The Mathematics of Voting and Apportionment: An Introduction (Compact Textbooks in Mathematics) [1st ed. 2019] 3030147673, 9783030147679

Table of contents :
Preface
Acknowledgments
Contents
1 Social Choice
1.1 Introduction
1.1.1 Plurality vs Majority
1.1.2 The Plurality Procedure
1.1.3 The 2000 US Presidential Election and the Shortcomings of the Plurality Procedure
1.2 Elimination Procedures
1.2.1 Plurality with Run-Off
1.2.2 The Hare Procedure
1.2.3 The Coombs Procedure
1.2.4 Monotonicity
1.2.5 Dealing with Non-monotonicity
1.3 Condorcet Ideas and Related Procedures
1.3.1 The Condorcet Tournament
1.3.2 Condorcet Winner and Condorcet Loser
1.3.3 Condorcet Winner vs Majority Candidate
1.3.4 The Condorcet Criteria
1.3.5 Agenda Voting
1.3.6 Weak Pareto Efficiency
1.3.7 Killer Amendment
1.3.8 The Copeland Procedure
1.3.9 Monotonicity Revisited
1.3.10 McGarvey's Theorem
1.4 Scoring Procedures: Borda Count
1.4.1 Borda Count
1.4.2 Shortcomings of the Borda Count Procedure
1.4.3 More on Borda Count
1.4.4 Hare-Scoring Procedures
1.4.5 Borda Meets Condorcet
1.5 A Glimpse Into Social Welfare Theory
1.5.1 Rankings and Semi Rankings
1.5.2 Independence of Irrelevant Alternatives (IIA)
1.5.3 Transitivity
1.5.4 Black's Theorem
1.5.5 May's Theorem
1.5.6 Dictatorship and Arrow's Theorem
1.5.7 Oligarchy and Gibbard's Theorem
1.6 Social Choice Procedures: Indifference and Ties Allowed
1.6.1 Standard Social Choice Procedures
1.6.2 More Condorcet-Type Classifications of the Alternatives
1.6.3 The Social Choice Set of the Hare–Borda Procedure
1.7 Manipulability of Social Choice Procedures: Indifference and Ties Allowed
1.7.1 Comparing Sets of Alternatives
1.7.2 Anonymity and Neutrality
1.7.3 Gärdenfors' Theorem
1.7.4 Manipulability of Social Choice Functions Not Covered by Gärdenfors' Theorem
Exercises
2 Yes-No Voting
2.1 Introduction
2.1.1 The Basics
2.1.2 The +/- Table
2.1.3 Parity of Banzhaf Scores
2.2 Quantification of Power in a Yes-No Voting System
2.2.1 The Banzhaf Index of Power
2.2.2 The Felsenthal–Machover Example
2.2.3 The Shapley–Shubik Index of Power
2.2.4 Banzhaf vs Shapley–Shubik Computations
2.3 Some Combinatorics
2.3.1 Permutations and Combinations
2.3.2 Cartesian Products
2.4 Banzhaf and Shapley–Shubik Indices in One View
2.4.1 Computing the Shapley–Shubik Indices from Winning Coalitions
2.4.2 Veto-Powered Voters, Dominant Voters and Dictators
2.5 Weightable Yes-No Voting Systems
2.5.1 Trades Among Coalitions
2.5.2 Trade-Robustness and the Taylor–Zwicker Theorem
2.5.3 The Magic Square Voting System
Exercises
3 Apportionment
3.1 Introduction
3.2 Axioms of Apportionment
3.3 Quota Procedures
3.3.1 Hamilton's Procedure
3.3.2 Lowndes' Procedure
3.3.3 The Alabama Paradox
3.3.4 The Population Paradox
3.3.5 The Balinski-Young Theorem
3.4 Divisor Procedures
3.4.1 The General Framework of Divisor Procedures
3.4.2 Jefferson's Procedure
3.4.3 Adams' Procedure
3.4.4 Webster's Procedure
3.4.5 The Hill–Huntington Procedure
3.5 Equity Criteria of Divisor Procedures
3.5.1 Measures of Inequity
3.5.2 Postulates of a Measure of Inequity
3.5.3 Webster's Procedure and the δ-Inequity
3.5.4 The Hill–Huntington Procedure and the ρ-Inequity
3.5.5 The φ-Inequity Measure
3.5.6 The Harmonic Mean
3.5.7 Dean's Procedure
3.5.8 A Collective View of Divisor Procedures
3.5.9 The Threshold Divisors of Dean's Procedure
3.6 Apportionment Paradoxes
3.6.1 Monotonicity and the Divisor Procedures
3.6.2 Consistency
3.7 Applications of Priority Formulas
3.7.1 One-By-One Seat Apportionment
3.7.2 The Quota-Divisor Procedures
Exercises
Answers to Selected Exercises
Chapter 1
Chapter 2
Chapter 3
References
Index
Answers to Selected Exercises
References
Index

Citation preview

Compact Textbooks in Mathematics

Sherif El-Helaly

The Mathematics of Voting and Apportionment An Introduction

Compact Textbooks in Mathematics

Compact Textbooks in Mathematics This textbook series presents concise introductions to current topics in mathematics and mainly addresses advanced undergraduates and master students. The concept is to offer small books covering subject matter equivalent to 2- or 3-hour lectures or seminars which are also suitable for self-study. The books provide students and teachers with new perspectives and novel approaches. They may feature examples and exercises to illustrate key concepts and applications of the theoretical contents. The series also includes textbooks specifically speaking to the needs of students from other disciplines such as physics, computer science, engineering, life sciences, finance. • • •

compact: small books presenting the relevant knowledge learning made easy: examples and exercises illustrate the application of the contents useful for lecturers: each title can serve as basis and guideline for a semester course/lecture/seminar of 2–3 hours per week.

More information about this series at http://www.springer.com/series/ 11225

Sherif El-Helaly

The Mathematics of Voting and Apportionment An Introduction

Sherif El-Helaly Department of Mathematics The Catholic University of America Washington, DC, USA

ISSN 2296-4568 ISSN 2296-455X (electronic) Compact Textbooks in Mathematics ISBN 978-3-030-14767-9 ISBN 978-3-030-14768-6 (eBook) https://doi.org/10.1007/978-3-030-14768-6 Library of Congress Control Number: 2019935503 Mathematics Subject Classification: 91-01, 91B12, 91F10 © Springer Nature Switzerland AG 2019 This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors, and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, express or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This book is published under the imprint Birkhäuser, www.birkhauser-science.com by the registered company Springer Nature Switzerland AG. The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

To my wife Afaf; my late mother, Safeyya; and my late father, Taha

vii

Preface Until recently, the mathematical content of the social sciences undergraduate curriculum was mostly limited to computations and routine manipulation of symbols. The undergraduate social sciences student was left with little or no exposure to the more sophisticated and elegant conceptual side of mathematics that involves the logical deduction of results from axioms and hypotheses. In the 1990s, mathematical applications involving deductive reasoning started to find their way into the social sciences undergraduate curriculum. The main topics introduced included social choice, yes-no voting, apportionment, game theory, and fair division. Most of the textbooks that covered this new material targeted the freshman and sophomore levels, while junior and senior students received the attention of only a few authors. Audience Based on two decades of teaching the above-mentioned topics at the Catholic University of America, I wrote this book to serve the upper-level audience as follows: 1. The book can be used as a textbook for a rigorous one-semester course in the mathematics of voting and apportionment for upper-level undergraduate and beginning graduate students majoring in economics, political science, and philosophy. 2. It can be used for self-study, especially by graduate students in economics and political science who need a straightforward and selfcontained introduction to the topics therein before they get started on their dissertations. 3. It can also be used in an elective course for mathematics majors in the applications of mathematics in the social sciences. 4. The book is also a suitable read for any curious mathematician who wishes to get acquainted with these unpublicized applications of mathematics. Developed by economists and political scientists with contributions by mathematicians, the theories of voting and apportionment (social choice and social welfare theory in particular) are genuine branches of advanced mathematics. For example, Arrow’s theorem on dictatorships (considered by many to be the cornerstone of social welfare theory) and the subsequent theorem by Gibbard on oligarchies are deep and original mathematical results whose only prerequisite is an elementary knowledge of the mathematical concept of “ranking” (or “order relation”) which I introduce in  Chap. 1 along with complete and detailed proofs of these two theorems. Prerequisites No political science prerequisites are needed to use this book. The mathematical prerequisites are minimal, and they do not include elementary algebra, calculus, statistics, or geometry. There is no need for

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Preface

computer skills; a simple calculator that can perform the basic arithmetical operations and compute square roots and includes a memory function will suffice. The only mathematical requirement (included in the book in order to make it self-contained) is some acquaintance with elementary concepts in combinatorics, graph theory, and order relations, in addition to the geometric and harmonic means. A more important requirement is the level of maturity that enables the student to think logically, derive results from axioms and hypotheses, and intuitively grasp logical notions such as “contrapositive” and “counterexample.” This level of maturity can be expected of junior and senior undergraduate students in economics, political sciences, philosophy, and mathematics. Book Contents The book consists of three mutually independent chapters which can be covered in any of the six possible orders, with one caveat: An instructor who wishes to start with  Chap. 1 should cover  Sect. 2.3 (on combinatorics) before starting  Sects. 1.5 (on social welfare functions) and 1.7 (on the manipulation of social choice functions). I included a detailed Table of Contents that should give the reader a good idea about the topics covered in each chapter. The following are some highlights:  Chapter 1, on social choice, starts with a discussion of the shortcomings of the widely used plurality procedure of elections in  Sect. 1.1. Elimination procedures are introduced in  Sect. 1.2 as attempts to avert the shortcomings of the plurality procedure. The Condorcet tournament (commonly known as the round-robin tournament) with its graph representation and related concepts are introduced in  Sect. 1.3; and special attention is given to the Condorcet paradox and how it can be used in a parliamentary procedure to manipulate the voting outcome by creating a killer amendment (aka poison pill amendment). I end Sect. 3 with the mathematically elegant but often overlooked theorem proved in 1953 by David McGarvey. This theorem is useful in the construction of examples and counterexamples as it guarantees the existence of a scenario of voting that can be represented by any given complete directed graph. In  Sect. 1.4, the vulnerability of Borda count to the influence of irrelevant alternatives and its frequent failure to elect an existing Condorcet winner are demonstrated. It is interesting that bringing an elimination principle to the Borda count (similar to that of Hare’s procedure) results in a procedure that satisfies the Condorcet winner criterion (the Hare-Borda procedure).  Section 1.5 is a brief introduction to social welfare theory, including the theorems of Duncan Black, Kenneth May, Kenneth Arrow, and Allen Gibbard. In  Sect. 1.6, I provide natural generalizations of some standard social choice procedures and Condorcet’s concepts to the case where indifference in the voter’s rankings and ties are allowed. The focus in  Sect. 1.7 is on the manipulability of social choice functions that satisfy the conditions of anonymity and neutrality and allow indifference and ties. I also present the 1976 theorem of Peter Gärdenfors

ix Preface

on the manipulability of social choice functions that satisfy the Condorcet winner criterion, in addition to the aforementioned conditions.  Chapter 2, on yes-no voting, starts with an elementary view of the Banzhaf and Shapley-Shubik indices of power in  Sects. 2.1 and 2.2, including a comparison between the concepts and the computations of the two indices.  Section 2.3 includes the combinatorics needed to understand the self-checking method of computing both indices in the same setting in  Sect. 2.4. The focus in  Sect. 2.5 is on the question of when a given yesno voting system can be represented by weights and a quota. The theorem of Taylor and Zwicker is partially proven, and their novel example of the magic square voting system is presented.  Chapter 3, on apportionment, starts with two introductory sections that demonstrate the complexity of the apportionment problem. In  Sect. 3.3, I present quota procedures with a focus on the procedures of Hamilton and Lowndes, the Alabama paradox, the population paradox, and the Balinski-Young theorem. Divisor procedures are then discussed in detail in  Sect. 3.4, together with the process of using the threshold divisors to find the interval of proper divisors for each of the procedures of Jefferson, Adams, Webster, and Hill-Huntington. The equity criteria and the measures of inequity associated with the procedures of Webster and Hill-Huntington are discussed in  Sect. 3.5. Motivated by a different measure of inequity, Dean’s procedure and the associated harmonic mean are introduced, and then the biases of the different divisor procedures are heuristically discussed.  Section 3.6 compares the quota procedures and the divisor procedures with regard to the apportionment paradoxes.  Section 3.7 is devoted to the quotadivisor procedures introduced by Balinski and Young, which reconcile house monotonicity with the quota rule. Pedagogy The pedagogical approach can be highlighted in the following points: 1. Each chapter begins with an elementary introduction and several examples, gradually leading to the more advanced material. (This is particularly useful to readers with no earlier exposure to the material at the freshman level.) 2. New concepts and definitions are motivated by examples before being formally defined. 3. The topics treated in this book are plagued by highly complicated and nonstandard notations. To help overcome this problem, I tried to make notations as simple and intuitive as possible, and, in some cases, I opted to use words (instead of subscripts and superscripts) in order to keep symbols simple and uncluttered. I also avoided using acronyms and abbreviations to denote mathematical objects and concepts. (The use of the standard acronym IIA instead of “Independence of Irrelevant Alternatives” is one of the few exceptions.)

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4. Landmark theorems are presented in this book, with fully detailed, rigorous, and streamlined proofs. Complex proofs, such as those of Arrow’s theorems, Gibbard’s theorem, and Gärdenfors’ theorem, are broken down into propositions and lemmas in order to make them easier to grasp. 5. Definitions of the form: “A scenario of voting for a set V of voters and a set A of alternatives is a member the set RV where R is the set of all order relations on A” and “A social welfare function is a function λ : RV → R” would be too alien to most of the audience of this book. Nevertheless, it is possible to convey the exact meanings of such definitions, in a workable manner that can be used in rigorous proofs, in a more lucid and less formal manner; this is what I opted to do. 6. Computations are kept as simple as possible in order to help the student focus on the concepts. 7. Each chapter ends with a set of exercises that vary from routine exercises (to help the student master concepts and techniques) to exercises in which the student is asked to prove or disprove a given statement. 8. Answers to most exercises are given at the end of the book. Washington, DC, USA

Sherif El-Helaly

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Acknowledgments I would like to express my gratitude to: • Christopher Tominich at Birkhäuser for his guidance throughout the preparation of the manuscript • The anonymous reviewers of the manuscript for their valuable feedback • My students at the Catholic University of America over the past two decades whose questions and comments have contributed to the development of this book I am also grateful to my wife Afaf for her patience and support.

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Contents 1

Social Choice . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Sherif El-Helaly 1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1.1 Plurality vs Majority . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1.2 The Plurality Procedure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1.3 The 2000 US Presidential Election and the Shortcomings of the Plurality Procedure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Elimination Procedures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2.1 Plurality with Run-Off . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2.2 The Hare Procedure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2.3 The Coombs Procedure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2.4 Monotonicity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2.5 Dealing with Non-monotonicity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Condorcet Ideas and Related Procedures . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3.1 The Condorcet Tournament . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3.2 Condorcet Winner and Condorcet Loser . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3.3 Condorcet Winner vs Majority Candidate . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3.4 The Condorcet Criteria . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3.5 Agenda Voting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3.6 Weak Pareto Efficiency . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3.7 Killer Amendment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3.8 The Copeland Procedure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3.9 Monotonicity Revisited . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3.10 McGarvey’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4 Scoring Procedures: Borda Count . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4.1 Borda Count . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4.2 Shortcomings of the Borda Count Procedure . . . . . . . . . . . . . . . . . . . . . . . . . 1.4.3 More on Borda Count . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4.4 Hare-Scoring Procedures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4.5 Borda Meets Condorcet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5 A Glimpse Into Social Welfare Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5.1 Rankings and Semi Rankings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5.2 Independence of Irrelevant Alternatives (IIA) . . . . . . . . . . . . . . . . . . . . . . . . . 1.5.3 Transitivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5.4 Black’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5.5 May’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5.6 Dictatorship and Arrow’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5.7 Oligarchy and Gibbard’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.6 Social Choice Procedures: Indifference and Ties Allowed . . . . . . . . . . . 1.6.1 Standard Social Choice Procedures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.6.2 More Condorcet-Type Classifications of the Alternatives . . . . . . . . . . . . . .

1 1 2 3 4 6 7 9 11 18 20 21 21 23 24 26 28 30 32 35 36 37 41 41 44 46 48 50 52 53 55 59 61 65 67 74 79 79 84

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The Social Choice Set of the Hare–Borda Procedure . . . . . . . . . . . . . . . . . . . Manipulability of Social Choice Procedures: Indifference and Ties Allowed . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.7.1 Comparing Sets of Alternatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.7.2 Anonymity and Neutrality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.7.3 Gärdenfors’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.7.4 Manipulability of Social Choice Functions Not Covered by Gärdenfors’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

102 105

1.6.3 1.7

88 92 94 95 98

Yes-No Voting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Sherif El-Helaly 2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1.1 The Basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1.2 The +/− Table . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1.3 Parity of Banzhaf Scores . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Quantification of Power in a Yes-No Voting System . . . . . . . . . . . . . . . . . 2.2.1 The Banzhaf Index of Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.2 The Felsenthal–Machover Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.3 The Shapley–Shubik Index of Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.4 Banzhaf vs Shapley–Shubik Computations . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Some Combinatorics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.1 Permutations and Combinations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.2 Cartesian Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4 Banzhaf and Shapley–Shubik Indices in One View . . . . . . . . . . . . . . . . . . 2.4.1 Computing the Shapley–Shubik Indices from Winning Coalitions . . . . . 2.4.2 Veto-Powered Voters, Dominant Voters and Dictators . . . . . . . . . . . . . . . . 2.5 Weightable Yes-No Voting Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5.1 Trades Among Coalitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5.2 Trade-Robustness and the Taylor–Zwicker Theorem . . . . . . . . . . . . . . . . . . 2.5.3 The Magic Square Voting System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

115 116 118 122 123 123 126 128 132 133 133 136 137 138 139 144 146 147 150 153

3

Apportionment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

159

3.1 3.2 3.3 3.3.1 3.3.2 3.3.3 3.3.4 3.3.5 3.4 3.4.1

Sherif El-Helaly Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Axioms of Apportionment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Quota Procedures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Hamilton’s Procedure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Lowndes’ Procedure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The Alabama Paradox . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The Population Paradox . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The Balinski-Young Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Divisor Procedures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The General Framework of Divisor Procedures . . . . . . . . . . . . . . . . . . . . . . . .

159 162 166 166 168 169 172 172 175 175

2

xv Contents

3.4.2 Jefferson’s Procedure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4.3 Adams’ Procedure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4.4 Webster’s Procedure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4.5 The Hill–Huntington Procedure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5 Equity Criteria of Divisor Procedures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5.1 Measures of Inequity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5.2 Postulates of a Measure of Inequity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5.3 Webster’s Procedure and the δ-Inequity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5.4 The Hill–Huntington Procedure and the ρ-Inequity . . . . . . . . . . . . . . . . . . 3.5.5 The ϕ-Inequity Measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5.6 The Harmonic Mean . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5.7 Dean’s Procedure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5.8 A Collective View of Divisor Procedures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5.9 The Threshold Divisors of Dean’s Procedure . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6 Apportionment Paradoxes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6.1 Monotonicity and the Divisor Procedures . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6.2 Consistency . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.7 Applications of Priority Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.7.1 One-By-One Seat Apportionment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.7.2 The Quota-Divisor Procedures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

177 185 192 199 207 208 209 211 213 215 216 217 219 222 225 226 227 233 233 234 242

Answers to Selected Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

251

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

259

Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

261

1

Social Choice Sherif El-Helaly © Springer Nature Switzerland AG 2019 Sherif El-Helaly, The Mathematics of Voting and Apportionment, Compact Textbooks in Mathematics, https://doi.org/10.1007/978-3-030-14768-6_1

1.1

Introduction

In this chapter, we study situations in which a society needs to choose from two or more alternatives available. Different members of the society may have different choices; and the problem is how to aggregate the different choices of all members of the society to make a choice for the entire society. The choice of the society will be called the social choice set (or the social choice) and may consist of one alternative or more. An alternative will be called a winner if it is a member of the social choice set; and will be called a loser if it is not. When the social choice set consists of more than one alternative, we shall say that its members are tied winners. The case of a social choice set consisting of more then one alternative will be mostly avoided in the first four sections of this chapter. The procedure used to produce the social choice set will be called a social choice procedure or a social choice function. A Note on the Use of the Terms “Function” and “Procedure” Many authors use the term “function” only when the alternatives and the voters are specified. This is to stay close to the standard mathematical concept of a function that specifies the domain in addition to the rule (or the “procedure”) that describes the “action” of the function. We find it more convenient to use the two terms interchangeably. The most important instance of a social choice situation is when the voters in a society need to elect a candidate for political office. Another example is that of a hiring committee in a company or a university trying to pick one out of several applicants for a vacant position. If necessary, a tie may be broken by some rule to be agreed upon before voting starts. Each of the social choice procedures in  Sects. 1.1, 1.2, 1.3 (except Sects. 1.3.5 and 1.3.7), 1.4, 1.6, and 1.7 is 1. anonymous, i.e., treats all voters equally. 2. neutral, i.e., treats all alternatives equally. Precise definitions of these two concepts will be given in  Sect. 1.7.

1

2

Chapter 1 • Social Choice

1 1.1.1 Plurality vs Majority To motivate the discussion, assume that a hiring committee of 19 voters is trying to pick one of the three candidates a, b, and c to fill a position in a certain company. Consider the following two possible scenarios: Scenario 1: a receives 9 votes, b receives 7 votes, and c receives 3 votes. Scenario 2: a receives 10 votes, b receives 6 votes, and c receives 3 votes. The above two scenarios do not look very different but, in fact, they are: • In both scenarios a has more votes than each individual opponent. • In Scenario 2, a goes one big step farther: He has more votes than all his opponents combined. Of course, this means that a has more than half of the total vote. This difference between the above two scenarios is of great importance, as we shall see. Definition 1.1.1 Suppose the entire set of voters is partitioned into subsets A1 , A2 , · · · , An such that each voter is included in one, and only one, of these subsets. 1. A subset Ai is said to be a plurality of the vote if the number of voters in Ai is greater than or equal to the number of voters in each Aj , j = 1, 2, · · · , n. A candidate who is voted for by a plurality set of voters is said to be a plurality candidate.   2. A subset Ai is said to be a majority of the vote if any of the following two equivalent conditions holds: The number of voters in Ai is greater than half of the total vote. The number of voters in Ai is greater than the total number of voters in all of the other subsets. A candidate who is voted for by a majority set of voters is said to be a majority candidate.  

In Scenario 1 above, a is a plurality candidate but not a majority candidate. In Scenario 2, a is both a plurality candidate and a majority candidate. (In general, a majority candidate is necessarily a plurality candidate but a plurality candidate need not be a majority candidate.) Note that in every social choice problem: • There is at least one plurality candidate. • There is at most one majority candidate. (It is possible to have no majority candidate, as in Scenario 1 above.) • If there are only two candidates and the number of voters is odd, then one of the two candidates must be a majority candidate.

1

3 1.1 · Introduction

Example 1.1.2 Identify the plurality candidates and the majority candidates in the following examples: 1. a has 29 votes, b has 26 votes, c has 9 votes, and d has 7 votes. 2. a has 22 votes, b has 13 votes, c has 6 votes, and d has 3 votes. 3. a has 17 votes, b has 17 votes, c has 14 votes, d has 12 votes, and e has 5 votes. 4. a has 27 votes, b has 14 votes, c has 5 votes, and d has 3 votes.  Solution 1. a is a plurality candidate. The total number of voters is 29 + 26 + 9 + 7 = 71. Since 71 2 = 35.5, one needs at least 36 votes to qualify as a majority candidate. Therefore, there is no majority candidate.   2. a is a plurality candidate. The total number of voters is 22 + 13 + 6 + 3 = 44. Since 44 2 = 22, one needs at least 23 votes to qualify as a majority candidate. Therefore, there is no majority candidate.   3. a and b are plurality candidates. The total number of voters is 17+17+14+12+5 = 65. Since 65 2 = 32.5, one needs at least 33 votes to qualify as a majority candidate. Therefore, there is no majority candidate.   4. a is a plurality candidate. The total number of voters is 27 + 14 + 5 + 3 = 49. Since 49 2 = 24.5, one needs at least 25 votes to qualify as a majority candidate. Therefore, a is, in fact, a majority candidate.  

1.1.2 The Plurality Procedure Definition 1.1.3 The social choice procedure which declares the plurality candidate (or candidates) as winner(s) is called the plurality procedure.  

Why can’t we replace the word “plurality” with the word “majority” in the above definition to define what we might call “the majority procedure?” The reason, which can be seen from the examples in Example 1.1.2 above, is that a majority candidate is not guaranteed to exist in every case. The plurality procedure has long been challenged by many political scientists. The trouble lies in electing a plurality candidate a who is not a majority candidate. In fact, • The number of voters who did not vote for a is greater than the number of voters who voted for a. • The victory of a depends, in part, on the way the rest of the vote was distributed among her opponents, which has nothing to do with the merits of a or the strength of her support among voters.

4

Chapter 1 • Social Choice

1 • As we shall see in  Sect. 1.3, the plurality procedure can pick a winner who happens to be the weakest candidate, in some “very reasonable” sense. In the first of the Examples 1.1.2, candidate a is the plurality procedure winner without having a majority of the total vote. Candidates c and d in this example could be regarded as minor candidates with no realistic chance of winning. Had these two candidates not been in the race, their combined vote of 9 + 7 = 16 votes would have been up for grabs by the two main candidates, namely a and b. In one possible scenario, 10 or more of the 16 votes of c and d could have gone to b and 6 or less would have gone to a. This way, a would have ended up with 35 or less votes and b would have ended up with 36 or more votes and won the election. Assume that, in this example, a is a left of center candidate, b is a right of center candidate while c and d are to the right of b. It can be argued that had c and d not been in the race, b would have picked up enough of their vote to win the election. The following is a real-life example.

1.1.3 The 2000 US Presidential Election and the Shortcomings of the Plurality Procedure In the 2000 presidential election, the Florida results were delayed due to a dispute between the two main contenders, George W. Bush and Al Gore, caused by a razor thin vote margin. Based on the election returns from all states and the District Of Columbia (with the exception of the delayed Florida returns), no candidate was able to secure the 270 electoral votes needed to win, and it was up to Florida with its 25 electoral votes to decide who would be the next occupant of the White House. In Florida, as in the vast majority of the states, the plurality procedure is used to pick one winner, who takes all the electoral votes of the state. That said, the new president was, for all practical purposes, being elected by the Florida voters using the plurality procedure; and the complicating factor of the Electoral College was neutralized. After a lengthy process of counting and recounting (and intervention by the Supreme Court), the final Florida election returns were as shown in the following table. George W. Bush won Florida with a plurality (but not a majority) of the vote and the difference between Bush and his nearest opponent, Al Gore, was only 537 votes. The rest of the candidates combined received 138,067 votes, enough to change the election outcome had those minor candidates (or some of them) been out of the race.

5 1.1 · Introduction

Candidate

Total vote

%

Party

George W. Bush (W)

2,912,790

48.847

Republican

Al Gore

2,912,253

48.838

Democratic

Ralph Nader

97,488

1.635

Green

Patrick J. Buchanan

17,484

0.293

Reform

Harry Browne

16,415

0.275

Libertarian

John Hagelin

2281

0.038

Natural law/reform

Howard Phillips

1371

0.023

Constitution

Other

3028

0.051

Total

5,963,110

100

The two main candidates who had realistic chances of winning the Florida vote were, of course, George W. Bush (R) and Al Gore (D). Nader’s vote came, in the most part, from Democratic voters while the rest of the vote received by the minor candidates came from conservative Republican voters. Had the election been between George W. Bush and Al Gore only, and assuming that Al Gore received only 71.1% of Nader’s votes while the rest of Nader’s votes combined with the votes of the other minor candidates went to George W. Bush, Al Gore would have won the 25 electoral votes of Florida and the presidency of the United States. Some Computations Half of the total vote is 2,981,555 and therefore, had the election been between George W. Bush and Al Gore only, the number of votes needed to win would have been 2,981,556. To reach this number, Al Gore would have needed 2, 981, 556 − 2, 912, 253 = 69, 303 more votes to win Florida. It would have been easy for Al Gore to get this number of votes which is only 71.1% of Nader’s supporters who numbered 97,488. (It is assumed in these computations that, in the absence of all minor candidates, all of the 5,963,110 voters would still cast their votes.) In cases where the plurality procedure winner is also a majority candidate, we have a very stable outcome that does not depend on the distribution of the opposing vote among the opponents, nor does it depend on who might have dropped out of the race or stayed in it. For over two centuries, political scientists have been proposing ideas to reform the plurality procedure (and some offered entirely new approaches) to avoid the shortcomings of the plurality procedure discussed above. However, due to its simplicity, the plurality procedure has persisted and continued to be by far the most widely used election procedure in the United States. The plurality procedure is also believed to be one of the main reasons behind the perpetuation of the two-party system in the United States, as it discourages voters from voting for minor (third party) candidates. In the 2000 presidential election, for example, many voters who voted for Al Gore would have voted for Ralph Nader but voted for Al Gore instead because they knew that Nader had no realistic chance of winning. The wise thing to do, in their opinion, was to vote for their second choice, Al Gore, in

1

6

Chapter 1 • Social Choice

1 order to prevent their most undesirable candidate, George W. Bush, from winning. A similar strategy was followed by many conservative republican voters who voted for their second choice (George W. Bush) instead of their first choice (Patrick Buchanan). Another vivid example of this phenomenon was in the presidential election of 1992 in which the two main candidates were Bill Clinton (D) and George H. W. Bush (R) with Ross Perot as a third party candidate, but the analysis is more complicated because of the electoral college.

1.2

Elimination Procedures

Next, we study some of the attempts to reform the plurality procedure (commonly known as elimination procedures). These reforms are based on one central idea: If a majority candidate exists, she is declared the sole winner. If not, we eliminate one (or more) of the weaker candidate(s) and take a second vote. (Elimination procedures differ in the way they define the term “weaker candidates”.) If a “majority” candidate emerges as a result of the second vote, he is the sole winner. If not, we eliminate one (or more) of the weaker candidate(s) and take a third vote. The process is continued until one of the following two mutually exclusive and exhaustive possibilities materializes: 1. A “majority” candidate emerges. This candidate is declared the sole winner. 2. The rule of elimination of weaker candidate(s) equally applies to all of the remaining candidates. In this case, we declare all of the remaining candidates as tied winners. The reason for enclosing the word “majority” above between quotation marks is that the candidate who emerges with a majority after the second (or later) vote is not a true majority candidate. A true majority candidate is a candidate who has a majority as a result of the first vote, without any elimination. We shall, however, drop those quotation marks from now on. The idea behind the elimination procedures is that elimination helps concentrate the vote in fewer hands, thus helping a majority to emerge. Next, we present three of the better known elimination procedure: Plurality with run-off, Hare’s procedure and Coombs’ procedure. A general illustration of elimination procedures is given in ⊡ Fig. 1.1 below.

Winner(s) Found?

Data Eliminate Weaker Candidate(s) ⊡ Fig. 1.1 Elimination procedures

No

Yes

End

1

7 1.2 · Elimination Procedures

1.2.1 Plurality with Run-Off Definition 1.2.1 The plurality with run-off procedure: Step 1. Look for a majority candidate. If one exists, declare her the winner. If not, go to Step 2. Step 2. Two candidates with the highest vote counts compete in a second run-off round of voting. (The rest of the candidates are eliminated.) In the run-off round, one of those two candidates wins by a majority of the vote or they are tied.   A technical difficulty may arise and will be addressed in the third of the following examples.

Example 1.2.2 Assuming that the plurality with run-off procedure is used, which two candidates qualify for the run-off round, if needed, in the following examples? 1. a has 17 votes, b has 14 votes, c has 11 votes, d has 7 votes, and e has 4 votes. 2. a has 17 votes, b has 17 votes, c has 11 votes, d has 7 votes, and e has 4 votes. 3. a has 17 votes, b has 14 votes, c has 14 votes, d has 7 votes, and e has 4 votes. 4. a has 38 votes, b has 14 votes, c has 11 votes, d has 7 votes, and e has 4 votes.  Solution 1. No majority candidate exists. a and b qualify for the run-off round.   2. No majority candidate exists. a and b qualify for the run-off round.   3. No majority candidate exists. a qualifies for the run-off round, joined with one of b and c. The tie between b and c is to be broken by some process (to be agreed upon before the voting begins) to determine which of b and c should join a in the run-off round. Difficulties of this kind are extremely unlikely in mass elections such as city-wide or state-wide elections, but they are not so rare in small committees.   4. There is no run-off round in this case. a is a majority candidate and is declared an “immediate” winner.  

Instant Run-Off In the case of no majority candidate, the voters are invited to come back after a reasonable period of time (usually 2 or 3 weeks) for the run-off round of voting. This is how the plurality with run-off procedure is usually carried out. The whole process can be completed in just one trip to the polling stations if the instant run-off idea is implemented. In this trip, each voter is asked to fill out a preference ballot (a complete preference list) showing a first choice, second choice, third choice, etc. If no majority candidate exists, two candidates with the highest first choice votes are kept and all others are eliminated from the preference ballots.

8

Chapter 1 • Social Choice

1 Note that 1. When preference ballots are used, the term “majority candidate” refers to a candidate who is the first choice of a majority of voters. Preference ballots will be used in the rest of this chapter. 2. In  Sects. 1.2, 1.3, and 1.4, indifference will not be allowed in the preference ballots, that is, given two alternatives x and y, a voter must either prefer x to y or y to x but cannot be indifferent. Indifference will be allowed in  Sects. 1.6 and 1.7. The following example makes the instant run-off procedure very clear. Example 1.2.3 Fifty three voters are to choose from the set of five candidates a, b, c, d and e. Identical preference ballots are grouped together and the number of voters in each group is indicated on the top. Process the ballots using 1. The plurality procedure. 2. The plurality with run-off procedure. 10

7

8

6

6

5

7

a

a

b

b

c

c

d

4 e

e

d

c

d

b

d

c

a

d

e

e

c

d

e

b

d

c

c

d

e

e

a

a

c

b

b

a

a

a

b

e

b

(This may be considered the same as the first of Example 1.2.2 after the voters have been instructed to submit complete preference ballots instead of “single pick” ballots.)  Solution 1. a is the first choice of 10 + 7 = 17 voters, b is the first choice of 8 + 6 = 14 voters, c is the first choice of 6 + 5 = 11 voters, d is the first choice of 7 voters, and e is the first choice of 4 voters. Candidate a is the sole plurality procedure winner.   = 26.5, one needs at least 27 first choice votes to qualify as a majority 2. Since 53 2 candidate. There is no majority candidate; and a, b qualify for the run-off round. Eliminating c, d, and e and updating the ballots we get: 10

7

8

6

6

5

7

4

a

a

b

b

b

a

b

a

b

b

a

a

a

b

a

b

Now b has 8 + 6 + 6 + 7 = 27 votes (a majority) and wins the plurality with run-off procedure.  

1

9 1.2 · Elimination Procedures

Example 1.2.4 Twenty five voters are to choose from the set of four candidates a, b, c, and d. The grouped ballots are given below. Process the ballots using 1. The plurality procedure. 2. The plurality with run-off procedure. 8

6

4

2

5

a

a

b

b

c

c

b

c

d

d

b

d

a

a

a

d

c

d

c

b

 Solution 1. a is the first choice of 14 voters, b is the first choice of 6 voters, c is the first choice of 5 voters, and d is the first choice of 0 voters. Candidate a is the sole plurality procedure winner.   2. Since 25 = 12.5, one needs at least 13 first choice votes to qualify as a majority 2 candidate; therefore a is a majority candidate. Candidate a wins the plurality with runoff. No run-off round in this case.  

1.2.2 The Hare Procedure Introduced in 1861 by Thomas Hare (a British law scholar), this procedure is a refinement of the instant run-off procedure. The version presented here is intended for electing a single winner. A multiple-winner version of it is used in The Republic of Ireland, Australia and in some local jurisdictions in the United States and the United Kingdom. Definition 1.2.5 Hare Procedure: Step 1. If a majority candidate exists, he is the sole winner. If all candidates have equal numbers of top choice votes, they are all tied for winning. If neither of these two events occurs, go to Step 2. Step 2. Eliminate the candidate(s) with the lowest first choice vote, update the ballots and go back to Step 1.  

Example 1.2.6 Process the data in Example 1.2.3 using the Hare procedure.



10

Chapter 1 • Social Choice

1 Solution The 53 preference ballots are: 10

7

8

6

6

5

7

a

a

b

b

c

c

d

4 e

e

d

c

d

b

d

c

a

d

e

e

c

d

e

b

d

c

c

d

e

e

a

a

c

b

b

a

a

a

b

e

b

Step 1. Since 53 2 = 26.5, a candidate needs at least 27 first choice votes to qualify as a majority candidate. a is the first choice of 10 + 7 = 17 voters, b is the first choice of 8 + 6 = 14 voters, c is the first choice of 6 + 5 = 11 voters, d is the first choice of 7 voters, and e is the first choice of 4 voters. We must go to Step 2. Step 2. Eliminate e since he has the lowest first choice vote. Here are the updated preference ballots: 10

7

8

6

6

5

7

a

a

b

b

c

c

d

4 a

d

d

c

d

b

d

c

d

c

c

d

c

d

a

b

c

b

b

a

a

a

b

a

b

Go to Step 1. Step 1. a is the first choice of 10 + 7 + 4 = 21 voters, b is the first choice of 8 + 6 = 14 voters, c is the first choice of 6 + 5 = 11 voters, and d is the first choice of 7 voters. We must go to step 2. Step 2. Eliminate d since he has the lowest first choice vote. Here are the updated preference ballots: 10

7

8

6

6

5

7

4

a

a

b

b

c

c

c

a

c

c

c

c

b

a

b

c

b

b

a

a

a

b

a

b

Go to Step 1. Step 1. a is the first choice of 10 + 7 + 4 = 21 voters, b is the first choice of 8 + 6 = 14 voters, and c is the first choice of 6 + 5 + 7 = 18 voters. We must go to step 2. Step 2. Eliminate b since he has the lowest first choice vote. Here are the updated preference ballots: 10

7

8

6

6

5

7

4

a

a

c

c

c

c

c

a

c

c

a

a

a

a

a

c

1

11 1.2 · Elimination Procedures

Step 1. a is the first choice of 10 + 7 + 4 = 21 voters and c is the first choice of 8 + 6 + 6 + 5 + 7 = 32 voters. Now c emerges with a majority and qualifies as the Hare procedure sole winner.   Example 1.2.7 Process the following data for 24 voters and 4 candidates using the Hare procedure. 7

5

5

3

4

a

a

b

b

c

c

b

c

d

d

b

d

a

a

b

d

c

d

c

a

 Solution Step 1. Since 24 2 = 12, one needs at least 13 first choice votes to qualify as a majority candidate. a is the first choice of 12 voters, b is the first choice of 8 voters, c is the first choice of 4 voters, and d is the first choice of zero voters. We must go to Step 2. Step 2. Eliminate d. Here are the updates ballots. 7

5

5

3

a

a

b

b

4 c

c

b

c

a

b

b

c

a

c

a

Step 1. a is the first choice of 12 voters, b is the first choice of 8 voters, and c is the first choice of 4 voters. We must go to Step 2. Step 2. Eliminate c. Here are the updated ballots. 7

5

5

3

4

a

a

b

b

b

b

b

a

a

a

Step 1. Each of a and b is the first choice of 12 voters. We have two tied winners: a and b.  

1.2.3 The Coombs Procedure This procedure was introduced by Clyde Coombs (a twentieth century American scholar in the field of mathematical psychology). It is based on a principle similar to that of Hare’s procedure, as we shall soon see.

12

Chapter 1 • Social Choice

1 Definition 1.2.8 Coombs Procedure: Step 1. If a majority candidate exists, she is the sole winner. If no majority candidate exists and all candidates have equal numbers of last choice vote, all candidates are tied for winning. If neither of these two events occurs, we go to Step 2. Step 2. Eliminate the candidate(s) with the highest last choice vote, update the ballots and go back to Step 1.  

Example 1.2.9 Use Coombs’ procedure to process the data in Example 1.2.3.



Solution The 53 preference ballots are: 10

7

8

6

6

5

7

a

a

b

b

c

c

d

4 e

e

d

c

d

b

d

c

a

d

e

e

c

d

e

b

d

c

c

d

e

e

a

a

c

b

b

a

a

a

b

e

b

Step 1. 27 first choice votes are necessary to qualify as majority candidate. No majority candidate exists, as seen in Example 1.2.6. Further, a is the last choice of 8+6+6 = 20 voters, b is the last choice of 10 + 7 + 5 + 4 = 26 voters, c is the last choice of 0 voters, d is the last choice of 0 voters, and e is the last choice of 7 voters. Last choice vote is not equally distributed among all candidates. We go to Step 2. Step 2. Eliminate b since he has the highest number of last choice votes. Here are the updated preference ballots: 10

7

8

6

6

5

7

4

a

a

c

d

c

c

d

e

e

d

e

c

d

d

c

a

d

e

d

e

e

e

a

d

c

c

a

a

a

a

e

c

Go to Step 1. Step 1. There is no majority candidate since a is the first choice of 10 + 7 = 17 voters, c is the first choice of 8 + 6 + 5 = 19 voters, d is the first choice of 6 + 7 = 13 voters, and e is the first choice of 4 voters. Also, a is the last choice of 8 + 6 + 6 + 5 = 25 voters, c is the last choice of 10 + 7 + 4 = 21 voters, d is the last choice of 0 voters, and e is the last choice of 7 voters. Last choice vote is not equally distributed among all candidates, so we go to step 2.

1

13 1.2 · Elimination Procedures

Step 2. Eliminate a since he has the highest number of last choice votes. Here are the updated preference ballots: 10

7

8

6

6

5

7

e

d

c

d

c

c

d

4 e

d

e

e

c

d

d

c

d

c

c

d

e

e

e

e

c

Go to Step 1. Step 1. There is no majority candidate since c is the first choice of 8 + 6 + 5 = 19 voters, d is the first choice of 7 + 6 + 7 = 20 voters, and e is the first choice of 10 + 4 = 14 voters. Further, c is the last choice of 10 + 7 + 4 = 21, d is the last choice of 8 voters, and e is the last choice of 6 + 6 + 5 + 7 = 24 voters. Last choice vote is not equally distributed among all voters. We go to step 2. Step 2. Eliminate e since he has the highest number of last choice votes. Here are the updated preference ballots: 10

7

8

6

6

5

7

d

d

c

d

c

c

d

4 d

c

c

d

c

d

d

c

c

Step 1. c is the first choice of 8 + 6 + 5 = 19 voters and d is the first choice of 10 + 7 + 6 + 7 + 4 = 34 voters. Now d qualifies as majority candidate and is declared the sole winner of Coombs’ procedure.   Example 1.2.10 Process the following data for 27 voters and 5 candidates using 1. The plurality procedure. 2. The plurality with run-off procedure. 3. Hare’s procedure. 4. Coombs’ procedure. 8

7

6

5

a

b

c

d

1 e

c

e

d

c

c

b

c

b

e

d

e

d

e

b

b

d

a

a

a

a

 Solution 1. Plurality procedure: a is the first choice of 8 voters, b is the first choice of 7 voters, c is the first choice of 6 voters, d is the first choice of 5 voters, and e is the first choice of one

14

Chapter 1 • Social Choice

1 voter. Candidate a is the plurality procedure winner. Note that a, who won according to the plurality procedure because he has the highest number of first choice votes, is ranked as last choice by a majority of 19 out of the 27 voters! We shall have more to say about this in  Sect. 1.3.   2. Plurality with run-off : Since 27 = 13.5, one needs at least 14 first choice votes to qualify 2 as a majority candidate. There is no majority candidate. The top two candidates are a and b. Eliminating c, d, and e and updating the ballots we get: 8

7

6

5

1

a

b

b

b

b

b

a

a

a

a

Now b has 7 + 6 + 5 + 1 = 19 votes (a majority) and wins the plurality with run-off procedure.   3. Hare: Step 1. There is no majority candidate. Go to Step 2. Step 2. Eliminate e since he has the lowest first choice vote. Here are the updated preference ballots: 8

7

6

5

a

b

c

d

1 c

c

c

d

c

d

b

d

b

b

b

d

a

a

a

a

Go to Step 1. Step 1. a is the first choice of 8 voters, b is the first choice of 7 voters, c is the first choice of 6 + 1 = 7 voters, and d is the first choice of 5 voters. There is no majority candidate. We go to Step 2. Step 2. Eliminate d since he has the lowest first choice votes. Here are the updated preference ballots: 8

7

6

5

1

a

b

c

c

c

c

c

b

b

b

b

a

a

a

a

Go to Step 1. Step 1. a is the first choice of 8 voters, b is the first choice of 7 voters, and c is the first choice of 6 + 5 + 1 = 12 voters. There is no majority candidate. We go to Step 2.

1

15 1.2 · Elimination Procedures

Step 2. Eliminate b since he has the lowest first choice votes. Here are the updated preference ballots: 8

7

6

5

1

a

c

c

c

c

c

a

a

a

a

Go to Step 1. Step 1. a is the first choice of 8 voters and c is the first choice of 7 + 6 + 5 + 1 = 19 voters. Now c emerges as the sole winner of Hare’s procedure.   4. Coombs: Step 1. There is no majority candidate. Further, a is the last choice of 19 voters, b is the last choice of 0 voters, c is the last choice of 0 voters, d is the last choice of 8 voters, and e is the last choice of 0 voters. The last choice vote is not equally distributed among the voters. We go to Step 2. Step 2. Eliminate a since she has the highest number of last choice votes. Here are the updated preference ballots: 8

7

6

5

c

b

c

d

1 e

b

e

d

c

c

e

c

b

e

d

d

d

e

b

b

Go to Step 1. Step 1. b is the first choice of 7 voters, c is the first choice of 14 voters, d is the first choice of 5 voters, and e is the first choice of one voter. c emerges with a majority and is declared the sole winner of Coombs’ procedure.   Example 1.2.11 Process the following data for 27 voters and 5 candidate using 1. Hare’s procedure 2. Coombs’ procedure 8

7

6

5

a

e

c

d

1 b

b

d

b

b

d

d

b

d

c

e

c

c

e

a

c

e

a

a

e

a



16

Chapter 1 • Social Choice

1 Solution 1. Hare: Step 1. a is the first choice of 8 voters, b is the first choice of 1 voter, c is the first choice of 6 voters, d is the first choice of 5 voters, and e is the first choice of 7 voters. One needs at least 14 first choice votes to qualify as a majority candidate. There is no majority candidate. Step 2. Eliminate b since he has the lowest first choice voter. Here are the updated ballots: 8

7

6

5

a

e

c

d

1 d

d

d

d

c

e

c

c

e

a

c

e

a

a

e

a

Step 1. a is the first choice of 8 voters, c is the first choice of 6 voters, d is the first choice of 6 voters, and e is the first choice of 7 voters. There is no majority candidate. Step 2. Eliminate c and d since they have the lowest first choice vote. Here are the updated preference ballots: 8

7

6

5

1

a

e

e

a

e

e

a

a

e

a

Step 1. a is the first choice of 13 voters and e is the first choice of 14 voters. Now e emerges with a majority and is declared the sole winner of Hare’s procedure.  

2. Coombs: Step 1. There is no majority candidate. Further, a is the last choice of 14 voters, e is the last choice of 13 voters, while each of b, c, and d is the last choice of 0 voters. Step 2. Eliminate a since he has the highest number of last choice votes. Here are the updated preference ballots: 8

7

6

5

b

e

c

d

1 b

d

d

b

b

d

c

b

d

c

e

e

c

e

e

c

Step 1. b is the first choice of 9 voters, c is the first choice of 6 voters, d is the first choice of 5 voters, and e is the first choice of 7 voters. There is no majority

17 1.2 · Elimination Procedures

candidate. Further, c is the last choice of 8 voters, e is the last choice of 19 voters, b is the last choice of 0 voters, and d is the last choice of 0 voters. We go to Step 2. Step 2. Eliminate e since he has the highest number of last choice votes. Here are the updated ballots: 8

7

6

5

b

d

c

d

1 b

d

b

b

b

d

c

c

d

c

c

Step 1. b is the first choice of 9 voters, c is the first choice of 6 voters, and d is the first choice of 12 voters. No candidate has a majority yet. Further, b is the last choice of 0 voters, c is the last choice of 21 voters, and d is the last choice of 6 voters. We go to Step 2. Step 2. Eliminate c since he has the highest number of last choice votes. Here are the updated ballots: 8

7

6

5

1

b

d

b

d

b

d

b

d

b

d

Step 1. b is the first choice of 15 voters and d is the first choice of 12 voters. b is the sole winner of Coombs’ procedure.   Note 1.2.12 In elimination procedures, it is important to stop the process as soon as a winner emerges (or winners emerge). If we do not realize that candidate a has already acquired a majority and should be declared the Coombs procedure winner in the following example, we will have to eliminate a because he has the largest number of last choice votes; this would lead to a wrong result. (The point here is that it is possible for a candidate to have, simultaneously, a majority of the first choice vote and a plurality of the last choice vote. It is impossible to have a majority of both, of course.) 3

2

2

2

a

a

b

c

c

b

c

b

b

c

a

a

Note that the Hare procedure is not affected by mistakenly continuing the process. Can you verify this?   Note 1.2.13 All of the four social choice procedures discussed so far are reasonable and are founded on sensible ideas, yet they may produce different results as you have noticed. (In Examples 1.2.3, 1.2.6, and 1.2.9 the four procedures delivered four different winners for the same data.) Do some of these procedures pick the “right” winner and others do not?

1

18

Chapter 1 • Social Choice

1 This question presumes that we already have a correct way of identifying the “right” winner. If this were the case, then there would be no need for those procedures. The point we are making here is that social choice (made by several individual voters) is far more complex than individual choice, and is plagued with anomalies and paradoxes to be studied in this chapter. In fact, all social choice procedures (including those presented above) are mere imperfect attempts to capture that elusive concept of “the choice of the society.” We shall study more social choice procedures in the rest of this chapter, after we take a critical look into the procedures presented above.  

1.2.4 Monotonicity In general, we say that a process is monotone if an increase in the input makes the output increase or stay the same. Studying ought to be a monotone process because studying harder should not lead to a lower grade. A healthy car engine is also monotone because the car does not slow down when you apply more gas. If your grades drop when you study harder, then there is something wrong with the way you study. If your car slows down when you apply more gas, you should take it to the shop because it is obviously malfunctioning. The following two examples show that the above three elimination procedures are all non-monotone in the sense that (albeit rarely) improving your positions in the preference ballots can turn you from a winner into a loser when those elimination procedures are used. Example 1.2.14 Twenty one voters are using the Hare procedure to choose from the available three candidates a, b, and c. Process the following two possible voting scenarios and comment on the results.

Scenario 1:

8 a b c

5 b c a

2 b a c

6 c a b

Scenario 2:

8 a b c

5 b c a

2 a b c

6 c a b

=

10 a b c

5 b c a

6 c a b



Solution In Scenario 1, a is the first choice of 8 voters, b is the first choice is 7 voters, and c is the first choice of 6 voters. To qualify as a majority candidate, one needs at least 11 first choice votes. Eliminate c. The updated ballots are 8

5

2

6

a

b

b

a

b

a

a

b

and a emerges as the winner. Before we process Scenario 2, we note that the only difference between the two scenarios is in the way the two voters in the third column rank a and b. In Scenario 1, b is the first choice of these two voters and a is their second choice. In Scenario 2, a is the first choice of these two voters and b is their second choice. This slight difference makes Scenario 2 more favorable to a than Scenario 1. (Think of the improvement of a’s position in Scenario 2 as

1

19 1.2 · Elimination Procedures

being the result of working harder on his campaign.) Since a is the winner in Scenario 1, one would expect a to win in “the more favorable” Scenario 2. Let us now process Scenario 2: a is the first choice of 10 voters, b is the first choice of 5 voters, and c is the first choice of 6 voters. Since there is no majority candidate, we eliminate b. The updated ballots are 8

5

2

6

a

c

a

c

c

a

c

a

and c (not the expected a) emerges as the winner. This is certainly quite ironic and paradoxical. (Think of a car that slows down when the gas pedal is pushed harder!)  

Note The plurality with run-off procedure coincides with the Hare procedure in this example. Therefore, the above comment would still apply had we used the plurality with run-off procedure instead. Example 1.2.15 Twenty one voters are using Coombs’ procedure to choose from the available three candidates a, b, and c. Process the following two possible voting scenarios and comment on the results.

Scenario 1:

8 a b c

7 c a b

2 c b a

4 b c a

Scenario 2:

8 a b c

7 c a b

2 c a b

4 b = c a

8 a b c

9 c a b

4 b c a



Solution In Scenario 1, a is the first choice of 8 voters, b is the first choice is 4 voters, and c is the first choice of 9 voters. To qualify as a majority voter, one needs at least 11 first choice votes. a is the last choice of 6 voters, b is the last choice of 7 voters, and c is the last choice of 8 voters. We eliminate c. The updated ballots are 8

7

2

4

a

a

b

b

b

b

a

a

and a emerges as the winner. In Scenario 2, a is the first choice of 8 voters, b is the first choice of 4 voters, and c is the first choice of 9 voters. There is no majority candidate. a is the last choice of 4 voters, b is the last choice of 9 voters, and c is the last choice of 8 voters. We eliminate b. The updated ballots are 8

7

2

4

a

c

c

c

c

a

a

a

20

Chapter 1 • Social Choice

1 and c emerges as the winner. The only difference between the two scenarios is in the way the two voters in the third column rank a and b. In Scenario 1, b is the second choice of these two voters and a is their third choice. In Scenario 2, a is the second choice of these two voters and b is their third choice. This slight difference makes Scenario 2 more favorable to a than Scenario 1, yet a loses in Scenario 2 and wins in Scenario 1. This is similar to what happened in Example 1.2.14.  

The paradoxical and flawed behavior exhibited by some of the social choice procedures in Examples 1.2.14 and 1.2.15 deserves further consideration. In the next definition we give a special name to the social choice procedures that do not suffer from this flaw. Definition 1.2.16 A social choice procedure is said to be monotone if a winner (i.e., a member of the social choice set) x always maintains his membership in the social choice set when some voters change their preference ballots in his favor by switching him with a specific candidate y who was originally placed immediately above x in their preference ballots, and all else stays the same. If the definition is modified so that the change is made by only one voter, we say that the social choice procedure is singly monotone.  

Examples 1.2.14 and 1.2.15 show that none of the procedures of plurality with runoff, Hare and Coombs is monotone. (They are non-monotone.) The student should not conclude from this that those elimination procedures should be abandoned. In fact, in most cases, a change of the kind described in the above definition will not turn a winner into a loser. Further, as we shall see, every social choice procedure has its own flaws. This will be studied in more detail in the course of the present chapter. We stated in the beginning of this chapter that the elimination procedures were introduced as reforms to the plurality procedures. Ironically, the plurality procedure outperforms the elimination procedures when it comes to monotonicity: The plurality procedure is, in fact, monotone. (You will be asked to prove this fact in Exercise 14 below.) This irony may be explained if we think of monotonicity as a sign of consistency and so the plurality procedure is monotone in the sense of being consistent, regardless of the quality of its choices.

1.2.5 Dealing with Non-monotonicity A lesson to be learned from the above discussion is that, if you are a candidate for office in a situation where an elimination procedure is used, you need to make some difficult calculations. For example, if the Hare procedure is used and you are not sure of receiving a majority at some round, you need to reconcile the following two events:

21 1.3 · Condorcet Ideas and Related Procedures

1. Receiving enough first choice votes to proceed to the next round. 2. Getting rid of “dangerous” opponents who can can beat you even if this requires conceding some first choice votes to another opponent. For example, it would be wise for a in Example 1.2.14 to concede to b those two top choice votes to allow b to proceed to the next round instead of the “dangerous” c, as in Scenario 1. We shall revisit this issue in the next section.

1.3

Condorcet Ideas and Related Procedures

Consider the case of only two alternatives x and y. There are exactly two possible events: 1. One of the two alternatives, say x, receives a majority of the vote. 2. The two alternates receive equal numbers of votes. The two-alternative majority rule is the social choice procedure for two alternatives that declares x the winner in the first event and declares a tie in the second. (For simplicity, we may say, in the first case, that x beats y in the one-on-one vote or x beats y one-on-one.) The following principle is basic to all social choice procedures: When applied to the case of only two alternatives, all of the social choice procedures that are designed for the several-alternative case must agree with the two-alternative majority rule. It is therefore possible to view those procedures as attempts to extend the majority rule from the two-alternative case to the general case of several alternatives. Before considering any proposed procedure for social choice, one must first verify that it satisfies this basic principle. The student can easily verify this for the social choice procedures studied here such as plurality, plurality with run off, Hare, Coombs, and Borda count ( Sect. 1.4).

1.3.1 The Condorcet Tournament Marquis de Condorcet, an eighteenth century French philosopher, mathematician, and political scientist (and a close friend of Thomas Jefferson when the latter was the United States ambassador to France) sought to extend the majority rule from the two-alternative case to the case of several alternatives in a different direction. His approach was to take the one-on-one vote for each pair of alternatives and use the two-alternative majority rule to determine the result of each vote. We shall call this process the Condorcet tournament (instead of its more common name: the Round-Robin tournament). To motivate the discussion, we present the following example. Example 1.3.1 The following are four different scenarios for the voting data of nine voters and four candidates a, b, c, and d.

1

22

Chapter 1 • Social Choice

1

Scenario 1:

4 a c b d

3 b c d a

2 d c a b

Scenario 3:

4 a c b d

3 b c d a

2 d c b a

Scenario 2:

4 d b c a

3 a d c b

2 b a c d

Scenario 4:

4 b a d c

3 c b d a

2 a d c b

Let us run the Condorcet tournament for each of the above scenarios. Since each vote involves only two candidates and the number of voters in our example is odd, there will be a sole winner in each of the one-on-one votes (no ties). We have six one-on-one votes to take in each scenario: a vs b, a vs c, a vs d, b vs c, b vs d, and c vs d. We conduct these one-on-one votes in Scenario 1 and leave the rest to the student. 

Scenario 1: • a beats b 6 to 3 since 6 voters rank a higher than b while 3 voters rank b higher than a. • c beats a 5 to 4 since 5 voters rank c higher than a while 4 voters rank a higher than c. • d beats a 5 to 4 since 5 voters rank d higher than a while 4 voters rank a higher than d. • c beats b 6 to 3 since 6 voters rank c higher than b while 3 voters rank b higher than c. • b beats d 7 to 2 since 7 voters rank b higher than d while 2 voters rank d higher than b. • c beats d 7 to 2 since 7 voters rank c higher than d while 2 voters rank d higher than c. In Graph theory (a branch of mathematics), a finite set of vertices some or all of its pairs are joined by unidirectional or bidirectional arrows is called a directed graph. A directed graph is said to be complete if each pair of its vertices is joined by an arrow (unidirectional or bidirectional). It will be useful to represent the Condorcet tournament of a given voting scenario by a complete directed graph in which each candidate is represented by a vertex and the result of a two-alternative majority vote is indicated by an arrow directed from the winner to the loser or, in case of a tie, a bidirectional arrow. This representation of a Condorcet tournament will be referred to as the graph representation of the Condorcet tournament or simply the graph representation of the given voting scenario.

1

23 1.3 · Condorcet Ideas and Related Procedures

x beats y: x −→ y. x and y are tied: x ←→ y. Before we continue the discussion of Example 1.3.1 we define two important concepts.

1.3.2 Condorcet Winner and Condorcet Loser Definition 1.3.2 A winner of the Condorcet tournament, called a Condorcet winner, is a candidate who beats all her opponents in the Condorcet tournament. A Condorcet loser is a candidate who loses to all her opponents in the Condorcet tournament.  

The graph representations for the four scenarios above are shown in ⊡ Fig. 1.2 below. In the graph representation of a Condorcet tournament, a Condorcet winner “shoots” arrows toward all his opponents and a Condorcet loser “receives” arrows from all his opponents. In any particular scenario of voting, one and only one of the following four cases is possible: • A Condorcet winner exists but a Condorcet loser does not. (Scenario 1 in Example 1.3.1.)

a

d

a

b

Scenario 1 Condorcet winner: c Condorcet loser: None c

d Scenario 2 Condorcet winner: None Condorcet loser: c

a

d Scenario 3 Condorcet winner: c Condorcet loser: a

b

c a

b

c

⊡ Fig. 1.2 Graph representations for Example 1.3.1

d Scenario 4 Condorcet winner: None Condorcet loser: None

b

c

24

Chapter 1 • Social Choice

1 • A Condorcet loser exists but a Condorcet winner does not. (Scenario 2 in Example 1.3.1.) • A Condorcet winner and a Condorcet loser both exist. (Scenario 3 in Example 1.3.1.) • Neither a Condorcet winner nor a Condorcet loser exists. (Scenario 4 in Example 1.3.1.)   The above example shows that the Condorcet approach to extend the two-alternative majority rule to the more general case of several candidates does not always succeed in finding a winner (which would be the Condorcet winner defined above).

1.3.3 Condorcet Winner vs Majority Candidate ⓘ Remark 1.3.3 It is important to note that 1. There is at most one Condorcet winner in any voting scenario. (The same holds for a majority candidate as noted earlier.)   2. There is at most one Condorcet loser in any voting scenario.   3. The election of a Condorcet winner would be a very stable and desirable outcome in any social choice situation because she cannot be challenged by any individual opponent. Since the existence of a Condorcet winner is not guaranteed, we cannot regard the Condorcet tournament as a social choice procedure and we have to resort to procedures that are guaranteed to deliver a winner (or tied winners) such as plurality, Hare, Coombs, etc. The problem is further exacerbated by the fact that those procedures do not necessarily pick the Condorcet winner when it exists!   4. The election of a Condorcet loser would be the worst possible outcome since he loses one-on-one to each of his opponents.   5. A majority candidate is necessarily a Condorcet winner. In fact, being the first choice of a majority of voters enables a majority candidate to beat each opponent in a majority of the ballots. However, a Condorcet winner need not be a majority candidate. (In Scenarios 1 and 3 (Example 1.3.1), c is a Condorcet winner without being a majority candidate.) To explain the difference between a majority candidate and a Condorcet winner, let a be a majority candidate and let b be a Condorcet winner (in two different scenarios, of course). • There is one and the same majority set of voters, call it S, who prefer a over all his opponents. The set of voters who place a on the top of their preference ballots is such a set. • For each opponent x of b there exists a majority set Sx of voters who prefer b over x. The sets Sx can be different and the set of all voters who prefer b over all opponents (which is the set of all voters who place b on the top of their ballots) need not be a majority set and may even be empty.

1

25 1.3 · Condorcet Ideas and Related Procedures

⊡ Fig. 1.3 The condorcet paradox

a

c

b

Consider, for example, the following preference ballots submitted by five individual voters. aa c bb b caa c c bb ca Clearly a is a Condorcet winner who is not a majority candidate. If we name the voters v, w, x, y, z from left to right, then the majority set {v, w, x} prefers a over b and the majority set {v, w, y} prefers a over c but the set of voters who prefer a over all of his opponents is {v, w} which is not a majority set.   6. Note that in Scenario 1 (Example 1.3.1), a beats b and b beats d, but instead of a beating d (as intuition would have lead us to conclude), d does beat a! This is indicated in the graph by the 3-cycle formed by the three arrows from a to b, from b to d, and from d to a. There are no 3-cycles in Scenario 3. Scenarios 2 & 4 have 3-cycles that should be identified by the student. The simplest form of these cycles was discovered by Condorcet in his example with three voters and three candidates a, b, and c, known as the Condorcet Paradox (⊡ Fig. 1.3). ab c b ca cab

Note that a beats b, b beats c, and c beats a. This will be studied further in  Sect. 1.5.   The student should find the plurality procedure winner, the Hare winner, and the Coombs winner in each of the four scenarios in Example 1.3.1 and verify the information in the following table. Scenario

Condorcet W

Condorcet L

Plurality W

Hare W

Coombs W

1

c

None

a

a

c

2

None

c

d

a

d

3

c

a

a

b

c

4

None

None

d

a

d

26

Chapter 1 • Social Choice

1 Again, it is highly undesirable for any social choice procedure to 1. Fail to pick a Condorcet winner when one exists (e.g., plurality and Hare in Scenario 1 and Scenario 3). 2. Pick a Condorcet loser (e.g, plurality in Scenario 3). This motivates the following definitions.

1.3.4 The Condorcet Criteria Definition 1.3.4 1. A social choice procedure is said to satisfy the Condorcet winner criterion if it always makes the Condorcet winner, when one exists, the sole member of its social choice set.   2. A social choice procedure is said to satisfy the Condorcet loser criterion if it never makes a Condorcet loser a member of its social choice set.  

The plurality and Hare procedures both failed to elect an existing Condorcet winner in Scenarios 1 and 3 of Example 1.3.1. Therefore, neither satisfies the Condorcet winner criterion. The plurality procedure does not satisfy the Condorcet loser criterion since it elected a Condorcet loser in Scenario 3. The following example shows that the Coombs procedure does not satisfy the Condorcet winner criterion. Example 1.3.5 The following are the grouped preference ballots for 21 voters and 5 candidates a, b, c, d, and e. 3

3

4

5

6

b

c

d

e

b

e

e

e

b

a

c

d

b

a

c

d

a

a

c

d

a

b

c

d

e

Show that 1. e is a Condorcet winner. 2. e does not win the Coombs procedure. 

27 1.3 · Condorcet Ideas and Related Procedures

Solution 1. e beats a 15 to 6; e beats b 12 to 9; e beats c 12 to 9, and e beats d 11 to 10. Therefore, e is a Condorcet winner. 2. A quick examination of the preference ballots shows that no majority candidate exists and that the last choice votes are not equally distributed among the candidates, therefore e must be eliminated since he has the highest number of last choice votes.  

The following two examples show that neither Hare nor Coombs satisfies the Condorcet loser criterion. Example 1.3.6 Consider the following data for 7 voters and 3 candidates a, b, and c. 3

2

a

b

2 c

b

c

b

c

a

a

There is no majority candidate. The Condorcet loser a wins the Hare procedure after eliminating b and c simultaneously because they both have the lowest first choice vote.  Example 1.3.7 Consider the following data for 11 voters and 3 candidates a, b, and c. 1

1

2

3

4

a

b

c

c

b

c

c

b

a

a

b

a

a

b

c

There is no majority candidate. The Condorcet loser a wins the Coombs procedure after eliminating b and c simultaneously because they both have the highest last choice vote.  Note 1.3.8 In Examples 1.3.6 and 1.3.7, Hare and Coombs elected a Condorcet loser, as a sole winner, after eliminating more than one opponent simultaneously. This is the only way a Condorcet loser can be the sole winner in either of these two procedures. If we modify the Hare and Coombs procedures so that only one candidate is eliminated at each elimination step, a Condorcet loser could never be the sole winner (but could still be a member of the social choice set). What do we do if two or more candidates “qualify” for elimination at one step? A rule to determine the eliminated candidate should be decided upon prior to voting. This will be addressed in Exercise 15 below.  

1

28

Chapter 1 • Social Choice

1 1.3.5 Agenda Voting This social choice procedure is not neutral (i.e., does not treat the available alternatives equally) and, therefore, cannot be used when the alternatives are candidates for office. It is mainly used by parliaments and committees when the alternatives to choose from are two or more policies. This is how it works: First, the available alternatives are listed in a certain order called the agenda. This may be done by the committee chair or imposed by the bylaws of the voting body. If the alternatives are a, b, c, d, and e, then bcead (in this specific order) is a possible agenda. Vote is taken, using a modified version of the two-alternative majority rule (that rules out ties), in the following sequence: 1. The first vote is taken on the pair b and c. 2. The winner in the first vote goes against e in the second vote. 3. The winner in the second vote goes against a in the third vote. 4. In the fourth vote, the winner in the third vote goes against d to determine the final winner. Should the two-alternative majority rule produce a tie in any of the above votes, between the ith alternative and the j th alternative in the agenda, where i < j , it will be broken in favor of the j th alternative. For example, if c beats b in the first vote above and beats e in the second vote then ties with a in the third vote, a will be considered the winner of the third vote and will proceed to face d in the fourth and final vote. Consequently, agenda voting is always won by a single alternative. In the case of only two alternatives a and b, the agenda ba, for example, means that b wins if and only if supporters of b > supporters of a; and a wins if and only if supporters of a ≥ supporters of b. This is typical of a situation in which b is a motion and a is the status quo. Of course, this added tie breaker sets the agenda voting for two alternatives apart from the two-alternative majority rule. Example 1.3.9 A committee of nine voters is voting on the alternatives a, b, c, and d. The grouped preference ballots are: 4

3

a

c

2 b

d

b

a

c

d

d

b

a

c

Find the winning alternative if the agenda is: 1. cbad 2. cabd 3. dabc 4. abcd. 

1

29 1.3 · Condorcet Ideas and Related Procedures

⊡ Fig. 1.4 Agenda cbad

c

a

b

a

c d

b

c

d

a a

d

a Winner

Solution It is very helpful to draw the graph representation of the Condorcet tournament first. We may then get the results of the one-on-one votes from the graph representation. 1. c beats b in the first vote then loses to a in the second vote. a beats d in the third and final vote. a is the winner. The voting process according to this agenda is shown to the right of the graph representation.   2. a beats c in the first vote then loses to b in the second vote. b beats d in the third and final vote. b is the winner.   3. a beats d in the first vote then loses to b in the second vote. c beats b in the third and final   vote. c is the winner. 4. b beats a in the first vote then loses to c in the second vote. d beats c in the third and final vote. d is the winner.  

ⓘ Remark 1.3.10 It is clear from Example 1.3.9 that the agenda can play a very

1. 2.

3. 4.

important role in picking the winner when agenda voting is implemented. You should also be able to verify the following facts about agenda voting. The left-most alternative in an agenda wins if and only if it is a Condorcet winner.   A Condorcet winner wins agenda voting regardless of the specific agenda used. Therefore, agenda voting satisfies the Condorcet winner criterion. It is also true that an alternative x that always wins agenda voting regardless of the agenda used must be a Condorcet winner. (Consider an agenda that places x in the left-most position.)   A Condorcet loser loses agenda voting regardless of the specific agenda. Therefore, agenda voting satisfies the Condorcet loser criterion.   An alternative that always loses agenda voting irrespective of the agenda need not be a Condorcet loser. An obvious example is when a Condorcet winner exists because the Condorcet winner will win regardless of the agenda, leaving no saving agenda for any of the competing alternatives. The following scenario for three voters and five alternatives demonstrates the validity of this claim even in the absence of a Condorcet winner. a

b

c

b

c

e

d

d

a

e

a

d

c

e

b

30

Chapter 1 • Social Choice

1 ⊡ Fig. 1.5 Graph representation for Example 1.3.11

a

d

b

c

The student should be able to see from the graph representation of this scenario that no Condorcet winner exists and that d is not a Condorcet loser, but no agenda can make d win.   5. Agenda voting is monotone. You will be asked to prove this fact in Exercise 21 below.   Example 1.3.11 A committee of fifteen voters is using agenda voting to pick one out of the four alternatives a, b, c, and d. Find an agenda for each of the alternatives to win, whenever possible. 5

4

3

2

1

a

c

d

d

c

c

b

a

c

d

d

d

b

b

a

b

a

c

a

b

 Solution One simple way, thought not the only way, to construct an agenda for a specific alternative to win (if this is at all possible) is to place this alternative at the end of the agenda and proceed backward, following the arrows on the graph representation, until each alternative has been visited exactly once. See ⊡ Fig. 1.5 below. Alternative a wins if the agenda is bdca. No agenda can make alternative b win, since it is a Condorcet loser. Alternative c wins if the agenda is badc. Alternative d wins if the agenda is bcad.

 

1.3.6 Weak Pareto Efficiency One criterion that agenda voting fails to satisfy is the Weak Pareto efficiency, named after the nineteenth to twentieth century Italian economist Vilfredo Pareto.

1

31 1.3 · Condorcet Ideas and Related Procedures

Definition 1.3.12 1. An alternative x is said to be Pareto superior to an alternative y (and y is Pareto inferior to x) if x is ranked higher than y by each voter, i.e., if x is unanimously preferred over y.   2. An alternative is said to be Pareto dominated if it is Pareto inferior to some other alternative.   3. A social choice procedure is said to be Weakly Pareto efficient if it never admits a Pareto dominated alternative into its social choice set.   The following example shows that agenda voting is not Weakly Pareto efficient.

Example 1.3.13 Three voters are picking one out of the four alternatives a, b, c, and d. Show that d is a Pareto dominated alternative and find an agenda that makes d win. a

b

d

c

c a

b

a

d

c

d

b

 Solution Alternative d is Pareto dominated since it is Pareto inferior to a. We indicate this on the graph representation of the Condorcet tournament by a double-headed arrow from a to d instead of the usual single-headed arrow.   You can easily verify that alternative d wins if the agenda is acbd (⊡ Fig. 1.6).

ⓘ Remark 1.3.14 It is important to note the difference between a Condorcet loser alternative and a Pareto dominated one. To be a Condorcet loser, an alternative must lose, by a majority, to each opponent. To be Pareto dominated, an alternative needs to lose, unanimously, to just one opponent. Alternative d in Example 1.3.13 is Pareto dominated without being a Condorcet loser. Alternative c in Example 1.3.1 (Scenario 2) is a Condorcet loser without being Pareto dominated. • We saw in Remarks 1.3.10 that agenda voting satisfies the Condorcet loser criterion. Example 1.3.13, on the other hand, shows that agenda voting is not Weakly Pareto efficient.   • The plurality procedure is Weakly Pareto efficient but does not satisfy the Condorcet loser criterion. These two facts will be addressed in Exercises 16 and 17 below.  

• • • •

32

Chapter 1 • Social Choice

1 ⊡ Fig. 1.6 Graph representation of Example 1.3.13

a

d

b

c

1.3.7 Killer Amendment When a motion is made in a voting body, such as a committee or a parliament, the voting body finds itself in a position of having to choose between two alternatives: the motion b and the status quo a. Unless an amendment to the motion is made, an agenda vote is taken according to the agenda ba to choose one of the two alternatives a and b. If an amendment to the motion is proposed, a new situation arises in which the voting body has to use agenda voting, with agenda cba, to choose one of the three alternatives a, b, and c, where c is the amended motion. If the amended motion c is further amended to get a twice-amended motion d, an agenda vote is taken to choose one of the four alternatives a, b, c, and d according to the agenda dcba. In general, if several amendments are made, an agenda vote is taken according to an agenda in which a later amendment comes before an earlier one. In other words, the agenda in this case is a listing of all alternatives in a chronological order in which the latest amendment is placed in the left-most position and the status quo is placed in the right-most position. This agenda is usually included in the parliamentary bylaws. The Condorcet paradox is sometimes utilized to manipulate the agenda voting processes for the purpose of defeating (or killing) a motion that would have otherwise been successful. Consider, for example, a situation in which a committee of eleven voters is to vote on a motion b versus the status quo a. Two of the eleven voters are in favor of the status quo while the remaining nine voters are in favor of the motion. Therefore, the grouped preference ballots are: 2

9

a

b

b

a

Clearly, if a vote is immediately taken, the status quo a will be defeated. The two pro status quo voters can prevail if they introduce a killer amendment (aka poison pill amendment) to the motion b so that the amended motion c would defeat b in the first vote, then lose to the status quo a in the second vote; thus forming a 3-cycle, or a Condorcet paradox. (See ⊡ Fig. 1.7 below.) To successfully create a killer amendment, the pro status quo voters should design it so that it divides the pro motion voters into two equal (or close to equal) groups. Members of one group like the amended motion c to the

1

33 1.3 · Condorcet Ideas and Related Procedures

⊡ Fig. 1.7 Killer Amendment

a Status quo: a Motion: b Killer Amendment: c

c

b

point of preferring it over the motion b, while members of the other group dislike c to the point of placing it below the status quo a. The pro status quo voters themselves should place the amended motion c in the middle, between a and b. The grouped preference ballots (with the amended motion c included), are: 2

4

5

a

c

b

c

b

a

b

a

c

According to the standard parliamentary procedure, the agenda is cba. In the first vote, c defeats b by a 6 to 5 vote. In the second vote, a defeats c by a 7 to 4 vote and wins the entire voting procedure. The status quo a survives against the wishes of a majority of the voters! Example 1.3.15 Five students meet in the study lounge in the dorms every evening to study and they order a vegetarian pizza to eat during the break. One evening, while they were studying, a member of the group made a motion to add pepperoni to the pizza. Discussions revealed that only one member of the group wanted to keep the status quo (vegetarian pizza) while the remaining four members were in favor of the motion (adding pepperoni). Knowing that two of the four pro motion members loved hot pepper while the other two strongly disliked it, the pro status quo member amended the motion so as to add hot pepper with the pepperoni. The following three alternatives ended up being on the table: The status quo a: Vegetarian pizza. The motion b: Vegetarian pizza with pepperoni. The amended motion c: Vegetarian pizza with pepperoni and hot pepper. The five students agreed to apply the standard parliamentary procedure of agenda voting with the agenda cba as they learned in class. The grouped preference ballots are: 1

2

2

a

c

b

c

b

a

b

a

c

34

Chapter 1 • Social Choice

1 In the first vote of c vs b, c defeats b by a 3 to 2 vote. In the second vote of c vs a, a defeats c by a 3 to 2 vote. The voting process is now complete with the status quo a being the winner, despite the fact that four out of the five voters were in favor of the motion! 

It is important to note that the killer amendment trick does not work by convincing any of the pro-motion voters that the status quo is better. In fact, the relative ranking by each voter of the status quo a and the motion b is the same before and after the introduction of the killer amendment c. Killer amendments are not easy to design and, in fact, many of them fail. Example 1.3.16 In a committee of 23 voters, 17 voters support a motion b and 6 voters want to maintain the status quo a. The pro status quo voters propose an amended motion c to kill the motion and keep the status quo. Assuming that all the pro status quo voters will rank c between a and b, will the amended motion c succeed in keeping the status quo if: 1. 6 of the 17 pro motion voters rank c as their top choice and 11 of them rank it as their last choice. 2. 11 of the 17 pro motion voters rank c as their top choice and 6 of them rank it as their last choice. 3. 5 of the 17 pro motion voters rank c as their top choice and 12 of them rank it as their last choice. 4. 12 of the 17 pro motion voters rank c as their top choice and 5 of them rank it as their last choice. 

(Of course, the relative ranking of a and b by each voter will remain unchanged.) Solution The preferences in the four cases above are as follows: 6 a 1. c b

6 c b a

11 b a c

6 a 2. c b

11 c b a

6 b a c

6 a 3. c b

5 c b a

12 b a c

6 a 4. c b

12 c b a

5 b a c

The graph representations of the Condorcet tournaments are as follows: a

a

b c

c 1

a

b c 2

a

bc 3

b 4

As you can easily see, the killer amendment succeeds in saving the status quo in the first and second scenarios. In the third and fourth scenarios, the amended motion c fails to

35 1.3 · Condorcet Ideas and Related Procedures

accomplish the goal of saving the status quo. In the third scenario, too many of the pro motion voters ranked c as their last choice and the motion b ended up being a Condorcet winner while, in the fourth scenario, too many of the pro motion voters ranked c as their top choice which made c a Condorcet winner. (In the fourth scenario, the killer turned “monster!”)   Example 1.3.17 (1.3.16 Generalized) Let us generalize the results of Example 1.3.16. Assume m voters are in support of the status quo a while n voters are in support of the motion b, with m < n. Assume also that • The pro status quo voters all rank their amended motion c between a and b. • n1 of the pro motion voters rank c as their top choice. • n2 of the pro motion voters rank c as their last choice. (n1 + n2 = n.) The preferences are now as follows. m

n1

n2

a

c

b

c

b

a

b

a

c

For the killer amendment to accomplish the goal of saving the status quo a, the amended motion c must beat the motion b in the first vote then lose to a in the second vote, that is, m + n1 > n2 and m + n2 > n1 . It follows that, under the stated conditions, the killer amendment succeeds in keeping the status quo if and only if |n1 − n2 | < m. The student should be able to see how this result agrees with the intuition that larger support for the status quo makes it easier for the killer amendment to succeed, and vice versa. 

1.3.8 The Copeland Procedure This procedure is based on the Condorcet tournament. We simply run all the one-onone votes and get the results from the two-alternative majority rule. In each of these votes, the winner receives 1 point, the loser receives 0 points and, in case of a tie, each of the two alternatives receives half a point. The alternative(s) with the highest point sum wins (win). You can easily see that this procedure never fails to elect an existing Condorcet winner and never elects a Condorcet loser. In other words, it satisfies both the Condorcet winner criterion and the Condorcet loser criterion. The main problem with this procedure is indecisiveness. Ties are quite likely in the absence of a Condorcet winner.

1

36

Chapter 1 • Social Choice

1 Example 1.3.18 Process each of the four scenarios in Example 1.3.1 using the Copeland procedure.



Solution We use the graph representations in ⊡ Fig. 1.2. In Scenario 1: The Condorcet winner c has 3 points; and each of a, b, and d has one point. Therefore, c is the sole Copeland winner.   In Scenario 2: Each of a, b, and d has 2 points; and the Condorcet loser c has zero points. We therefore have a three-way tie involving a, b, and d.   In Scenario 3: The Condorcet winner c has 3 points, b has 2 points, d has 1 point, and the Condorcet loser a has zero points. Therefore, c is the sole Copeland winner.   In Scenario 4: Each of a and b has 2 points; and each of c and d has one point. We have a two-way tie involving a and b.   Example 1.3.19 Seventeen voters are using the Copeland procedure to pick a candidate for a position in a company. 5

7

3

2

a

c

b

e

b

d

d

b

c

e

e

c

d

b

a

a

e

a

c

d

Draw the graph representation and find the winner(s) (⊡ Fig. 1.8).



Solution a has zero points (Condorcet loser), b has 3 points, c has 3 points, d has 2 points, and e has 2 points. Candidates b and c are tied for winning. Can you suggest a tie breaker?  

1.3.9 Monotonicity Revisited ⓘ Remark 1.3.20 Let us view Example 1.2.14 from a different angle.

Scenario 1:

8 a b c

5 b c a

2 b a c

6 c a b

Scenario 2:

8 a b c

5 b c a

2 a b c

6 c a b

=

10 a b c

5 b c a

6 c a b

Of these two scenarios, we assume Scenario 2 to be the representation of the true and sincere opinions of the voters. Both scenarios have the same graph representation (⊡ Fig. 1.9) which is a Condorcet paradox.

1

37 1.3 · Condorcet Ideas and Related Procedures

⊡ Fig. 1.8 Graph representation of Example 1.3.19

a e

b d c ⊡ Fig. 1.9 Remark 1.3.20

a

c

b

As we saw in Example 1.2.14, a would not win the Hare (or plurality with run-off) procedure in scenario 2 (the sincere scenario). The reason, as seen from the graph representation of the Condorcet tournament, is that b (whom a can beat one-on one) is eliminated by Hare’s procedure and a has to face c (who defeats a one-on-one). This is obviously an undesirable outcome for the voters who rank a as their top choice. Can those voters get a better outcome for themselves by a manipulative trick? Yes they can. If two of a’s supporters change their preferences from a first and b second to b first and a second (which appears to hurt a’s chances), the graph representation remains the same but the preference ballots change to Scenario 1. Now c is eliminated in the first round; and a defeats b in the second round. As you can see, the voters whose top choice is a get a better outcome for themselves by seemingly acting against their own interest!

1.3.10 McGarvey’s Theorem The representation of a Condorcet tournament of a given voting scenario by a complete directed graph is quite useful and intuitive. It is natural to ask if any given complete directed graph represents the Condorcet tournament of some voting scenario. The answer, in the affirmative, was given [14] by David McGarvey. (See also [13].)

Theorem 1.3.21 (McGarvey 1953) Every complete directed graph represents a Condorcet tournament for some voting scenario.  

38

Chapter 1 • Social Choice

1 The idea of the proof is quite original and simple at the same time but the notations may be complicated. To simplify matters, we illustrate the proof in the following two examples of complete directed graphs with 5 vertices. The general case of n vertices follows exactly the same path. Example 1.3.22 In Example 1.3.19 above we constructed a complete directed graph (⊡ Fig. 1.8) for the given voting scenario of 17 voters and 5 candidates. Making the graph in ⊡ Fig. 1.8 our starting point, we construct a voting scenario whose Condorcet tournament is representable by this graph. 

The candidates and the voters in our scenario are introduced as follows: • For each vertex in the given complete directed graph we introduce a candidate with the same name as the vertex, and so we have the five candidates a, b, c, d, and e. • For each unidirectional arrow in the given graph we introduce two voters (with two preference ballots) in the manner we explain next. Let us start with the arrow pointing from b to a in ⊡ Fig. 1.8. On account of this arrow, we introduce two voters with preferences e b d a and c c b d a e Note that in these two preference ballots (combined) b beats a while all of the remaining nine pairs are tied. How did we accomplish this? • We placed b first and a second in the first preference ballot and we placed b second to last and a last in the second preference ballot. • We placed c, d, and e in an arbitrary order in the remaining spots in the first ballot and in the reverse order in the remaining spots in the second preference ballot. Similarly, we account for the arrow pointing from e to b, for example, by introducing two new voters with the following preference ballots: d e c b and a a e c b d

1

39 1.3 · Condorcet Ideas and Related Procedures

Graph representation for Example 1.3.23

a e

b d c

In these two preference ballots (combined) e beats b while all of the remaining nine pairs are tied. Continuing with this process of introducing two new voters for each unidirectional arrow we get the following voting scenario for five candidates and twenty voters, which is representable by the given complete directed graph. b→a b e a d c c d b e a

c→a c e a d b b d c e a

d→a d e a c b b c d e a

e→a e d a c b b c e d a

b→c b e c d a a d b e c

b→d b e d c a a c b e d

e→b e d b c a a c e d b

c→d c e d b a a b c e d

c→e c d e b a a b c d e

d→e d c e b a a b d c e

As explained above, the result of the vote on a certain pair of candidates is enforced only by the two preference ballots associated with that pair (the result of the remaining eighteen preference ballots is a tie for that particular pair).   Example 1.3.23 Construct a voting scenario whose Condorcet tournament is representable by the following complete directed graph. 

Solution The only difference between the present graph and the one in Example 1.3.22 is that the arrow between c and d, the arrow between c and e, and the arrow between d and e are all bidirectional in the present graph. For a bidirectional arrow we need not introduce any voters or, if we wish, we may introduce two voters whose two preference ballots are exact opposite of one another. Choosing the first option we get the scenario b→a b e a d c c d b e a

c→a c e a d b b d c e a

d→a d e a c b b c d e a

e→a e d a c b b c e d a

b→c b e c d a a d b e c

b→d b e d c a a c b e d

e→b c↔d c↔e d↔e e d b c a a c e d b

40

Chapter 1 • Social Choice

1 or b→a b e a d c c d b e a

c→a c e a d b b d c e a

d→a d e a c b b c d e a

e→a e d a c b b c e d a

b→c b e c d a a d b e c

b→d b e d c a a c b e d

e→b e d b c a a c e d b

McGarvey’s original proof treats the case of a complete directed graph with n vertices whose all arrows are unidirectional. The proof uses the construction explained in Example 1.3.22 but, as expected, the notations are considerably more complicated in the general case. Since (by Proposition 2.3.1 in the next chapter) the number of pairs of vertices, which is   the same as the number of arrows, is n2 and two voters are introduced for each unidirectional arrow we see that there are   n n(n − 1) 2· =2· = n(n − 1) 2 2·1 voters in the original proof of McGarvey’s theorem.

 

ⓘ Remark 1.3.24 1. The voting scenario constructed by McGarvey’s theorem for the complete directed graph in ⊡ Fig. 1.8 is obviously different from the voting scenario from which this graph originated in Example 1.3.19. This should not come as a surprise because any given complete directed graph represents an infinitude of voting scenarios but each given voting scenario is represented by only one complete directed graph.   2. The McGarvey voting scenario created in Example 1.3.22 is obviously not the most compact one as it involves a large number of voters whose preference ballots are all distinct. (Just compare McGarvey’s voting scenario with the voting scenario 5

7

3

2

a

c

b

e

b

d

d

b

c

e

e

c

d

b

a

a

e

a

c

d

in Example 1.3.19.) This is typical of “existence theorems” in mathematics, whose only purpose is to prove existence, leaving the refinements for future efforts.   3. McGarvey’s theorem can be quite useful in proving the existence of a voting scenario with certain features that depend only on its graph representation. In such case, one can proceed directly with the construction of the graph, without having to worry about the more difficult construction of the voting scenario since its existence is guaranteed by the theorem.  

1

41 1.4 · Scoring Procedures: Borda Count

1.4

Scoring Procedures: Borda Count

In this section, we study a different family of social choice procedure with main emphasis on the most important procedure in this family: the Borda count. A scoring procedure assigns to each candidate a score from each preference ballot, depending on her position in that ballot. This score is a real number such that 1. The higher the position in the preference ballot, the higher the score. Precisely, if the j th position is higher than the ith position, then the score received by the occupant of the j th position ≥ the score received by the occupant of the ith position. 2. For each i, the ith positions in all preference ballots are assigned the same score. 3. The score assigned to the top position must be greater than the score assigned to the bottom position. This condition guarantees that a scoring procedure agrees with the two-alternative majority rule in the case of only two alternatives. The total score of each candidate, from all ballots, is tallied up and the candidate(s) with the highest score wins (win).

1.4.1 Borda Count This procedure was proposed by Jean-Charles de Borda, a French mathematician and political scientist who was contemporary with Marquis de Condorcet. Here is how it works: The number of points a candidate receives from each preference ballot is equal to the number of opponents who are placed lower than him in that ballot. In other words, a candidate receives zero points from a preference ballot in which he is placed as last choice, one point from a preference ballot in which he is placed second to last choice, two points from a ballot in which he is placed third to last choice, and so on. This description of Borda count rules out indifference in the voters’ individual preference ballots; in  Sect. 1.6, we shall generalize it to the case where indifference is allowed. Example 1.4.1 Process the following scenario for seventeen voters and five alternatives using the Borda count. 5

7

3

4

a

c

b

e

3

b

d

d

b

2

c

e

e

c

1

d

b

a

a

0

e

a

c

d

(This is the same as the data in Example 1.3.19.)

2



42

Chapter 1 • Social Choice

1 Solution We added a column on the left indicating the Borda scores of different positions. Let us compute the total score of the candidates one candidate at a time. Total score of a = 5(4) + 7(0) + 3(1) + 2(1) = 25 points. Total score of b = 5(3) + 7(1) + 3(4) + 2(3) = 40 points. Total score of c = 5(2) + 7(4) + 3(0) + 2(2) = 42 points. Total score of d = 5(1) + 7(3) + 3(3) + 2(0) = 35 points. Total score of e = 5(0) + 7(2) + 3(2) + 2(4) = 28 points.  

It follows that c is the sole Borda count winner.

You may, if you wish, do the above calculations in the following table. 5

7

3

2

Total

a

× 4 = 20

× 0 = 00

× 1 = 03

× 1 = 02

= 25

b

× 3 = 15

× 1 = 07

× 4 = 12

× 3 = 06

= 40

c

× 2 = 10

× 4 = 28

× 0 = 00

× 2 = 04

= 42

d

× 1 = 05

× 3 = 21

× 3 = 09

× 0 = 00

= 35

e

× 0 = 00

× 2 = 14

× 2 = 06

× 4 = 08

= 28

Note The correct sum of all scores should be known to us ahead of time and can be used to check on our computations. In fact, each of the 17 voters contributes 0 + 1 + 2 + 3 + 4 = 10 to the “pool” of points. The total scores should therefore be 17 × 10 = 170 points. The sum of the scores calculated above is 25 + 40 + 42 + 35 + 28 = 170 which is the correct sum. This is a strong indication that our calculations are all correct unless we have made two (or more) mistakes that balanced one another! This is quite unlikely, of course.

Our next scoring procedure is one that you are very familiar with. Plurality Procedure The plurality procedure itself is a scoring procedure in which the first choice in a preference ballot receives one point and the rest of the candidates receive zero points each. The total score of a candidate will therefore be the number of voters who place her as first choice. It follows that the candidate(s) with the highest score is (are) the same as the plurality winner(s). Having found the plurality procedure to be a scoring procedure, it is natural to ask if any of the other procedures we studied is (or can be viewed as) a scoring procedure too. We shall have the answers shortly. Proposition 1.4.2 Scoring procedures are monotone. Proof Assume that candidate a is a winner in some given scenario when an arbitrary scoring procedure is used. If, in some ballots, a is switched with the candidate just above him (say b), then a’s score will increase or stay the same, while b’s score will decrease or stay the same;

43 1.4 · Scoring Procedures: Borda Count

and the scores of the rest of the candidates will stay the same. Therefore, a will continue to be a winner in the new scenario that resulted from the switch.  

Being a scoring procedure, the Borda count is monotone. ⓘ Corollary 1.4.3 None of the procedures of Hare, Coombs, and plurality with run-off is a scoring procedure. Proof Examples 1.2.14 and 1.2.15 demonstrate the non-monotonicity of the procedures named in the Corollary statement. We also see from Proposition 1.4.2 that scoring procedures are all monotone. From these two results, together, we see that none of the named procedures is a scoring procedure.  

Theorem 1.4.4 The Borda count satisfies the Condorcet loser criterion but fails the Condorcet winner criterion.

Proof The first assertion is proved in Theorem 1.4.11 below. As for the second assertion, we present the following example for seven voters and three alternatives: 4

3

2

c

b

1

b

a

0

a

c

Candidate c is clearly a Condorcet winner. (In fact, c is more than just a Condorcet winner; it is a majority candidate.) a receives 3 points. b receives 10 points. c receives 8 points. The above scores indicate that the Condorcet winner c is not a Borda count winner. (Note that the sum of scores is 21, which is the correct sum since each of the 7 preference ballots contributes 0 + 1 + 2 = 3 points.)  

1

44

Chapter 1 • Social Choice

1 1.4.2 Shortcomings of the Borda Count Procedure In addition to its failure to satisfy the Condorcet winner criterion (Theorem 1.4.4 above), the Borda count suffers from more serious issues, as shown by the following two examples. Example 1.4.5 A committee of seven voters is using Borda count to pick one out of the four candidates a, b, c, and d to fill a position in a company. The grouped preference ballots are as follows: 2

2

2

1

3

a

b

c

d

2

b

c

d

a

1

c

d

a

b

0

d

a

b

c

a receives 2(3) + 2(0) + 2(1) + 1(2) = 10 points. b receives 2(2) + 2(3) + 2(0) + 1(1) = 11 points. c receives 2(1) + 2(2) + 2(3) + 1(0) = 12 points. d receives 2(0) + 2(1) + 2(2) + 1(3) = 09 points. Minutes before declaring c the winner and sending him the job offer, the committee received news that candidate d, who finished last, had included forged documents in his application file and should have therefore been disqualified. Two opinions arose: Opinion 1. The disqualification of d should have no effect on the above result and c should still be the winner. Opinion 2. A fresh Borda count procedure for the three candidates a, b, and c should be conducted after excluding d. The preference ballots and the scores, after the exclusion of d, would be as follows: 2

2

2

2

a

b

c

1 a

1

b

c

a

b

0

c

a

b

c

a receives 2(2) + 2(0) + 2(1) + 1(2) = 8 points. b receives 2(1) + 2(2) + 2(0) + 1(1) = 7 points. c receives 2(0) + 2(1) + 2(2) + 1(0) = 6 points. This would lead to declaring a as the winner, not c. Opinion 2 is the more sound one as it prevents the undue influence of the irrelevant alternative d on the selection of a winner. If you look more thoroughly into the above results you would notice that • In the presence of d, Borda count would place c first, b second, and a third.

45 1.4 · Scoring Procedures: Borda Count

• In the absence of d, Borda count would place a first, b second, and c third, which is the exact reverse of the order of these three alternatives in the presence of d.  We advise the student to eliminate the candidates in Example 1.4.5, one at a time, and compare the Borda count ranking of the remaining three candidates to their Borda count ranking obtained above when all of the four candidates are present.  Example 1.4.6 Five voters are using Borda count to elect one candidate out of a, b, c, and d. The main contenders are a and b. The “honest” preference ballots are as follows: 2

3

3

a

b

2

b

a

1

c

c

0

d

d

a receives 12 points, b receives 13 points, c receives 5 point, and d receives 0 points and so b is the sole winner. This is the best possible outcome for the voters in the right column since b is their first choice, but it is not the best outcome for the voters in the left column since b is their second choice. The voters in the left column can get a better outcome for themselves if they vote strategically by placing c and d between their favorite candidate a and their less favored b so that the preference ballots become as follows: 2

3

3

a

b

2

c

a

1

d

c

0

b

d

The new scores are as follows: a receives 12 points, b receives 9 points, c receives 7 points, and d receives 2 points. Now a is the sole winner; and the two voters in the left column manage to get their favorite candidate elected by voting dishonestly! Next, suppose that the three voters in the right column also vote strategically and place c and d between their favorite candidate b and their less favored a. The preference ballots become 2

3

3

a

b

2

c

c

1

d

d

0

b

a

1

46

Chapter 1 • Social Choice

1 In this case, a receives 6 points, b receives 9 points, c receives 10 points, and d receives 5 points. Now c emerges as the sole winner and the two main contenders lose due to the “mutual destruction” by their supporters! 

For a possible real-life situation in which the scenarios described in the above example can happen, think of the five voters here as the voters in a miniature predominantly Republican district, candidates a, b as two Republican candidates and candidates c, d as Democrats. Naturally, a and b will be the top two choices of all voters, but some voters might have slight preference for a over b (the two voters in the left column) and others (the three voters in the right column) might have a slight preference for b over a. If each voter ranks the candidates according to his/her honest opinion, the Republican b wins. By misrepresenting their preferences as described above, the two voters who slightly prefer a over b can make a win, which is a better outcome for them. If all voters follow this insincere strategy, they end up with a Democrat representing their Republican district, which is a highly undesirable outcome for all voters. The above examples highlight the biggest flaw with Borda count, namely, its vulnerability to the effect of irrelevant alternatives. In Example 1.4.5, the presence of the irrelevant alternative d affected the societal ranking of the three other candidates. Example 1.4.6 goes further to show how irrelevant alternatives (c, d in this case) can be used to manipulate the Borda count procedure. A detailed study of this topic will follow in  Sect. 1.5.

1.4.3 More on Borda Count Theorem 1.4.4 above states that the Borda count procedure satisfies the Condorcet loser criterion. This simply means that a Condorcet loser cannot receive the highest score in the Borda count. The following includes a proof of this fact, together with a proof of the analogous fact that a Condorcet winner cannot obtain the lowest score in the Borda count. First we need to adopt some notations. Notation 1.4.7 In the sequel, we shall be dealing with n preference ballots numbered 1, 2, · · · , n; and k candidates named x, y, · · · , a, b, · · · . For ballot number i and candidates a, x we use #(a ↓ x, i) to denote the number of times a is placed higher than x in ballot number i which is, obviously, 0 or 1. More precisely, #(a ↓ x, i) = 1 if and only if a is placed higher than x in ballot number i and, #(a ↓ x, i) = 0 if and only if a is placed lower than x in ballot number i or x = a.

 

 It is clear that for a fixed a, the sum x #(a ↓ x, i) is the Borda score a receives from ballot number i and therefore the total Borda score of a (from all ballots) is given by the equation The Borda score of a

=

 i

x

#(a ↓ x, i) =

 x

i

#(a ↓ x, i).

1

47 1.4 · Scoring Procedures: Borda Count

 But, for a specific x, the sum i #(a ↓ x, i) is the number of ballots in which a is placed higher than x (the number of voters who prefer a to x). We denote this sum by #(a ↓ x) and we call it the score of a against x. We now have The Borda score of a

=

 i

#(a ↓ x, i) =



x

x

#(a ↓ x, i) =

i



#(a ↓ x).

x

(Note that by our notation, #(a ↓ a) = 0 so we do not have to rule out x = a in the above sum.) The last equation provides an alternative way to compute the Borda score of a candidate, say a: Simply add the scores of a against all of his opponents in the one-on-one votes. The sum of all these scores is the Borda score of a. (Compare this to Copeland procedure where we ignore the actual scores and give a one point for each win in the one-on-one votes, zero for each loss and half a point for each draw.) Proposition 1.4.8 In a society of n voters, if a one-on-one vote is taken on two candidates x and y then, in the two-alternative majority tule, x beats y if and only if #(x ↓ y) > n2 and x loses to y if and only if #(x ↓ y) < n2 Proof This follows easily since in a one-on-one vote, x beats y if and only if more than half of the voters place x higher than y and loses to it if less than half of the voters place x higher than y.  

ⓘ Corollary 1.4.9 In the case of n voters and k candidates: If c is a Condorcet winner, then The Borda score of c =



#(c ↓ x) >

x

n x

2

=

(k − 1)n . 2

=

(k − 1)n .   2

On the other hand, if c is a Condorcet loser, then The Borda score of c =

 x

#(c ↓ x)


(k − 1)n · 2

If, further, x has the lowest Borda score, then (in contradiction to Lemma 1.4.10) we get Sum of the Borda scores of all candidates

>

k(k − 1)n · 2

2. The proof of this part is similar to the proof of Part 1.  

1.4.4 Hare-Scoring Procedures The Hare-scoring procedure derived from a given scoring procedure consists of a finite sequence of applications (or rounds) of the given scoring procedure where, at each round, we eliminate the candidate(s) with the lowest score. We stop the process when all remaining candidates have the same score, and we declare those candidates to be tied winners. Viewing the plurality procedure as a scoring procedure as in Sect. 1.4.1 above (a candidate receives one point from a ballot if he occupies its top and receives zero points otherwise), the student should be able to verify that The Hare-plurality procedure is simply the Hare procedure studied in earlier sections of the present chapter. (Hence the term “Hare-scoring procedures.”) The Hare–Borda count procedure will simply be called the Hare–Borda procedure. Example 1.4.12 Process the following voting scenario for 21 voters and 5 candidates a, b, c, d, e using

1

49 1.4 · Scoring Procedures: Borda Count

3

3

4

5

4

b

c

d

e

6 b

3

e

e

e

b

a

2

c

d

b

a

c

1

d

a

a

c

d

0

a

b

c

d

e

1. The Borda count procedure. 2. The Hare–Borda procedure.  Solution 1. Borda count: Score of a = 3(0) + 3(1) + 4(1) + 5(2) + 6(3) = 35 points. Score of b = 3(4) + 3(0) + 4(2) + 5(3) + 6(4) = 59 points. Score of c = 3(2) + 3(4) + 4(0) + 5(1) + 6(2) = 35 points. Score of d = 3(1) + 3(2) + 4(4) + 5(0) + 6(1) = 31 points. Score of e = 3(3) + 3(3) + 4(3) + 5(4) + 6(0) = 50 points. The winner of Borda count is b. 2. Hare–Borda procedure: We eliminate d since he has the lowest Borda count in the first round and we update the ballots. Second round: 3

3

4

5

6

3

b

c

e

e

b

2

e

e

b

b

a

1

c

a

a

a

c

0

a

b

c

c

e

Score of a = 3(0) + 3(1) + 4(1) + 5(1) + 6(2) = 24 points. Score of b = 3(3) + 3(0) + 4(2) + 5(2) + 6(3) = 45 points. Score of c = 3(1) + 3(3) + 4(0) + 5(0) + 6(1) = 18 points. Score of e = 3(2) + 3(2) + 4(3) + 5(3) + 6(0) = 39 points.

We eliminate c since he has the lowest Borda count in the second round and we update the ballots.

50

Chapter 1 • Social Choice

1 Third round: 3

3

4

5

6

2

b

e

e

e

b

1

e

a

b

b

a

0

a

b

a

a

e

Score of a = 3(0) + 3(1) + 4(0) + 5(0) + 6(1) = 9 points. Score of b = 3(2) + 3(0) + 4(1) + 5(1) + 6(2) = 27 points. Score of e = 3(1) + 3(2) + 4(2) + 5(2) + 6(0) = 27 points. We eliminate a since he has the lowest Borda count in the third round and we update the ballots. Fourth round: 3

3

4

5

6

1

b

e

e

e

b

0

e

b

b

b

e

Score of b = 3(1) + 3(0) + 4(0) + 5(0) + 6(1) = 9 points. Score of e = 3(0) + 3(1) + 4(1) + 5(1) + 6(0) = 12 points. We eliminate b since he has the lowest Borda count in the fourth round and we are left with e as winner of the Hare–Borda procedure.  

1.4.5 Borda Meets Condorcet In the above example, e is a Condorcet winner that was overlooked by the Borda count procedure. However, the Hare–Borda procedure was able to detect and elect this Condorcet winner. This is no coincidence, as established by the following theorem.

Theorem 1.4.13 The Hare–Borda procedure satisfies the Condorcet winner criterion.

Proof If x is a Condorcet winner in the preference ballots submitted by the voters, she will continue to be a Condorcet winner at all rounds of the Hare–Borda procedure since the elimination of some candidates does not affect the relative rankings of any pair of the remaining candidates. From Theorem 1.4.11 we then see that x will not be eliminated at any round of the process and will be the sole winner of the Hare–Borda procedure.  

1

51 1.4 · Scoring Procedures: Borda Count

⊡ Fig. 1.10 Graph representation for Example 1.4.14

a e

b d c

The above discussion of the procedures of Copeland, Borda count, and Hare–Borda gives rise to the following two inquiries: 1. We proved in Theorem 1.4.11 that a Condorcet winner, which is also the sole Copeland procedure winner, cannot obtain the lowest score in Borda count. In the absence of a Condorcet winner, can a Copeland winner obtain the lowest score in Borda Count? 2. Since both of the Copeland and the Hare–Borda procedures pick the Condorcet winner (if it exists), it is natural to ask: Do these two procedures continue to pick the same winner(s) in the absence of a Condorcet winner? The following example addresses these two inquiries. Example 1.4.14 Thirteen voters are to pick one of the five candidates a, b, c, d, and e. Show that 1. There is no Condorcet winner (⊡ Fig. 1.10). 2. b is the Copeland procedure winner. 3. b has the lowest Borda count score. 4. The winners of the Hare–Borda procedure and the Copeland procedure are different. 2

5

6

4

b

c

e

3

a

b

a

2

d

d

d

1

e

e

c

0

c

a

b

 Solution 1 and 2. From the graph we see that there is no Condorcet winner and the Copeland scores are a

b

c

d

e

2

3

1

2

2

52

Chapter 1 • Social Choice

1 and so b is the sole Copeland procedure winner. 3. The Borda scores are:

 

a: 2(3) + 5(0) + 6(3) = 24 b: 2(4) + 5(3) + 6(0) = 23 c: 2(0) + 5(4) + 6(1) = 26 d: 2(2) + 5(2) + 6(2) = 26 e: 2(1) + 5(1) + 6(4) = 31 and so, the Copeland procedure winner, b, has the lowest Borda count score.   4. The Copeland procedure winner, b, is the first candidate to be eliminated in the Hare–Borda procedure, and therefore cannot win it. This is sufficient for our purposes, but if you wish to continue with the Hare–Borda procedure, you will find d to be the sole winner.  

1.5

A Glimpse Into Social Welfare Theory

In the earlier sections of this chapter we used social choice functions to aggregate the preference ballots provided by all voters in a given voting scenario and produce a non-empty set of winning alternatives (commonly known as the social choice set) that, presumably, is the collective choice of the society of voters. Examples include plurality, plurality with run-off, Hare, Coombs, Borda count, Copeland, and Hare–Borda. A procedure that aggregates the preference ballots provided by all voters in a given scenario and delivers a semi ranking (i.e., pairwise rankings of the alternatives) will be called a semi social welfare function. Given any two alternatives x and y, a semi social welfare function is able, after aggregating the preference ballots of the voters, to tell us whether the society prefers one over the other or is indifferent. A procedure that assigns a numerical value to each alternative can naturally compare x to y based on the numerical values assigned to them. Examples include the Copeland procedure and scoring procedures (such as plurality and Borda count). The Condorcet tournament is another example as it compares x to y using the two-alternative majority rule. If the pairwise rankings provided by a semi social welfare function fit together properly (in a sense that will be made clear soon), we obtain a ranking of the alternatives into first tier, second tier, third tier, and so on; and we speak of a social welfare function instead of a semi social welfare function. Examples include all procedures that assign a numerical value to each voter such as Copeland, Borda count, and plurality. (Those procedures can naturally list the alternatives in the decreasing order of their numerical values.) Every social welfare function is a semi social welfare function but the converse is not true.

53 1.5 · A Glimpse Into Social Welfare Theory

1.5.1 Rankings and Semi Rankings The following is a more precise definition of a ranking: Definition 1.5.1 Let A be a set of at least two alternatives. A ranking on A by a certain entity (which may be an individual voter or a society) is a partition of A into subsets A1 , A2 , · · · , An so that each alternative is in one and only one of these sets; and for each pair x, y of alternatives in A, • The entity thinks that each of these two alternatives is as good as the other if and only if x and y are in the same Ai . • The entity prefers y to x if and only if x is in some Ai and y is in some Aj with i < j. An An−1 . A ranking may be presented in a column: . . A2 A1 An is called the first tier (or top tier), An−1 is called the second tier, ..., and A1 is called the bottom tier.  

For example, if A = {q, r, s, t, u, v, w, x, y, z} is the set of alternatives, then {s, v} {q, u} {x} {r, w} {t, y, z} is a ranking with the first (or top) tier A5 = {s, v}, the second tier A4 = {q, u}, the third tier A3 = {x}, the fourth tier A2 = {r, w}, and the fifth (or bottom) tier A1 = {t, y, z}. To explain what it means for the pairwise rankings of a semi ranking to fit together properly (see the underlined phrase above), we examine the four scenarios in Example 1.3.1. We present the graph representations of their Condorcet tournaments one more time here for easy reference (⊡ Fig. 1.11). In Scenario 3, we can place the Condorcet winner c above all of its opponents (in the first tier) because it beats each one of them one-on-one. Next, b beats each of a and d one-one-one which entitles it be placed in the second tier. We then place d in the third tier and a in the fourth tier since d beats a one-on-one. Thus the Condorcet tournament

1

54

Chapter 1 • Social Choice

1 a

d

a

b

Scenario 1 Condorcet winner: c Condorcet loser: None c

d Scenario 2 Condorcet winner: None Condorcet loser: c

a

d Scenario 3 Condorcet winner: c Condorcet loser: a

c a

b

c

b

d Scenario 4 Condorcet winner: None Condorcet loser: None

b

c

⊡ Fig. 1.11 Rankings and semi rankings

c b works as a social welfare function and provides the collective ranking , in which the d a first tier consists of c only, the second tier consists of b only, the third tier consists of d only, and the fourth tier consists of a only. Looking at Scenario 1, we can place the Condorcet winner c in the first tier as in Scenario 3 but then we are left with three alternatives (a, b, and d) with no possible way of sorting them into tiers since a beats b, b beats d, and d beats a (a Condorcet paradox). This makes it impossible to proceed any further; and the semi ranking obtained from the Condorcet tournament falls short of being a ranking. c a→b→d→a In Scenario 2 we can place c in the bottom tier but then, as in Scenario 1, it is impossible to proceed because a, b, and d are involved in a Condorcet paradox. Again, the semi ranking obtained from the Condorcet tournament falls short of being a ranking. a→d→b→a c

55 1.5 · A Glimpse Into Social Welfare Theory

The situation is even worse in Scenario 4 as you can see: a → d → c → b → a. Only in Scenario 3 did the pairwise rankings provided by the Condorcet tournament fit together properly and the semi ranking became a ranking! For another example where the Condorcet tournament succeeds in providing a ranking consider the following simple profile with two voters and four alternatives: a

a

b

c

c

b

d

d

You can easily see that the Condorcet tournament places a in the first tier, b and c in the second tier, and d in the bottom tier: a bc d Why are we interested in generating a ranking out of the Condorcet tournament when Borda count is readily available to secure one for us with very little effort?

1.5.2 Independence of Irrelevant Alternatives (IIA) The relative ranking of two alternatives x and y by Borda count can be altered while keeping the relative ranking of x and y by each voter unchanged. To illustrate, consider the following (first) scenario for three voters and two alternatives a and b: a

a

b

b

b

a

a is ranked higher than b by Borda count since a and b receive 2 points and 1 point, respectively. If we add a third alternative c as in the following (second) scenario, a

a

b

b

b

c

c

c

a

the relative ranking of a and b by Borda count becomes a tie with 4 points for each of these two alternatives. Adding a fourth alternative d as in the following (third) scenario, the relative ranking by Borda count changes again as b is now ranked higher than a with 7 points for b and 6 points for a. Obviously, withdrawing d and c will bring back the original Borda count ranking when the contest involved only a and b. If, in the third

1

56

Chapter 1 • Social Choice

1 a

a

b

b

b

c

c

c

d

d

d

a

scenario with four alternatives, the second voter switches b and c and the third voter switches a and d, we get the (fourth) scenario a

a

b

b

c

c

c

b

a

d

d

d

in which the Borda count relative ranking of a and b changes in favor of a with 7 points for a and 6 points for b. What is significant here is that all these changes in the Borda count relative ranking of a and b occurred while none of the three voters changed his/her opinion as to whether a is better than b or b is better than a. This phenomenon, known as the dependence on irrelevant alternatives, is a serious issue with Borda count and social welfare functions in general, that has generated considerable interest and debate in the literature. Why is dependence on irrelevant alternative seen as a problem? In the first scenario above (involving a and b only), a is 1 point higher than b in each of the first and the second ballots and is 1 point lower than b in the third ballot and so a scores 1 point higher than b in Borda count. In the third scenario (involving a, b, c, and d) a continues to be 1 point higher than b in the first and the second ballots but is 3 points lower than b in the third ballot and so a scores 1 point lower than b in Borda count. No voter changed his/her relative ranking of a and b but the wider gap in the third ballot in the third scenario, created by the presence of c and d, is what changed the outcome in favor of b. Presumably, the Borda score difference between two alternatives in a certain ballot is a measure of the intensity of preference for one over the other, but this is questionable as we briefly explain in the following two points: 1. It is not clear how to quantify and compare the intensities of preference of different voters, not to mention combining them. In fact, it is doubtful that this makes any sense at all. 2. Even if this were possible, why should the presence of c and d in the third scenario increase the third voter’s intensity of preference for b over a? What if c and d were purposely brought in to manipulate the result in favor of b? For a concrete example, why should the intensity of your preference for coke b over seven up a increase if we make available to you a can of pepsi c and a can of ginger ale d as in the third scenario? For further investigation of social welfare theory, we need some definitions and notations.

1

57 1.5 · A Glimpse Into Social Welfare Theory

Notation 1.5.2 We shall denote voting scenarios by capital Greek letters , , , · · · . Social welfare functions and semi social welfare functions shall be denoted by small Greek letters λ, μ, ν, · · · . The ranking (or semi ranking) of the alternatives obtained from a scenario  by a social welfare function (or a semi social welfare function) λ will be denoted λ. The voters will be labelled 1st, 2nd, 3rd, . . . . The restriction of a scenario or a ranking (or a semi ranking) to a subset B of the set of A of all alternatives will be indicated by the suffix /B.  

These notation and definitions are made clear in the following example. Example 1.5.3 The restriction of the scenario 1st c = a b d

2nd d c a b

3rd b d c a

to the two alternatives a and b is 1st 2nd 3rd /{a, b} = a a b b b a in which the voters are asked to choose between a and b only. If the social welfare function λ is the Borda count, then c d λ = b a as you can easily verify. (Borda points: 3 points for a, 4 points for b, 6 points for c, and 5 points for d.) We therefore have (λ)/{a, b} =

b a

λ(/{a, b}) =

a b

Note also that

with 2 Borda points for a and 1 Borda point for b.



58

Chapter 1 • Social Choice

1 Definition 1.5.4 A semi social welfare function λ is said to satisfy the independence of irrelevant alternative (IIA) if, for any given scenario  and any two alternatives x and y, (λ)/{x, y} = λ(/{x, y}).

 

In other words, the relative ranking of any two alternative by λ depends only on their relative rankings by the voters without any interference from the rest of the alternatives. This leads to the following formulation of the same definition: Definition 1.5.4 A semi social welfare function λ is said to satisfy the independence of irrelevant alternatives (IIA) if for every two given scenarios  and  and two alternatives x and y with /{x, y} = /{x, y} we have (λ)/{x, y} = (λ)/{x, y}.

 

Example 1.5.5 As a social welfare function, the Borda count does not satisfy IIA. This is clear form Example 1.5.3 since (λ)/{a, b} = λ(/{a, b}) for the given scenario  with λ being the Borda count. 

Theorem 1.5.6 As a semi social welfare function, the Condorcet tournament satisfies IIA.

 

Theorem 1.5.6 needs no proof since the IIA property is built into the definition of the Condorcet tournament. Example 1.5.7 As a social welfare function, the Copeland procedure does not satisfy the IIA. This is shown by the following voting scenario. 1st d c = b e a

2nd e a d c b

3rd c b e a d 

1

59 1.5 · A Glimpse Into Social Welfare Theory

Graph representation for Example 1.5.7

a e

b d c

The Copeland point distribution is a

b

c

d

e

1

2

3

2

2

Let μ be the Copeland social welfare function then c μ = b d e a

and therefore (μ)/{c, d} =

c d

while 1st 2nd 3rd /{c, d} = d d c c c d

and therefore μ(/{c, d}) =

d c

and so (μ)/{c, d} = μ(/{c, d}).   The failure of Copeland social welfare function to inherit the IIA from the Condorcet tournament is worth noticing since the first is directly derived from the second!

1.5.3 Transitivity Next, we investigate how the pairwise rankings can “fit together properly” so that a semi ranking becomes a ranking and a semi social welfare function (in particular, the Condorcet tournament) becomes a social welfare function. Notation 1.5.8 Let A be a set of at least two alternatives on which a certain semi ranking is defined. Let x and y be two alternatives in A. • We write x → y (or y ← x) if x is preferred over y.   • We write x ↔ y or y ↔ x if each of x and y is as good as the other.   • We write x  y if x is at least as as good as y, that is, x → y or x ↔ y. Clearly, x  y and y  x together mean that x ↔ y.  

60

Chapter 1 • Social Choice

1 (→ denotes preference, ↔ denotes indifference and  denotes weak preference.)   • If x → y we say that x is superior to y; if x ← y we say that x is inferior to y; and if x ↔ y we say that x is a match for y (and y is a match for x).   • When several rankings are involved, we shall use a subscript to represent the particular voter (or the semi social welfare function) making the ranking. For example, we shall write x →v y to indicate that voter v prefers x over y.   x →λ y to indicate that the society of voters with voting scenario  prefer x over y when the semi social welfare function λ is used to aggregate the voters’ rankings.  

Definition 1.5.9 A semi ranking on a set A of alternatives is said to be transitive if for every x, y and z in A with x  y and y  z we have x  z.  

Clearly, a semi ranking is transitive if and only if its restriction to every set of three alternatives is transitive. The following corollary is clear. ⓘ Corollary 1.5.10 A semi ranking is transitive if and only if the following four conditions are all satisfied for every set of three alternatives x, y, and z. Condition 1. If x ↔ y and y ↔ z then x  z. Condition 2. If x ↔ y and y → z then x  z. Condition 3. If x → y and y ↔ z then x  z. Condition 4. If x → y and y → z then x  z.

 

The above four conditions can be made more precise as in the following proposition. Proposition 1.5.11 A semi ranking is transitive if and only if the following four conditions are all satisfied for every set of three alternatives x, y, and z. Condition 1. If x ↔ y and y ↔ z then x ↔ z. Condition 2. If x ↔ y and y → z then x → z. Condition 3. If x → y and y ↔ z then x → z. Condition 4. If x → y and y → z then x → z. Proof Condition 1: Assume x ↔ y and y ↔ z then by Condition 1 in Corollary 1.5.10 we have x  z. Similarly, from z ↔ y and y ↔ x we get z  x. From x  z and z  x we then get x ↔ z.   Condition 2: Assume x ↔ y and y → z. If it were not true that x → z then (from Condition 2 in Corollary 1.5.10) we have x ↔ z. From z ↔ x and x ↔ y we conclude by Condition 1 (in this proposition) that z ↔ y, a contradiction. Therefore, x → z.  

1

61 1.5 · A Glimpse Into Social Welfare Theory

Condition 3. Similar to Condition 2.   Condition 4. Assume x → y and y → z. If it were not true that x → z then (from Condition 4 in Corollary 1.5.10) we have x ↔ z. From z ↔ x and x → y we conclude by Condition 2 (in this proposition) that z → y, a contradiction. Therefore, x → z.  

We see from Condition 1 of Proposition 1.5.11 that the indifference ↔ of a transitive semi ranking on a set A of alternatives partitions A into subsets (tiers) A1 , A2 , · · · , An such that all alternatives in the same tier are equally favored and if the alternatives x, y are in two different tiers then one of them is preferred over the other. Further, Conditions 2,3,4 of Proposition 1.5.11 make it possible to list those tiers in a column such that (after renumbering if necessary), y → x if and only if x is in some Ai and y is in some Aj with i < j . This proves the “if” part of the following theorem. The “only if” part is clear. Theorem 1.5.12 A semi ranking on a set of alternatives is a ranking if and only if it is transitive.

 

1.5.4 Black’s Theorem The lack of transitivity of the pairwise ranking created by the Condorcet tournament is the reason behind its failure to be a social welfare function. This is what we learn from Theorem 1.5.12. In Theorem 1.5.16 below we provide a sufficient condition on the voting scenarios that brings about this missing transitivity. It is a special case of both of the theorems of Black [3] and Sen [21]. Those theorems bring transitivity to the Condorcet tournament semi ranking by restricting the domain, i.e, restricting the allowed scenarios to certain types. (More on this topic can be found in [5, 16, 19, 20, 24].) Proposition 1.5.13 Assume we have a semi ranking on a set A of alternatives such that for every set of three alternatives: one of the three is superior to each of the other two or one of the three is inferior to each of the other two, Then the semi ranking is transitive. Proof Let a, b, and c be three alternatives. If, say, b → a and b → c, then the restriction of our semi ranking to a, b, and c is one of the following three rankings:

62

Chapter 1 • Social Choice

1 b b b a , c or ac c a  

and therefore the semi ranking is transitive. The other case is similar.

ⓘ Remark 1.5.14 1. In the sequel, we shall consider scenarios where indifference in the voters’ rankings (preference ballots) occurs. This will require a slight adaptation of the two-alternative majority rule as follows: Given two alternatives x and y, let m be the number of voters who prefer x over y and let n be the number of voters who prefer y over x. • x beats y in the two-alternative majority rule if and only if m > n. (Note that x may or may not have a majority of the total vote, but must have a majority of the non-indifferent vote, in order to beat y.) • x and y are tied in the two-alternative majority rule if and only if m = n. 2. To conform with the principle that the winner in the two-alternative majority rule should receive a majority of the total vote, we add half of the indifferent vote to each of m and n above. 3. In the case where voters are allowed to express indifference, ties in the two-alternative majority rule may occur regardless of whether the number of voters is even or odd. In the case where indifference is not allowed, ties may occur only when the number of voters is even.

ⓘ Remark 1.5.15 In some applications, the alternatives and the voters can naturally be positioned on a straight line and each voter ranks the alternatives based on their distances from his/her position: the nearer the better. For example, consider the case of several candidates running for political office and let a, b, and c be three of these candidates whose positions on the political spectrum are as follows:

a

x

b

y

c

b is a centrist, a is a leftist, and c is a rightist. x and y are not candidates: x is the midpoint between a and b, and y is the midpoint between b and c on the political spectrum. Under ideal circumstances: a • A voter to the left of x should rank the candidates as b . c ab • A voter exactly at x should rank the candidates as . c

63 1.5 · A Glimpse Into Social Welfare Theory

b b b • A voter between x and y should rank the candidates as a , or c . ac c a • A voter exactly at y should rank the candidates as

bc . a

c • A voter to the right of y should rank the candidates as b . a Therefore, the general form of a voting scenario (restricted to three alternatives) is h

i

a

j

ab b

l

a

n c

bc c

ac c

m

b b

c c

k

b

b a

a

a

where h, i, j, k, l, m, and n are the corresponding numbers of voters. For example, the bc number of voters who ranked the candidates as is m. Note that some rankings such a a as c will not occur in the present situation. (A voter whose most preferred of the three b candidates is the leftist a is not expected to prefer the rightist c to the centrist b.)  

Theorem 1.5.16 If the alternatives and the voters are as in Remark 1.5.15 and the number of voters is odd, then the Condorcet tournament is a social welfare function.

1

64

Chapter 1 • Social Choice

1 Proof Assume that the alternatives and voters are as in Remark 1.5.15 and let μ be the Condorcet tournament semi social welfare function. Let  be a scenario of voting whose restriction to arbitrary three alternatives a, b, and c is as in Remark 1.5.15. By Theorem 1.5.12, all we need to prove is that μ is transitive. h i j k l m n a /{a, b, c} =

b ab

b

b a

c c

b

c bc

c ac

c

b a

a

a

We have three cases to consider: Case 1. a →μ c: In this case we have h + i + j > l + m + n and therefore h + i + j + k + l > n which proves that b →μ c. The transitivity of μ now follows from Proposition 1.5.13 in this case. Case 2. a ↔μ c: In this case we have h + i + j = l + m + n. The total number of voters is then (h + i + j ) + k + (l + m + n) = 2(h + i + j ) + k which is assumed to be an odd number and so k is a positive odd integer. In particular, k > 0. It follows that h + i + j + k > l + m + n and therefore

h+i+j +k+l >n

which proves that b →μ c. We also get k + l + m + n > h + i + j and therefore

j +k+l+m+n>h

which proves that b →μ a. The transitivity of μ follows from Proposition 1.5.13 in this case too. Case 3. c →μ a: By symmetry, this case is similar to Case 1.

 

From Definition 1.5.4 we see that a semi social welfare function can satisfy the IIA if it uses, as building blocks, the pairwise rankings resulting from some twoalternative rule (which need not be the two-alternative majority rule. (As a social

1

65 1.5 · A Glimpse Into Social Welfare Theory

welfare function, the Borda count is not built in this manner since it creates a ranking for the available alternative all at once.) If the consequent semi ranking turns out to be transitive, we obtain a social welfare function that satisfies the IIA. The Condorcet tournament satisfies the IIA simply because it is built in this manner (with the pairwise rankings obtained from the two-alternative majority rule). The problem with the Condorcet tournament is that it does not, in general, produce a transitive semi ranking. Theorem 1.5.16 presents a way to bring transitivity to the Condorcet tournament by limiting our consideration to certain types of scenarios. A simple but obviously unattractive way to secure both transitivity and IIA together, without limiting the scenarios considered, is to endow one specific voter with the sole authority to rank the alternatives on behalf of the entire society of voters. The precise definition is as follows: Definition 1.5.17 A social welfare function δ is said to be a strong dictatorship if a specific voter v (called the strong dictator) is designated such that for every scenario , δ is the ranking provided by v, that is, given any two alternatives x and y, x →δ y if and only if x →v y and x ↔δ y if and only if x ↔v y.

 

1.5.5 May’s Theorem In 1952 Kenneth May proved a landmark theorem about the two-alternative majority rule which we discuss next. We first explain the notations to be used throughout this subsection (Sect. 1.5.5) then give formal definitions of some of the desirable properties satisfied by the two-alternative majority rule. Notation 1.5.18 1. The two alternatives being compared are a and b.   2. We have n voters labeled vi , i = 1, 2, · · · , n; and the ranking of a and b provided by voter vi is labeled with the subscript i. For example, a →i b means that voter vi prefers a over b. Scenarios will involve only those n voters and the two alternatives a, b.   3. The number of voters who prefer a over b is n(a → b), the number of voters who prefer b over a is n(b → a), and the number of voters who are indifferent is n(a ↔ b) so that n(a → b) + n(b → a) + n(a ↔ b) = n.   4. The two-alternative rule used to aggregate the rankings of a and b in a scenario  for the n voters and the two alternatives is denoted by μ and, as usual, the ranking that results from this aggregation is denoted by μ.  

66

Chapter 1 • Social Choice

1 5. For any given scenario , one and only one of a →μ b, b →μ a and a ↔μ b is true.   6. A scenario obtained by redistributing the same rankings in  among the voters (i.e., permuting them as discussed in  Sect. 2.3 according to a certain permutation p) will be denoted by p .   7. The inverse of a scenario , denoted by  is the scenario obtained from  by replacing every a →i b with b →i a and every b →i a with a →i b while leaving every a ↔i b unchanged. Likewise, the inverse of μ, denoted by μ is obtained from μ by replacing a preference of a over b with a preference of b over a and vice versa, while leaving an indifference unchanged.   8. If, in scenario , a single voter, say vi , changes her ranking from a ←i b to a ↔i b or a →i b OR from a ↔i b to a →i b, the new scenario, which is obviously more favorable to a than , will be denoted  + .  

Definition 1.5.19 A two-alternative rule μ is said to satisfy 1. Anonymity if μp = μ for every scenario  and every permutation p of the voters’ rankings.   2. Neutrality if μ = μ for every scenario .     3. Monotonicity if a →μ + b whenever a ↔μ b or a →μ b. (Anonymity means that the outcome depends only on the numbers n(a → b), n(b → a) and n(a ↔ b) regardless of the identities of the voters who provided each of the three types of rankings.)

Neutrality means that if a and b exchange votes then the outcome is reversed (if it was in favor of either one) and stays the same if they were originally tied. Monotonicity means that an alternative that is already tied or winning will be winning if it receives additional support. The two-alternative majority rule obviously satisfies all of the three properties in Definitions 1.5.19. The significance of May’s theorem is that it singles out the two-alternative majority rule as the only two-alternative rule that satisfies those three properties. In other words, if we start building a two-alternative rule with those three properties in mind, we shall find ourselves face-to-face with the two-alternative majority rule and no other! Proposition 1.5.20 If the scenario  is such that n(a → b) = n(b → a) and the twoalternative rule μ satisfies anonymity and neutrality, then a ↔μ b.

1

67 1.5 · A Glimpse Into Social Welfare Theory

Proof Assuming that n(a → b) = n(b → a) we see that  = p for some permutation p of the voters’ rankings in  in which the voters who prefer a over b exchange their rankings with those who prefer b over a. By anonymity of μ we have μp = μ and by neutrality of μ we have μ = μ. It follows that μ = μ = μp = μ. From Notations 1.5.18 we see that the only way that μ and μ could be equal is that μ is an indifference, that is a ↔μ b.   Proposition 1.5.21 If the scenario  is such that n(a → b) > n(b → a) and the twoalternative rule μ satisfies anonymity, neutrality, and monotonicity, then a →μ b. Proof Assume n(a → b) > n(b → a) and let k = n(a → b) − n(b → a), then k > 0. Let us construct a scenario  from  by “convincing” k of the voters who prefer a over b to change their rankings to indifference. Clearly the conditions of Proposition 1.5.20 apply to  and so a ↔μ b. Now let us recover  from  by making those k voters switch their rankings back from indifference to preference for a over b, one at a time. By monotonicity of μ we   get a →μ b.

Theorem 1.5.22 (Kenneth May, 1952) The two-alternative majority rule is the only two-alternative rule that satisfies anonymity, neutrality, and monotonicity.

Proof The theorem follows immediately from Propositions 1.5.20 and 1.5.21.

 

1.5.6 Dictatorship and Arrow’s Theorem Perhaps the most important result in social welfare theory is the theorem known as Arrow’s Impossibility Theorem proven by the 1972 Nobel Prize winner Kenneth Arrow in his monograph Social Choice and Individual Values published in 1951. We present here a stronger version of the theorem, proven by Arrow in the second edition [1] of the same monograph in 1963. Arrow puts forward a seemingly innocuous set of axioms that a social welfare function should satisfy and proves that the only social welfare function that satisfies those reasonable axioms is a form of dictatorship somewhat similar to (though slightly weaker than) the strong dictatorship in Definition 1.5.17.

68

Chapter 1 • Social Choice

1 Arrow’s Axioms 1.5.23 A semi social welfare function λ needs to satisfy the following axioms: 1. Unrestricted Domain. This means that all scenarios are allowed, in contrast to Black’s theorem (Theorem 1.5.16). which limits the allowable scenarios. 2. Independence of Irrelevant Alternatives (IIA). 3. Weak Pareto Condition: If, in a scenario , every voter prefers alternative x to alternative y then x →λ y. 4. Collective Rationality. This axiom means that for every scenario , λ is transitive. (This axiom is already included in our definition of a social welfare function.)   Before we state and prove Arrow’s theorem, we need the following definition.

Definition 1.5.24 A semi social welfare function δ is said to be dictatorial (or a dictatorship) if a specific voter v (called the dictator) is designated such that for every scenario  and every pair of alternatives x, y, x →v y implies x →δ y.

 

A strong dictator (as in Definition 1.5.17) fully enforces her ranking (both preferences and indifferences) of the alternatives on the society of voters, while a dictator (as in Definition 1.5.24) enforces only his preferences. For example, if the ranking of voter a a a a b v is b c , then the societal ranking δ is b c if v is a strong dictator; but can be b c , c d d d d a c or if v is only a dictator. b d Arrow’s Impossibility Theorem 1.5.25 (Arrow [1]) For a set A of at least three alternatives and a set V of at least two voters, a semi social welfare function satisfies Arrow’s axioms if and only if it is a dictatorship.  

If we add to Arrow’s axioms a fifth axiom that rules out dictatorship, it would be impossible to find a semi social welfare function (or a social welfare function) that satisfies all the five axioms together, hence the “impossibility” part in the theorem’s title. The proof of Arrow’s theorem will follow as a culmination of several partial results. It is a truly original, deep, and pioneering mathematical work.

1

69 1.5 · A Glimpse Into Social Welfare Theory

Notation 1.5.26 1. In the sequel, V will denote a set of at least two voters, A will denote a set of at least three alternatives, and λ will denote a semi social welfare function. Further, D will denote a nonempty subset of V .   2. The set of voters not in D will be denoted by V \D.   3. If all the voters in D prefer alternative x over alternative y we shall write x →D y. Similarly, x →V \D y will denote a concerted vote by voters in V \D for x over y.   4. If  is a scenario in which x →D y and y →V \D x, we may write D V \D /{x, y} = x y · y x  

Definition 1.5.27 D is said to be 1. nearly λ-decisive if for every x, y in A and every scenario  such that x →D y and y →V \D x we have x →λ y.   2. λ-decisive if for every x, y in A and every scenario  such that x →D y we   have x →λ y.

ⓘ Remark 1.5.28 1. Members of a nearly λ-decisive set D of voters are able to force their concerted preference for x over y against the concerted opposite preference (y over x) by the voters not in D. Members of a λ-decisive set E of voters, on the other hand, are able to force their concerted preference for x over y against any configuration of rankings by the voters not in E, including their concerted opposite preference (y over x).   2. It may seem intuitive that a nearly λ-decisive set should also be λ-decisive since it is able to prevail against the most challenging opposition possible, but we work with axioms, not with intuition, and we have nothing available in our axioms to lend an immediate support to this intuition. It will take some elaborate arguments involving a very interesting interplay among Arrow’s axioms to prove the validity of this intuition, which constitutes a significant part of the proof of Arrow’s theorem.   The following corollary follows immediately from the definitions, with no need to invoke any of Arrow’s axioms. (See the first Remark in 1.5.28.) ⓘ Corollary 1.5.29 A λ-decisive set is nearly λ-decisive.  

70

Chapter 1 • Social Choice

1 Proposition 1.5.30 Assume we have two specific alternatives a, b in A and a specific scenario  that satisfies D V \D Condition I: /{a, b} = a b b a and a Condition II: (λ)/{a, b} = b where λ is a semi social welfare function that satisfies IIA. Then every scenario  that satisfies Condition I must also satisfy Condition II, that is D V \D a implies (λ)/{a, b} = . /{a, b} = a b b b a Proof This is an immediate consequence of the IIA (Definition 1.5.4 ).

 

The student should note the relation between Proposition 1.5.30 and the definition of near λ-decisiveness in 1.5.27. Informally speaking, Proposition 1.5.30 assumes just one single occurrence of the defining condition of near λ-decisiveness for D (involving a specific pair of alternatives a and b, and a specific scenario ) and, using only the IIA, the proposition extends this single occurrence to all scenarios that satisfy Condition I for these two specific alternatives. Using the axioms of Arrow, combined, the following lemma will further extend this to all pairs of alternatives x and y, thus proving the near λ-decisiveness of D. ⓘ Lemma 1.5.31 Assume we have two specific alternatives a, b in A, and a specific scenario  with Condition I:

D V \D /{a, b} = a b b a

and

Condition II:

(λ)/{a, b} =

a b

and assume further that λ satisfies Arrow’s axioms. Then D is nearly λ-decisive. Proof By Proposition 1.5.30, every scenario that satisfies Condition I will necessarily satisfy Condition II.

71 1.5 · A Glimpse Into Social Welfare Theory

Step 1: Pick an arbitrary alternative y that is different from each of a and b. (This is possible since there are at least three alternatives in A.) By the axiom of unrestricted domain, there is a scenario such that D V \D a b

/{a, b, y} = . b y y a It follows from Proposition 1.5.30 that a →λ b and from the weak Pareto axiom that b →λ y. Collective rationality now implies a Condition II: (λ )/{a, y} = · y Since we also have D V \D Condition I: /{a, y} = a y , y a it follows from Proposition 1.5.30 and the arbitrariness of y that D V \D a implies (λ)/{a, y} = /{a, y} = a y y y a for every scenario  and every alternative y that is different from a. (Notice how the arbitrary y has replaced the specific b we started with. In the next step, an arbitrary x will replace the specific a.) Step 2: Pick an arbitrary alternative x that is different from each of a and y. (This is possible since there are at least three alternatives in A.) By the axiom of unrestricted domain, there is a scenario  such that D V \D x y /{a, x, y} = . a x y a It follows from Proposition 1.5.30 that a →λ y and from the weak Pareto axiom that x →λ a. Collective rationality now implies x Condition II: (λ)/{x, y} = · y Since we also have D V \D Condition I: /{x, y} = x y , y x it follows from Proposition 1.5.30 and the arbitrariness of x and y that D V \D x implies (λ)/{x, y} = /{x, y} = x y y y x

1

72

Chapter 1 • Social Choice

1 for every scenario  and every pair of different alternatives x and y. This completes the proof of our lemma.  

ⓘ Lemma 1.5.32 Assume that λ satisfies Arrow’s axioms. Then D is λ-decisive if and only if it is nearly λ-decisive. Proof We only need to prove that near λ-decisiveness implies λ-decisiveness, since the converse has already been established in Corollary 1.5.29. Let x and y be two different arbitrary alternatives. Let  be a scenario in which every voter in D prefers x over y. Partition V \D into three sets E, F, G such that DEF G /{x, y} = x x y x y . y y x Pick an arbitrary alternative z that is different from each of x and y. (This is possible since we have at least three alternatives.) By the axiom of unrestricted domain, we can find a scenario such that D x

/{x, y, z} = z y

E x y z

F G y xy x z z

It follows from the near λ-decisiveness of D that z →λ y and from the weak Pareto axiom that x →λ z. Collective rationality now implies x →λ y. Since /{x, y} = /{x, y} (as you can easily verify), it follows from the axiom of IIA that (λ)/{x, y} = (λ )/{x, y} =

x · y

Since  is an arbitrary scenario with the only restriction that all voters in D prefer x over y (with no restrictions as to how the individual voters in V \D rank x and y relative to one another), and since x and y are arbitrary alternatives, it follows that D is λ-decisive. This completes the proof of the lemma.  

ⓘ Corollary 1.5.33 Assume we have two specific alternatives a, b in A, and a specific scenario  such that D V \D a /{a, b} = a b and Condition II: (λ)/{a, b} = b b a are both satisfied, and assume further that λ satisfies Arrow’s axioms. Then D is λ-decisive. Condition I:

1

73 1.5 · A Glimpse Into Social Welfare Theory

Proof This follows immediately from Proposition 1.5.30, Lemma 1.5.31, and Lemma 1.5.32.

 

ⓘ Lemma 1.5.34 Assume that λ satisfies Arrow’s axioms and that D is a λ-decisive set of at least two voters. Partition D into two non-empty subsets D1 and D2 . Then one (and only one) of D1 and D2 is λ-decisive. Proof Let x, y, and z be three different alternatives. (Again, this is possible since we have at least three alternatives.) By the axiom of unrestricted domain, we can find a scenario  such that D1 x /{x, y, z} = y z If (λ)/{y, z} = (λ)/{y, z} =

D2 V \D z y . x z y x

z then, D2 is λ-decisive (by Corollary 1.5.33) and D1 is not. If y

z , then λ ranks y at least as high as z that is, y y λ z.

But from the λ-decisiveness of D we have x →λ y. It then follows from Collective rationality that x →λ z and therefore D1 is λ-decisive (by Corollary 1.5.33) and D2 is not. The proof is now complete.   Proof 1.5.35 (Arrow’s Impossibility Theorem) Starting off with the set V of all voters (which is evidently λ-decisive by the weak Pareto axiom) we shall reach, after a finite sequence of applications of Lemma 1.5.34, a λ-decisive set with only one voter. This voter is a dictator as easily seen from Definitions 1.5.24 and 1.5.27. This will complete the proof. In detail: 1. Partition V into two nonempty sets (this is possible since there are at least two voters in V ). One and only one of those two sets is λ-decisive by Lemma 1.5.34. Denote this set by D1 .

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Chapter 1 • Social Choice

1 2. If D1 contains two or more voters, we can partition it into two nonempty subsets, one and only one of which is λ-decisive. Denote this set by D11 . Obviously, D11 has fewer members than D1 . Continuing in this fashion, we must reach, after finitely many applications of Lemma 1.5.34, a λ-decisive set with just one voter. This voter is our dictator or, more precisely, λ-dictator.  

1.5.7 Oligarchy and Gibbard’s Theorem In Sect. 1.5.4 above, we saw how restricting the allowable scenarios can bring transitivity to the Condorcet tournament (Black’s theorem), thus making it into a social welfare function. It can be easily verified that the social welfare function so obtained satisfies the Weak Pareto axiom in addition to the inbuilt IIA and the Collective rationality that has been brought into it. This is a good example of how relaxing one or more of Arrow’s axioms (the axiom of unrestricted domain in this case) can replace Arrow’s dictatorship with a more appealing outcome. In an unpublished manuscript [8, 23], Allen Gibbard replaced the collective rationality (included here in our definition of a social welfare functions) with the weaker property in the following definitions. (Compare to Definition 1.5.9, Corollary 1.5.10, Proposition 1.5.11, and Theorem 1.5.12 above.) Definition 1.5.36 1. A semi ranking on a set A of alternatives is said to be quasi transitive if for every x, y, and z in A with x → y and y → z we have x → z.   2. A quasi transitive semi ranking is said to be a quasi ranking.   3. A semi social welfare function λ is said to be a quasi social welfare function if for every scenario , λ is quasi transitive. We also say, in this case that λ satisfies the axiom of collective quasi-rationality.  

ⓘ Corollary 1.5.37 A semi ranking on a set A of alternatives is quasi transitive if and only if it satisfies the following two conditions for every x, y, z in A: 1. If x → y and y  z then x  z. 2. If x  y and y → z then x  z. Proof First, assume that the semi ranking at hand is quasi transitive. If Condition 1 is not satisfied, then there are x, y, z in A such that x → y, y  z and z → x. From z → x and x → y and the assumed quasi transitivity we get z → y, a contradiction. This proves condition 1. The proof of Condition 2 is similar. Next, assume the above two conditions and let x, y, z be alternatives in A such that x → y and y → z then, from Condition 1 we see that x  z (i.e., x ↔ z or x → z). If

1

75 1.5 · A Glimpse Into Social Welfare Theory

z ↔ x then, this together with x → y and Condition 2 yields z  y, a contradiction. Hence x → z which proves quasi transitivity.   It is clear from Proposition 1.5.11 that transitivity implies quasi transitivity. The converse is not true: Consider, for example, the case where A consists of three alternatives x, y, z with x → y, y ↔ z and x ↔ z.  

Following the notations in 1.5.26 we have Definition 1.5.38 A non-empty subset D of the set V of voters is said to be 1. nearly λ-blocking if for every x, y in A and every scenario  such that x →D y and y →V \D x we have x λ y.   2. λ-blocking if for every x, y in A and every scenario  such that x →D y we have x λ y.   The use of the adjective “blocking” can be explained by noting that D is λ-blocking if and only if the members of D can block the collective preference of y over x by their concerted preference for x over y (but may allow a tie between x and y). If {v} is a nearly λ-blocking (or λ-blocking) set, we say that v is a nearly λ-blocking (or a λ-blocking) voter.

Definition 1.5.39 A semi social welfare function λ is said to be oligarchic if there exists a non-empty set D of voters, called oligarchy (or λ-oligarchy, to be more precise) such that D is λ-decisive and each member of D is λ-blocking.  

A λ-dictator is automatically λ-blocking, as you can easily verify. It is therefore possible to think of a dictatorship as an oligarchy consisting of only one member. Gibbard’s Theorem 1.5.40 If λ is a semi social welfare function that satisfies the following axioms (henceforth called Gibbard’s axioms): 1. Unrestricted Domain, 2. Independence of Irrelevant Alternatives, 3. Weak Pareto, 4. Collective quasi-rationality then λ is oligarchic.

 

The proof of Gibbard’s theorem will come as a culmination of the following partial result (Corollary 1.5.41 through Proposition 1.5.43). First, we note that Corollary 1.5.33 above was arrived at with the help of collective quasi-rationality, not the full strength of collective rationality. This establishes Part 1 of the following corollary. We ask the

76

Chapter 1 • Social Choice

1 student to prove Part 2 by revising the statements and proofs of Proposition 1.5.30, Lemmas 1.5.31, and 1.5.32. ⓘ Corollary 1.5.41 Assume that the semi social welfare function λ satisfies Gibbard’s axioms and let D be a nonempty subset of the set V of voters. 1. If two specific alternatives a, b in A, and a specific scenario  exist such that D V \D a Condition I: /{a, b} = a b and Condition II: (λ)/{a, b} = b b a are both satisfied, then D is λ-decisive.   2. If two specific alternatives a, b in A, and a specific scenario  exist such that D V \D a Condition I: /{a, b} = a b and Condition II: (λ)/{a, b} = or ab b b a are both satisfied, then D is λ-blocking.  

A student who is not familiar with the set-theoretic notations used below can refer to 1.7.A for an explanation of these notations. ⓘ Lemma 1.5.42 Assume that the semi social welfare function λ satisfies Gibbard’s axioms and let D and E be two λ-decisive subsets of the set V of voters. Then 1. D ∩ E is nonempty. 2. D ∩ E is λ-decisive. Proof 1. Assume to the contrary that D ∩ E = ∅ and let a and b be two different alternatives. By the axiom of unrestricted domain, we can find a scenario  such that a →D b and b →E a· The λ-decisiveness of both D and E now yields the two contradictory conclusions a →λ b and b →λ a. Hence D ∩ E = ∅. 2. Let x, y, z be three different alternatives. By the axiom of unrestricted domain we can find a scenario  such that D ∩ E D\E E\D V \(D ∪ E) x z y z /{x, y, z} = y x z y z y x x Since D is λ-decisive, we have x →λ y; and since E is also λ-decisive, we have y →λ z. By the axiom of collective quasi-rationality we then have x →λ z. It follows from Corollary 1.5.41 that D ∩ E is λ-decisive.  

77 1.5 · A Glimpse Into Social Welfare Theory

Proposition 1.5.43 Assume that the semi social welfare function λ satisfies Gibbard’s axioms. Then there exists a unique subset D of V such that 1. D is a nonempty λ-decisive subset of V . 2. If E is a λ-decisive subset of V , then D is a subset of E. (D is termed the smallest λ-decisive subset of V .) 3. If D is a singleton, that is D = {v} for some voter v, then v is a dictator (more precisely, a λ-dictator). Proof 1. The weak Pareto axiom guarantees the existence of at least one λ-decisive subset of the set V of all voters, namely V itself. Let D be the intersection of all λ-decisive subsets of V . Finitely many applications of Lemma 1.5.42 prove that D is a nonempty λ-decisive subset of V . 2. This follows since E is one of the sets whose intersection is D. 3. This is clear since D is λ-decisive. The uniqueness of D is evident because if G is another smallest λ-decisive subset of V , then D ⊆ G and G ⊆ D and so G = D.   Proof 1.5.44 (Gibbard’s Theorem) We show that the nonempty subset D of V , obtained in Proposition 1.5.43, is an oligarchy. If D is a singleton, then λ is both oligarchic and dictatorial as pointed out above. It remains to consider the case when D consists of more than one voter, which we do next. Since D is already known to be λ-decisive, we only need to show that each voter v in D is λ-blocking. Let x, y, z be three different alternatives and let v be an arbitrary voter in D. By the axiom of unrestricted domain we can find a scenario  such that D\{v} z /{x, y, z} = x y

{v} V \D x y y z z x

If z →λ y, then D\{v} is λ-decisive by Part 1 of Corollary 1.5.41, contrary to the assumption that D is the smallest λ-decisive subset of V . We must therefore have y λ z; but since D is λ-decisive we must also have x →λ y. Using Corollary 1.5.37, it now follows from x →λ y and y λ z that x λ z. Finally, it follows from Part 2 of Corollary 1.5.37 that voter v is λ-blocking.  

1

78

Chapter 1 • Social Choice

1 A Note on the Proofs of Arrow’s Theorem and Gibbard’s Theorem In the proof of Lemma 1.5.34, the stronger property of transitivity enabled us to partition a given λdecisive set D of at least two voters into two nonempty subsets D1 and D2 , one of which is λ-decisive. Finitely many applications of this result led to a λ-decisive set with just one voter: Arrow’s Dictator. The weaker property of quasi transitivity (if used in the proof of Lemma 1.5.34 instead of transitivity) will fall short of establishing that one of D1 and D2 is λ-decisive. It will, instead, establish that either D2 is λ-decisive or D1 is λ-blocking (Verify!). If D2 is not λ-decisive, we stop. If D2 is λ-decisive, it must contain all members of the oligarchy. We partition D2 into D21 and D22 . We can continue with the partitioning process till further partitioning results in two parts none of which is λ-decisive. The last λ-decisive set so obtained must contain Gibbard’s Oligarchy. It will be exactly that oligarchy if we are so lucky that all our partitions leave the entire oligarchy in one of the two parts. Of course, if the oligarchy happens to consist of only one voter (a dictator), it will always fall into one of the two parts and will be found at the end of the partitioning process. Example 1.5.45 Assume that λ satisfies Arrow’s axioms and consider the following scenario s a  = c b

t u a b b ac c

for the set V = {s, t, u} of voters and the set A = {a, b, c} of alternatives, and assume that a →λ b. 1. How does λ rank a and c? 2. Find all the possibilities for λ and identify the dictator associated with each possibility.  Solution 1. Let D = {s, t}. Since a →D b, b →V \D a and a →λ b, it follows from Corollary 1.5.33 that D is λ-decisive and so a →λ c since we also have a →D c.   a 2. From Lemma 1.5.34 we see that either s is the dictator and therefore λ = c , or t is the b a dictator and therefore λ = b . c

 

79 1.6 · Social Choice Procedures: Indifference and Ties Allowed

1.6

Social Choice Procedures: Indifference and Ties Allowed

In our study of social choice procedures in the first four sections of this chapter, we did not allow indifference in the voters’ ballots and, with few exception including Example 1.3.23 on McGarvey’s theorem, we avoided ties in the outcome. In this section, we generalize the standard social choice procedures so that voters can submit rankings of the kind introduced in Definition 1.5.1 instead of mandating them to submit strict preferences. The crucial point here is that when no indifference occurs in the voters’ ballots, the proposed generalization of a certain social choice procedure should agree with the original version of the procedure. Recall that when two or more alternatives are tied for winning, we speak of a winning set or a social choice set. For example, if a, b, and c are tied for winning, the social choice set will be {a, b, c} and, if a is the sole winner, the social choice set will be {a}.

1.6.1 Standard Social Choice Procedures In  Sect. 1.4 we viewed the plurality procedure as a scoring procedure in which every voter is in possession of exactly one point to be given, undivided, to the one candidate he prefers most. The most natural generalization, if we are to allow indifference in voters’ rankings, is to equally divide each voter’s point among the candidates in his first tier. For example, if a voter picks 3 candidates as her top choice, each of those candidates receives one third of a point from her. (A voter whose ranking consists of only one tier equally distributes his point among all candidates, or we may just drop that voter’s ballot from our calculations.) The total number of points received by all candidates remains equal to the number of voters (as it should). As usual, a majority candidate is a candidate who receives more than half of the total number of points. The generalizations of the plurality with run-off and the Hare procedures use the present manner of point calculation in implementing Definitions 1.2.1 and 1.2.5 of the original versions of these two procedures. As for the Coombs procedure, we only need to mention that in Definition 1.2.8, the last choice votes are calculated in the same way as the top choice votes: If a voter’s last tier consists of 3 candidates, each of those candidates get one third of a last choice vote from him. In the procedures of plurality, plurality with run-of, Hare and Coombs, one-tier ballots are to be dropped from our calculations. If all ballots are one-tier ballots, the social choice set consists of all candidates. To explain our generalization of Borda count to the case where a tier may include more than one alternative, consider the following example of two rankings of seven alternatives: The first ranking does not include indifference and the second ranking does.

1

80

Chapter 1 • Social Choice

1 a

6

b

5

Tiers

Points in a tier

Total point in a tier

Points per alternative

c

4

a, b

5,6

11

11/2 = 5.5

d

3

c

4

4

4/1 = 4

e

2

d, e, f

1,2,3

6

6/3=2

f

1

g

0

0

0/1 =0

g

0

In the second ranking, alternatives in the same tier equally share the sum of points that they would have earned if they were in different tiers (as in the first ranking). As shown, g earns 0 points, each of d, e, f earns 2 points, c earns 4 points, and each of a, b earns 5.5 points. Following the same pattern, the student should be able to verify the point-worth of being in each tier in the following ranking of 10 alternatives: Tier

Points per alternative

a, b

8.5

c, d, e

6

f

4

g, h

2.5

i, j

0.5

Here is a unified definition of the generalized Borda count and the standard Borda count: An alternative x receives from a ranking ballot one point per each alternative in lower tiers, half a point per each alternative (other than x itself) in the same tier, and zero points per each alternative in higher tiers. To justify the above definition, assume that alternative x is in a tier that has p alternatives (including x) and the total number of alternatives in the tiers lower than x’s tier is q. Then the total score of the alternatives in x’s tier is q+(q+1)+(q+2)+· · ·+(q+p−1) = pq+(1+2+· · ·+(p−1)) = pq+

p(p − 1) . 2

Since this total is equally shared by p alternatives (including x), we see that x receives   p(p − 1) 1 1 pq + = q + (p − 1), p 2 2 in agreement with the above generalized Borda count definition. As in Lemma 1.4.10, each voter has k(k − 1) 2

1

81 1.6 · Social Choice Procedures: Indifference and Ties Allowed

points to contribute to the pool of points; and the sum of the Borda scores of all alternatives is k(k − 1)n points. 2 The two-alternative majority rule was defined in the second remark in 1.5.14 in a generalized form that allows indifference. Based on this generalization of the two-alternative majority rule, the generalizations of the following concepts follow immediately: Copeland procedure, Condorcet tournament, Condorcet winner, and Condorcet loser. Example 1.6.1 The following scenario shows the rankings of five alternatives a, b, c, d, and e by five voters. 1st

2nd

3rd

4th

5th

ab

c

cd e

a

bd

c

d

a

bc

ac

de

abe

b

d

e

e

The corresponding graph representation is shown in ⊡ Fig. 1.12. 1. There is no Condorcet winner and e is a Condorcet loser. 2. The plurality procedure scores are as follows:



a: 12 + 0 + 0 + 1 + 0 = 96 b: 12 + 0 + 0 + 0 + 12 = 66 c: 0 + 1 + 13 + 0 + 0 = 86 d: 0 + 0 + 13 + 0 + 12 = 56 e: 0 + 0 + 13 + 0 + 0 = 26 The sum of plurality scores is equal to the number of voters, as it should. The plurality procedure choice set is {a}.  ⊡ Fig. 1.12 Graph representation of Example 1.6.1

a e

b d c

82

Chapter 1 • Social Choice

1 3. If indifference was not allowed, a majority would have meant ≥ 3 votes since we have 5 voters. With indifference allowed, votes are not necessarily integers so we must follow the formal definition of a majority as greater than half of the number of voters ( 52 in the present example). 

There is no majority alternative to win the Hare procedure in the first round. We eliminate e since it has the lowest top choice score. The updated scenario is 1st

2nd

3rd

4th

5th

ab

c

cd

a

bd

c

d

a

bc

ac

d

ab

b

d

and the top choice scores are a: 12 + 0 + 0 + 1 + 0 = 96 b: 12 + 0 + 0 + 0 + 12 = 66 c: 0 + 1 + 12 + 0 + 0 = 96 d: 0 + 0 + 12 + 0 + 12 = 66 The sum checks correctly. No majority. b and d are eliminated and we are left with a and c which are tied as seen from the graph representation. The Hare procedure choice set is therefore {a, c}.   4. There is no majority candidate to win the Coombs procedure in the first round. The bottom choice distribution is a: 0 + 13 + 0 + 0 + 0 = 26 b: 0 + 13 + 1 + 0 + 0 = 86 c: 0 + 0 + 0 + 0 + 0 = 0 d: 12 + 0 + 0 + 0 + 0 = 36 e: 12 + 13 + 0 + 1 + 1 = 17 6 We eliminate e since it has the highest last choice score. The updated scenario is 1st

2nd

3rd

4th

5th

ab

c

cd

a

bd

c

d

a

bc

ac

d

ab

b

d

There is no majority candidate in this scenario. The bottom choice scores are a: 0 + 12 + 0 + 0 + 12 = 66 b: 0 + 12 + 1 + 0 + 0 = 96 c: 0 + 0 + 0 + 0 + 12 = 36 d: 1 + 0 + 0 + 1 + 0 = 12 6

1

83 1.6 · Social Choice Procedures: Indifference and Ties Allowed

Here is the updated scenario after the elimination of d: 1st

2nd

3rd

4th

5th

ab

c

c

a

b

a

bc

ac

c ab

b

The top choice scores are a: 12 + 0 + 0 + 1 + 0 = 96 b: 12 + 0 + 0 + 0 + 1 = 96 c: 0 + 1 + 1 + 0 + 0 = 12 6 Again, no majority candidate exists. The last choice scores are a: 0 + 12 + 0 + 0 + 12 = 66 b: 0 + 12 + 1 + 12 + 0 = 12 6 c: 1 + 0 + 0 + 12 + 12 = 12 6 b and c are eliminated and the Coombs social choice set is {a}.

 

5. The Borda count scores are: a: 72 + 1 + 1 + 4 + 32 = 22 2 b: 72 + 1 + 0 + 52 + 72 = 21 2 c: 2 + 4 + 3 + 52 + 32 = 26 2 d: 12 + 3 + 3 + 1 + 72 = 22 2 e: 12 + 1 + 3 + 0 + 0 = 92 The total sum of scores of all candidates is 50, which is the correct sum. The Borda count social choice set is {c}.   6. To proceed with the Hare–Borda procedure, we eliminate e. The updated scenario is

1st

2nd

3rd

4th

5th

ab

c

cd

a

bd

c

d

a

bc

ac

d

ab

b

d

and the new Borda scores scores are a: 52 + 12 + 1 + 3 + 12 = 15 2 b: 52 + 12 + 0 + 32 + 52 = 14 2 c: 1 + 3 + 52 + 32 + 12 = 17 2 d: 0 + 2 + 52 + 0 + 52 = 14 2 We eliminate b and d, and the Hare–Borda social choice set is {a, c} as can be seen from the graph representation.  

84

Chapter 1 • Social Choice

1 1.6.2 More Condorcet-Type Classifications of the Alternatives Example 1.6.2 The following are the rankings of five alternatives a, b, c, d, and e by four voters, and the corresponding graph representation. 1st

2nd

3rd

4th

a

e

d

abe

b

c

c

cd

cd

ab

ab

e

d

e

There is no Condorcet winner or loser in this scenario. With close examination, you can see that in the Condorcet tournament: 1. a beats each of b, d, and e; and ties with c. 2. b beats each of d and e; ties with c; and loses to a. 3. c ties with all of its opponents. 4. d ties with each of c and e; and loses to each of a and b. 5. e ties with each of c and d; and loses to each of a and b. Motivated by the above five observations, we make the following definitions that lie between  the two extremes of a Condorcet winner and a Condorcet loser (⊡ Fig. 1.13).

Definition 1.6.3 An alternative is said to be 1. a weak Condorcet winner if it loses to no opponent in the Condorcet tournament, beats at least one and ties with at least one. (a in Example 1.6.2.) 2. a Condorcet neutralizer if it ties with all its opponents in the Condorcet tournament. (c in Example 1.6.2.)

 

 

(continued ) ⊡ Fig. 1.13 Graph representation of Example 1.6.2

a e

b d c

1

85 1.6 · Social Choice Procedures: Indifference and Ties Allowed

Definition 1.6.3 (continued) 3. a strong Condorcet loser if it beats no opponent in the Condorcet tournament, loses to at least one and ties with at least one.   (d and e in Example 1.6.2.) (b in Example 1.6.2 is “none of the above.”)

Proposition 1.6.4 No weak Condorcet winner or Condorcet neutralizer can exist in the presence of a Condorcet winner. Proof This is clear since a Condorcet winner must beat each of his opponents, leaving no possibility for a weak Condorcet winner or a Condorcet neutralizer to exist.   Proposition 1.6.5 1. Let j be the number of weak Condorcet winners in a scenario with k ≥ 3 alternatives. Then the only possible values for j are 0, 1, 2, · · · , k − 1. 2. A weak Condorcet winner cannot exist in the case of exactly two alternatives. Proof 1. The requirement that a weak Condorcet winner must beat at least one opponent rules out the possibility of j = k. A complete directed graph with k vertices all of its arrows are bidirectional proves the case j = 0. For the case of exactly one weak Condorcet winner, construct a graph with 3 of the k vertices, say a, b, c, are such that a → b, b → c, and c ↔ a, then draw a unidirectional arrow from each of a, b, and c to each of the remaining k − 3 vertices. This makes a the only weak Condorcet winner. For 2 ≤ j ≤ k − 1, construct a complete directed graph with j vertices all of its arrows are bidirectional, then augment this graph with k − j vertices and unidirectional arrows from each of the j vertices to each of the k − j vertices. Each of the graphs constructed in this proof can be realized by a scenario of voting, thanks to McGarvey’s theorem (Theorem 1.3.21).   2. This is clear.   Proposition 1.6.6 Let a and b be two weak Condorcet winners in a given scenario, then a ↔ b in the Condorcet tournament. Proof Neither one can lose to the other and therefore we must have a ↔ b.

 

Next, we shall adapt the results in Sects. 1.4.3, 1.4.4, and 1.4.5 to accommodate the case of indifference by voters and cover the Condorcet-type concepts of Definition 1.6.3.

86

Chapter 1 • Social Choice

1 ⓘ Remark 1.6.7 Given two alternatives a and x we introduced the notation #(a ↓ x, i) in 1.4.C to represent the number of times a beats x in the ith voter’s ballot. To accommodate the above generalization of Borda count that allows indifference, we need to modify the meaning of this notation as follows: #(a ↓ x, i) = 0 if the ith voter prefers x to a or x = a. #(a ↓ x, i) = 12 if the ith voter ranks a and x in the same tier. #(a ↓ x, i) = 1 if the ith voter prefers a to x. You can verify that, as in Notations 1.4.7, we still have The Borda score of a

=

 i

#(a ↓ x, i) =

x

 x

#(a ↓ x, i) =

i



#(a ↓ x).

x

 where #(a ↓ x) = i #(a ↓ x, i) is the score of a against x in the two-alternative majority rule (as in the second Remark in 1.5.14).

 

As noted in Proposition 1.4.8, x beats y in the Condorcet tournament if and only if #(x ↓ y) > n2 where n is the number of voters. For this to happen, #(x ↓ y) must be at least 12 more than n2 . Similarly, x loses to y in the Condorcet tournament if and only if #(x ↓ y) is at least 12 less than n2 . This leads to the following (sharper) version of Proposition 1.4.8. Proposition 1.6.8 Given two alternatives x and y in an n-voter scenario, 1. x beats y in the Condorcet tournament if and only if #(x ↓ y) ≥ n+1 2 · 2. x loses to y in the Condorcet tournament if and only if #(x ↓ y) ≤ n−1 2 · 3. x and y are tied in the Condorcet tournament if and only if #(x ↓ y) = n2 ·

ⓘ Corollary 1.6.9 In a scenario with k alternatives and n voters, 1. If c is a Condorcet winner, then The Borda score of c ≥

(k − 1)n k − 1 + · 2 2

2. If c is a weak Condorcet winner, then The Borda score of c ≥

(k − 1)n 1 + · 2 2

3. If c is a Condorcet neutralizer, then The Borda score of c =

(k − 1)n · 2

4. If c is a strong Condorcet loser, then The Borda score of c ≤

(k − 1)n 1 − · 2 2

1

87 1.6 · Social Choice Procedures: Indifference and Ties Allowed

5. If c is a Condorcet loser, then The Borda score of c ≤

(k − 1)n k − 1 − · 2 2

Proof 1. If c is a Condorcet winner, it follows from Proposition 1.6.8 that c must score at least n+1 2 points against each of its k − 1 opponents in the Condorcet tournament. Now from Remark 1.6.7 we have The Borda score of c ≥

(k − 1)(n + 1) (k − 1)n k − 1 = + · 2 2 2

  2. To be a weak Condorcet winner, c must beat at least one opponent in the Condorcet tournament and tie with the rest. Therefore, from Proposition 1.6.8 and Remark 1.6.7 we have The Borda score of c =



#(c ↓ x) ≥

x

(k − 1)n 1 (k − 2)n n + 1 + = + · 2 2 2 2

  3. To be a Condorcet neutralizer, c must tie with each of its k−1 opponents in the Condorcet tournament. The result follows from Proposition 1.6.8 and Remark 1.6.7.   4. To be a strong Condorcet loser, c must lose to at least one opponent in the Condorcet tournament and tie with the rest. Therefore, from Proposition 1.6.8 and Remark 1.6.7 we have The Borda score of c =

 x

#(c ↓ x) ≤

(k − 1)n 1 (k − 2)n n − 1 + = − · 2 2 2 2

  5. If c is a Condorcet loser, it follows from Proposition 1.6.8 that c must score at most n−1 2 points against each of its k − 1 opponents in the Condorcet tournament. Now from Remark 1.6.7 we have The Borda score of c ≤

(k − 1)(n − 1) (k − 1)n k − 1 = − · 2 2 2  

Theorem 1.6.10 In a scenario with k alternatives and n voters 1. Neither a Condorcet winner nor a weak Condorcet winner can obtain the lowest score in the Borda count. (There will always be an opponent with a lower score.)

(continued )

88

Chapter 1 • Social Choice

1 Theorem 1.6.10 (continued) 2. If a Condorcet neutralizer c obtains the lowest score in Borda count, then each alternative obtains the same score as c, which is (k − 1) n2 points. 3. If a Condorcet neutralizer c obtains the highest score in Borda count, then each alternative obtains the same score as c, which is (k − 1) n2 points. 4. Neither a Condorcet loser nor a strong Condorcet loser can obtain the highest score in the Borda count. (There will always be an opponent with a higher score.)

Proof 1. This follows from a similar argument to the one used in Theorem 1.4.11.   2. Let c be a Condorcet neutralizer. By Corollary 1.6.9, the Borda score of c is (k − 1) n2 . If each of the k − 1 opponents of c scores at least as high as c and at least one opponent scores strictly higher than c, the sum of scores of all alternatives will be greater than   k(k − 1) n2 , in contradiction to Lemma 1.4.10. 3. This is analogous to Part 2. 4. This follows from a similar argument to the one used in Theorem 1.4.11.  

1.6.3 The Social Choice Set of the Hare–Borda Procedure The following remark is an immediate consequence of the independence of irrelevant alternatives, satisfied by the Condorcet tournament. ⓘ Remark 1.6.11 In the course of eliminations in the Hare–Borda procedure 1. 2. 3. 4.

A Condorcet winner maintains its Condorcet winner status. A Condorcet neutralizer maintains its Condorcet neutralizer status. A Condorcet loser maintains its Condorcet loser status. A weak Condorcet winner will turn into a Condorcet neutralizer if the opponents it defeats in the Condorcet tournament are eliminated, otherwise, it remains a weak Condorcet winner, or becomes a Condorcet winner. 5. A strong Condorcet loser will turn into a Condorcet neutralizer if the opponents it loses to in the Condorcet tournament are eliminated, otherwise, it remains a strong Condorcet loser, or becomes a Condorcet loser.

Theorem 1.6.12 In the Hare–Borda procedure 1. If a Condorcet winner exists, it is the sole member of the choice set.

(continued )

1

89 1.6 · Social Choice Procedures: Indifference and Ties Allowed

Theorem 1.6.12 (continued) 2. The social choice set includes all Condorcet neutralizers and weak Condorcet winners that may exist. 3. A strong Condorcet loser may or may not be included in the social choice set. 4. A Condorcet loser cannot be included in the social choice set.

Proof 1. This is a restatement of Theorem 1.4.13.   2. This is a consequence of Parts 2 and 4 of Remark 1.6.11 and Parts 1 and 2 of Theorem 1.6.10.   3. This will be demonstrated by Examples 1.6.13 and 1.6.14 below.   4. This is a consequence of Part 3 of Remark 1.6.11 and Part 3 of Theorem 1.6.10.   Example 1.6.13 The following scenario shows the rankings of three alternatives a, b, and c by five voters. 1st

2nd

3rd

4th

5th

a

c

b

a

ac

b

a

c

b

b

c

b

a

c

(The corresponding graph representation is shown in ⊡ Fig. 1.14.) Find the Hare–Borda social choice set. It is clear that a is a weak Condorcet winner and c is a strong Condorcet loser. The Borda count scores are: a: 2 + 1 + 0 + 2 + 32 = 13 2 b: 1 + 0 + 2 + 1 + 0 = 82 c: 0 + 2 + 1 + 0 + 32 = 92 ⊡ Fig. 1.14 Graph representation of Example 1.6.13

a

c

b

90

Chapter 1 • Social Choice

1 ⊡ Fig. 1.15 Graph representation of Example 1.6.14

a e

b d c

The elimination of b will turn both a and c into Condorcet neutralizers; and the Hare–Borda social choice set is {a, c}. Note how the strong Condorcet loser c ended up in the social choice set of the Hare–Borda procedure after turning into a Condorcet neutralizer (⊡ Fig. 1.15).  Example 1.6.14 The following scenario shows the rankings of five alternatives a, b, c, d, and e by four voters, and the corresponding graph representation. 1st

2nd

3rd

4th

a

e

d

abe

b

c

c

cd

cd

d

a

e

ab

be

You can see that a is a weak Condorcet winner. b is a strong Condorcet loser. c is a Condorcet neutralizer. d is a Condorcet neutralizer. e is a strong Condorcet loser. We find the Hare–Borda social choice set. 1. The Borda count scores are Alternative

a

b

c

d

e

Borda score

9.5

7

8

8

7.5

1

91 1.6 · Social Choice Procedures: Indifference and Ties Allowed

2. The updated scenario after the elimination of b is 1st

2nd

3rd

4th

a

e

d

ae

c

c

cd

cd

d

a

e

a

e

and the new Borda scores are Alternative

a

c

d

e

Borda score

6.5

6

6

5.5

The graph representation is in ⊡ Fig. 1.16(1). a is a weak Condorcet winner. c is a Condorcet neutralizer. d is a Condorcet neutralizer. e is a strong Condorcet loser. 3. The updated scenario after the elimination of e is 1st

2nd

3rd

4th

a

c

d

a

d

c

cd

a

a

cd

and the new Borda scores are Alternative

a

c

d

Borda score

4

4

4

The graph representation is in ⊡ Fig. 1.16(2). a turns from a weak Condorcet winner into a Condorcet neutralizer. c is a Condorcet neutralizer. d is a Condorcet neutralizer. The Hare–Borda procedure social choice set is {a, d, c}. Note that the strong Condorcet losers b and e did not make it to the Hare–Borda social choice set in this example. 

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Chapter 1 • Social Choice

1 ⊡ Fig. 1.16 Graph representations of the elimination steps in Example 1.6.14

a a

e

c

d

c

d

(1)

(2)

Example 1.6.15 We leave it to the student to verify that the Hare–Borda social choice set for the following scenario is {a, b, c, d} and that a is a Condorcet neutralizer, while none of b, c, d falls under any of our Condorcet-type classifications in Definition 1.6.3. 1st

2nd

3rd

4th

5th

6th

b

ad

c

ab

d

ac

c

b

d

c

b

d

ad

c

ab

d

ac

b



1.7

Manipulability of Social Choice Procedures: Indifference and Ties Allowed

It is often possible for a voter to get a more favorable outcome for herself by misrepresenting her preference. For example, we saw in the course of our discussion of the Florida 2000 presidential election (in Sect. 1.1.3). that, when the plurality procedure is used, a voter whose top choice is a minor candidate with a slim chance of winning may very well get a better outcome for herself if she bypasses her top choice and votes for her most preferred candidate among those with a realistic chance of winning. In Example 1.2.14 and Remark 1.3.20 we saw an instance of how the Hare procedure can be manipulated by voters through insincere reporting of preferences to bring a better outcome. Example 1.2.15 can also be transformed as in Remark 1.3.20 to illustrate the possibility of manipulating the Coombs procedure. The manipulability of Borda count is illustrated in Example 1.4.6. A social choice procedure is said to be resolute if its social choice set is always a singleton, i.e., includes only one alternative. Most of the study of manipulability deals with resolute social choice procedures, with only one manipulator among the voters. The best known result in this regard is the Gibbard–Satterthwaite theorem, which was independently proven by Gibbard in 1973 and Sattarthwaite in 1975. Different

1

93 1.7 · Manipulability of Social Choice Procedures: Indifference and Ties. . .

versions of this theorem, with different proofs, are abundantly available in the literature and we shall therefore bypass it for the less known theorem proven [7] by Peter Gärdenfors in 1976, which deals with non-resolute social choice procedures and only one manipulator. (See [17] for more on Gärdenfors’ theorem; and [6, 10, 16] for the Gibbard–Satterthwaite theorem.) Notation 1.7.1 1. We shall follow Notations 1.5.2 with some changes. The small Greek letters λ, μ, ν, · · · will denote social choice functions (procedures) instead of social welfare and semi social welfare functions. Given a scenario  and a social choice function λ, λ will denote the social choice set that results from applying the social choice function λ to the scenario . 2. Given two alternatives x and y we shall write • x →t y if voter t prefers x to y. • x ↔t y if the voter t places x and y in the same tier. (To avoid complicating the notations, the scenario in which this occurs will be verbally specified instead of adding it to the notations.) 3. Given a scenario  and a specific voter t,  t will denote any scenario such that • The ranking of the alternatives by voter t in  t is different from his ranking of the alternatives in . • For s = t, the rankings by voter s in  and  t are the same. 4. When the rankings by voter t in different scenarios are considered, we shall speak of the voter t -ranking, the voter t -ranking, and so on.

Definition 1.7.2 A social choice function μ is said to be 1. manipulable by voter t at scenario  if there exists a scenario  t (as in Notations 1.7.1) such that voter t prefers μ t to μ, based on his “initial” -ranking. 2. manipulable if there exists a scenario  at which μ is manipulable by some voter. 3. stable or Strategy-proof if it is not manipulable.

     

If a stable social choice function is used, a voter has no incentive to submit an insincere ranking of the alternatives since the ranking he reveals will get him the best possible outcome according to that ranking, under the circumstances imposed by the other voters’ rankings.

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Chapter 1 • Social Choice

1 1.7.1 Comparing Sets of Alternatives The definition of manipulability in Sect. 1.7.2 can be directly applied to resolute social choice functions without any ambiguity involved. This is because each of the outcomes μ i and μ is a single alternative; and the comparison of alternatives by the ith voter is readily available. The case of non-resolute social choice functions is different because μ and μ i are not necessarily singletons and we do not have a concept of comparing sets of alternatives that are not singletons. Gärdenfors [7] used an existing ranking on a set of alternatives to induce a comparison of subsets of alternatives which we present next. We shall find it convenient to use some standard set-theoretic notations. Let A and B be sets of objects (elements). 1. x ∈ A means that x is a member of A. (∈ stands for “in” or “is in.”) 2. A ⊂ B (A is a proper subset of B) means that every member of A is also a member of B but there is at least one member of B that is not in A. 3. A ∩ B (the intersection of A and B) is the set of all x such that x ∈ A and x ∈ B. 4. Given a finite collection of sets Ai , i = 1, 2, · · · , k, the set of all x such that x ∈ Ai for all i = 1, 2, · · · , k is denoted ∩ki=1 Ai . 5. A ∪ B (the union of A and B) is the set of all x such that x ∈ A or x ∈ B (or both). 6. Given a finite collection of sets Ai , i = 1, 2, · · · , k, the set of all x such that x ∈ Ai for at least one i, i = 1, 2, · · · , k, is denoted ∪ki=1 Ai . 7. A\B is the set of all x in A but not in B. 8. A = B means that every x in A is also in B and every x in B is also in A. 9. A ⊆ B (A is a subset of B) means that A ⊂ B or A = B. 10. The set that does not have any member, called the empty set, will be denoted ∅. 11. We say that A and B are disjoint if A ∩ B = ∅.

Definition 1.7.3 Let A, B be two subsets of a set S of alternatives on which a ranking is defined, and assume that A = B. We say that A is preferred over B, and we write A  B, in any of the three following cases: Case 1. A ⊂ B and for every x ∈ A and y ∈ B\A we have x  y, with x → y for at least one such x and one such y. Case 2. B ⊂ A and for every x ∈ A\B and y ∈ B we have x  y, with x → y for at least one such x and one such y. Case 3. Neither A\B nor B\A is empty and for every x ∈ A\B and y ∈ B\A we   have x  y, with x → y for at least one such x and one such y.

1

95 1.7 · Manipulability of Social Choice Procedures: Indifference and Ties. . .

B

A

A

c

A

B

f

c

c

B

f

f a

b g

g Case 1

e Case 2

Case 3

⊡ Fig. 1.17 Example 1.7.4

Example 1.7.4 We start with the following ranking ab cd e fg h According to Definition 1.7.3, A  B in each of the following three cases (⊡ Fig. 1.17): Case 1. A = {c, f } and B = {c, f, g}. Case 2. A = {c, f, e} and B = {c, f }. Case 3. A = {a, c, f } and B = {b, c, f, g}. The intuition behind Definition 1.7.3 is that A  B when, on “average,” a member of A is preferred over a member of B. Note that not every A and B are comparable by the concept introduced in Definition 1.7.3. Consider, for example, A = {b, c, f, g} and B = {a, c, f, e}. We only need to check Case 3 since none of the two sets is a subset of the other. We have • b ∈ A\B, e ∈ B\A and b → e. • a ∈ B\A, g ∈ A\B and a → g. and therefore neither set is preferred over the other (⊡ Fig. 1.17).



1.7.2 Anonymity and Neutrality In Notations 1.5.18 we used p to denote a scenario  after the rankings of the voters have been permuted. We find it more convenient here to replace the notation p with p; and we shall also use q to denote  after a certain permutation q has been applied to the alternatives. If, in a certain scenario, alternatives a, b, c, d are permuted so that, in all rankings, c replaces a, a replaces b, d replaces c, and b replaces d, we write q(a) = c, q(b) = a, q(c) = d and q(d) = b. Given a set of alternatives such as

96

Chapter 1 • Social Choice

1 A = {a, b} we shall write q(A) = q({a, b}) = {q(a), q(b)} = {c, a} (or {a, c}). We can also represent q as follows: a b q= c d

→ → → →

c a d b

We shall always use p to denote a permutation of the voters and q to denote a permutation of the alternatives. Definition 1.7.5 A social choice function μ is said to be 1. anonymous if μ(p) = μ for every scenario  and every permutation p of the voters’ rankings. (See Notations 1.5.18 and Definitions 1.5.19.)   2. neutral if μ(q) = q(μ) for every scenario  and every permutation q of the alternatives.

In the sequel, all social choice functions are non-resolute and voters are allowed to express indifference in their rankings. Further, in Proposition 1.7.6 through 1.7.11, we shall have only three alternatives (named a, b, c) and three voters (named s, t, u). Proposition 1.7.6 Consider the following scenario for the voters s, t, u and the alternatives a, b, c s t u b ab a = c c c a b and assume that the social choice function μ is anonymous and neutral. 1. If one of a and b is in μ, then so is the other. 2. μ is {c}, {a, b, c} or {a, b}.

97 1.7 · Manipulability of Social Choice Procedures: Indifference and Ties. . .

Proof 1. Let p be the permutation of the voters’ rankings and q be the permutation of the alternatives given by s →  u p= t →  t u→  s

a→  b and q = b →  a· c→  c

It is easy to check that  = pq and therefore, μ = μ(pq) = μ(q) = q(μ) where the second and the third equalities follow from the anonymity and neutrality of μ, respectively. If a ∈ μ, then by the above equality we see that both a and b are in μ.   2. From Part 1, μ includes both a and b or neither of the two.  

ⓘ Lemma 1.7.7 The social choice set from a Condorcet paradox scenario by a neutral and anonymous social choice function consists of all the three alternatives involved. Proof Consider the Condorcet paradox scenario s a = b c

t b c a

u c · a b

Let μ be a neutral and anonymous social choice function and let p be the permutation of the voters’ rankings and q be the permutation of the alternatives given by s →  u p= t →  s u→  t

a→  b and q = b →  c· c→  a

It is easy to check that  = pq and therefore, μ = μ(pq) = μ(q) = q(μ) where the second and the third equalities follow from the anonymity and neutrality of μ, respectively. If a ∈ μ, then by the above “triple” equality we see that b ∈ μ as well. Next, starting with b ∈ μ and applying the above argument a second time we get c ∈ μ. We have thus shown that the presence of a in μ implies the presence of both b and c in μ. In a similar fashion, we can start with b ∈ μ or c ∈ μ and conclude that μ = {a, b, c}.  

1

98

Chapter 1 • Social Choice

1 1.7.3 Gärdenfors’ Theorem The following sequence of four propositions constitute Gärdenfors’ proof of his 1976 theorem on the manipulability of social choice functions [7]. The central idea of the proof is to consider a certain scenario (denoted below by ) for three voters and three alternatives, and examine all possible choice sets μ where μ is a social choice function that satisfies anonymity, neutrality, and the Condorcet winner criterion. For each of those μ possibilities, the manipulability of μ is demonstrated. Proposition 1.7.8 Assume that the social choice function μ satisfies the Condorcet winner criterion and assume that c ∈ μ where s t u c ab a = · b c c a b Then μ is manipulable. Proof t t a If voter t changes his -ranking from a b to , we get the new scenario b c c s c  = b a t

t a b c

u a c b

whose choice set by μ is μt = {a} since μ satisfies the Condorcet winner criterion. Clearly, t in the -ranking of t, namely a b , {a} is preferred to any set containing c (Definition 1.7.3), c and therefore μ is manipulable by t at scenario .

 

Proposition 1.7.9 Assume that the social choice function μ is neutral and anonymous. Let  be as in Proposition 1.7.8, that is, s t u c ab a = · b c c a b

99 1.7 · Manipulability of Social Choice Procedures: Indifference and Ties. . .

If μ = {a, b}, then μ is manipulable. Proof Consider the Condorcet paradox scenario s c = b a

t b a c

u a · c b

t t b By Lemma 1.7.7, we have μ = {a, b, c}. If voter t changes his -ranking from to a b , a c c t the resulting scenario will be identical with scenario  whose choice set by μ is {a, b}. Clearly, in the -ranking of t, {a, b}  {a, b, c} and therefore μ is manipulable by voter t at scenario .   Proposition 1.7.10 Assume that the social choice function μ satisfies anonymity, neutrality, and the Condorcet winner criterion. Let  be as in Proposition 1.7.8, that is, s t u c ab a = · b c c a b If μ = {b}, then μ is manipulable. Proof u u a c If voter u changes her -ranking from to , we get the new scenario c a b b s t u c ab c  = b c a a b u

whose choice set by μ is μu = {c} since μ satisfies the Condorcet winner criterion. It follows that μ is manipulable by voter u at scenario  since {c}  {b} in the -ranking of voter u.  

1

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Chapter 1 • Social Choice

1 Proposition 1.7.11 Assume that the social choice function μ is anonymous and neutral. Let  be as in Proposition 1.7.8, that is, s t u c ab a = · b c c a b If μ = {a}, then μ is manipulable. Proof s s c b If voter s changes her -ranking from to , the resulting s scenario will be identical b c a a with scenario  of Proposition 1.7.6. By Part 2 of Proposition 1.7.6 we then have μs = {c}, μs = {a, b, c} or μs = {a, b}, each of which is preferred to {a} in the -ranking of voter s. Therefore, μ is manipulable by voter s at scenario .  

Theorem 1.7.12 (Gärdenfors [7]) A neutral, anonymous social choice function that satisfies the Condorcet winner criterion and is defined for at least three voters and at least three alternatives is manipulable.

Proof In Propositions 1.7.8 through 1.7.11 we considered all possible social choice sets for a certain scenario of three alternatives and three voters, denoted , by an arbitrary social choice function μ that satisfies anonymity, neutrality, and the Condorcet winner criterion. In each case of a possible social choice set, we demonstrated the manipulability of μ. This proves the theorem for three alternatives and three voters. The following table shows all possibilities for the social choice set μ and the Proposition that demonstrates the manipulability of μ for each possibility. μ

Proposition

{a}

1.7.11

{b}

1.7.10

{c}

1.7.08

{a, b}

1.7.09

{a, c}

1.7.08

{b, c}

1.7.08

{a, b, c}

1.7.08

101 1.7 · Manipulability of Social Choice Procedures: Indifference and Ties. . .

The case of m alternatives and n voters with m > 3 or n > 3 (or both) can be established by augmenting the scenarios in Propositions 1.7.6 through 1.7.11 as in the following two steps: Step 1: If m > 3, we add m − 3 dummy alternatives, x4 , x5 , · · · , xm , all tied (i.e., in one tier), below a, b, c in the rankings of voters s, t, u. If m = 3, no dummy alternatives will be added. Step 2: If n > 3, we add n − 3 dummy voters, v4 , v5 , · · · , vm , such that each of these dummy voters places a, b, c, x4 , x5 , · · · , xm (a, b, c if no dummy alternatives are added in Step 1) all in one tier. If n = 3, no dummy voters are added. We leave it to the student to verify that the changes in Steps 1 and 2 do not affect the validity of the proofs of Propositions 1.7.6 through 1.7.11. This completes the proof.  

To illustrate Steps 1 and 2 in the proof of Theorem 1.7.12, we present the augmented version  of scenario  in the case of five voters (s, t, u, v, w) and six alternatives (a, b, c, x, y, z): s t u v w c ab a abcxyz abcxyz = b c c a b xyz xyz xyz To better understand the proof of Gärdenfors’ theorem, we show in the next example how it works in the case of the Copeland social choice procedure, which obviously satisfies anonymity, neutrality, and the Condorcet winner criterion. Example 1.7.13 Part 2 of Proposition 1.7.6 presents all possibilities for μ for the given scenario  by an arbitrary anonymous and neutral social choice function μ. You can easily verify that the actual possibility for μ is {a, b} when μ is the Copeland social welfare function. You can also verify that for scenario  considered in Propositions 1.7.8, 1.7.9, 1.7.10, and 1.7.11, μ = {a} as in Proposition 1.7.11. The suggested change of the -ranking of s s c b voter s from the sincere to the insincere results in scenario  above, whose Copeland b c a a choice set is {a, b}. Clearly, {a, b} is better than {a} from the point of view of someone whose c sincere and true ranking is b .  a

1

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Chapter 1 • Social Choice

1 ⓘ Remark 1.7.14 1. In the case of a Condorcet paradox scenario for three alternatives, Lemma 1.7.7 formally delivers the intuitive and natural choice by a neutral and anonymous social choice function, which is the set of all three alternatives. Similarly, Proposition 1.7.6 implies that a neutral and anonymous social choice function cannot pick one of a or b without picking the other too. The indispensability of anonymity and neutrality in most social choice problems thus weighs in favor of Gärdenfors’ theorem as it deals with social choice functions that are neutral, anonymous, and non-resolute. On the other hand, the Gibbard– Satterthwaite theorem deals only with resolute social choice functions and therefore does not postulate anonymity or neutrality. A common version of the Gibbard–Satterthwaite theorem reads as follows:

»

Assume that indifference is not allowed in the voters’ rankings and that there are at least three alternatives and three voters. Then, the only resolute, weakly Pareto efficient and Strategy-proof social choice functions are dictatorships.

Note When indifference is not allowed, a social choice function δ is said to be dictatorial or a dictatorship if there exists a voter v, called the dictator, such that for every scenario , δ = {a} where a the unique alternative on the top of the dictator’s ranking. 2. The scenario  whose manipulability is established in the proof of Gärdenfors’ theorem includes voters’ indifference. Therefore, the theorem does not cover the case where indifference is not permitted. In the same paper [7] of 1976, Gärdenfors includes the following theorem which we state without proof.

Theorem 1.7.15 Assume that 1. There are at least three alternatives and three voters. 2. Scenarios are restricted to those that prohibit indifference in the voters’ rankings. We define the social choice function κ as follows: κ = {c} if a majority alternative c exists in , and κ is the entire set of alternatives otherwise. Then κ is a strategy-proof social choice function that satisfies anonymity, neutrality, and the Condorcet winner criterion.  

1.7.4 Manipulability of Social Choice Functions Not Covered by Gärdenfors’ Theorem Two of the social choice functions studied in this chapter, namely the Copeland and the Hare–Borda social choice functions, satisfy all of the conditions of Gärdenfors’ theorem and their manipulability is thus established. The following examples demonstrate the

103 1.7 · Manipulability of Social Choice Procedures: Indifference and Ties. . .

manipulability of some standard social choice functions that do not meet all of the postulates of the theorem. Example 1.7.16 Let λ be the plurality social choice function and consider the following “sincere” scenario for the 5 voters s, t, u, v, w and the three alternatives a, b, c. s a  = b c

t b a c

u b a c

v c a b

w c a b

s a Clearly, λ = {b, c}, but if voter s changes his ranking from the sincere to the insincere b c s b we get the scenario a c

s

s b = a c

t b a c

u b a c

v c a b

w c a b

with λ s = {b}. Voter s thus benefits from voting insincerely since {b} is preferred to {b, c} according to the sincere ranking of voter s.  Example 1.7.17 Let λ be the Hare social choice function and consider the following “sincere” scenario for three alternatives a, b, c and 21 voters. 9 a  = b c

5 b c a

6 c a b

s a b c

The last column represents the sincere ranking by voter s which will be our manipulator. You s a can easily verify that λ = {c}, but if voter s changes his ranking from the sincere to the b c

1

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Chapter 1 • Social Choice

1 s b insincere we get the scenario a c

s

9 a = b c

5 b c a

6 c a b

s b a c

with social choice set λ s = {a} which is preferred to {c} according to the sincere ranking of s. 

Note that Example 1.7.17 is actually a modification that conforms with Definitions 1.7.2 (which involves only one manipulator) of the example with two manipulators discussed in Remark 1.3.20. Example 1.7.18 Consider the following sincere Condorcet paradox scenario for 3 alternatives a, b, c and 3 voters s, t, u. Let λ be the agenda voting with agenda abc. s a  = b c

t b c a

u c a b

Clearly λ = {c}, but if voter s switches a and b in his ranking we get the manipulated scenario



s

s b = a c

t b c a

u c a b

with λ s = {b}. As you can easily see, voter s prefers {b} to {c}.



Note that both plurality and Hare are neutral and anonymous but fail the Condorcet winner criterion; while agenda voting (with a fixed agenda) satisfies anonymity and the Condorcet winner criterion but does not meet the neutrality condition.

105 Exercises

Exercises In each of Scenarios 1, 2, 3, 4, 5, and 6, 1. Draw the graph representation of the Condorcet tournament and identify the Condorcet winner and the Condorcet loser that may exist. 2. Process the given scenario using the plurality procedure. 3. Process the given scenario using the plurality with run-off procedure. 4. Process the given scenario using Hare’s procedure 5. Process the given scenario using Coombs’ procedure 6. Identify the procedures that elected a Condorcet loser or failed to elect an existing Condorcet winner. Scenario 1 3

2

2

2

a

c

c

d

b

b

d

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d

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Scenario 2 5

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Scenario 3

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Chapter 1 • Social Choice

1 Scenario 4 3

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Scenario 5 6

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Scenario 6 8

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7. Show that in the following incomplete scenario (for 13 voters and 5 candidates a, b, c, d, e) there is a Condorcet winner that loses in all of the standard procedures of plurality, plurality with run off, Hare, Coombs, and Borda count. 1st

2nd

3rd

4th

5th

6th

7th

8th

9th

10th

11th

12th

13th

a

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107 Exercises

8. In the following incomplete scenario for 11 voters and 4 candidates a, b, c, and d,

1. 2. 3. 4.

1st

2nd

3rd

4th

5th

6th

7th

8th

9th

10th

11th

b

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Identify a Condorcet winner. Identify a plurality procedure winner. Identify a Hare procedure winner. Identify a Coombs procedure loser.

9. The following is a voting scenario for 59 voters and 5 candidates a, b, c, d, and e. 14

13

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1. Draw the graph representation of the Condorcet tournament and search for a Condorcet winner and a Condorcet loser. 2. Rather than using any of the standard social choice procedures, they invented their own procedure, call it the top-averaging procedure, which we now explain. Since there are 59 voters and five candidates, the average number of top choice votes for each candidate is 59/5 = 11.8. Each candidate with less than the threshold of 11.8 top choice votes is eliminated and the process is repeated with the threshold adjusted for the remaining number of candidates. Continue until all the remaining candidates have equal numbers of top choice votes. These candidates constitute the social choice set of the top-averaging procedure. Process the given scenario using this procedure. 3. Show that the top-averaging procedure coincides with the two-alternative majority rule in the case of only two alternatives. 4. Does the top-averaging procedure satisfy the Condorcet winner criterion? 10. Use Hare’s procedure to process the voting scenario in Exercise 9. 11. Use the Coombs’ procedure to process the voting scenario in Exercise 9. 12. The following two voting scenarios for 21 voters and 3 candidates are given. Note that the only difference between the two scenarios is that, in scenario II, the last two voters improved the position of candidate a by switching his position with candidate c. This makes

1

108

Chapter 1 • Social Choice

1 scenario II more favorable to candidate a. Find the Hare winner in each scenario and compare the results. Do you notice a violation of monotonicity by Hare in this example?

Scenario I:

8 a . .

6 b a c

5 c b a

2 c a b

8 a . .

Scenario II:

6 b a c

5 c b a

2 a c b

13. The following two voting scenarios for 21 voters and 3 candidates are given. Note that the only difference between the two scenarios is that, in scenario II, the last two voters improved the position of candidate a by switching his position with candidate c. This makes scenario II more favorable to candidate a. Find the Coombs winner in each scenario and compare the results. Do you notice a violation of monotonicity by Coombs in this example?

Scenario I:

8 a c b

7 b a c

4 c b a

2 b c a

8 a c b

Scenario II:

7 b a c

4 c b a

2 b a c

14. Prove that the plurality procedure is monotone. 15. Prove that the procedures of Hare and Coombs cannot make a Condorcet loser the sole winner if they are modified so that only one candidate is eliminated at each elimination step, but can still include a Condorcet loser in the social choice set. 16. Prove that the plurality procedure is Weakly Pareto efficient. 17. Prove that the plurality procedure does not satisfy the Condorcet loser criterion. 18. Draw the graph representation of the Condorcet tournament for the following 15 ballots and 4 alternatives, and (whenever possible) find an agenda for each alternative to win. If this is not possible for one or more alternatives, explain the reason. 4

4

3

2

1

1

d

c

d

a

b

a

a

b

b

b

d

d

c

a

a

c

a

b

b

d

c

d

c

c

19. Draw the graph representation of the Condorcet tournament for the following 15 ballots and 4 alternatives, and (whenever possible) find an agenda for each alternative to win. If this is not possible for one or more alternatives, explain the reason.

109 Exercises

4

4

3

2

1

1

d

c

b

a

b

a

c

b

d

b

d

d

a

a

a

c

a

b

b

d

c

d

c

c

20. Draw the graph representation of the Condorcet tournament for the following 15 ballots and 4 alternatives, and (whenever possible) find an agenda for each alternative to win. If this is not possible for one or more alternatives, explain the reason. 4

4

3

2

1

1

d

c

b

c

a

d

a

b

a

b

d

a

b

d

a

c

a

b

d

b

c

d

c

c

21. Prove that agenda voting is monotone. 22. 1. Which of the 5 alternatives in the following scenario can win agenda voting by a suitable choice of the agenda? a

b

c

b

c

e

d

d

a

e

a

d

c

e

b

2. As stated in Remarks 1.3.10, a Condorcet loser loses agenda voting regardless of the agenda adopted. Is the converse of this fact true? More precisely, is it always possible to find a winning agenda for a given non-Condorcet loser. 23. Suppose that an alternative x wins all agendas in which it is placed in the last position and the preceding alternatives are permuted in all possible ways. Is x necessarily a Condorcet winner? 24. 1. A committee of 21 members is to vote on a motion b vs the status quo a. Discussions reveal that b would win with ballots being as follows:

1

110

Chapter 1 • Social Choice

1 5

16

a

b

b

a

Where should a killer amendment c be placed in the ballots in order to save the status quo a? 2. Show that the pro b voters can amend c so that the new motion d succeeds in making b win. 3. Show that the pro a voters can amend d so that the new motion e succeeds in making a win. 25. Can a Pareto dominated alternative occupy the top of any preference ballot? Explain. 26. Identify all Pareto dominated alternatives in the following ballots and, whenever possible, find an agenda for each one of these alternatives to win. 4

3

2

d

a

b

c

d

a

e

c

c

b

e

e

a

b

d

27. Identify all Pareto dominated alternatives in the following ballots and, whenever possible, find an agenda for each one of these alternatives to win. 4

3

2

d

a

c

c

d

e

e

c

a

a

e

b

b

b

d

28. Show that once an agenda is adopted and fixed, agenda voting is not neutral (as in Definitions 1.7.5). 29. Use the Borda count to process the ballots in each of Exercises 1, 2, 3, 4, 5, 6. 30. Use the Hare–Borda procedure to process the ballots in each of Exercises 1, 2, 3, 4, 5, 6.

111 Exercises

31. Theorem 1.4.11 states that, in Borda count, a Condorcet winner cannot obtain the lowest score and a Condorcet loser cannot obtain the highest score. The two scenarios in this exercise demonstrate the “limits” of these two statements.

Scenario 1 :

7 a b c d e

1 b c d e a

5 d c b e a

Scenario 2 :

7 e d c b a

1 a e d c b

5 a e b c d

1. In Scenario 1, show that there is a Condorcet winner which ranks second to last in Borda count. 2. In Scenario 2, show that there is a Condorcet loser which ranks second in Borda count. 3. Process Scenario 1 using the Hare–Borda procedure and confirm Theorem 1.4.13. 32. Use the Borda count procedure to process the voting scenario in Exercise 09 above. Can you predict the outcome of the Hare–Borda procedure in advance? 33. 1. Draw the graph representation of the Condorcet tournament for the following scenario. 2

2

3

2

a

a

d

c

d

d

b

d

c

b

a

b

b

c

c

a

2. Show that the semi ranking of the Condorcet tournament is actually a ranking in this example. Does this ranking agree with the Borda count ranking? 34. 1. Prove that the Copeland social welfare function satisfies the weak Pareto axiom. 2. We showed by example (Example 1.5.7 above) that the Copeland social welfare function does not satisfy IIA. Can you obtain this result from Arrow’s theorem? 35. Allowing indifference in voters’ rankings, we introduce a new scoring social choice procedure in this exercise, call it the top-bottom procedure, as follows: From each voter’s ranking, alternatives in the top tier equally share 1 point, and alternatives in the bottom tier equally share −1 points. (One-tier ballots are dropped out.) All voters in the tiers in between receive 0 points each. The points are tallied and the social choice set consists of the alternatives with the highest score. 1. Show that the top-bottom social choice procedure coincides with the two-alternative majority rule in the case of only two alternatives.

1

112

Chapter 1 • Social Choice

1 2. Process the following voting scenario for 7 voters and 4 alternatives using the top-bottom procedure, Hare’s procedure and Coombs’ procedure. 2

2

2

1

b

c

d

b

a

a

a

c

d

b

c

d

c

d

b

a

3. Does the top-bottom social choice procedure satisfy the Condorcet winner criterion? 36. Let μ be the Hare–Borda social choice function. Demonstrate the manipulability of μ by comparing μ and μs in Proposition 1.7.11. 37. Show that a social choice function that is not singly monotone (as in Definition 1.2.16) is manipulable. 38. Let s and t be two specific members of the set V of voters. Let λ be a semi social welfare function such that: Given a scenario  and two alternatives x, y we have x →λ y if and only if one of the following three events occurs in : x →s y and x ↔t y, x ↔s y and x →t y or, x →s y and x →t y. Otherwise, x ↔λ y. Show that λ satisfies Collective quasi-rationality and that the set {s, t} is an oligarchy. 39. Consider the following voting scenario for 3 alternatives a, b, c and 3 voters s, t, u. s t u  = a bc ab bc a c and let λ be a social welfare function that satisfies Arrow’s axioms. Is it possible to have c λ = ? ab

113 Exercises

40. Let {a, b, c, d} be the set of alternatives and {s, t, u} be the set of voters. Consider the following two scenarios: s a  = b cd

t u b cd ad ab c

s t a abd  = c c bd

and

u c bd a

1. Find all pairs of alternatives x, y such that /{x, y} = /{x, y}. 2. Assuming that λ is the Borda count, find λ and λ. 3. Detect all violations of IIA 41. Assuming that μ is the Condorcet tournament, 1. Find μ and μ where  and  are as in Exercise 40. 2. Detect all violations of Collective rationality. In each of Exercises 42 and 43 you are asked to do the following: 1. Draw the graph representation of the Condorcet tournament of the given voting scenario and find any existing Condorcet winner, weak Condorcet winners, Condorcet neutralizers, strong Condorcet losers or Condorcet loser. 2. Using Theorem 1.6.12, determine the alternatives that have a place “reserved in advance” in the Hare–Borda social choice set. 3. Process the given voting scenario using Hare–Borda social choice procedure and find the actual Hare–Borda social choice set. 42. a

cde

a

bde

b

ade

b

ace

d

b

a

c

a

c

b

d

b

c

d

cde

b

bde

c

ade

c

ace

d

abe

c

43. 1

1

1

1

1

5

a

c

b

a

ac

d

b

a

c

b

b

c

b

a

c

d

d

d

d

d

abc

abe

1

114

Chapter 1 • Social Choice

1 44. Let us define the Hare–Copeland procedure following the same iteration steps as in the Hare–Borda procedure. At each step we eliminate the alternative(s) with the lowest point sum; and we continue till we have a set of alternatives with equal point sums. Show that Theorem 1.6.12 holds with the Hare–Copeland procedure replacing the Hare–Borda procedure. 45. Replace the Hare–Borda procedure with the Hare–Copeland procedure in Exercises 42, 43; and compare the social choice sets of the two procedures.

115

Yes-No Voting Sherif El-Helaly © Springer Nature Switzerland AG 2019 Sherif El-Helaly, The Mathematics of Voting and Apportionment, Compact Textbooks in Mathematics, https://doi.org/10.1007/978-3-030-14768-6_2

2.1

Introduction

This kind of voting is quite different from the voting studied in Social Choice. The main differences are summarized in the following points: 1. The number of alternatives to choose from: While the voter in the topic of social choice has several (usually more than two) alternatives to choose from, the voter in the context of yes-no voting has only two available alternatives: YES and NO. Examples of yes-no voting systems include committees and all instances in which a body of voters has to vote YES or NO on a motion, such as referendums. (For simplicity, abstention is ruled out in our study.) 2. The most important application of social choice theory is the election of a candidate to occupy a political office or to fill a position in a company. The alternatives in this case are the candidates who should be treated in a neutral fashion. In yes-no voting, the two alternatives (YES and NO) need not be treated with neutrality. For example, voting to amend the constitution of a nation is usually biased in favor of NO by requiring two thirds or three quarters of the voters to vote YES before the constitutional amendment can be adopted. On the other hand, to discuss a motion in a committee, only two YES votes are needed: one from the member who submits the motion and another from a member who seconds it. The committee, of course may consist of any number of members. 3. When applied to elections, social choice procedures need to be anonymous. This means that they treat the voters equally: one person one vote. In contrast, a yesno voting system may discriminate among the voters. The United Nations Security Council (UNSC), for example, is a yes-no voting system with fifteen member nations five of which (China, France, Russia, UK and USA) are endowed with the veto power which enables any of those five nations to block a motion by casting a NO vote on it. Another form of discrimination (which is not of concern to us here) in favor of those five nations is that their membership in the UNSC is permanent, while each of the remaining ten members serves for a 2-year term.

2

116

2

Chapter 2 • Yes-No Voting

2.1.1 The Basics Definition 2.1.1 A set consisting of some (or all) of the voters in a yes-no voting system is called a coalition. Voters are usually denoted by lowercase letters a, b, c, . . . , x, y, . . . and a coalition whose members are the voters a, b, c is denoted {a, b, c}. Notice that the order in which voters are listed in a coalition has no significance, for example, {a, b, c}, {a, c, b}, {b, a, c}, {b, c, a}, {c, a, b}, and {c, b, a} denote the same coalition. (Please note that using the term “coalition” here does not necessarily mean that its members have actually joined forces to support or oppose a certain motion. In our study, a coalition is simply a subset of the set of all voters, whether or not they agreed to vote in a certain manner. We may also denote a coalition by an uppercase letter. For instance, we may write A = {a, b, c}.) If each voter in A is also in B, we say that A is a subcoalition of B and B is a supercoalition of A. If, in addition, there is a voter in B that is not in A, we say that A is a proper subcoalition of B and B is a proper supercoalition of A. A coalition is said to be winning if a YES vote by all its members is sufficient to pass a motion, and is said to be losing if it is not winning.  

Our yes-no voting systems will be monotone: All supercoalitions of a winning coalition are winning. (Consequently, all subcoalitions of a losing coalition are losing.) The coalition of all voters in the system will always be winning. Example 2.1.2 Four people, a, b, c, and d established a new business. Their contributions to the capital of the business were as follows: a contributed 4 million dollars, b contributed 3 million dollars, c contributed 2 million dollars, and d contributed 1 million dollars. (The total is 10 million dollars.) They decided to run their business based on a weighted yes-no voting system in which every 1 million dollars of capital is equivalent to one point of weight. This means that a’s vote is worth four points of weight, b’s vote is worth three points of weight, c’s vote is worth two points of weight, and d’s vote is worth one point of weight. They decided to set the quota (the minimum weight required to pass a motion) at six points of weight. List all the winning coalitions.  Solution The one-voter coalitions are {a} with a weight of four points, {b} with a weight of three points, {c} with a weight of two points, and {d} with a weight of one point. None of these meets the quota of six. Two-voter coalitions: How can we go through all of these coalitions without missing any of them? We can start by fixing a and pairing each of b, c, d with it one at a time. Next, we fix b and pair each of c, d with it one at a time. We then fix c and pair d with it. Results are:

2

117 2.1 · Introduction

Coalition

Weight

Status

{a, b}

7

W

{a, c}

6

W

{a, d}

5

L

{b, c}

5

L

{b, d}

4

L

{c, d}

3

L

Three-voter coalitions: We can list those by leaving out one voter at a time. Results are: Coalition

Weight

Status

{a, b, c}

9

W

{a, b, d}

8

W

{a, c, d}

7

W

{b, c, d}

6

W

Finally, the coalition {a, b, c, d} of all voters is certainly winning with a weight of 10. We may now list all the winning coalitions as follows: {a, b} {a, c} {a, b, c} {a, b, d} {a, c, d} {b, c, d} {a, b, c, d} Definition 2.1.3 A member in a winning coalition is said to be critical in that coalition if its withdrawal causes the coalition to become losing. The number of winning coalitions in which a voter is critical is called the Banzhaf score of the voter.

 

The Banzhaf score and the Banzhaf index (to be discussed in detail later) were introduced by attorney John F. Banzhaf III in connection with a lawsuit involving the board of Nassau county, New York in the 1960s. Being critical in a winning coalition is a source of power since a critical voter in a winning coalition can turn the coalition into a losing one by withdrawing from it. A non-critical voter in a winning coalition does not have this power since his withdrawal does not turn the winning coalition into a losing one.

118

2

Chapter 2 • Yes-No Voting

Example 2.1.4 Identify all critical voters in each of the winning coalitions in Example 2.1.2 and find the Banzhaf score of each voter.  Solution Examine, for instance, the winning coalition {a, b, c}. Voter c is a member of this coalition but is not critical since its withdrawal results in the coalition {a, b} which also appears in the same list of all winning coalitions. Likewise, b is not critical in {a, b, c} since its withdrawal results in the coalition {a, c} which is also winning, being in the same list of all winning coalitions. On the other hand, a is critical in {a, b, c} since its withdrawal results in the coalition {b, c} which is losing since it is not in the list of all winning coalitions. The following table shows the critical voters in each of the winning coalitions: Winning coalition

Critical voters

{a, b}

a, b

{a, c}

a, c

{a, b, c}

a

{a, b, d}

a, b

{a, c, d}

a, c

{b, c, d}

b, c, d

{a, b, c, d}

None

The following table shows the Banzhaf score of each voter: Voter

a

b

c

d

Banzhaf score

5

3

3

1

2.1.2 The +/− Table If we are not interested in knowing which voter is critical in which coalition, but only in the Banzhaf score of each voter, we can use the following table (Taylor [24].) Each time a voter is present in a coalition, we enter a plus sign, otherwise, we enter a minus sign. (We do not check whether or not the voter is critical in the coalition.) The net of the plus and minus signs in the column of a voter yields the Banzhaf score of that voter.

Winning coalition

a

b

c

d

{a, b}

+

+

−

−

{a, c}

+

−

+

−

{a, b, c}

+

+

+

−

{a, b, d}

+

+

−

+

{a, c, d}

+

−

+

+

{b, c, d}

−

+

+

+

{a, b, c, d}

+

+

+

+

Net

5

3

3

1

2

119 2.1 · Introduction

How does this table work? Let us think of the plus sign a voter gets for being present (but not critical) in a winning coalition as “unmerited.” Let us also think of the plus sign a voter gets for being present (and critical) in a winning coalition as “merited.” • To see how the +/− table handles a voter that is present (but not critical) in a winning coalition, consider the case of voter c in the winning coalition {a, b, c}. Since c is not critical in {a, b, c}, the withdrawal of c from {a, b, c} leaves us with another winning coalition, namely {a, b}. Voter c gets a minus sign in the line of the winning coalition {a, b} which cancels the “unmerited” plus sign c gets in the line of the winning coalition {a, b, c}. • To see how the +/− table handles a voter that is present (and is critical) in a winning coalition, consider the case of voter c in the winning coalition {a, c, d}. Since c is critical in {a, c, d} the withdrawal of c from {a, c, d} leaves us with a losing coalition, namely {a, d}. Since {a, d} is losing, it does not have a line in the +/− table to enter a minus sign in. Therefore, the “merited” plus sign c gets in the line of the winning coalition {a, c, d} persists. Consequently, the net of plus signs and minus signs in the column of a voter (obtained by subtracting the number of minus signs from the number of plus signs) yields the Banzhaf score of the voter. Example 2.1.2 is an example of a kind of yes-no voting system which we now define: Definition 2.1.5 A yes-no voting system is said to be weighted if each voter is assigned a positive number called weight so that a coalition is winning if and only if the total weight of its members (called the weight of the coalition) is greater than or equal to a specific given number called the quota.  

Example 2.1.6 Find the Banzhaf score of each voter in the weighted system consisting of the four voters a, b, c, d with weights 6,4,2,1, respectively, if 1. The quota is 7. 2. The quota is 8. 3. The quota is 9.  Solution 1. Quota q = 7. There are eight winning coalitions.

120

Chapter 2 • Yes-No Voting

2

Winning coalition

Weight

a

b

c

d

{a, b}

10

+

+

−

−

{a, c}

8

+

−

+

−

{a, d}

7

+

−

−

+

{a, b, c}

12

+

+

+

−

{a, b, d}

11

+

+

−

+

{a, c, d}

9

+

−

+

+

{b, c, d}

7

−

+

+

+

{a, b, c, d}

13

+

+

+

+

Banzhaf score

N/A

6

2

2

2

2. Quota q = 8. The table for this part is the same as in Part 1 except that the two lines for coalitions {a, d} and {b, c, d} are eliminated since these two coalitions will no longer be winning when q = 8. There are six winning coalitions. Winning coalition

Weight

a

b

c

d

{a, b}

10

+

+

−

−

{a, c}

8

+

−

+

−

{a, b, c}

12

+

+

+

−

{a, b, d}

11

+

+

−

+

{a, c, d}

9

+

−

+

+

{a, b, c, d}

13

+

+

+

+

Banzhaf score

N/A

6

2

2

0

3. Quota q = 9. The table for this part is the same as in Part 2 except that the line for coalition {a, c} is eliminated since this coalition will no longer be winning when q = 9. There are five winning coalitions.

• •

Winning coalition

Weight

{a, b}

10

+

+

−

−

{a, b, c}

12

+

+

+

−

{a, b, d}

11

+

+

−

+

{a, c, d}

9

+

−

+

+

{a, b, c, d}

13

+

+

+

+

Banzhaf score

N/A

5

3

1

1

a

b

c

d

In Part 2 of the Example 2.1.6 we notice that Voter a has no minus signs in his/her column. This means that no coalition can be winning without a. Voter d has a Banzhaf score of zero. This means that d is never critical in any winning condition.

This motivates the following definitions.

121 2.1 · Introduction

Definition 2.1.7 1. A voter who is critical in every winning coalition is said to have veto power.   (Note that the word “critical” can be replaced with the word “present” in this definition. Can you see the reason?) 2. A voter who is never critical in any winning coalition (i.e., with a Banzhaf score of zero) is said to be a dummy.  

ⓘ Remark 2.1.8 1. A yes-no voting system need not be defined by weights and quota. All we need is to determine whether any given coalition is winning or losing. This can be done in various ways including • weights and quota as in weighted systems, • a verbal definition of winning coalitions, • an explicit and complete list of all winning coalitions. 2. In the axiomatic approach of the yes-no voting theory, a list of coalitions in a yes-no voting system for a set V of voters is regarded as a complete list of winning coalitions, if and only if it satisfies the following three axioms: Axiom 1: V is in the list. Axiom 2: A coalition in the list must include at least one voter. Axiom 3: Monotonicity: If coalition A is in the list and B is a supercoalition of A, then B is also in the list. The reader should always verify for himself that all the yes-no voting systems presented here satisfy the above three axioms. 3. Axiom 3 implies that if a yes-no voting system is modified by removing a coalition from the list of all winning coalitions, all subcoalitions of the removed coalition must also be removed. On the other hand, if a coalition is added to the list, all its supercoalitions must also be added too. Example 2.1.9 A yes-no voting system consists of four voters a, b, c, and d. A coalition in this system is winning if and only if it contains at least two voters including a or b (or both.) Find the Banzhaf score of each voter and identify the dummies and the voters with veto power.  Solution We construct the +/− table. There are no dummies since no voter has a Banzhaf score of zero; and there are no voters with veto power since minuses exist in every voter’s column.  

2

122

2

Chapter 2 • Yes-No Voting

Winning coalition

a

b

c

d

{a, b}

+

+

−

−

{a, c}

+

−

+

−

{a, d}

+

−

−

+

{b, c}

−

+

+

−

{b, d}

−

+

−

+

{a, b, c}

+

+

+

−

{a, b, d}

+

+

−

+

{a, c, d}

+

−

+

+

{b, c, d}

−

+

+

+

{a, b, c, d}

+

+

+

+

Banzhaf score

4

4

2

2

2.1.3 Parity of Banzhaf Scores If you closely examine the above examples you will notice that: • The Banzhaf scores are non-negative integers. • In all cases where the number of winning coalitions was odd, all the Banzhaf scores were odd. • In all cases where the number of winning coalitions was even, all the Banzhaf scores were even. The above remarks are true in general and can be put together in the following theorem. Theorem 2.1.10 In a yes-no voting system 1. The Banzhaf scores of all voters are non-negative integers. 2. The Banzhaf scores of all voters have the same parity, which is the parity of the number of winning coalition. 3. No dummies can exist in cases where the number of winning coalitions is odd.

Proof Let x be a voter and let m be the number of minuses and n be the number of pluses that x receives in the +/− table, then the Banzhaf score of x is n − m and the total number of winning coalitions is m + n. 1. Each minus earned by x in the +/− table corresponds to a certain winning coalition A that does not include x. Adding x to A, we get a winning coalition Ax that includes x, from which x earns a plus. If B is a winning coalition (different from A) that does not include x, then Bx is a winning coalition (Different from Ax ) that includes x, from which

123 2.2 · Quantification of Power in a Yes-No Voting System

x earns a plus. This shows that for each minus earned by x there corresponds a plus also earned by x; and different minuses correspond to different pluses. It follows that n ≥ m and therefore n − m ≥ 0.   2. If m + n is even, then m and n have the same parity (both even or both odd) and therefore n − m is even. If m + n is odd, then m and n have different parities and therefore n − m is odd.   3. To have n − m = 0 we must have m = n and therefore m + n is even.  

2.2

Quantification of Power in a Yes-No Voting System

We now focus on the way “power” is distributed in a yes-no voting system. More precisely, we would like to quantify the power of each voter by a non-negative integer. It may seem that in a weighted system, the weight of the voters can be used for this purpose. For a simple example that dramatizes the inadequacy of weights as measures of power in a weighted system, consider a weighted system with three voters a, b, and c with weights 49, 48, and 3, respectively; and a quota of 51. As you can easily verify, the winning coalitions in this system are {a, b}, {a, c}, {b, c}, and {a, b, c}. Simply stated: All we need, to pass a motion, is to get any two of the three voters to vote YES on it. Intuitively, this means that “power” in this system is equally distributed among the three voters despite the large difference in weight between c and each of a and b.

2.2.1 The Banzhaf Index of Power The Banzhaf scores developed above can give an idea about the distribution of power within the same yes-no voting systems, but cannot be used to compare the distributions of power of two systems for the same voters. For example, it may seem that voter a in Example 2.2.6 (below) has more power in Part 1 than he has in Part 2 since his Banzhaf score in Part 1 is 9 and his Banzhaf score in Part 2 is only 8. This, however, ignores the contexts in which those scores are obtained. Definition 2.2.1 The Banzhaf index of a voter in a yes-no voting system is defined as the Banzhaf score of this voter divided by the sum of the Banzhaf scores of all voters in the system. The Banzhaf distribution of power in a yes-no voting system is a statement of the Banzhaf indices of all voters.  

If we denote the Banzhaf score of voter x by b(x) and the Banzhaf index by bi(x), and if the voters in a yes-no voting system are a, b, c, d, then bi(a) =

b(a) . b(a) + b(b) + b(c) + b(d)

2

124

2

Chapter 2 • Yes-No Voting

Note that the Banzhaf index of a voter is always non-negative (by Theorem 2.1.10) and the sum of the Banzhaf indices of all voters in a yes-no voting system is one. The intuition behind the concept of Banzhaf index is that the power of a voter resides in being critical in winning coalitions. Therefore, it makes sense to say that in a system with four voters a, b, c, d, power is distributed according to the ratio b(a) : b(b) : b(c) : b(d) If we think of the total power in a system as a “pie,” then the portion voter a gets out of it is bi(a) defined above. Example 2.2.2 Find the Banzhaf distribution of power for the yes-no voting system in Example 2.1.4.



Solution We have b(a) = 5, b(b) = 3, b(c) = 3, b(d) = 1. Therefore bi(a) =

5 5 = , 5+3+3+1 12

bi(b) =

3 3 = , 5+3+3+1 12

bi(c) =

3 3 = , 5+3+3+1 12

bi(d) =

1 1 = . 5+3+3+1 12

Example 2.2.3 Find the Banzhaf distribution of power for Example 2.1.6.

 



Solution 1. When the quota is 7, b(a) = 6, b(b) = 2, b(c) = 2, and b(d) = 2. The common denominator for the Banzhaf indices is therefore 6 + 2 + 2 + 2 = 12 and bi(a) =

6 3 2 1 2 1 2 1 = , bi(b) = = , bi(c) = = , bi(d) = = .  12 6 12 6 12 6 12 6

2. When the quota is 8, b(a) = 6, b(b) = 2, b(c) = 2, and b(d) = 0. The common denominator for the Banzhaf indices is therefore 6 + 2 + 2 + 0 = 10 and bi(a) =

3 2 1 2 1 0 6 = , bi(b) = = , bi(c) = = , bi(d) = = 0.  10 5 10 5 10 5 10

125 2.2 · Quantification of Power in a Yes-No Voting System

3. When the quota is 9, b(a) = 5, b(b) = 3, b(c) = 1, and b(d) = 1. The common denominator for the Banzhaf indices is therefore 5 + 3 + 1 + 1 = 10 and bi(a) =

5 3 1 1 , bi(b) = , bi(c) = =, bi(d) = .  10 10 10 10

Example 2.2.4 Compute the Banzhaf distribution of power for the yes-no voting system in Example 2.1.9.  Solution b(a) = 4, b(b) = 4, b(c) = 2, and b(d) = 2. The common denominator for the Banzhaf indices is therefore 4 + 4 + 2 + 2 = 12 and bi(a) =

2 4 2 2 1 2 1 4 = , bi(b) = = , bi(c) = = , bi(d) = = .  12 6 12 6 12 6 12 6

Example 2.2.5 A yes-no voting system consists of five voters a, b, c, d, and e with the following defining rule for the winning coalitions: A coalition is winning if and only if it consists of at least three voters including a. Compute the Banzhaf distribution of power and identify the dummies and the voters with veto power.  Solution We have not dealt with a system with five voters before, but the presence of a in every winning coalition will help identify all winning coalitions. Since a winning coalition must contain at least three voters including a, we can list all winning coalitions as follows: Three-voter winning coalitions: Following the argument used earlier in the solution of the Example 2.1.2 to find all possible ways of picking two out of b, c, d, e to join a, the threevoter winning coalitions are {a, b, c}, {a, b, d}, {a, b, e}, {a, c, d}, {a, c, e}, {a, d, e}. Four-voter winning coalitions: We can list those by leaving out one of b, c, d, e at a time. This yields {a, b, c, d}, {a, b, c, e}, {a, b, d, e}, {a, c, d, e}. Of course, we should not forget the coalition of all voters: {a, b, c, d, e}. There are exactly 11 winning coalitions, an odd number. Therefore, all Banzhaf scores must be odd (by Theorem 2.1.10.) Next, we set up the +/− table.

Winning coalition

a

b

c

d

e

{a, b, c}

+

+

+

−

−

{a, b, d}

+

+

−

+

−

{a, b, e}

+

+

−

−

+

{a, c, d}

+

−

+

+

−

{a, c, e}

+

−

+

−

+

{a, d, e}

+

−

−

+

+

{a, b, c, d}

+

+

+

+

−

{a, b, c, e}

+

+

+

−

+

{a, b, d, e}

+

+

−

+

+

{a, c, d, e}

+

−

+

+

+

{a, b, c, d, e}

+

+

+

+

+

Net

11

3

3

3

3

2

126

2

Chapter 2 • Yes-No Voting

The sum of the Banzhaf scores is 11 + 3 + 3 + 3 + 3 = 23 and therefore bi(a) =

11 3 , bi(b) = bi(c) = bi(d) = bi(e) = . 23 23  

There are no dummies and voter a has veto power.

It can be difficult to identify all the winning coalitions in a yes-no voting system with more than four voters. The following is a list of all coalitions in a five-voter yes-no voting system with voters a, b, c, d, and e. Given any particular yes-no voting system with five voters, the student can examine each coalition in the list to determine whether it is winning or not. Systems with more than five voters shall be treated later, after we present some needed combinatorial concepts. One-member coalitions: {a}, {b}, {c}, {d}, {e} Two-member coalitions: {a, b}, {a, c}, {a, d}, {a, e}, {b, c}, {b, d}, {b, e}, {c, d}, {c, e}, {d, e} Three-member coalitions: {c, d, e}, {b, d, e}, {b, c, e}, {b, c, d}, {a, d, e}, {a, c, e}, {a, c, d}, {a, b, e}, {a, b, d}, {a, b, c} Four-member coalitions: {b, c, d, e}, {a, c, d, e}, {a, b, d, e}, {a, b, c, e}, {a, b, c, d} Five-member coalitions: {a, b, c, d, e}

2.2.2 The Felsenthal–Machover Example Example 2.2.6 This example was discovered by Felsenthal and Machover in 1994. 1. The weights of voters in a five-voter weighted yes-no voting system are given in the following table. The quota is 8.

Voter

a

b

c

d

e

Weight

5

3

1

1

1

127 2.2 · Quantification of Power in a Yes-No Voting System

Compute the Banzhaf distribution of power and identify the dummies and the voters with veto power. 2. Voter a ceded one point of weight to voter b. The quota remained the same as in Part 1. The new weights are: Voter

a

b

c

d

e

Weight

4

4

1

1

1

Compute the Banzhaf distribution of power and identify the dummies and the voters with veto power. 3. Comment on the results in Parts 1 and 2.  Solution 1. The +/− table for Part 1 is Coalition

a

b

c

d

e

{a, b}

+

+

−

−

−

{a, b, c}

+

+

+

−

−

{a, b, d}

+

+

−

+

−

{a, b, e}

+

+

−

−

+

{a, b, c, d}

+

+

+

+

−

{a, b, c, e}

+

+

+

−

+

{a, b, d, e}

+

+

−

+

+

{a, c, d, e}

+

−

+

+

+

{a, b, c, d, e}

+

+

+

+

+

Banzhaf score

9

7

1

1

1

There are no dummies and a has veto power. The sum of Banzhaf scores is 9 + 7 + 1 + 1 + 1 = 19 and so the Banzhaf indices are as in the following table. Voter

a

b

c

d

e

Banzhaf index

9/19

7/19

1/19

1/19

1/19

2. The +/− table for Part 2 is Coalition

a

b

c

d

{a, b}

+

+

−

−

−

{a, b, c}

+

+

+

−

−

{a, b, d}

+

+

−

+

−

e

{a, b, e}

+

+

−

−

+

{a, b, c, d}

+

+

+

+

−

{a, b, c, e}

+

+

+

−

+

{a, b, d, e}

+

+

−

+

+

{a, b, c, d, e}

+

+

+

+

+

Banzhaf score

8

8

0

0

0

2

128

2

Chapter 2 • Yes-No Voting

c, d, and e are dummies; a and b have veto power. The sum of Banzhaf scores is 8 + 8 + 0 + 0 + 0 = 16 and so the Banzhaf indices are as in the following table. Voter

a

b

c

d

e

Banzhaf index

8/16 = 1/2

8/16 = 1/2

0

0

0

3. Comments: 7 • Voter b gained veto power and her Banzhaf index increased (from 19 = 0.368 to 0.5) as a result of receiving an additional point of weight from voter a. This seems reasonable. 9 = 0.473 to 0.5) after ceding • It is ironic that a’s Banzhaf index increased (from 19 one point of weight to b. How did this happen? A careful comparison of the lists of winning coalitions in Part 1 and Part 2 shows that the only difference between the two lists is that coalition {a, c, d, e} (which was winning in Part 1) is losing in Part 2. The removal of this coalition from the list of winning coalitions in Part 2 made each of a, c, d, e drop a plus and b drop a minus. The drop of the Banzhaf score of a from 9 to 8 was outweighed by the drop of the denominator of the Banzhaf indices from 19 to 16. The 9 8 result was an increase in the Banzhaf index of a from 19 = 0.473 to 16 = 0.500. This paradoxical behavior may not sound as dramatic if we observe that coalition {a, c, d, e} is the only winning coalition in Part 1 where b is not present and, therefore, removing it from the list of winning coalitions (as in Part 2) is equivalent to endowing b with veto power. (Note that by the third remark in Remark 2.1.8, the removal of {a, c, d, e} from the list of winning coalitions necessitates the removal of all of its subcoalitions but, in fact, none of those subcoalition is in that list to begin with.) Rephrasing the change as “endowing b with veto power” instead of “a ceding one point of weight to b” makes the increase of a’s Banzhaf index seem less paradoxical. • Voters c, d, and e, who were bystanders in the deal between a and b, lost the little power they had and became dummies!

2.2.3 The Shapley–Shubik Index of Power This power index is an application of an important game theoretic notion known as the Shapley value which is beyond the scope of this book. We shall therefore take a direct path to the Shapley–Shubik power index and refer the interested reader to [4] and [9] for information on the more general and basic notion of Shapley value. Consider the yes-no voting system in Example 2.2.2 with four voters a, b, c, d with weights 4, 3, 2, 1, respectively, and a quota of 6. Let us examine, for example, the arrangement or permutation adcb of the voters in which a comes first, d comes second, c comes third, and b comes fourth in this specific chronological order. Voter c possesses special power in this permutation because it is the first voter to form a winning coalition

129 2.2 · Quantification of Power in a Yes-No Voting System

together with the preceding voters. (Note that both {a} and {a, d} are losing but {a, d, c} is winning.) This leads to the following concept, which is central to the development of the Shapley–Shubik index. Definition 2.2.7 The first voter to form a winning coalition together with the preceding ones in a permutation of all voters is said to be the pivotal voter in this particular permutation.  

Example 2.2.8 In Example 2.2.2, identify the pivotal voter in each of the permutations abcd, dcba, and bcda.  Solution The pivotal voter in abcd is b since coalition {a} is losing but coalition {a, b} is winning. Similarly, b is the pivotal voter in dcba and d is the pivotal voter in bcda.  

Let us count the permutations of four voters a, b, c, and d. The first position in a permutation can be filled by any of the four voters. Once the occupant of the first position has been chosen, we have three possible occupants for the second position. Therefore we have 4 × 3 possible ways to fill both the first and the second positions in a permutation. For each one of these 4 × 3 possible ways there are two possible occupants for the third position then only one choice left for the fourth position. Therefore, there are 4 × 3 × 2 × 1 = 24 permutations for any system with four voters. Similarly, in a system with five voters we have 5 × 4 × 3 × 2 × 1 = 120. In general, the number of permutations of n voters is n(n − 1)(n − 2) · · · 2 · 1. Here is a complete list of the 24 permutations of four voters a, b, c, and d: a a a a a a

b b c c d d

c d b d b c

d c d b c b

b b b b b b

a a c c d d

c d a d a c

d c d a c a

c c c c c c

a a b b d d

b d a d a b

d b d a b a

d d d d d d

a a b b c c

b c a c a b

c b c a b a

This is how we composed the list: 1. It seems plausible that the first place should be equally shared among the four voters, we therefore created four columns, each contains six permutations: the first column is for the permutations that start with a, the second column is for the permutations that start with b, and so on.

2

130

2

Chapter 2 • Yes-No Voting

2. In the first column, in which a occupies the first place, the second place will be equally shared by the b, c, and d. We therefore fill the second place in this column twice with b, twice with c, and twice with d. We fill the second spot in each of the permutation in the other columns in a similar fashion. 3. The first and second permutations in the first column are now filled with a in the first spot and b in the second spot. We complete one of these two permutations by placing c third and d fourth, and the other by placing d third and c fourth. We follow the same pattern in filling the third and fourth spots in the remaining 22 permutation. Definition 2.2.9 The Shapley–Shubik score of a voter in a yes-no voting system is the number of permutations in which this voter is pivotal. The Shapley–Shubik index of a voter is its Shapley–Shubik score divided by the total number of permutations. The Shapley–Shubik distribution of power in a yes-no voting system is a statement of the Shapley–Shubik indices of all voters.  

The Shapley–Shubik score and the Shapley–Shubik index of voter x will be denoted by s(x) and si(x), respectively. Example 2.2.10 Compute the Shapley–Shubik distribution of power for the yes-no voting system in Example 2.2.2.  Solution We list all the 24 permutations of the voters a, b, c, d. The pivotal voter in each permutation is in boldface. a a a a a a

b b c c d d

c d b d b c

d c d b c b

b b b b b b

a a c c d d

c d a d a c

d c d a c a

c c c c c c

a a b b d d

b d a d a b

d b d a b a

d d d d d d

a a b b c c

b c a c a b

c b c a b a

s(a) = 10, s(b) = 6, s(c) = 6, and s(d) = 2. The total number of permutations is s(a) + s(b) + s(c) + s(d) = 10 + 6 + 6 + 2 = 24. The above sum of 24 should have been known to us in advance since the total number of permutations for four voters is 4×3×2×1 = 24 as shown above. However, it is recommended that you compute this sum for double checking. The Shapley–Shubik distribution of power now follows: si(a) =

10 5 6 3 6 3 2 1 = , si(b) = = , si(c) = = , si(d) = = . 24 12 24 12 24 12 24 12

2

131 2.2 · Quantification of Power in a Yes-No Voting System

ⓘ Remark 2.2.11 1. A dummy in a yes-no voting system must have zero Banzhaf index and zero Shapley– Shubik index. Conversely, a voter with zero Banzhaf index or a zero Shapley–Shubik index must be a dummy. Verify!   2. As noted earlier, a voter has veto power if and only if her column in the +/- table has no minus signs. It is also clear that a voter has veto power if and only if she is pivotal in all permutations in which she is placed in the last position.   3. Comparing the results in Example 2.2.2 and Example 2.2.10 we see that the Banzhaf distribution of power and the Shapley–Shubik distribution of power (for the same yesno voting system) coincided. This is not always the case as we shall see in some of the examples below. In fact, the processes of computing the Banzhaf and the Shapley–Shubik distributions of power are not two different methods to arrive at the same result. Rather, they are two different attempts to interpret and quantify the vague concept of power in a yes-no voting system.   Example 2.2.12 Find the Shapley–Shubik distribution of power for the weighted system consisting of the four voters a, b, c, d with weights 6,4,2,1, respectively, if 1. The quota is 7. 2. The quota is 8. 3. The quota is 9. 

This is the same voting system in Example 2.2.3. Solution 1. Quota=7. a a a a a a

b b c c d d

c d b d b c

d c d b c b

b b b b b b

a a c c d d

c d a d a c

d c d a c a

c c c c c c

a a b b d d

b d a d a b

d b d a b a

d d d d d d

a a b b c c

b c a c a b

c b c a b a

s(a) = 12, s(b) = 4, s(c) = 4, s(d) = 4. Therefore, si(a) =

3 4 1 4 1 4 1 12 = , si(b) = = , si(c) = = , si(a) = = · 24 6 24 6 24 6 24 6

This is the same as the Banzhaf Distribution of power, obtained in Example 2.2.3.

132

2

Chapter 2 • Yes-No Voting

2. Quota=8. a a a a a a

b b c c d d

c d b d b c

d c d b c b

b b b b b b

a a c c d d

c d a d a c

d c d a c a

c c c c c c

a a b b d d

b d a d a b

d b d a b a

d d d d d d

a a b b c c

b c a c a b

c b c a b a

s(a) = 16, s(b) = 4, s(c) = 4, s(d) = 0. Therefore, si(a) =

16 4 4 1 4 1 0 = , si(b) = = , si(c) = = , si(d) = = 0· 24 6 24 6 24 6 24

This is different from the Banzhaf Distribution of power, obtained in Example 2.2.3. 3. Quota=9. a a a a a a

b b c c d d

c d b d b c

d c d b c b

b b b b b b

a a c c d d

c d a d a c

d c d a c a

c c c c c c

a a b b d d

b d a d a b

d b d a b a

d d d d d d

a a b b c c

b c a c a b

c b c a b a

s(a) = 14, s(b) = 6, s(c) = 2, s(d) = 2. Therefore, si(a) =

14 7 6 3 2 1 2 1 = , si(b) = = , si(c) = = , si(d) = = · 24 12 24 12 24 12 24 12

This is different from the Banzhaf distribution of power, obtained in Example 2.2.3.

 

ⓘ Remark 2.2.13 Suppose you are voter b in the yes-no voting system in Part 2 of Examples 2.2.3 and 2.2.12. You have a Banzhaf index of 15 which is higher than your Shapley–Shubik index of 16 . Does this mean that you get more power if the Banzhaf index is adopted? Not so. In fact, neither of the two indices gives you any power. They only try to quantify and estimate the power you have. You should be able to think of situations where the lower of the two indices better serves your interest and other situations where the higher index is more preferable to you. Exercise 5 at the end of this chapter is related to this remark.

2.2.4 Banzhaf vs Shapley–Shubik Computations ⓘ Remark 2.2.14 1. While the Banzhaf computations identify critical voters in an already formed winning coalition, the Shapley–Shubik computations observe voters as they join a coalition in the process of formation, one at a time, and identifies the voter who turns the coalition from a losing to a winning one.  

133 2.3 · Some Combinatorics

2. As we saw, permutations and pivotal voters play a role in the development of the Shapley–Shubik index analogous to the role played by winning coalitions and critical voters in the development of the Banzhaf index. There are, however, some differences that make the Shapley–Shubik approach more appealing (at least mathematically) than the Banzhaf approach: • A permutation has one and only one pivotal voter (this is clear from the definition) and, as a consequence, the sum of the Shapley–Shubik scores in a given yes-no voting system is equal to the pre-known number of permutations of the voters. On the other hand, a winning coalition of m voters can have any number of critical voters from 0 to m. (In Example 2.1.4, both a and c are critical in {a, c, d} but d is not, all voters in {b, c, d} are critical, and no voter is critical in {a, b, c, d}.) We can therefore think of the Shapley–Shubik index of a voter as the ratio of her Shapley–Shubik score to the total available score (just as we do to evaluate a student’s performance in a test.) No analogous observation holds in the Banzhaf computations since the sum of scores of all voters is, in general, different from the number of winning coalitions.   • We do not have an obvious relationship between the number of winning coalitions and the number of voters in a yes-no voting system. In fact, the number of winning coalitions depends on the defining condition of the winning coalitions in addition to the number of voters. For example, the different quotas in Example 2.1.6 lead to different numbers of winning coalitions. Permutations are much more orderly than this. In fact, the number of permutations depends only on the number of voters in the yes-no voting system at hand. (For n voters, there are n(n − 1)(n − 2) · · · 2 · 1 permutations.)  

2.3

Some Combinatorics

In this section, we present some combinatorial concepts and tools that will help us study the linkage between the Banzhaf and the Shapley–Shubik indices of power and deal with yes-no voting systems with larger numbers of voters.

2.3.1 Permutations and Combinations Suppose we want to pick a committee of three out of the five people a, b, c, d, and e. We study the following two different types of committees: 1. Committees in which the members serve in equal, indistinguishable, capacities: To denote a committee of this type, we list the members, (in any order) separated by commas and enclosed between two curly brackets. For instance, the committee whose members are a, b, and c may be denoted by any of the expressions {a, b, c}, {a, c, b}, {b, a, c}, {b, c, a}, {c, a, b}, {c, b, a}. We shall call a committee of this type a combination of 3 objects out of 5. 2. Committees in which the members are assigned different roles (whether or not some of these roles are superior to others.): A committee of this type is denoted by listing the members, separated by commas and enclosed between two round brackets. The order of listing the members

2

134

2

Chapter 2 • Yes-No Voting

is now significant: The expressions (a, b, c), (a, c, b), (b, a, c), (b, c, a), (c, a, b), (c, b, a) denote the various committees whose members are a, b, and c. The difference between any two of these committees is similar to the difference between the presidential ticket (Trump, Pence) in which Trump is president and Pence is vice president, and the ticket (Pence, Trump) in which Pence is president and Trump is vice president. We shall call each one of these committees a permutation of 3 objects out of 5. We have two questions to answer: Question 1. Question 2.

How many permutations of 3 objects out of 5 are there? How many combinations of 3 objects out of 5 are there?

Despite being apparently more complex, permutations are easier to count. There are five different choices of a first member (the president, say): It could be a, b, c, d, or e. Each of these five people may be paired with any of the remaining four (who would serve as vice president, say). We now have 5 × 4 possibilities for a (president, vice president) pair. Next, each of these 5 × 4 choices can be complemented with any of the remaining three people (who would serve, say, as secretary.) We now have the answer to Question 1: There are 5 × 4 × 3 = 60 permutations of 3 objects out of 5. To answer Question 2, we notice that fixing three out of the given five objects (a, b, and c for example) exactly one combination can be composed, but how many permutations? With the same argument used in answering Question 1, we can conclude that exactly 3 × 2 × 1 = 6 permutations can be composed from those three fixed objects. Therefore, each 3 × 2 × 1 permutations of 3 objects out of 5 correspond to one combination. Therefore, there are 5×4×3 3×2×1 combinations of 3 out of 5 objects. The above correspondence is shown in the following array. (a, b, c) (a, b, d) (a, c, d) (b, c, d) (a, b, e) (a, c, e) (b, c, e) (a, d, e) (b, d, e) (c, d, e) (a, c, b) (a, d, b) (a, d, c) (b, d, c) (a, e, b) (a, e, c) (b, e, c) (a, e, d) (b, e, d) (c, e, d) (b, a, c) (b, a, d) (c, a, d) (d, b, c) (b, a, e) (c, a, e) (c, b, e) (d, a, e) (d, b, e) (d, c, e) (b, c, a) (b, d, a) (c, d, a) (d, c, b) (b, e, a) (c, e, a) (c, e, b) (d, e, a) (d, e, b) (d, e, c) (c, a, b) (d, a, b) (d, a, c) (c, b, d) (e, a, b) (e, a, c) (e, c, b) (e, a, d) (e, b, d) (e, c, d) (c, b, a) (d, b, a) (d, c, a) (c, d, b) (e, b, a) (e, c, a) (e, b, c) (e, d, a) (e, d, b) (e, d, c) ↓

↓

↓

↓

↓

↓

↓

↓

↓

↓

{a, b, c} {a, b, d} {a, c, d} {b, c, d} {a, b, e} {a, c, e} {b, c, e} {a, d, e} {b, d, e} {c, d, e}

The general case can be handled in the same way.

2

135 2.3 · Some Combinatorics

Proposition 2.3.1 For integers k, n with 0 < k ≤ n, the number of permutations of k objects out of n objects is n(n − 1)(n − 2) · · · (n − k + 1) and the number of combinations of k objects out of n objects is n(n − 1)(n − 2) · · · (n − k + 1) · k(k − 1)(k − 2) · · · 2 · 1



Note that in the case k = n, the above formula yields the intuitive result that the number of combinations of n objects out of n objects is 1. Notation 2.3.2 For integers n, k with 0 < k ≤ n we denote the number of combinations of   k objects out of n by nk (read n choose k.) Definition 2.3.3 Given a positive integer n we define factorial n, denoted n!, by n! = n(n − 1)(n − 2) · · · 2 · 1 and we define factorial 0 by 0! = 1.

Proposition 2.3.4 For integers n, k with 0 < k ≤ n we have   n! n = · k (n − k)! k! Proof For 0 < k < n, this follows from   n n(n − 1)(n − 2) · · · (n − k + 1) n(n − 1)(n − 2) · · · (n − k + 1) (n − k)! = = k k(k − 1)(k − 2) · · · 2 · 1 k(k − 1)(k − 2) · · · 2 · 1 (n − k)! and for k = n this follows from our convenient definition 0! = 1.

ⓘ Corollary 2.3.5 For integers n, k with 0 < k < n we have     n n = · k n−k

 

136

2

Chapter 2 • Yes-No Voting

Proof From Proposition 2.3.4 we have     n n! n! n = = = · k (n − k)! k! k! (n − k)! n−k  

The above result is intuitively clear since picking k out of n given objects is the same as unpicking (n − k) out of those n objects. By entering k = n, we can also use this result to define the number of combinations of 0 objects out of n objects as follows:     n n = = 1· 0 n

2.3.2 Cartesian Products Next, suppose we have two sets: A with m ≥ 1 objects and B with n ≥ 1 objects. In how many ways can we pair an object from A with an object from B? The set of all such pairs is called the cartesian product of A and B and is denoted A × B. More precisely, A × B is the set of all ordered pairs (a, b) where a is from A and b is from B. In general, suppose we have nonempty sets A1 with m1 members, A2 with m2 members, · · · and Ak with mk members. The set of all k-tuples (a1 , a2 , · · · , ak ) where a1 is from A1 , a2 is from A2 , · · · and ak is from Ak is called the cartesian product of A1 , A2 , · · · Ak (in this order) and is denoted A1 × A2 × · · · × Ak . The proof of the following theorem is clear.

Theorem 2.3.6 If A and B are nonempty sets with m and n members, respectively, then A×B has m×n members (pairs (a, b) where a is from A and b is from B.) In general, with the same notations as above, the cartesian product A1 × A2 × · · · × Ak has m1 × m2 × · · · × mk members (k-tuples (a1 , a2 , · · · , ak ) as above.)

Finally, suppose we have a set A = {a1 , a2 , · · · , am } of m objects. How many subsets (or combinations) of all possible sizes can we pick from A? An effective way of answering this question is to try to construct a subset of A and count our choices as we pause at each member of A and choose whether to include it in our set or not. Let {y1 , n1 } denote the set of choices for a1 where y1 means “include a1 in our set” and n1 means “do not include a1 in our set.” In general, for each ai , i = 1, 2, · · · m let {yi , ni } be the set of choices for ai where yi means “include ai in our set” and ni means “do not include ai in our set.” Each subset of A corresponds to a member of the cartesian product {y1 , n1 } × {y2 , n2 } × · · · × {ym , nm }

137 2.4 · Banzhaf and Shapley–Shubik Indices in One View

which, by Theorem 2.3.6, contains 2 × 2 × · · · × 2 = 2m members, i.e., possible choices for a subset of A. Note that the empty set is included in this count as it corresponds to the m-tuple (n1 , n2 , · · · nm ). This proves the following theorem. Theorem 2.3.7 A set of m ≥ 0 members has 2m subsets, including the empty set. (Note that the empty set has exactly one subset and so the theorem holds true for m = 0.)  

The set of all subsets of a set A is called the power set of A and is denoted P (A).

2.4

Banzhaf and Shapley–Shubik Indices in One View

To motivate the discussion, assume that 1. V = {a, b, c, d, e, f } is the set of voters in a yes-no voting system. 2. The coalition C = {a, b, c} is winning. 3. c is a critical voter in C. Since c is critical in the coalition C = {a, b, c}, c is pivotal in all permutations in which it is preceded by a and b in any of the 2! possible orders (a, b) and (b, a), and succeeded by d, e, and f in any of the 3! possible orders (d, e, f ), (d, f, e), (e, d, f ), (e, f, d), (f, d, e), and (f, e, d). There are exactly 2! · 3! = 12 such permutations: (a, b, c, d, e, f ) (a, b, c, d, f, e) (a, b, c, e, d, f ) (a, b, c, e, f, d) (a, b, c, f, d, e) (a, b, c, f, e, d) (b, a, c, d, e, f ) (b, a, c, d, f, e) (b, a, c, e, d, f ) (b, a, c, e, f, d) (b, a, c, f, d, e) (b, a, c, f, e, d) Likewise, if c is critical in a winning coalition of five voters, say D = {a, b, c, e, f }, c will be pivotal in all of the 4! · 1! = 24 permutations in which it is preceded by a, b, e, and f (in all of the 4! possible orders) and succeeded by d (in 1! = 1 order.) The generalization of the above argument is straightforward. The number of members in a set S will be called the size of S and will be denoted |S|. Proposition 2.4.1 Let V be the set of voters in a yes-no voting system. Given a pair x, C, where C is a winning coalition and x is a critical voter in C, let P(x, C) be the set of

2

138

2

Chapter 2 • Yes-No Voting

all permutations (of the voters in V ) in which x is preceded by all of its “partners” in C. Then, 1. Every permutation in which x is pivotal belongs to P(x, C) for exactly one winning coalition C that includes x.   2. |P(x, C)| = (|C| − 1)! · (|V | − |C|)!.  

2.4.1 Computing the Shapley–Shubik Indices from Winning Coalitions The following theorem is a direct consequence of Proposition 2.4.1. Theorem 2.4.2 Let V be the set of voters in a yes-no voting system and let x be a member of V . Then the Shapley–Shubik score s(x) and the Shapley–Shubik index si(x) of x are given by s(x) =



|P(x, C)| =

C



(|C| − 1)! · (|V | − |C|)!

C

and  si(x) =

C

|P(x, C)| = |V |!



C (|C| − 1)! · (|V | − |C|)!

|V |!

·  

where C runs through all winning coalitions in which x is critical.

We may now abandon the cumbersome process of counting the permutations in which a voter is pivotal, and compute the Shapley–Shubik score and the Shapley–Shubik index of a voter from his critical membership in winning coalitions, as we do with the Banzhaf score and the Banzhaf index. Unfortunately, the +/− table will not be useful in the Shapley–Shubik computation. Definition 2.4.3 Let V be the set of voters in a yes-no system and let C be a winning coalition. We define the Banzhaf weight and the Shapley–Shubik weight of C (not to be confused with the weight of a voter or a coalition in a weighted system) as follows: The Banzhaf weight of C = 1. The Shapley–Shubik weight of C = (|C| − 1)! · (|V | − |C|)!.

   

(If a voter is critical in a winning coalition C, the Banzhaf weight of C counts toward the Banzhaf score of the voter and its Shapley–Shubik weight counts toward the voter’s Shapley–Shubik score. Neither weight counts for the voter if he is not present in C or is present but not critical in it. The Banzhaf weights of all winning coalitions are equal but the Shapley–Shubik weights are obviously not.)

2

139 2.4 · Banzhaf and Shapley–Shubik Indices in One View

2.4.2 Veto-Powered Voters, Dominant Voters and Dictators The following diagram shows the Shapley–Shubik weight of a winning coalition C in a yes-no voting system with eight voters (⊡ Fig. 2.1). |C| = 1, 2, · · · , 8 is represented on the horizontal axis and the Shapley–Shubik weight of C is represented on the vertical axis. The shape of the diagram can be explained by taking the ratio of the Shapley–Shubik weight of some coalition C to the Shapley–Shubik weight of another coalition of size |C| + 1. |V | − |C| (|C| − 1)! · (|V | − |C|)! = (|C| + 1 − 1)! · (|V | − |C| − 1))! |C| which is greater than 1 if |C| < |V2 | and is less than 1 if |C| > |V2 | . It follows that the Shapley–Shubik weight of a coalition assumes its highest value of (|V | − 1)! when |C| = 1 (a one-voter coalition) and when |C| = |V | (the coalition of all voters) and takes the shape of a symmetric open-up bell between these two extremes. This is intuitively and logically appealing since 1. A voter x who is critical in the coalition V of all voters is particularly strong because her YES vote is necessary for a motion to pass. 2. A voter y such that {y} is a winning coalition is particularly strong because his YES vote is sufficient for a motion to pass. Clearly, x is a veto-powered voter as in Definition 2.1.7(1). What about y?

⊡ Fig. 2.1 Shapley-Shubik weight of a coalition in a yes-no voting system with 8 voters

5040

5040

720

720 240

1

2

3

144 144 240

4

5

6

7

8

140

2

Chapter 2 • Yes-No Voting

Definition 2.4.4 Let x be a voter in a yes-no voting system with V as the set of all voters. 1. x is said to be a veto-powered voter if her membership is necessary for a coalition to be winning.   (This is a restatement of Definition 2.1.7(1).) 2. x is said to be a dominant voter if her membership is sufficient for a coalition to be winning.   3. x is said to be a dictator if it is dominant and veto-powered.  

You can easily see that a yes-no voting system may have more than one vetopowered voter or more than one dominant voter but cannot simultaneously have a veto-powered voter and a dominant voter unless they are the same voter (which would then be a dictator) since the sufficiency of some voter and the necessity of another voter for a coalition to win will contradict one another. Both of the Banzhaf and the Shapley–Shubik concepts have probabilistic roots. The Banzhaf concept is based on the assumption that all winning coalitions are equally likely to form and, therefore, the power of a voter is proportional to the number of winning coalitions this voter is critical in. The Shapley–Shubik concept, on the other hand, is based on the assumption that the permutations of the set of voters (or the sequences in which the voters line up to support a motion) are equally likely. Therefore, the power of a voter is proportional to the number of permutations this voter is pivotal in. The probabilistic basis of the Shapley–Shubik index seems more sound than that of the Banzhaf index since permutations are exactly similar in size and construction while winning coalitions are not. In our first application of the above results, we redo Example 2.2.6 (due to Felsenthal and Machover) in a setting that enables us to compute both of the Banzhaf and Shapley– Shubik distributions of power together. We shall be able to determine whether or not the unexpected behavior of the Banzhaf index in this example is shared by the Shapley– Shubik index. Example 2.4.5 1. The weights of voters in a five-voter weighted yes-no voting system are given in the following table. The quota is 8. Voter

a

b

c

d

e

Weight

5

3

1

1

1

Compute the Banzhaf and the Shapley–Shubik distributions of power and identify the dummies and the voters with veto power. 2. Voter a ceded one point of weight to voter b. The quota remained the same as in Part 1. The new weights are: Voter

a

b

c

d

e

Weight

4

4

1

1

1

2

141 2.4 · Banzhaf and Shapley–Shubik Indices in One View

Compute the Banzhaf and the Shapley–Shubik distributions of power and identify the dummies and the voters with veto power. 3. Is the paradoxical behavior of the Banzhaf index discussed in Example 2.2.6 shared by the Shapley–Shubik index?  Solution Computations are done in the following table. The check mark  in the column of a voter x and the row of a winning coalition C is an indication that x is critical in C and therefore receives both its Banzhaf and Shapley–Shubik weights. The absence of  means that x is not critical in C (or is not in C to begin with) and therefore receives neither weight. 1. Winning coalition

SS weight

Banzhaf weight

a

b

{a, b}

1! · 3! = 6

1





{a, b, c}

2! · 2! = 4

1





{a, b, d}

2! · 2! = 4

1





{a, b, e}

2! · 2! = 4

1





{a, b, c, d}

3! · 1! = 6

1





{a, b, c, e}

3! · 1! = 6

1





{a, b, d, e}

3! · 1! = 6

1





{a, c, d, e}

3! · 1! = 6

1



{a, b, c, d, e}

4! · 0! = 24

1



c

d

e







Shapley–Shubik computations: s(a) = 6 + 4 + 4 + 4 + 6 + 6 + 6 + 6 + 24 = 66. s(b) = 6 + 4 + 4 + 4 + 6 + 6 + 6 = 36. s(c) = 6. s(d) = 6. s(e) = 6. Banzhaf computations b(a) = 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 9. b(b) = 1 + 1 + 1 + 1 + 1 + 1 + 1 = 7. b(c) = 1. b(d) = 1. b(e) = 1. Note that the sum of the Shapley–Shubik scores of all voters is 66 + 36 + 6 + 6 + 6 = 120 which is the correct number of permutations of the five voters (5 × 4 × 3 × 2 × 1). This verifies both of our Shapley–Shubik and Banzhaf computations. The Banzhaf and Shapley–Shubik distributions of power are shown in the following table: Index a b c d e si 66/120 = 0.550 36/120 = 0.300 6/120 = 0.050 6/120 = 0.050 6/120 = 0.050 bi 9/19 ≈ 0.473 7/19 ≈ 0.368 1/19 ≈ 0.053 1/19 ≈ 0.053 1/19 ≈ 0.053

142

2

Chapter 2 • Yes-No Voting

2. Winning coalition {a, b} {a, b, c} {a, b, d} {a, b, e} {a, b, c, d} {a, b, c, e} {a, b, d, e} {a, b, c, d, e}

SS weight 1! · 3! = 6 2! · 2! = 4 2! · 2! = 4 2! · 2! = 4 3! · 1! = 6 3! · 1! = 6 3! · 1! = 6 4! · 0! = 24

Banzhaf weight 1 1 1 1 1 1 1 1

a        

b        

c

d

e

Shapley–Shubik computations: s(a) = 6 + 4 + 4 + 4 + 6 + 6 + 6 + 24 = 60. s(b) = 6 + 4 + 4 + 4 + 6 + 6 + 6 + 24 = 60. s(c) = 0. s(d) = 0. s(e) = 0. Banzhaf computations b(a) = 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 8. b(b) = 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 8. b(c) = 0. b(d) = 0. b(e) = 0. The sum of the Shapley–Shubik scores of all voters is 60 + 60 + 0 + 0 + 0 = 120 which is the correct number of permutations of the five voters. This, again, verifies both of our Shapley–Shubik and Banzhaf computations as in Part 1. The Banzhaf and Shapley– Shubik distributions of power are shown in the following table: Index a b c d e si 60/120 = 0.500 60/120 = 0.500 0/120 = 0.000 0/120 = 0.000 0/120 = 0.000 bi 8/16 = 0.500 8/16 = 0.500 0/16 = 0.000 1/16 = 0.000 0/16 = 0.000

3. Comments: Interestingly, the unexpected increase in the Banzhaf index of voter a upon ceding one point of weight to b was not paralleled by a similar increase in the Shapley– Shubik index of a. As shown from Parts 1 and 2, the Shapley–Shubik index of a decreased from 0.55 to 0.50. (See the comments in Example 2.2.6.)   Example 2.4.6 The set of voters in an eight-voter yes-no voting system is V = {a, b, c, d, e, f, g, h} and a coalition is winning if and only if it consists of four or more voters including a and b (both a and b have veto power.) Compute the Banzhaf and Shapley–Shubik distributions of power. 

143 2.4 · Banzhaf and Shapley–Shubik Indices in One View

Solution From Theorem 2.3.7 we can see that there are 28 − 1 = 255 coalitions in this case; and if we restrict our consideration to coalitions of four or more voters then by Proposition 2.3.4 we have to examine           8 8 8 8 8 + + + + = 70 + 56 + 28 + 8 + 1 = 163 4 5 6 7 8 coalitions to find the winning ones. We can avoid this formidable task by utilizing the combinatorial tools presented in the last section. Obviously, a and b play interchangeable roles in the system, and the same can be said about c, d, e, f, g, and h. Therefore, it is enough to compute the Banzhaf and Shapley– Shubik indices for one from each group, say a and c.

Let us sort out the winning coalitions in which a is critical by size (number of voters.) •

•

•

•

•

Coalitions of size 4 in which a is critical: A coalition of this size must  include a, b and any two of the remaining six voters which can be chosen in 62 = 15 ways. Therefore, there are 15 such coalitions each has a Banzhaf weight of 1 and a Shapley–Shubik weight of 3! · 4! = 144. In conclusion, a receives 15 × 1 = 15 Banzhaf points and 15 × 144 = 2160 Shapley–Shubik points from those coalitions. Coalitions of size 5 in which a is critical: A coalition of this size must  include a, b and any three of the remaining six voters which can be chosen in 63 = 20 ways. Therefore, there are 20 such coalitions each has a Banzhaf weight of 1 and a Shapley–Shubik weight of 4! · 3! = 144. In conclusion, a receives 20 × 1 = 20 Banzhaf points and 20 × 144 = 2880 Shapley–Shubik points from those coalitions. Coalitions of size 6 in which a is critical: A coalition of this size must  include a, b and any four of the remaining six voters which can be chosen in 64 = 15 ways. Therefore, there are 15 such coalitions each has a Banzhaf weight of 1 and a Shapley–Shubik weight of 5! · 2! = 240. In conclusion, a receives 15 × 1 = 15 Banzhaf points and 15 × 240 = 3600 Shapley–Shubik points from those coalitions. Coalitions of size 7 in which a is critical: A coalition of this size must include  a, b and any five of the remaining six voters which can be chosen in 65 = 6 ways. Therefore, there are 6 such coalitions each has a Banzhaf weight of 1 and a Shapley– Shubik weight of 6! · 1! = 720. In conclusion, a receives 6 × 1 = 6 Banzhaf points and 6 × 720 = 4320 Shapley–Shubik points from those coalitions. Coalitions of size 8 in which a is critical: A coalition of this  size must include a, b and all of the remaining six voters which can be chosen in 66 = 1 way. Therefore, there is 1 such coalition that has a Banzhaf weight of 1 and a Shapley–Shubik weight of 7! · 0! = 5040. In conclusion, a receives 1 × 1 = 1 Banzhaf point and 1 × 5040 = 5040 Shapley–Shubik points from this coalition.

It follows that b(b) = b(a) = 15 + 20 + 15 + 6 + 1 = 57, and s(b) = s(a) = 2160 + 2880 + 3600 + 4320 + 5040 = 18,000.

2

144

2

Chapter 2 • Yes-No Voting

Next, let us sort out the winning coalitions in which c is critical by size (number of voters.) Only in winning coalitions of size 4 can c be critical. (This should be clear.) In a coalition of this size, a, b, and c must be  included together with any one of the remaining five voters, which can be chosen in 51 = 5 ways. Therefore, there are 5 such coalitions each has a Banzhaf weight of 1 and a Shapley–Shubik weight of 3! · 4! = 144. In conclusion, a receives 5 × 1 = 5 Banzhaf points and 5 × 144 = 720 Shapley–Shubik points from those coalitions. It follows that b(h) = b(g) = b(f ) = b(e) = b(d) = b(c) = 5, and s(h) = s(g) = s(f ) = s(e) = s(d) = s(c) = 720. We may now verify our computations by adding the Shapley–Shubik scores of all voters: 2(18000) + 6(720) = 40320 = 8! which is the correct number of permutations of the eight voters. Therefore, the Shapley– Shubik indices of the voters are = 25 ≈ 0.4464285. si(b) = si(a) = 18000 40320 56 si(h) = si(g) = si(f ) = si(e) = si(d) = si(c) =

720 40320

=

1 56

≈ 0.0178571.

The common denominator of the Banzhaf indices is 2(57)+6(5) = 144 and the Banzhaf indices are 57 bi(b) = bi(a) = 144 =≈ 0.3958333. bi(h) = bi(g) = bi(f ) = bi(e) = bi(d) = bi(c) =

2.5

5 144

≈ 0.0347222.

 

Weightable Yes-No Voting Systems

We close this chapter with a brief investigation of the possibility of describing a yes-no voting system by means of weights and a quota. For a deeper presentation we refer the reader to Taylor [24]. An advanced treatment of this topic is available in Taylor and Zwicker [25]. To motivate the discussion, we consider the following two examples. Example 2.5.1 A yes-no voting system has four voters a, b, c, and d. A coalition is this system is winning if and only if it contains two or more voters including a. The following is the list of winning coalitions: {a, b} {a, c} {a, d} {a, b, c} {a, b, d} {a, c, d} {a, b, c, d}

145 2.5 · Weightable Yes-No Voting Systems

Can we find weights for the voters, and a quota, so that the complete list of winning coalitions of the new weighted system is exactly as above? It makes sense to assign the same weight, say λ, to each of b, c, and d; and a different weight, say μ, to a. If the quota is q then, since {b, c, d} is losing and {a, b} is winning we have 3λ < q ≤ μ + λ· We have infinitely many solutions: Assign a value to λ, then choose a value for μ such that μ > 2λ and a value for q so that the above inequality holds. For example, λ = 1, μ = 3, q = 4 is a solution, and λ = 1, μ = 6, q = 5 is another.  Example 2.5.2 A yes-no voting system has four voters: two women a, b and two men c, d. A coalition is winning if and only if it contains two or more voters with both genders represented. The following is the list of winning coalitions. {a, c} {a, d} {b, c} {b, d} {a, b, c} {a, b, d} {a, c, d} {b, c, d} {a, b, c, d} It is impossible to find weights for the voters and a quota so that the new weighted system has the above list as a complete list of winning coalitions. To see this, assume to the contrary that such weights: w(a), w(b), w(c), w(d) and a quota q were found. Since {a, c} and {b, d} are both winning we have w(a) + w(c) ≥ q, w(b) + w(d) ≥ q and therefore w(a) + w(b) + w(c) + w(d) ≥ 2q. If these two winning coalitions trade b for c, we get the two coalitions {a, b} and {c, d} which are both losing. It follows that w(a) + w(b) < q, w(c) + w(d) < q and therefore w(a) + w(b) + w(c) + w(d) < 2q· The above two inequalities (w(a) + w(b) + w(c) + w(d) ≥ 2q and w(a) + w(b) + w(c) + w(d) < 2q) are contradictory, and this proves the non-existence of a weighted yes-no voting system with the same list of winning coalitions as the given system. 

2

146

2

Chapter 2 • Yes-No Voting

Definition 2.5.3 1. Two yes-no voting systems (for the same set of voters) are said to be equivalent if they have the same list of winning coalitions. (We may also say that each of the two systems is equivalent to the other.)   2. A yes-no voting system is said to be weightable if it is equivalent to a weighted system. A weighted system that is equivalent to a given system is said to be a weighted representation for it.   (According to Definition 2.5.3, the voting system in Example 2.5.1 is weightable while the system in Example 2.5.2 is not.)

2.5.1 Trades Among Coalitions Example 2.5.4 Let V = {a, b, c, d, e, f, g, h} be the set of voters in a yes-no voting system and consider the following four coalitions: A1 = {a, d, f, h}, A2 = {d, g}, A3 = {a, c, e} and A4 = {d, g, h}

(A)

V A1 A2 A3 A4 

a 1 0 1 0 2

b 0 0 0 0 0

c 0 0 1 0 1

d 1 1 0 1 3

e 0 0 1 0 1

f 1 0 0 0 1

g 0 1 0 1 2

h 1 0 0 1 2

(B)

V B1 B2 B3 B4 

a 0 0 1 1 2

b 0 0 0 0 0

c 0 1 0 0 1

d 1 1 1 0 3

e 0 0 1 0 1

f 0 0 1 0 1

g 0 1 0 1 2

h 1 0 1 0 2

In Table (A), the top row represents the set V of all voters in the system. The second, third, fourth, and fifth rows represent coalitions A1 , A2 , A3 , and A4 , respectively. (The membership in a coalition is indicated by placing 1 in the columns of the voters who are in that coalition, and placing 0 in the columns of the voters who are not.) The bottom row sums up the entries in each column to find the number of occurrences of each voter in all of the coalitions. Table (B) is derived from Table (A) by moving some of the 1’s and the 0’s vertically (thus keeping the numbers of 1’s and 0’s in each column unchanged.) This results in the new coalitions:  B1 = {d, h}, B2 = {c, d, g}, B3 = {a, d, e, f, h} and B4 = {a, g}. Definition 2.5.5 Let V be the set of voters in a yes-no voting system. 1. Given a coalition A in the system and a voter x in V , we define the membership index of x in A, denoted [x; A], by [x; A] = 1 if x is in A and [x; A] = 0 if x is not in A.  

(continued )

2

147 2.5 · Weightable Yes-No Voting Systems

Definition 2.5.5 (continued) 2. Given a set of coalitions {Ai : i = 1, 2, · · · , k} and a voter x in V , we define the membership index of x in the given set of coalitions, denoted [x; {Ai : i = 1, 2, · · · , k}] by [x; {Ai : i = 1, 2, · · · , k}] =

k 

[x; Ai ].

 

i=1

3. For k ≥ 2, a set of coalitions {Ai : i = 1, 2, · · · , k} is said to be trade-equivalent to another set of coalitions {Bi : i = 1, 2, · · · , k} (some Ai ’s or Bi ’s may be empty) if for every voter x in V , [x; {Ai : i = 1, 2, · · · , k}] = [x; {Bi : i = 1, 2, · · · , k}] ·

 

The membership index of a voter in a set of coalitions is simply the number of occurrences of that voter in the coalitions of the set.

ⓘ Remark 2.5.6 1. Clearly, if the set {Ai : i = 1, 2, · · · , k} is trade-equivalent to the set {Bi : i = 1, 2, · · · , k}, then {Bi : i = 1, 2, · · · , k} is trade-equivalent to {Ai : i = 1, 2, · · · , k} and we may therefore say that the two sets are trade-equivalent. It is also clear that if {Ai : i = 1, 2, · · · , k} is trade-equivalent to the set {Bi : i = 1, 2, · · · , k} and the set {Bi : i = 1, 2, · · · , k} is trade-equivalent to the set {Ci : i = 1, 2, · · · , k} then the set {Ai : i = 1, 2, · · · , k} is trade-equivalent to the set {Ci : i = 1, 2, · · · , k} (In fact, tradeequivalence is an “equivalence relation.”) All sets of coalitions that are trade-equivalent to one another form a trade-equivalence class.   2. In Example 2.5.4, the two given sets of coalitions are trade-equivalent. All tradeequivalent sets to a given set of coalitions can be obtained by representing the given set by a table as in Example 2.5.4 then redistributing the 1’s and 0’s among the cells in each column in all possible ways. This redistribution of the 1’s and 0’s amounts to allowing the voters to move freely among the coalitions in the set {Ai : i = 1, 2, · · · , k}. This gives a more intuitive understanding of the formal definition of trade equivalence in 2.5.05(3). The movement of the voters (in this manner) is called a trade and we may say that the set {Bi : i = 1, 2, · · · , k} of coalitions is derived from the set {Ai : i = 1, 2, · · · , k} by   a trade among the Ai ’s.

2.5.2 Trade-Robustness and the Taylor–Zwicker Theorem ⓘ Remark 2.5.7 Assuming that the yes-no voting system in Example 2.5.4 is weighted, let us replace each of the 1s in Table (A) and Table (B) with the weight of the voter in its column and keep the 0s. Note that 1. The weight of a coalition can be obtained by adding up the entries in its row.   2. The sum of all entries in either table yields the sum of weights of the coalitions in the table.  

148

2

Chapter 2 • Yes-No Voting

3. Since the entries in Table (B) are the same as the entries in Table (A) (albeit in different cells), the sum of all entries in Table (B) is equal to the sum of all entries in Table (A).   4. It follows from 2 &3 that the sum of weights of the coalitions in Table (B) is the same as the sum of weights of the coalitions in Table (A).   5. The above remarks hold in general in any weighted yes-no voting system. More precisely, Proposition 2.5.8 If {Ai : i = 1, 2, · · · , k} and {Bi : i = 1, 2, · · · , k} are two-trade equivalent sets of coalitions in a weighted yes-no voting system, then k  i=1

w(Ai ) =

k 

w(Bi )

i=1

 

where w(X) denotes the weight of coalition X. Definition 2.5.9 A (not necessarily weighted) yes-no voting system is said to be 1. n-trade-robust for a certain positive integer n ≥ 2 provided that given two arbitrary trade-equivalent sets of coalitions {Ai : i = 1, 2, · · · , k} and {Bi : i = 1, 2, · · · , k} (with 2 ≤ k ≤ n) such that the coalitions Ai , i = 1, 2, · · · , k are all winning, at least one of the coalitions Bi , i = 1, 2, · · · , k is winning. 2. trade-robust if it is n-trade-robust for all positive integer n ≥ 2.

   

Trade-robustness is an expression of the fact that weight does not dissipate. More precisely, if an open exchange of voters takes place among coalitions, the weight lost by a coalition due to the departure of a voter is gained, in the exact same amount, by the coalition this voter joins. We may compare this to the exchange of players among football teams in the open season, but the analogy is not exact since the value a player adds to his new team is not necessarily equal to the value he takes away from his old team, due to issues of compatibility and ability to work with other players. Theorem 2.5.10 (Taylor & Zwicker, 1992) A yes-no voting system is weightable if and only if it is trade-robust.

Proof Let V be the set of voters in a yes-no voting system. Assuming that the system is weightable, each voter x in V can be assigned a weight w(x), and a quota q can be found so that the resulting weighted system have the same list of winning coalitions as the given system. For an arbitrary integer k ≥ 2, let {Ai : i = 1, 2, · · · , k} and {Bi : i = 1, 2, · · · , k} be two trade-equivalent sets of coalitions such that each Ai is winning,

2

149 2.5 · Weightable Yes-No Voting Systems

i = 1, 2, · · · , k. It follows that k 

w(Bi ) =

i=1

k 

w(Ai ) ≥ kq·

i=1

From this we conclude that w(Bi ) ≥ q for at least one i, for otherwise we would have k i=1 w(Bi ) < kq. This proves that every weightable yes-no voting system is trade-robust. The proof that every trade-robust yes-no voting system is weightable is too involved to be presented here. We refer the interested reader to Taylor and Zwicker [25] for a proof.  

Using the “if” part of Theorem 2.5.10 to prove that a given yes-no voting system is weightable by demonstrating its trade-robustness is often impractical because one will have to examine all pairs of trade-equivalent sets of coalitions. In fact, it is much easier to try to find weights and a quota that generate the list of winning coalitions of the given system. The “only if” part, on the other hand, comes in handy when we embark on proving that a given system is not weightable. To this end, we only need to find a set of allwinning coalitions and a trade among its members that results in another set of all-losing coalitions. Example 2.5.11 A yes-no voting system has the set of voters V = {a, b, c, d, e, f, g}. A coalition in this system is winning if and only if it consists of three or more voters including at least one of a and b. Show that this system is not weightable.  Solution Consider the set {A1 , A2 } where A1 = {a, c, d} and A2 = {b, e, f, g}; and the set {B1 , B2 } where B1 = {a, b} and B2 = {c, d, e, f, g}. The two sets are trade-equivalent as shown in the following two tables.

Table (A)

V A1 A2 

a 1 0 1

b 0 1 1

c 1 0 1

d 1 0 1

e 0 1 1

f 0 1 1

g 0 1 1

Table (B)

V B1 B2 

a 1 0 1

b 1 0 1

c 0 1 1

d 0 1 1

e 0 1 1

f 0 1 1

g 0 1 1

(You can see that {B1 , B2 } was derived from {A1 , A2 } by a trade between A1 and A2 in which c and d moved from A1 to A2 and b moved from A2 to A1 .) Note that A1 and A2 are both winning while B1 and B2 are both losing and so the system is not trade-robust. It follows from Theorem 2.5.10 that it is not weightable.   Example 2.5.12 Let us keep the same set of voters as in Example 2.5.11 but modify the defining condition of a winning coalition by requiring both a and b to be present. The new defining condition for winning coalitions is: A coalition is winning if and only if consists of three or more voters including both a and b.

150

2

Chapter 2 • Yes-No Voting

Given a set {A1 , A2 , · · · , Ak } of winning coalitions, we can see that each Ai , i = 1, 2, · · · , k includes both a and b in addition to one or more of c, d, e, f , and g. If a set {B1 , B2 , · · · , Bk } is derived from the given set by a trade then, obviously, each Bi must include both a and b. If all of the Bi ’s are losing, then none of them can include any of c, d, e, f, g which is impossible. Therefore our system is trade-robust and, by Theorem 2.5.10, is weightable. This is one of the rare occasions where Theorem 2.5.10 can be used to show that a given yes-no system is weightable. To find weights and a quota for this system, let us assign a weight of 1 to each of c, d, e, f , and g, a weight of μ to each of a and b and set the quota at q. Since {a, b, c} is winning and {b, c, d, e, f, g} is losing we have μ + 5 < q ≤ 2μ + 1· We obviously have infinitely many possibilities to choose a pair q, μ. One of those choices is q = 11 and μ = 5. Another choice is q = 16 and μ = 10. 

We close with the following ingenious example by Taylor and Zwicker [24, 26] of a yes-no voting system that is 2-trade-robust but not 3-trade-robust and, therefore, not trade-robust.

2.5.3 The Magic Square Voting System For an integer n ≥ 3, a magic square of order n is an array with n rows and n columns filled with the numbers 1, 2, · · · , n2 so that the sum of each row, columns or diagonal is the same. Since 1 + 2 + · · · + n2 =

1 2 2 n (n + 1) 2

we can see that this common sum (called the magic sum) is 12 n(n2 + 1). Magic squares exist for all n ≥ 3 (but not for n = 2) and have been a subject of interest since antiquity. The following is an example of a magic square of order 3. The magic sum for a magic square of order 3 is 12 × 3(32 + 1) = 15. 4

3

8

9

5

1

2

7

6

The set of voters in the magic square voting system of order 3 consists of 9 voters a, b, c, d, e, f, g, h, i, each occupies one of the cells of the above magic square. Let us think of the number in each cell as a dollar amount owned by the voter in that cell. The dollar amount owned by a coalition X, denoted $(X), is the sum of the dollar amounts owned by its members; and the number of voters in X is denoted #(X). a|$4

b|$3

c|$8

d|$9

e|$5

f |$1

g|$2

h|$7

i|$6

151 2.5 · Weightable Yes-No Voting Systems

Winning coalitions: Let X be a coalition in the above magic square yes-no voting system. By definition, the winning/losing status of X is determined as follows: 1. X is winning if #(X) ≥ 4 and is losing if #(X) ≤ 2. 2. Let #(X) = 3. X is winning if $(X) ≥ 16 and is losing if $(X) ≤ 14. 3. If #(X) = 3 and $(X) = 15, then the members of X occupy a row, a column, or a diagonal in the magic square. (This will be shown in Proposition 2.5.13 below.) Rows are declared winning, while columns and diagonals are declared losing. Proposition 2.5.13 Let S be a coalition in the magic square voting system of order 3, with #(S) = 3 and $(S) = 15, then the members of S occupy a row, a column, or a diagonal in the magic square. Proof If a is in S and if the other two members of S are x and y then, since $(a) = 4, we have $(x) + $(y) = 11. The only possible values for $(x) and $(y) are 9&2, 8&3 and 6&5; and therefore X = {a, d, g} (a column) or X = {a, b, c} (a row) or X = {a, e, i} (a diagonal.) Similarly, if e is in S and the other two members of S are x and y then, since $(e) = 5, we have $(x) + $(y) = 10. The only possible values for $(x) and $(y) are 9&1, 8&2, 7&3 and 6&4; and therefore X = {d, e, f } (a row), X = {c, e, g} (a diagonal), X = {b, e, h} (a column), or X = {a, e, i} (a diagonal.) We leave it to the student to consider the membership of each of the remaining seven voters in S and show that, in each case, S is a row, a column, or a diagonal.   Proposition 2.5.14 The magic square voting system of order 3 is 2-trade-robust. Proof Let {A1 , A2 } be a set of winning coalitions and assume that the set {B1 , B2 } was derived from {A1 , A2 } by a trade between A1 and A2 involving unequal numbers of voters (the number of voters moving from A1 to A2 is different from the number of voters moving from A2 to A1 .) Since #(A1 ) ≥ 3 and #(A2 ) ≥ 3, we must have #(B1 ) > 3 or #(B2 ) > 3 and therefore at least one of B1 and B2 is winning. To complete the proof, we consider trades between A1 and A2 that involve equal numbers of voters (the number of voters moving from A1 to A2 is the same as the number of voters moving from A2 to A1 .) We have three possibilities: 1. One of A1 and A2 , say A1 , is such that #(A1 ) ≥ 4. In this case, we have #(B1 ) ≥ 4 and therefore B1 is winning. 2. #(A1 ) = 3 and #(A2 ) = 3; and one of A1 and A2 , say A1 , is such that $(A1 ) ≥ 16. In this case, we have #(B1 ) = 3 and #(B2 ) = 3. Further, since $(A1 ) ≥ 16 and $(A2 ) ≥ 15, at least one of B1 and B2 , say B1 , will be such that $(B1 ) ≥ 16 and therefore B1 is winning. 3. #(A1 ) = 3, #(A2 ) = 3, $(A1 ) = 15 and $(A2 ) = 15. (Both A1 and A2 must be rows.) In this case, we have #(B1 ) = 3 and #(B2 ) = 3. If one of B1 and B2 , say B1 , is such that $(B1 ) > 15, then it is winning. If $(B1 ) = 15 and $(B2 ) = 15 then, by Proposition 2.5.13, each of B1 and B2 is a row, a column, or a diagonal. But a column or

2

152

2

Chapter 2 • Yes-No Voting

diagonal must include members from all three rows, which is not the case with B1 or B2 whose members come only from the two rows of A1 and A2 . It follows that B1 and B2 are rows ({B1 , B2 } is the same as {A1 , A2 }) and therefore both B1 and B2 are winning.   Proposition 2.5.15 The magic square voting system of order 3 is not 3-trade-robust. Proof Each of the three following coalitions is winning since its members occupy a row of the magic square A2 = {d, e, f }, A3 = {g, h, i}, A1 = {a, b, c}, while each of the three following coalitions is losing since its members occupy a column of the magic square B1 = {a, d, g}, B2 = {b, e, h}, B3 = {c, f, i}. It can be easily verified that the two sets {A1 , A2 , A3 } and {B1 , B2 , B3 } are tradeequivalent and so the magic square voting system is not 3-trade-robust.  

Propositions 2.5.14 and 2.5.15 amount to the following theorem.

Theorem 2.5.16 The magic square voting system of order 3 is 2-trade-robust but not 3-trade-robust.  

The following more general theorem, also by Taylor and Zwicker, can be found in their paper [26].

Theorem 2.5.17 For each integer k ≥ 3, there exists a yes-no voting system that is (k − 1)-trade-robust but not k-trade-robust.  

We close the present chapter with a brief note on a different direction of the study of the structure of a yes-no voting system. Weighted (or weightable) systems have the simplest structure, in the sense that an arbitrary yes-no voting system can be reconstructed from a set of weighted systems. To explain, consider the system in Example 2.5.2. One can think of this system as an “intersection” of two weighted systems: • In the first system, each of a and b has a weight of 1, each of c and d has a weight of zero, and the quota is 1. • In the second system, each of a and b has a weight of zero, each of c and d has a weight of 1, and the quota is 1. You can easily see that a coalition in Example 2.5.2 is winning if and only if it is winning in both of the above two weighted systems. In other words, the system in

153 Exercises

Example 2.5.2 is the intersection of the above two weighted systems, and can be represented as follows: a

b

c

d

q

1

1

0

0

1

0

0

1

1

1

It turns out that every yes-no voting system is representable as an intersection of finitely many weighted yes-no voting system. The dimension of a given yes-no voting system is the smallest number of weighted systems whose intersection yields that given system [24]. (see also Exercise 28 below.)

Exercises 1. In a weighted yes-no voting system, voter a has weight of 6, voter b has weight of 3 and b(a) = 8. What are the possible values of b(b)? 2. In a weighted yes-no voting system, voter a has weight of 7, voter b has weight of 4 and b(a) = 7. What are the possible values of b(b)? 3. In a weighted yes-no voting system, b(a) = 8. Which of the following is a possible value for the total number of winning coalition: 6, 7, 8, 9, 10? 4. In a weighted yes-no voting system, b(a) = 9. Which of the following is a possible value for the total number of winning coalition: 7, 8, 9, 10, 11? 5. Suppose you are a voter in a yes-no voting system and your Banzhaf index is different from your Shapley–Shubik index. • Can you think of a situation where the lower index better serves your interest? • Can you think of a situation where the higher index better serves your interest? 6. Show that the following three Yes-No systems for voters a, b, c, d, e, f, g, h are equivalent. 1. A coalition is winning if and only if it contains at least four voters including a, b. 2. Each of a and b is assigned a weight of 5, each of c, d, e, f, g, and h is assigned a weight of 1, and a quota of 12 is adopted. 3. Each of a and b is assigned a weight of 100, each of c, d, e, f, g, and h is assigned a weight of 1, and a quota of 202 is adopted. 7. A yes-No voting system assigns weights of 5,4,3,2,1 to the voters a, b, c, d, e, respectively. Compute the Banzhaf distribution of power and identify the dummies and the voters with veto power in each of the following three cases: 1. Quota is 8. 2. Quota is 9. 3. Quota is 10.

2

154

2

Chapter 2 • Yes-No Voting

8. A Yes-No voting system consists of six voters named a, b, c, d, e, f together with the following rule: A coalition is winning if and only if it contains at least four voters whose names are four consecutive letters. List all winning coalitions and compute the Banzhaf distribution of power. Identify the dummies and the voters with veto power that may exist. 9. A hypothetical country consists of five states a, b, c, d, e whose populations, in millions, are given in the following table: State

a

b

c

d

e

Population

5

5

1

1

1

In the National Supreme Council that runs the country, a coalition of states is winning if and only if the total population of the states in the coalition is at least 8 million citizens. List all the winning coalitions and compute the Banzhaf distribution of power. Identify the dummies and the voters with veto power. 10. In Exercise 9, the three smaller states c, d, and e demanded that a winning coalition of states must contain at least three states, in addition to the condition of at least 8 million citizens. List all the winning coalitions and compute the Banzhaf distribution of power. Show how this additional condition shifts some Banzhaf power from the larger states to the smaller ones. 11. The following are the weights of the voters in a yes-no voting system: a

b

c

d

e

6

4

2

2

1

Find the Banzhaf distribution of power if the quota is 10 and identify the dummies and the voters with veto power if they exist. 12. It was suggested that increasing the quota from 10 to 11 in Exercise 11 might be helpful to voter e. Find the Banzhaf distribution of power and identify the dummies and the voters with veto power if this change is implemented. Did voter e actually gain from this change? 13. Instead of increasing the quota as in Exercise 12, let us bring the quota back to 10 and add the condition that a winning coalition must include three or more voters. Find the Banzhaf distribution of power and identify the dummies and the voters with veto power if they exist. Did this help voter e? Which of the two modifications is better for e: the one in Exercise 12 or this one? 14. In a yes-no voting system for seven voters a, b, c, d, e, f , and g, a coalition is winning if and only if it includes four or more voters including both a and b. Mark the pivotal voter in each of the following permutations: acbf d eg acgf deb gcaf bed ecgf dba

155 Exercises

15. In a yes-no voting system for seven voters a, b, c, d, e, f , and g, a coalition is winning if and only if it includes four or more voters including at least one of a and b. Mark the pivotal voter in each of the permutations in Exercise 14. acbf d eg acgf deb gcaf bed ecgf dba 16. Is the yes-no voting system in Exercise 14 weightable? 17. Is the yes-no voting system in Exercise 15 weightable? 18. A yes-No voting system assigns weights of 5,4,1,1 to the voters a, b, c, d, respectively. Compute the Banzhaf and Shapley–Shubik distributions of power and identify the dummies and the voters with veto power in the following three cases: 1. Quota is 6. 2. Quota is 7. 3. Quota is 8. 19. A Yes-No voting system consists of four voters named a, b, c, d together with the following rule: A coalition is winning if and only if it includes two voters whose names are consecutive letters. Compute the Banzhaf and Shapley–Shubik distributions of power. Identify the dummies and the voters with veto power, if they exist. 20. Is the yes-no voting system in Exercise 19 weightable? 21. A Yes-No voting system for five voters a, b, c, d, and e has the following defining rule for the winning coalitions: A coalition is winning if and only if it contains three or more voters including both a and b. 1. Find a weighted representation for the system. 2. Find the Banzhaf distribution of power. 3. Find the Shapley–Shubik distribution of power. 22. A Yes-No voting system for five voters a, b, c, d, and e has the following defining rule for the winning coalitions: A coalition is winning if and only if it contains three or more voters including at least one of a and b. 1. Find a weighted representation for the system. 2. Find the Banzhaf distribution of power. 3. Find the Shapley–Shubik distribution of power. 23. Consider a weighted yes-no voting system in which all voters have positive even integer weights except for one voter, say x, whose weight is 1; and assume that the quota is an even positive integer. Show that x is a dummy. 24. The European Union (EU) started in 1957 under the name The European Common Market which consisted of six members: France, West Germany, Italy, The Netherlands,

2

156

2

Chapter 2 • Yes-No Voting

Belgium, and Luxembourg. The decision making body was a weighted yes-no voting system with weights and quota as follows: Member Weight

France 4

Germany 4

Italy 4

Netherlands 2

Belgium 2

Luxembourg 1

and the quota was set at 12. Without listing all winning coalitions, show that Luxembourg was a dummy in the European Common Market yes-no voting system. 25. Find the Banzhaf and the Shapley–Shubik distributions of power for the European Common Market in Exercise 24. (Hint: Instead of sorting coalitions in which a voter is critical by the size, sort them by the weight.) 26. Without listing all winning coalitions, show that voters d, e, f are all dummies in the following weighted system if the quota is 16. Member

a

b

c

d

e

f

Weight

9

9

7

3

1

1

27. Let D be a coalition in a yes-no voting system. • D is said to be minimal winning if it is winning but every proper subcoalition of D is losing. (This is equivalent to the condition that every voter in D is critical.) • D is said to be maximal losing if it is losing but every proper supercoalition of D is winning. Prove or disprove 1. Every minimal winning coalition contains a maximal losing one. 2. Every maximal losing coalition is contained in a minimal winning one. 3. Every winning coalition contains a minimal winning one. 4. Every losing coalition is contained in a maximal losing one. 28. The following table shows the population data (in millions) of a federated nation of eight states with a total population of 22 million citizens. State

a

b

c

d

e

f

g

h

Population

8

8

1

1

1

1

1

1

The nation is governed by a yes-no voting system in which a coalition is winning if and only if it includes at least five states with a total population of at least 12 million people. Show that this system is not 2-trade robust, and is therefore not weightable. Note The yes-no voting system in Exercise 28 can be presented as a double-weighted system as follows:

State

a

b

c

d

e

f

g

h

Quota

w1

8

8

1

1

1

1

1

1

12

w2

1

1

1

1

1

1

1

1

5

157 Exercises

Such system can be viewed as a blend of two weighted systems. It is often used in federated nations in order to protect the smaller states from being dominated by the larger ones. The system in Exercise 10 is another example of a double weighted system; it can be presented as: State

a

b

c

d

e

Quota

w1

5

5

1

1

1

8

w2

1

1

1

1

1

3

The two weight assignments involved are denoted w1 and w2 . Note that the system in Exercise 10 is weightable (as we shall see next, in Exercise 29) while the system in Exercise 28 is not. 29. Show that the yes-no voting system in Exercise 10 is weightable. 30. How many sets {X1 , X2 , X3 , X4 } are there in the trade-equivalence class of the set {A1 , A2 , A3 , A4 } in Example 2.5.4? 31. The United Nations Security Council (UNSC) comprises 15 members, 5 of which are endowed with veto power. The following is a simplified version of the yes-no voting system used by the UNSC that precludes abstentions: A coalition is winning if and only if it includes four or more voters in addition to the five veto-powered members. 1. Show that the UNSC system is weightable. 2. Compute the Banzhaf and the Shapley–Shubik distributions of power for the UNSC system.

2

159

Apportionment Sherif El-Helaly © Springer Nature Switzerland AG 2019 Sherif El-Helaly, The Mathematics of Voting and Apportionment, Compact Textbooks in Mathematics, https://doi.org/10.1007/978-3-030-14768-6_3

3.1

Introduction

This chapter is devoted to the problem of apportioning the seats of the House of Representatives every 10 years after the census. Article I  Sect. 2.3 of the constitution states that “Representatives and direct taxes shall be apportioned among the several states which may be included within this union, according to their respective numbers . . . .” In other words, the representation of each state should be commensurate with its population size. To get started, consider the following two examples. Example 3.1.1 Suppose we have a federated nation of three states A, B, and C and a total population of 1000 people: 500 in state A, 300 in state B, and 200 in state C. The house size (which is the total number of seats in the house) is 10 seats. Intuitively, A should be represented by 5 seats, B should be represented by 3 seats, and C should be represented by 2 seats. The underlying computations are as follows. 

Start by computing the number of people to be represented by one seat (nationwide), called the natural divisor and denoted by d: Natural Divisor =

1000 Total Population of the Nation = = 100 person/seat. House Size 10

To find the number of seats that should represent a state (called the exact quota or natural quota of the state), we divide the natural divisor into its population. The results are shown in the following table:

3

160

Chapter 3 • Apportionment

3

State

Population

Natural quota

A

500

500/100 = 5

B

300

300/100 = 3

C

200

200/100 = 2

Total

1000

10

The exact apportionment of the house is the apportionment in which each state is represented by its natural quota. It is called exact because it guarantees that the number of people to be represented by one seat is same regardless of the state they live in. In other words, the ratio Population of a state number of seats apportioned to that state (called the district size in the state) is same for all states. The exact apportionment is the ideal that we should aim at despite its virtual impossibility. The study of apportionment is all about trying to get as close as possible to this ideal. In the present (unrealistically simple) example, the exact apportionment is 5 seats for A, 3 seats for B, and 2 seats for C with a district size of 100 people in each state. Example 3.1.2 In the above example, let us change the population data as in the following table and keep the house size at 10 seats.

The natural divisor is table.

State

A

B

C

Total

Population

378

346

276

1000

1000 10

= 100 person/seat. Calculations are shown in the following

State

Population

Natural quota

A

378

378/100 = 3.78

B

346

346/100 = 3.46

C

276

276/100 = 2.76

Total

1000

10

Therefore, the exact apportionment is 3.78 seats for A, 3.46 seats for B, and 2.78 seats for C. 

161 3.1 · Introduction

In contrast with the exact apportionment in Example 3.1.1 which assigned integer values of seats to each state, the exact apportionment in Example 3.1.2 involves noninteger values and is therefore impossible to implement in the real world since seats come in whole numbers. This is what makes apportionment a problematic, but rich and interesting field of study. We still have a practical problem waiting for a solution, namely, apportioning the house seats among the states in the last example. Obviously, a practical solution must replace the non-integer values with integer ones which means that we have to deviate from the exact apportionment. A state that receives a whole number of seats higher than its natural quota will be over-represented and a state that receives a whole number of seats lower than its natural quota will be under-represented. Perhaps the first thing that comes to mind is to round each non-integer natural quota to the nearest integer value. In other words, we use 0.5 as the cut-off point of rounding so that a fraction that is lower than 0.5 is rounded down, a fraction that is higher than 0.5 is rounded up and we agree to always round a fraction of exactly 0.5 either up (which we shall do) or down. This rounding is called the conventional rounding. Let us try conventional rounding in Example 3.1.2: State A’s natural quota of 3.78 rounds up to 4, state B’s natural quota of 3.46 rounds down to 3, and state C’s quota of 2.76 rounds up to 3. The total is 10 seats. Obviously, A is over-represented, B is under-represented, and C is over-represented. The following table summarizes the work done in this problem. State

Population

Natural quota

Rounded

A

378

378/100 = 3.78

4

B

346

346/100 = 3.46

3

C

276

276/100 = 2.76

3

Total

1000

10

10

Let us try the following two examples using the conventional rounding. Example 3.1.3 The population data of a certain country is shown in the following table. The house size is 10 seats. State

A

B

C

Total

Population

418

346

236

1000

3

162

Chapter 3 • Apportionment

The natural divisor is table:

3

1000 10

= 100 person/seat. Calculations are shown in the following 

State

Population

Natural quota

Rounded

A

418

418/100 = 4.18

4

B

346

346/100 = 3.46

3

C

236

236/100 = 2.36

2

Total

1000

10

9

Example 3.1.4 The population data of a certain country is shown in the following table. The house size is 10 seats.

The natural divisor is table:

State

A

B

C

Total

Population

378

356

266

1000

1000 10

= 100 person/seat. Calculations are shown in the following

State

Population

Natural quota

Rounded

A

378

378/100 = 3.78

4

B

356

356/100 = 3.56

4

C

266

266/100 = 2.66

3

Total

1000

10

11

It is clear from the above two examples that conventional rounding does not always help. (The rounded natural quotas did not add up to the correct house size of 10 seats.) There are two main approaches to the apportionment problem, neither is perfect. Each of these two approaches has its own shortcomings which we shall discuss in detail. These two approaches result in quota procedures and divisor procedures. 

3.2

Axioms of Apportionment

Now that exact apportionments have been seen to be virtually impossible, we should settle for lesser requirements (or axioms) that we should expect an apportionment to satisfy. We shall implicitly assume that the house size ≥ the number of states.

163 3.2 · Axioms of Apportionment

Axioms of Apportionment 3.2.1 Axiom 1. Axiom 2.

Axiom 3.

Axiom 4.

The number of seats apportioned to a state must be a non-negative integer. The total number of seats apportioned to all states must equal the house size; and if all the natural quotas are integers, then the apportionment obtained from a procedure must coincide with the exact apportionment. Any two states with equal populations must be apportioned the same number of seats; and if the population of state X is larger than the population of state Y , then the number of seats apportioned to Y must not exceed the number of seats apportioned to X. All apportionment procedures must satisfy the above three axioms. Quota procedures must also satisfy the following (fourth) axiom: A solution to the apportionment problem must satisfy the quota rule.

What does the quota rule mean? Let q be the natural quota of a state. If q is not an integer, the integer that is lower than q and is nearest to it is called the lower natural quota of the state, and the integer that is higher than q and is nearest to it is called the upper natural quota of the state. In other words, the lower natural quota and the upper natural quota are derived from the natural quota by rounding it to the nearest lower integer, or to the nearest higher integer, respectively. In the extremely rare case that q itself is an integer then, by definition, the lower natural quota and the upper natural quota are both equal to q.

A state that receives less seats than its lower natural quota is said to violate lower quota (otherwise it is said to satisfy lower quota), and a state that receives more seats than its upper natural quota is said to violate upper quota (otherwise it is said to satisfy upper quota). A state that satisfies both lower and upper quotas is said to satisfy quota or stay within quota. A solution to an apportionment problem in which all states stay within quota is said to satisfy the quota rule or stay within quota; it is also called a quota solution or a quota apportionment. The rationale behind the quota rule is that if the natural quota is unusable because it is not an integer, we should replace it with one of the two integers nearest to it: the lower natural quota or the upper natural quota. Note that the above axioms allow a state to receive zero seats. This goes against the constitutional requirement that each state must be represented by at least one seat in the House of Representatives. The law should provide a provision to deal with this issue if it materializes (increasing the house size is one possible solution). There are, however, other applications of apportionment where zero is allowed and even desirable as we shall see. Proportional Representation (common in European countries) is a good example of those applications. In a country that applies the proportional representation system, the entire country is one electoral district. Candidates for parliamentary seats do not run as individuals. Instead, each political party submits an ordered list of candidates. When a voter goes to the balloting station, she does not cast her vote for an individual candidate, but for a party list. The votes are then tallied and the parliament seats are apportioned among the political parties in accordance with their

3

164

3

Chapter 3 • Apportionment

respective numbers of votes. If, for example, the Social Democratic Party is apportioned 37 seats, the top 37 candidates in that party list occupy those seats. Typically, the parties in the parliament form two main blocs which balance one another. A minor political party with only a small number of seats can have an undue influence by holding the balance between the two main blocs. For this reason, it is undesirable to have minor parties represented in the parliament. In fact, countries that use the proportional representation system often set a threshold that a party must meet in order to participate in the seat apportionment process. For example, if that threshold is set at 5% of the total vote, then parties that receive below 5% of the total vote are denied access to the parliament and the seats will be apportioned only among parties that meet that threshold. The downside of this, of course, is that supporters of minor pairs are denied representation in the parliament and may be discouraged from voting according to their own opinions. We remind the student of the way the Plurality voting procedure (studied in  Chap. 1) discourages voters from voting to minor candidates. Example 3.2.2 Find all quota solutions for Example 3.1.3.

State

A

B

C

Total

Population

418

346

236

1000

 Solution In the initial allocation (IA) we give each state its lower natural quota.

State

Population

NQ

IA

(A)

(B)

(C)

A

418

418/100 = 4.18

4

5

4

4

B

346

346/100 = 3.46

3

3

4

3

C

236

236/100 = 2.36

2

2

2

3

Total

1000

10

9







There is one surplus seat left after the initial allocation. In the columns headed (A), (B), and (C) we try giving the surplus seat to each of A, B, and C, respectively. The check mark  in the bottom of a column indicates that the solution contained in the column conforms with the four axioms in Axiom 3.2.1 and is, therefore, acceptable. There are exactly three quota apportionments.

3

165 3.2 · Axioms of Apportionment

Example 3.2.3 Find all quota solutions for Example 3.1.4.

State

A

B

C

Total

Population

378

356

266

1000

 Solution In the initial allocation we give each state its lower natural quota.

State

Population

NQ

IA

(A, B)

(A, C)

(B, C)

A

378

378/100 = 3.78

3

4

4

3

B

356

356/100 = 3.56

3

4

3

4

C

266

266/100 = 2.66

2

2

3

3

Total

1000

10

8





X

There are two surplus seats left after the initial allocation. In the columns headed (A, B), (A, C), and (B, C) we try giving the two surplus seats to the pair appearing in the heading of the column. As above, the check mark  in the bottom of a column indicates that the solution contained in the column is acceptable because it conforms with the four axioms in Axiom 3.2.1, while “X” marks a solution that violates Axiom 3 and is, therefore, unacceptable. There are exactly two quota apportionments. Example 3.2.4 Find all quota solutions of the following problem in which 10 seats are to be apportioned among the four states A, B, C, and D.

State

A

B

C

D

Total

Population

4432

2537

1617

1414

10,000

 Solution The natural divisor is 10,000/10 = 1000. Work is shown in the following table. State

Population

NQ

IA

(A, B)

(A, C)

(A, D)

(B, C)

(B, D)

(C, D)

A

4432

4.432

4

5

5

5

4

4

4

B

2537

2.537

2

3

2

2

3

3

2

C

1617

1.617

1

1

2

1

2

1

2

D

1414

1.414

1

1

1

2

1

2

2

Total

10,000

10

8





X



X



166

Chapter 3 • Apportionment

There are exactly four quota solutions to this apportionment problem. Solutions to the above three examples can be outlined in the following two steps:

3

Step 1:

Step 2:

3.3

In the initial allocation, each state is awarded the whole number part of its natural quota (which is same as the lower natural quota). The number of seats assigned at this point will be less than or equal to the house size. The seats that remain unassigned after the initial allocation are called the surplus seats; and the number of surplus seats is denoted by s. Since s is the sum of the “dropped” fractional parts of the natural quotas, we see that s is always a nonnegative integer, smaller than the number of states. Now to the surplus seats. How should they be allocated? Since s is always less than the number of states as pointed out above, some states will receive surplus seats and others will not. We try all possible allocations of the surplus seats that allow no state to receive more than one surplus seat and choose, as valid solutions, those that satisfy Axiom 3 above. (Axioms 1, 2, 4 are automatically satisfied.)

Quota Procedures

In this section we present two commonly used quota procedures: the first was proposed by Alexander Hamilton to apportion the seats of the House of Representatives after the census of 1790. The second was proposed in 1822 by Representative William Lowndes of South Carolina. Step 1 above is exactly the same in all quota procedures. As for Step 2, the states are listed according to some priority rule that conforms with the above axioms and the top s states receive one surplus seat each. Quota procedures differ in the priority rules they use.

3.3.1 Hamilton’s Procedure Definition 3.3.1 (Hamilton’s Procedure) The states are listed according to the decreasing order of the fractional part of the natural quota and the top s states receive one seat each, in addition to its initial allocation.  

The rationale behind this rule is obvious: The larger the fractional part in the natural quota of a state, the more eligible it is for a surplus seat. The student should find it easy to verify that the quota solution picked by Hamilton’s procedure always satisfies Axiom 3 above. Example 3.3.2 Conventional rounding failed to apportion the 10 seats in Example 3.1.3 above. Let us try the procedure of Hamilton. In the initial allocation, A receives 4 seats, B receives 3 seats, and C receives 2 seats for a total of 9 seats. Therefore, s = 10 − 9 = 1. The priority list for surplus

3

167 3.3 · Quota Procedures

seats is BCA since 0.46 > 0.36 > 0.18. Since we have only one surplus seat (s = 1), we draw a vertical line between the first and second states in the list: B|CA to demonstrate the fact that only the top state will add a surplus seat to its initial allocation. The final Hamilton apportionment is therefore as follows: A receives 4 seats, B receives 4 seats, and C receives 2 seats. The complete solution of this problem is presented in the following table.

State

Population

Natural quota

Initial allocation

Surplus seats

Apportionment

A

418

418/100 = 4.18

4

0

4

B

346

346/100 = 3.46

3

1

4

C

236

236/100 = 2.36

2

0

2

Total

1000

10.00

9

1

10

 Example 3.3.3 Conventional rounding also failed to apportion the 10 seats in Example 3.1.4 above. Let us try the procedure of Hamilton. In the initial allocation, A receives 3 seats, B receives 3 seats, and C receives 2 seats for a total of 8 seats. Therefore, s = 10 − 8 = 2. The priority list for surplus seats is ACB since 0.78 > 0.66 > 0.56. Since we have two surplus seats (s = 2), the vertical line is drawn between the second and third states in the list: AC|B. The final Hamilton apportionment is therefore as follows: A receives 4 seats, B receives 3 seats, and C receives 3 seats. Solution of this problem is presented in the following table.

State

Population

Natural quota

Initial allocation

Surplus seats

Apportionment

A

378

378/100 = 3.78

3

1

4

B

356

356/100 = 3.56

3

0

3

C

266

266/100 = 2.66

2

1

3

Total

1000

10.00

8

2

10

 Note 3.3.4 Hamilton’s procedure is really a variant of conventional rounding in which the cut-off point of rounding is floating, instead of being fixed at 0.5. In Example 3.3.2, any cut-off point between the first and the second largest fractional parts (0.46 and 0.36), such as 0.40 will allow only B to receive a surplus seat. Likewise, in Example 3.3.3 any cut-off point between the second and the third largest fractional parts (0.66 and 0.56) such as 0.61 will allow only A and C to receive surplus seats.  

168

Chapter 3 • Apportionment

3.3.2 Lowndes’ Procedure

3

Definition 3.3.5 (Lowndes’ Procedure) The states are listed according to the decreasing order of the district size based on the initial allocation. The top s states in the list receive one seat each. The district size of a state with zero initial allocation is set at ∞.  

The rationale behind this rule is that if state X has a larger district size than state Y after the initial allocation, then X is poorly represented compared to Y and therefore has a stronger claim than Y to the next available seat. It is easy to verify that the quota solution picked by the procedure of Lowndes satisfies Axiom 3. Example 3.3.6 In Example 3.3.2 the district sizes after the initial allocation are 418/4 = 104.5 for A, 346/3 = 115.3 for B, and 236/2 = 118 for C. Therefore the Lowndes priority list is CBA. Since we have only one surplus seat, we draw the line between the first and the second states in the list: C|BA and the Lowndes apportionment is 4 seats for A, 3 seats for B, and 3 seats for C. (Compare to Hamilton’s solution of the same problem.)  Example 3.3.7 In Example 3.3.3, the district sizes after the initial allocation are 378/3 = 126 for A, 356/3 = 118.6 for B and 266/2 = 133 for C. Therefore the Lowndes priority list is CA|B and the Lowndes apportionment is 4 seats for A, 3 seats for B, and 3 seats for C which coincides with Hamilton’s solution of the same problem. 

ⓘ Remark 3.3.8 The following is an equivalent statement of the Lowndes priority rule presented in Definition 3.3.5: The states are listed according to the decreasing order of the ratio: fractional part of the natural quota entire natural quota and the top s states in the list receive one seat each. The equivalence of this statement to Definition 3.3.5 follows from the equality i δ i+δ + = = 1. q q q where q is the natural quota, i is the integer part of the natural quota (which is also the initial allocation), and δ is the fractional part of the natural quota. You will be asked to provide the details in the exercises.  

ⓘ Remark 3.3.9 The above alternative statement of the Lowndes priority rule highlights the difference between the priority rules of Hamilton and Lowndes. While Hamilton bases priority only on the absolute size of the fractional part of the natural quota,

3

169 3.3 · Quota Procedures

Lowndes puts the size of the fractional part in perspective by taking its ratio to the entire natural quota. Consider, for example, state X with a natural quota of 5.76 and state Y with a natural quota of 1.28. In the initial allocation, X receives 5 seats, and Y receives 1 seat. In Hamilton’s procedure, X’s claim to a surplus seat would be stronger than Y ’s claim since 0.76 > 0.28 while in Lowndes’ procedure Y ’s claim would be stronger than 0.76 X’s claim since 0.28 1.28 > 5.76 . Lowndes’ rationale here is that 0.28 is a more significant portion of Y ’s natural quota than 0.76 as a portion of X’s natural quota. Lowndes’ procedure is therefore more sympathetic to smaller states than Hamilton’s.  

3.3.3 The Alabama Paradox After the 1880 census, C.W. Seaton, the chief clerk of the U.S. Census Bureau, computed the Hamilton’s apportionments of all house sizes from 275 to 350 seats. (The house size was not fixed then since the population of the country and the number of states were both rapidly growing.) To his astonishment, Seaton found that Alabama receives 8 seats in a 299-seat house but receives only 7 seats in the larger house of 300 seats! This paradox became known in the literature as the Alabama Paradox. To help understand how this paradoxical behavior can occur, we present the following simple example. Example 3.3.10 Use Hamilton’s procedure to apportion 1. 10 seats 2. 11 seats among the following three states.

State

A

B

C

Total

Population

135

333

532

1000

 Solution 1. The natural divisor is 1000/10 = 100 person/seat. Calculations are presented in the following table:

State

Population

Natural quota

Initial allocation

Surplus seats

Apportionment

A

135

135/100 = 1.35

1

1

2

B

333

333/100 = 3.33

3

0

3

C

532

532/100 = 5.32

5

0

5

Total

1000

10.00

9

1

10

170

3

Chapter 3 • Apportionment

Only one surplus seat remains after the initial allocation and is taken by A since its natural quota has the largest fractional part. 2. The natural divisor is 1000/11 = 90.90 person/seat. Calculations are presented in the following table:

State

Population

Natural quota

Initial allocation

Surplus seats

Apportionment

A

135

135/90.90 = 1.485

1

0

1

B

333

333/90.90 = 3.663

3

1

4

C

532

532/90.90 = 5.852

5

1

6

Total

1000

11.00

9

2

11

Two surplus seats remain after the initial allocation. They are taken by C and B since there natural quotas have the largest two fractional parts. State A is the “Alabama” in this example. It received 2 seats in a 10-seat house but only 1 seat in an 11-seat house.

Let us try to understand how the calculations led to this anomaly. If p is the population of one of the three states, then its natural quota is

p 1000/10

=

10 p 1000

p 11 p in a 10-seat house and is 1000/11 = 1000 in an 11-seat house. Therefore, increasing the house size from 10 seats to 11 seats had the effect of enlarging the natural quota of each state by a factor of

11 p 10 p 11 ÷ = = 1.1 1000 1000 10 and therefore, Increase in the natural quota of A = (0.1)(1.35) = 0.135 Increase in the natural quota of B = (0.1)(3.33) = 0.333 Increase in the natural quota of C = (0.1)(5.32) = 0.532 Note 1 The increments added to the natural quotas all went to the fractional parts, leaving the initial allocations unchanged. Note 2 State A (which originally had the largest fractional part) had the smallest natural quota and therefore received the smallest increase in its natural quota. States B and C received larger increases in their natural quotas and the result (due to Note 1) was a reversal in priority for surplus seats from A|BC in a 10-seat house to CB|A in an 11seat house. Next, we present the formal definition. Definition of the Alabama Paradox 3.3.11 We say that the Alabama paradox has occurred when an increase in the house size causes the number of seats apportioned to a state to decrease. An apportionment procedure that never exhibits the Alabama paradox is said to be house-monotone.  

3

171 3.3 · Quota Procedures

The above discussion indicates that Hamilton’s procedure is not house-monotone. Lowndes’ procedure is not house-monotone either but this fact is not highlighted in the literature possibly because Lowndes’ procedure was never implemented in the House apportionment and is not widely used in other apportionment applications. The following example presents an occurrence of the Alabama paradox with the Lowndes procedure. Example 3.3.12 Use the Lowndes’ procedure to apportion 1. 10 seats 2. 11 seats among the following four states.

State

A

B

C

D

Total

Population

1262

1264

3736

3738

10,000

 Solution 1. The natural divisor is 10000/10 = 1000 person/seat. Calculations are shown in the following table:

State

Population

Natural quota

Initial allocation

Surplus seats

Apportionment

A

1262

1262/1000 = 1.262

1

1

2

B

1264

1264/1000 = 1.264

1

1

2

C

3736

3736/1000 = 3.736

3

0

3

D

3738

3738/1000 = 3.738

3

0

3

Total

10,000

10.000

8

2

10

1264 The district sizes based on the initial allocation are 1262 1 = 1262 for A, 1 = 1264 3738 = 1245.3 for C, and = 1246 for D. The two surplus seats go to A and for B, 3736 3 3 B since they have the two highest district sizes based on the initial allocation. 2. The natural divisor is 10000/11 = 909.09 person/seat. Calculations are shown in the following table.

172

3

Chapter 3 • Apportionment

State

Population

Natural quota

Initial allocation

Surplus seats

Apportionment

A

1262

1262/909.09 = 1.3882

1

0

1

B

1264

1264/909.09 = 1.3904

1

1

2

C

3736

3736/909.09 = 4.1096

4

0

4

D

3738

3738/909.09 = 4.1118

4

0

4

Total

10,000

11.0000

10

1

11

1264 The district sizes based on the initial allocation are 1262 1 = 1262 for A, 1 = 1264 3738 = 934 for C, and = 934.5 for D. We have only one surplus seat in for B, 3736 4 4 this case and it is picked up by B since it has the largest district size based on the initial allocation. The Alabama paradox has obviously occurred in this example since state A (the Alabama) lost a seat (going down from two seats to one) when one seat was added to the house.

3.3.4 The Population Paradox We alert the reader to the fact that this paradox has various definitions in the literature. We use the definition adopted by Balinski and Young in their monograph [2]. Definition of the Population Paradox 3.3.13 We say that the population paradox has occurred when a state that gained population loses one seat or more and a state that lost population gains one seat or more, while the house size is kept unchanged. An apportionment procedure that never exhibits the population paradox is said to be population-monotone.  

ⓘ Remark 3.3.14 Two events need to simultaneously take place for the population paradox to occur: 1. A state that gained population loses one seat or more. 2. Another state that lost population gains one seat or more. The occurrence of one of these two events without the other is not paradoxical or counter-intuitive.  

ⓘ Remark 3.3.15 1. To examine house-monotonicity, we need to fix the population data and allow the house size to change.   2. To examine population-monotonicity, we need to fix the house size and allow the population data to change.  

3.3.5 The Balinski-Young Theorem The following two examples, given by Young in [29] establish a very important fact about quota procedures in general.

3

173 3.3 · Quota Procedures

Example 3.3.16 The following is the 1990 census data of a nation. Find all quota apportionments of 7 seats among the states A, B, C, and D.

State

A

B

C

D

Total

Population

752

101

99

98

1050

 Solution The natural divisor is 1050/7 = 150 person/seat. There are exactly 2 surplus seats. Work is shown in the following table.

State

Population

NQ

IA

(A, B)

(A, C)

(A, D)

(B, C)

(B, D)

(C, D)

A

752

5.0133

5

6

6

6

5

5

5

B

101

0.6733

0

1

0

0

1

1

0

C

099

0.6600

0

0

1

0

1

0

1

D

098

0.6533

0

0

0

1

0

1

1

Total

1050

7

5



X

X



X

X

There are exactly two quota solutions to this apportionment problem: 6, 1, 0, 0 and 5, 1, 1, 0 for A, B, C, and D, respectively. (Both Hamilton and Lowndes pick the 5, 1 ,1, 0 solution.) Example 3.3.17 The following is the 2000 census data of the nation in Example 3.3.16. Find all quota apportionments of 7 seats among the states A, B, C, and D.

State

A

B

C

D

Total

Population

753

377

96

97

1323

 Solution The natural divisor is 1323/7 = 189 person/seat. There are exactly 3 surplus seats. Work is shown in the following table.

174

3

Chapter 3 • Apportionment

State

Population

NQ

IA

(A, B, C)

(A, B, D)

(A, C, D)

(B, C, D)

A

753

3.9841

3

4

4

4

3

B

377

1.9947

1

2

2

1

2

C

096

0.5079

0

1

0

1

1

D

097

0.5132

0

0

1

1

1

Total

1323

7

4

X







There are exactly three quota solutions to this apportionment problem. (Hamilton picks the 4, 2, 0, 1 solution while Lowndes picks the 3, 2, 1, 1 solution.)

The above two examples show that regardless of the particular quota procedure used: 1. State A whose population increased from 752 to 753 suffered a drop in representation from at least 5 seats to at most 4 seats. 2. State D whose population decreased from 98 to 97 gained in representation from 0 seats to 1 seat. This proves the following 1983 result due to Balinski and Young [28].

Theorem 3.3.18 No quota procedure is population-monotone.

 

In other words, a population-monotone procedure must allow the allotment of a state to go above the upper natural quota or below the lower natural quota. For example, a state with a natural quota of 7.3654 should be allowed to receive more than 8 seats or less than 7 seats. This happens to be the case with the divisor procedures which we study in the next section. Note 3.3.19 It can be problematic to use quota procedures to apportion a certain house size. This can happen in the case where two or more states have exactly the same priority for a surplus seat. For example, if we try to use Hamilton’s procedure to apportion ten seats among the three states in the following table,

State

A

B

C

Total

Population

522

339

139

1000

A gets 5 seats, B gets 3 seats, and C gets one seat in the initial allocation. Which state gets the remaining surplus seat? B or C?

175 3.4 · Divisor Procedures

We run into the same difficulty if we try to use Lowndes’ procedure to apportion ten seats among the three states in the following table: State

A

B

C

Total

Population

616

256

128

1000

Nine seats are apportioned in the initial allocation but which state gets the surplus seat? B or C? Mathematically, we can say that an apportionment of ten seats in the above two examples either does not exist or is not unique. Problems of this kind are extremely rare and can be resolved by provisions in the apportionment law that would break the ties for eligibility for surplus seats (thus choosing one out of available different solutions) or by changing the house size (in the grounds that a solution for this particular house size does not exist).  

3.4

Divisor Procedures

The earliest of these procedures was introduced in 1790 by Thomas Jefferson (president of the United States from 1801 to 1809). President George Washington vetoed Hamilton’s apportionment bill and Jefferson’s procedure was passed and was used to apportion the U.S. House of Representatives from 1790 to 1840. Later on, several other divisor procedures were introduced but they were all based on the principle originated by Thomas Jefferson in 1790. We shall study four different divisor procedures in this section and one more procedure will be developed in the next section. The four divisor procedures to be studied in this section are: 1. Jefferson’s procedure. 2. Adams’ procedure: Introduced by John Quincy Adams (president of the United States from 1825 to 1829) in 1832 when he was a representative from Massachusetts. It was never implemented. 3. Webster’s procedure: Introduced in 1840 by Senator Daniel Webster of New York. It was used on-and-off to apportion the U.S. House of Representatives from 1840 to 1940. 4. The Hill–Huntington: Introduced by Joseph Hill (Head of the Federal Bureau of Statistics) and E.V. Huntington (professor of mathematics at Harvard University). It was adopted in 1940 and has been in use ever since to apportion the U.S. House of Representatives.

3.4.1 The General Framework of Divisor Procedures The quotas of the states are computed by dividing their populations by one and the same divisor d > 0. Each divisor procedure adopts a certain rule for rounding the quotas to non-negative integers. More precisely, a quota whose value is q is rounded to one of

3

176

3

Chapter 3 • Apportionment

the two nearest non-negative integers, according to the rounding rule specified by the procedure. Each state is then apportioned a number of seats equal to its rounded quota. Each divisor procedure is characterized by its rounding rule. If the house size h is pre-specified, which is often the case, we adjust the divisor so that the apportioned seats add up to h. Note A rounding rule must be monotone in the sense that if α and β are quotas that are rounded to the integers m and n, respectively, and if α > β, then m ≥ n. The student should be able to verify that Axioms 1, 2, and 3 in 3.2.1 are all satisfied by these procedures. Preliminary Definitions and Notations 3.4.1 If a divisor d is used, a state with population p will have a quota of q = pd which is rounded, say, to n seats. As we gradually decrease the divisor, we will reach a value for the divisor called the critical or threshold divisor to add 1 seat, and denoted Th(+1), so that divisors just higher than Th(+1) apportion n seats and divisors just lower than Th(+1) apportion n + 1 seats to the state in question. If we continue to decrease the divisor we will reach, for every positive integer i, a divisor value Th(+i) (called the critical or threshold divisor to add i seats) so that divisors just higher than Th(+i) apportion n + i − 1 seats and divisors just lower than Th(+i) apportion n + i seats to the state in question.

On the other hand, if we gradually increase the divisor, we will reach a value for the divisor called the critical or threshold divisor to drop 1 seat, denoted Th(−1), so that divisors just higher than Th(−1) apportion n − 1 seats and divisors just lower than Th(−1) apportion n seats to the state in question. If we continue to increase the divisor we will reach, for i = 1, 2, · · · , n, a divisor value Th(−i) (called the critical or threshold divisor to drop i seats) so that divisors just higher than Th(−i) apportion n − i seats and divisors just lower than Th(−i) apportion n − i + 1 seats to the state in question. In an apportionment problem with a pre-specified house size h, a good strategy would be to make a first attempt using the natural divisor. If the rounded natural quotas add up to the pre-specified house size, we are done! If not, we replace the natural divisor with a modified divisor and the populations of the states are all divided by this modified divisor to produce the corresponding modified quotas. If the modified quotas add up to the pre-specified house size, the modified divisor used is said to be a proper divisor and the process is completed by giving each state its rounded modified quota. In general (barring some extremely rare situations) there is an interval (called the interval of proper divisors) that contains, exclusively, all proper divisors for a particular problem and they all deliver the same solution to the apportionment problem at hand. The threshold divisors defined above will play a crucial role in finding that interval. Lower modified quotas and upper modified quotas are the modified quotas rounded down and up, respectively.  

3

177 3.4 · Divisor Procedures

3.4.2 Jefferson’s Procedure In the light of the framework explained in Sect. 3.4.1, all we need to do in order to define a divisor procedure is to present its rounding rule. The Rounding Rule of Jefferson’s Procedure 3.4.2 All quotas are rounded down. More precisely, a state whose quota is q > 0 receives k seats where k is the unique non-negative integer such that k ≤ q < k + 1. Example 3.4.3 Use Jefferson’s procedure to apportion 10 seats among the states A, B, C, and D based on the following population data.

State

A

B

C

D

Total

Population

6514

2608

611

267

10,000

 Solution Let us first try the natural divisor d = shown in the following table.

10000 10

= 1000 person/seat. Calculations are

State

Population

NQ

Apportionment

A

6514

6.514

6

B

2608

2.608

2

C

0611

0.611

0

D

0267

0.267

0

Total

10,000

10

8

Obviously, the natural divisor of 1000 person/seat is not a proper divisor: it apportions only 8 seats, not 10. We need to replace the natural divisor with a smaller (modified) divisor d to produce new (modified) quotas, larger than the natural quotas. Hopefully, the resulting lower modified quotas will add up to the pre-specified house size of 10 seats. The choice of modified divisor is a delicate process because if it is too small, the resulting modified quotas will be too large and the sum of the lower modified quotas will be larger than the pre-specified house size. On the other hand, if the modified divisor is too large, the modified quotas will not be large enough and the sum of the lower modified quotas will be less than the pre-specified house size.

178

Chapter 3 • Apportionment

Let us try a modified divisor of 800 persons/seat.

3

State

Population

MQ

Apportionment

A

6514

6514/800 = 8.14 · · ·

8

B

2608

2608/800 = 3.26 · · ·

3

C

0611

0611/800 = 0.76 · · ·

0

D

0267

0267/800 = 0.33 · · ·

0

Total

10,000

12.5

11

Obviously, a modified divisor of 800 person/seat is not a proper divisor. It is too small since it apportions 11 seats while the pre-specified house size is 10 seats only. We need a divisor larger than 800 person/seat. Let us try a modified divisor of 900 person/seat. State

Population

MQ

Apportionment

A

6514

6514/900 = 7.23 · · ·

7

B

2608

2608/900 = 2.89 · · ·

2

C

0611

0611/900 = 0.67 · · ·

0

D

0267

0267/900 = 0.29 · · ·

0

Total

10,000

11.11 · · ·

9

A modified divisor of 900 person/seat is not a proper divisor either. It is too large since it apportioned 9 seats only. Therefore, we should confine our search to the interval 800 < d < 900. We won’t keep you waiting any longer: a divisor of 840 person/seat is a proper divisor, as shown in the following table. State

Population

MQ

Apportionment

A

6514

6514/840 = 7.75 · · ·

7

B

2608

2608/840 = 3.10 · · ·

3

C

0611

0611/840 = 0.72 · · ·

0

D

0267

0267/840 = 0.31 · · ·

0

Total

10,000

11.90 · · ·

10

The solution to the above problem by Jefferson’s procedure stays within the quota. In fact, State A whose natural quota is 6.514 ended up with its upper natural quota of 7 seats. State B whose natural quota is 2.608 ended up with its upper natural quota of 3 seats. State C whose natural quota is 0.611 ended up with its lower natural quota of 0 seats. State D whose natural quota is 0.267 ended up with its lower natural quota of 0 seats.  

3

179 3.4 · Divisor Procedures

In the following example Jefferson’s solution violates the quota rule. Example 3.4.4 Use Jefferson’s procedure to apportion 10 seats among the states A, B, C, and D based on the following population data.

State

A

B

C

D

Total

Population

5942

1684

1527

847

10,000

 Solution We may use trial and error to find a divisor that apportions the exact house size of 10 seats as we did in the previous example but this will not be easy in this example and may take several attempts because the interval of proper divisors is the very narrow interval 842 < d < 847, as we shall see later. Let us use the proper divisor d = 845. State

Population

MQ

Apportionment

A

5942

5942/845 = 7.031 · · ·

7

B

1684

1684/845 = 1.992 · · ·

1

C

1527

1527/845 = 1.807 · · ·

1

D

0847

0847/845 = 1.002 · · ·

1

Total

10,000

11.834 · · ·

10

Jefferson’s solution to this problem violates quota rule. To see this, note that the natural 5942 divisor is 10000 10 = 1000 person/seat and therefore state A’s natural quota is 1000 = 5.942. Jefferson’s procedure apportions 7 seats to state A which exceeds its upper natural quota of 6. You can verify that the apportionments of B, C, and D are within the quota.   Example 3.4.5 In Example 3.4.3, each of C and D was assigned zero seats by Jefferson’s procedure in a house of 10 seats. 1. Find the smallest house size that satisfies the constitutional requirement that each state should receive at least one seat and the corresponding number of seats apportioned to each state. 2. Detect all violations of the quota rule. 

180

3

Chapter 3 • Apportionment

Solution 1. We need to use a modified divisor d so that the modified quota of the smallest state rounds down to an integer ≥ 1, that is, 267 ≥ 1 and so d ≤ 267 d To make the house as small as possible, we choose the largest possible value for d allowed by the above inequality, that is, d = 267. Computations are shown in the following table. State

Population

MQ

Apportionment

A

6514

6514/267 = 24.39 · · ·

24

B

2608

2608/267 = 9.76 · · ·

9

C

0611

0611/267 = 2.28 · · ·

2

D

0267

0267/267 = 1

1

Total

10,000

37.45 · · ·

36

From the table we see that the smallest house size that enables each state to have at least one seat is 36 seats and the apportionment is shown in the last column of the table. 2. The natural divisor is 10000 36 = 277.7 and the natural quotas are 6514/277.7 = 23.45 · · · for A, 2608/277.7 = 9.38 · · · for B, 611/277.7 = 2.19 · · · for C and, 267/277.7 = 0.96 · · · for D. The following table shows the lower and upper natural quota of each state based on a house size of 36 seats and the quota rule violations.

State

Pop

NQ

LNQ

UNQ

Apportionment

Within quota?

A

6514

23.450

23

24

24

YES

B

2608

9.388

9

10

9

YES

C

0611

2.199

2

3

2

YES

D

0267

0.961

0

1

1

YES

Total

10,000

36

36

Jefferson’s apportionment for this problem is within quota.

 

Next, we proceed to determine the interval of proper divisors for Jefferson’s procedure.

3

181 3.4 · Divisor Procedures

Proposition 3.4.6 A state with population p receives n seats (n = 1, 2, 3, · · · ) by Jefferson’s procedure if and only if the divisor d satisfies the inequality p p p. Proof The state in question receives n seats by Jefferson’s procedure if and only if its quota q satisfies the inequality n ≤ q < n + 1, that is, n ≤ pd < n + 1 which is equivalent to d > p p < d ≤ pn when n = 0 and can be transformed with simple algebraic manipulations to n+1 for n = 0.  

ⓘ Corollary 3.4.7 If n seats are apportioned to a state with population p by Jefferson’s procedure, then Th(+i) =

p , i = 1, 2, 3, · · · n+i

Th(−i) =

p , i = 1, 2, 3, · · · , n · n − (i − 1)

 

Example 3.4.8 Consider a state with a population of 7648 persons. The interval of divisors that allow the state to have exactly 5 seats is 7648 7648 2p when n = 0, and can be transformed with simple algebraic manipulations to p p   1 < d ≤ 1 for n > 0. n+ 2

n− 2

ⓘ Corollary 3.4.26 If n seats are apportioned to a state with population p by Webster’s procedure, then Th(+i) = Th(−i) =

p n + (i − 12 ) p n − (i − 12 )

, i = 1, 2, · · · , i = 1, 2, · · · , n ·

 

Example 3.4.27 Use Webster’s procedure to apportion 10 seats among the states A, B, C, and D in Examples 3.4.3 and 3.4.13.

State

A

B

C

D

Total

Population

6514

2608

611

267

10,000

 Solution Let us first try the natural divisor d = shown in the following table.

10000 10

= 1000 person/seat. Calculations are

195 3.4 · Divisor Procedures

State

Population

NQ

Apportionment

A

6514

6.514

7

B

2608

2.608

3

C

0611

0.611

1

D

0267

0.267

0

Total

10,000

10

11

This is Case 3 in Examples 3.4.24 above. We need a modified divisor higher than the natural divisor of 1000. Since there is one excess seat, we need to compute, for each state, the threshold divisors to drop up to two seats. Calculations are shown in the following table:

State

Pop

d = 10

Th(−1)

A

6514

7

B

2608

3

C

0611

1

6514 6.5 = 1002.153 · · · 2608 2.5 = 1043.2 (2) 611 0.5 = 1222

D

0267

0

···

Th(−2) (1)

6514 5.5 2608 1.5

= 1184.363 · · · = 1738.666 · · ·

··· ···

The following diagram describes the gradual increase of the divisor and the interval of proper divisors. 1043.2 Do not go this far. We do not want to drop a second seat.  1002.153 · · · A drops a seat ↑ 1000 Natural divisor The interval of proper divisors for Webster’s procedure in this example is 1002.153 · · · < d < 1043.2. A drops 1 excess seat and ends up with 6 seats, B stays at 3 seats, C stays at 1 seat, and D stays at 0 seats. We may choose any number in the interval 1002.154 < d < 1043.2 as our modified divisor. Let us use the modified divisor d = 1030. State

Population

MQ

Apportionment

A

6514

6514/1030 = 6.32 · · ·

6

B

2608

2608/1030 = 2.53 · · ·

3

C

0611

0611/1030 = 0.59 · · ·

1

D

0267

0267/1030 = 0.25 · · ·

0

Total

10,000

9.69 · · ·

10

The student should verify that the solution provided by Webster’s procedure in this example is within quota.  

3

196

3

Chapter 3 • Apportionment

Example 3.4.28 Use Webster’s procedure to apportion 10 seats among the states A, B, C, and D in Examples 3.4.9 and 3.4.19. State

A

B

C

D

Total

Population

5942

1684

1527

0847

10,000

 Solution Let us first try the natural divisor d = shown in the following table.

10000 10

= 1000 person/seat. Calculations are

State

Population

NQ

Apportionment

A

5942

5.942

6

B

1684

1.684

2

C

1527

1.527

2

D

0847

0.847

1

Total

10,000

10

11

This is Case 3 in Examples 3.4.24 above. We need a modified divisor higher than the natural divisor of 1000. Since there is one excess seat, we need to compute, for each state, the threshold divisors to drop up to two seats. Calculations are shown in the following table: State

Pop

d = 10

Th(−1)

A

5942

6

B

1684

2

C

1527

2

D

0847

1

5942 5.5 = 1080.363 · · · 1684 1.5 = 1122.666 · · · 1527 1.5 = 1018 (1) 847 0.5 = 1694

Th(−2) (2)

5942 4.5 1684 0.5 1527 0.5

= 1320.444 · · · = 3368 = 3054

···

The following diagram describes the gradual increase of the divisor and the interval of proper divisors. 1080.363 · · · Do not go this far. We do not want to drop a second seat.  1018 C drops a seat ↑ 1000 Natural divisor The interval of proper divisors for Webster’s procedure in this example is 1018 < d < 1080.363 · · · . A stays at 6 seats, B stays at 2 seats, C drops 1 excess seat and ends up with 1 seat, and D stays at 1 seat. We may choose any number in the interval 1018 < d < 1080.363 as our modified divisor. Let us use the modified divisor d = 1050.

3

197 3.4 · Divisor Procedures

State

Population

MQ, d = 1050

Apportionment

A

5942

5942/1050 = 5.65 · · ·

6

B

1684

1684/1050 = 1.60 · · ·

2

C

1527

1527/1050 = 1.45 · · ·

1

D

0847

0847/1050 = 0.80 · · ·

1

Total

10,000

9.5 · · ·

10

 

Webster’s solution to this problem is within quota.

Webster’s procedure rarely violates the quota rule; the following is an example where it does. Example 3.4.29 Use Webster’s procedure to apportion 10 seats among the states A, B, C, and D whose population data is shown in the following table:

State

A

B

C

D

Total

Population

5942

1362

1351

1345

10,000

 Solution As always, we start with the natural divisor d = Calculations are shown in the following table.

10000 10

State

Population

NQ

Apportionment

A

5942

5.942

6

B

1362

1.362

1

C

1351

1.351

1

D

1345

1.345

1

Total

10,000

10

9

= 1000 person/seat.

This is Case 1 in Examples 3.4.24 above. We need a modified divisor lower than the natural divisor of 1000. Since there is one surplus seat, we need to compute, for each state, the threshold divisors to add up to two seats. Computations are shown in the following table:

198

Chapter 3 • Apportionment

State

3

Pop

d = 10

Th(+1) 5942 6.5 1362 1.5 1351 1.5 1345 1.5

A

5942

6

B

1362

1

C

1351

1

D

1345

1

Th(+2)

= 914.153 · · · (1) = 908 (2) = 900.666 · · · = 896.666 · · ·

5942 7.5 1362 2.5 1351 2.5 1345 2.5

= 792.266 · · · = 544.8 = 540.4 = 538

The following diagram describes the gradual reduction of the divisor and the interval of proper divisors. 1000 Natural divisor ↓ 914.153 · · · A gets the first surplus seat  908 Stop before you reach this point. We do not have a second surplus seat. The interval of proper divisors for Webster’s procedure in this example is 908 < d ≤ 914.153. A adds the surplus seat and ends up with 7 seats, B stays at 1 seat, C stays at 1 seat, and D stays at 1 seat. We may choose any number in the interval 908 < d < 914.153 as our modified divisor. Let us use the modified divisor d = 911. State

Population

MQ, d = 911

Apportionment

A

5942

5942/911 = 6.52 · · ·

7

B

1362

1362/911 = 1.49 · · ·

1

C

1351

1351/911 = 1.48 · · ·

1

D

1345

1345/911 = 1.47 · · ·

1

Total

10,000

10.97 · · ·

10

Webster’s solution for this problem violates the quota rule since state A whose natural quota is 5.942 ends up with 7 seats instead of 5 or 6.   Example 3.4.30 In Example 3.4.27, the smallest state, D, received 0 seats by Webster’s procedure.

State

A

B

C

D

Total

Population

6514

2608

611

267

10,000

Let us try to compute the smallest and the largest houses that allow D to receive exactly one seat. From Proposition 3.4.25 we see that D receives exactly one seat if and only if 267 1+

1 2

0 2 2 2 2

and so g(x, y) < m(x, y) which completes the proof of Part 1. Part 2 is clear.

 

The conventional rule of rounding, adopted by Webster’s procedure, uses the arithmetic mean of two consecutive integers as the cut-off point of rounding for all quotas that lie between those two consecutive integers. More precisely, given a quota q such that n ≤ q < n + 1, we compare it to the arithmetic mean m(n, n + 1) =

n + (n + 1) 1 =n+ . 2 2

If q < m(n, n + 1), we round it down to n; and if q ≥ m(n, n + 1), we round it up to n + 1. In the Hill–Huntington procedure, the geometric mean plays a rôle similar to the one played by the arithmetic mean in Webster’s procedure. In other words, to round a quota q, we find an integer n such that n ≤ q < n + 1 and we compare q to the geometric mean g(n, n + 1) =



n(n + 1).

If q ≤ g(n, n + 1) we round it down to n; and if q > g(n, n + 1) we round it up to n+1. We call this the geometric-mean rounding. (Note: q is—by definition—a “rational number,” while for n = 1, 2, 3, · · · , g(n, n + 1) is always an “irrational number” and therefore, the equality q = g(n, n + 1) is impossible.) The Rounding Rule of the Hill–Huntington Procedure 3.4.33 A state whose quota is q > 0 receives k seats where k is the unique positive integer such that

(k − 1)k < q <



k(k + 1)·

The following table shows the Webster cut-off point of rounding, m(n, n + 1), and Hill– Huntington cut-off point of rounding, g(n, n + 1), for each of the intervals n ≤ q < n + 1, n = 0, 1, 2, · · · , 19, 99, 199, 499, 999.

201 3.4 · Divisor Procedures

Cut Webster 0 to 1 0.5 1 to 2 1.5 2 to 3 2.5 3 to 4 3.5 4 to 5 4.5 5 to 6 5.5 6 to 7 6.5 7 to 8 7.5 8 to 9 8.5 9 to 10 9.5 10 to 11 10.5 11 to 12 11.5 12 to 13 12.5 13 to 14 13.5 14 to 15 14.5 15 to 16 15.5 16 to 17 16.5 17 to 18 17.5 18 to 19 18.5 19 to 20 19.5 99 to 100 99.5 199 to 200 199.5 499 to 500 499.5 999 to 1000 999.5 Interval

Off Point of Rounding Hill - Huntington √ 0×1=0 √ 1 × 2 = 1.4142135 √ 2 × 3 =2.4494897 √ 3 × 4 = 3.4641016 √ 4 × 5 = 4.4721359 √ 5 × 6 = 5.4772255 √ 6 × 7 = 6.4807406 √ 7 × 8 = 7.4833147 √ 8 × 9 = 8.4852813 √ 9 × 10 = 9.4868329 √ 10 × 11 = 10.488088 √ 11 × 12 = 11.489125 √ 12 × 13 = 12.489995 √ 13 × 14 = 13.490737 √ 14 × 15 = 14.491376 √ 15 × 16 = 15.491933 √ 16 × 17 = 16.492422 √ 17 × 18 = 17.492855 √ 18 × 19 = 18.493242 √ 19 × 20 = 19.493588 √ 99 × 100 = 99.498743 √ 199 × 200 = 199.49937 √ 499 × 500 = 499.49974 √ 999 × 1000 = 999.49987

Based on the above table, we can conjecture that for n = 0, 1, 2, · · · , 1. The Hill–Huntington cut-off point of rounding is always to the left of Webster’s one, that is, g(n, n+1) < m(n, n+1) = n+ 12 . This has already been proven in Proposition 3.4.32 above. 2. As n increases, g(n, n + 1) gets closer to m(n, n + 1). (See ⊡ Fig. 3.1.) This is to be proven in Part 1 of Proposition 3.4.34 below. 3. Except for n = 0, the first digit to the right of the decimal point in g(n, n + 1) is 4. This follows from Part 1 of Proposition 3.4.34 below and the fact that the first digit to the right √ of the decimal point in 1 × 2 is 4. 4. The difference m(n, n + 1) − g(n, n + 1) between the arithmetic mean and the geometric mean tends to zero as n → ∞. This is to be proven in Part 2 of Proposition 3.4.34.

3

202

Chapter 3 • Apportionment

⊡ Fig. 3.1 Positions of g(n, n + 1) relative to m(n, n + 1)

g

0

3

m

1

g

1

2

3

9

14

m g m g m g m g m

2

3

4

10

15

Proposition 3.4.34 1. For n = 0, 1, 2, · · · we have m(n, n + 1) − g(n, n + 1) > m(n + 1, n + 2) − g(n + 1, n + 2). 2. limn→∞ {m(n, n + 1) − g(n, n + 1)} = 0· Proof √ 1. From Proposition 3.4.32 we have g(n, n + 1) < m(n, n + 1) and so 2 n(n + 1) < 2n + 1. Adding 1 + n(n + 1) to both sides we get (1 +



n(n + 1))2 = 1 + 2 n(n + 1) + n(n + 1) < n2 + 3n + 2 = (n + 1)(n + 2)·

It follows that 1 +

√

n(n + 1)
1 = m(n + 1, n + 2) − m(n, n + 1) from which we get m(n + 1, n + 2) − g(n + 1, n + 2) < m(n, n + 1) − g(n, n + 1)

3

203 3.4 · Divisor Procedures

which proves Part 1. To Prove Part 2 we note that m(n, n + 1) − g(n, n + 1) = n + √ n(n + 1). Rationalizing the numerator we get

1 2

−

1

m(n, n + 1) − g(n, n + 1) =

n+

1 2

+

4 √

n(n + 1)

→ 0 as n → ∞·  

Note 3.4.35 It is now clear that 1. Any quota that is rounded up in Webster’s procedure is also rounded up in the Hill– Huntington procedure; and any quota that is rounded down in the Hill–Huntington procedure is also rounded down in Webster’s procedure.   2. Quotas less than 1 are all rounded up to 1 by the Hill–Huntington procedure. Therefore, the Hill–Huntington procedure (like Adams’ procedure) never gives zero seats to a state with non-zero population.   3. For a quota q with integer part n ≥ 1, let δ be the first digit to the right of the decimal point. If δ is 3 or lower, q is rounded down; if δ is 5 or higher, q is rounded up and if √ δ = 4 then we have to check q carefully against the n(n + 1) cut-off point.   Since some quotas are rounded down and others are rounded up, Remark 3.4.23 applies to the Hill–Huntington procedure as it does to Webster’s procedure. Example 3.4.36 We use the Hill–Huntington procedure to apportion ten seats among the states A, B, C, and D based on the following data.

State

A

B

C

D

Total

Population

6514

2608

611

267

10,000

(This was done in Examples 3.4.3, 3.4.13, and 3.4.27 using the procedures of Jefferson, Adams and Webster, respectively.) The natural divisor of 10000 10 = 1000 is not a proper divisor but d = 1100 is. Computations are shown in the following table. (We shall determine the interval of proper divisors later.)

State

Pop

NQ

Rounded NQ

MQ, d = 1100

Apportionment

A

6514

6.514

7

6514/1100 = 5.92 · · ·

6

B

2608

2.608

3

2608/1100 = 2.37 · · ·

2

C

0611

0.611

1

611/1100 = 0.55 · · ·

1

D

0267

0.267

1

267/1100 = 0.24 · · ·

1

Total

10,000

10

12

9.09 · · ·

10

The Hill–Huntington solution to this problem is within quota.



204

Chapter 3 • Apportionment

Proposition 3.4.37 A state with population p receives n seats, (n = 1, 2, 3, · · · ) by the Hill–Huntington procedure if and only if the divisor d satisfies the inequality

3

√

p p 0 such that max1≤j ≤r k = 1, 2, · · · , r we get mk −

mk − 12 xk

≤

1 xk 1 ≤ < mk + 2 d 2

1 d


0 so that mi − 12 ≤ xdii < mi + 12 but the di ’s may have to be different. Thanks to the δ-inequity pairwise minimization property we are able to find one common   value d so that di = d, i = 1, 2, · · · , r.

3.5.4 The Hill–Huntington Procedure and the ρ-Inequity The ρ-inequity is related to the Hill–Huntington procedure in the same way the δinequity is related to Webster’s procedure. This is to be established in the next theorem whose proof follows similar arguments to those in the proof of Theorem 3.5.12, with the algebra switching from addition and subtraction to multiplication and division.

Theorem 3.5.14 1. The Hill–Huntington procedure minimizes the pairwise ρ-inequities, in the sense that if the Hill–Huntington procedure is used, the resulting apportionment must minimize the pairwise ρ-inequities. 2. The Hill–Huntington procedure is the only apportionment procedure with this property, that is, an apportionment that minimizes the pairwise ρ-inequities must coincide with the Hill–Huntington apportionment.

Proof 1. Assume that the Hill–Huntington procedure apportioned m seats to state A and n seats to state B whose populations are x persons and y persons, respectively (note that m ≥ 1, n ≥ 1 by Notes 3.4.35). and let d be the divisor used. Then



x (m − 1)m < < m(m + 1) d

and



y (n − 1)n < < n(n + 1), m ≥ 1, n ≥ 1· d

Therefore, 1 m(m + 1) (m − 1)m < 2 < x2 d x2

and

(n − 1)n 1 n(n + 1) < 2 < · y2 d y2

214

3

Chapter 3 • Apportionment

n m m+1 From this we get n−1 y × y < x × x . Assuming pcr(B : n) ≥ pcr(A : m) (i.e., B is the rich state and A is the poor state) and using some simple algebraic manipulations we get

n m m+1 n−1 ÷ < ÷ = ρ(A : m + 1, B : n − 1) y x x y

ρ(A : m, B : n) =

as in Part 2 of Corollary 3.5.11. This completes the proof of Part 1. 2. Denote the states among which the seats are apportioned by Ai , i = 1, 2, · · · , r and for each i let xi be the population of Ai and mi be the number of seats apportioned to it (none m of the mi ’s is zero, by Proposition 3.5.10). Let Ai and Aj be two states with xjj > mxii (Aj is the rich state and Ai is the poor state). Assuming that the pairwise ρ-inequity minimization property holds we get mj mi mi + 1 mj − 1 ÷ < ÷ xj xi xi xj √ which, with simple algebraic manipulation, can be transformed to √

mi (mi +1) . xi



Since the last inequality trivially holds if

(mj − 1)mj xj

√


xj mj

using the ϕ-

< (without the aforementioned minimization property, but it is obviously true if property). Therefore, we can find a positive number d so that max

1≤j ≤r

h(mj − 1, mj ) 1 h(mi , mi + 1) < · ≤ min 1≤i≤r xj d xi

For every k = 1, 2, · · · , r we then have h(mk − 1, mk )
m(n + 1, n + 2) − h(n + 1, n + 2)· Proof 1. For n = 0, 1, 2, · · · we have m(n, n + 1) − h(n, n + 1) =

1 2n + 1 2n(n + 1) − = → 0 as n → ∞· 2 2n + 1 2(2n + 1)

2. This follows from m(n, n+1)−h(n, n+1) =

1 1 > = m(n+1, n+2)−h(n+1, n+2)· 2(2n + 1) 2(2n + 3)  

⊡ Figure 3.2 shows the cut-of points of rounding of the five divisor procedures we presented, in three different intervals. Note that the left endpoint of an interval is always the cut-off point of rounding for Adams’ procedure, the right endpoint of the interval is the cut-off point of rounding for Jefferson’s procedure and the midpoint is the cut-off point of Webster’s procedure. We also tried to demonstrate in the figure the fact that as n → ∞, the difference between the cut-off point of Webster and the cut-off point of each of Dean and Hill–Huntington tends to zero. Note also that since h(0, 1) = 0, h(1, 2) = 1.333 · · · , and h(2, 3) = 2.4 · · · , it follows from Part 2 of Proposition 3.5.22 that for n ≥ 2, the first digit to the right of the decimal point in

220

Chapter 3 • Apportionment

⊡ Fig. 3.2 Cut-off points of rounding for the standard divisor procedures

h

0

3

m

g g

1

m

h g

9

h m

1

2

10

h(n, n + 1) is always 4. This observation will be helpful in rounding quotas with less computations. The following table shows the cut-off points of rounding for our five divisor procedures in several intervals. Interval

Adams

Dean

H–H

Webster

Jefferson

0 to 1

0.0000000

0.0000000

0.0000000

0.5000000

1.0000000

1 to 2

1.0000000

1.3333333

1.4142135

1.5000000

2.0000000

2 to 3

2.0000000

2.4000000

2.4494897

2.5000000

3.0000000

3 to 4

3.0000000

3.4285714

3.4641016

3.5000000

4.0000000

4 to 5

4.0000000

4.4444444

4.4721359

4.5000000

5.0000000

5 to 6

5.0000000

5.4545454

5.4772255

5.5000000

6.0000000

6 to 7

6.0000000

6.4615384

6.4807406

6.5000000

7.0000000

7 to 8

7.0000000

7.4666666

7.4833147

7.6000000

8.0000000

8 to 9

8.0000000

8.4705882

8.4852813

8.5000000

9.0000000

9 to 10

9.0000000

9.4736842

9.4868329

9.5000000

10.000000

19 to 20

19.000000

19.487179

19.493588

19.500000

20.000000

99 to 100

99.000000

99.497487

99.498743

99.500000

100.00000

199 to 200

199.00000

199.49874

199.49937

199.50000

200.00000

999 to 1000

999.00000

999.49974

999.49987

999.50000

1000.0000

Since, in Dean’s procedure, some quotas are rounded up and others are rounded down, the three cases in Remark 3.4.23 occur in Dean’s procedure as they did in the Webster and Hill–Huntington procedures. Note also that, like the procedures of Adams and Hill–Huntington, Dean’s procedures never apportions zero seats to any state. Note 3.5.23 For n ≥ 1, the cut-off points of rounding of the five divisor procedures follow the inequality n < h(n, n + 1) < g(n, n + 1) < m(n, n + 1) < n + 1 and for n = 0 we have 0 = n = h(n, n + 1) = g(n, n + 1) < m(n, n + 1) =

1 < n + 1 = 1· 2

3

221 3.5 · Equity Criteria of Divisor Procedures

(with n being the cut-off point of rounding for Adams’ procedure and n + 1 being the cut-off point of rounding for Jefferson’s procedure).  

Obviously, the smaller the quota of a state, the more significant its fractional part to the state in question. This, together with Proposition 3.5.22 and ⊡ Fig. 3.2, explains why procedures get more favorable to smaller states as we proceed from right to left in the above inequality. The following is a list of the five standard divisor procedures with arrows pointing in the direction of favorability to smaller states. Adams ←− Dean ←− Hill–Huntington ←− Webster ←− Jefferson

 

Next, we discuss the relationships among the intervals of proper divisors for divisor procedures. First, let us agree to say that an interval I1 is greater than or equal to an interval I2 and write I1 ≥ I2 if for every x1 in I1 there is an x2 in I2 with x2 ≤ x1 and for every y2 in I2 there is a y1 in I1 with y1 ≥ y2 . For example, if I1 is 5 < t < 8 and I2 is 1 < t < 7, then I1 ≥ I2 . Proposition 3.5.24 Let Proc1 and Proc2 be two divisor procedures and in each interval [k, k + 1] (where k is a non-negative integer) assume that the cut-off points of rounding are such that cut-off point of rounding for Proc1 ≤ cut-off point of rounding for Proc2· If a house of h seats is to be apportioned, then I1 ≥ I2 where I1 and I2 are the intervals of proper divisors for Proc1 and Proc2, respectively. Proof 1. Due to the above relation between the cut-off points of rounding of the two procedures, we see that a divisor α1 that apportions h seats for Proc1 will apportion ≤ h seats for Proc2 and therefore α1 may have to be reduced to a value α2 ≤ α1 in order to apportion the h seats for Proc2. 2. Again, due to the above relation between the cut-off points of rounding of the two procedures, we see that a divisor β2 that apportions h seats for Proc2 will apportion ≥ h seats for Proc1 and therefore β2 may have to be increased to a value β1 ≥ β2 in order to apportion the h seats for Proc1.  

222

Chapter 3 • Apportionment

3.5.9 The Threshold Divisors of Dean’s Procedure

3

Proposition 3.5.25 A state with population p receives n seats, (n = 1, 2, 3, · · · ) by Dean’s procedure if and only if the divisor d satisfies the inequality 

p 2n(n+1) 2n+1

m1 , we get from the monotonicity of rounding rules that dx > xd11 and so d1 x1 > > 1· d x

3

227 3.6 · Apportionment Paradoxes

With a similar argument we also get the contradictory statement d1 y1 < < 1. d y  

The result follows from the contradiction obtained.

3.6.2 Consistency The following example, from Young’s book [29] demonstrates one more imperfect behavior by Hamilton’s procedure. Example 3.6.3 1. Use Hamilton’s procedure to apportion 21 seats among the three states A, B, and C with the following population data.

State

A

B

C

Total

Population

727

123

222

1072

2. Reapportion, between A and B, the combined seats apportioned to those two states in Part 1.  Solution 1. The natural divisor is 1072 21 = 51.047619 · · · person/seat. Calculations and solution are shown in the following table.

State

Population

NQ

Initial

Surplus seats

Apportionment

A

727

14.2416 · · ·

14

0

14

B

123

2.4095 · · ·

2

1

3

C

222

4.3488 · · ·

4

0

4

Total

1072

21

20

1

21

We now reapportion the 14 + 3 = 17 seats apportioned to A and B combined in Part 1 among those two states. The natural divisor is 727+123 = 50 person/seat. 17

228

Chapter 3 • Apportionment

3 • •

State

Population

NQ

Initial

Surplus seats

Apportionment

A

727

14.54

14

1

15

B

123

2.46

2

0

2

Total

850

21

16

1

17

Note that: In the presence of C, A receives 14 seats, and B receives 3 seats out of the 17 seats allotted to these two states combined. If those 17 seats are reapportioned between A and B (with C removed), A receives 15 seats, and B receives 2 seats.

This is obviously a type of inconsistency that bares resemblance to the dependence on irrelevant alternatives that afflicts many social welfare functions, studied in  Chap. 1.

Definition 3.6.4 An apportionment procedure is said to be consistent if the reapportionment of the combined shares of any set of states among the members of that set always agrees with the original apportionment. Precisely, if A1 , A2 , · · · Ar are some of the states in a nation and for each i, state Ai was apportioned ni seats, then the sum n1 + n2 + · · · + nr , when reapportioned among A1 , A2 , · · · Ar , state Ai receives again ni seats, i = 1, 2, · · · r. An apportionment procedure that is not consistent is said to be inconsistent.  

Example 3.6.5 1. Use Lowndes’ procedure to apportion 21 seats among the three states A, B, and C with the following population data.

State

A

B

C

Total

Population

717

133

222

1072

2. Reapportion, between A and C, the combined seats apportioned to those two states in Part 1.  Solution 1. The Lowndes apportionment for this problem is 14 seats for A, 3 seats for B, and 4 seats for C. 2. The combined population of A and C is 939 citizens and their combined share of seats in Part 1 is 18 seats. Reapportioning those 18 seats between A and C using the Lowndes

3

229 3.6 · Apportionment Paradoxes

procedure, A receives 13 seats, and C receives 5 seats. Compare to their allotments in the presence of B in Part 1.  

Computations are left to the student.

ⓘ Corollary 3.6.6 The procedures of Hamilton and Lowndes are inconsistent. (This follows from Example 3.6.3 and Example 3.6.5 above.)

 

Theorem 3.6.7 Divisor procedures are consistent.

Proof Assume a divisor procedure was used to apportion the house seats in a certain nation (and the proper divisor used was d). The seats apportioned to a certain set of states were then combined and reapportioned among the states in this set (and the proper divisor used was d1 ). Assume also that some member of that set, say state A (with population p) received m seats in the original apportionment and m1 seats in the above- mentioned reapportionment. If m > m1 , then p p > and so d1 > d· d d1 Note that some other member of the set, say state B, must have gained a seat or more in the reapportionment. A similar argument will show that d1 < d; a contradiction. Hence m1 = m.   Note 3.6.8 Theorems 3.6.1, 3.6.2, and 3.6.7 above establish some positive attributes of divisor procedures. The only shortcoming in these procedures we have come across so far is their possible quota violations. Divisor procedures, however, differ in the manner and the frequency of their quota violations: 1. Jefferson’s procedure can only violate upper quota, while Adams’ procedure can only violate lower quota. The procedures of Webster, Hill–Huntington, and Dean can violate both upper and lower quotas: Violations of upper quota can occur only when the divisor used is lower than the natural divisor; and violations of lower quota can occur only when the divisor used is higher than the natural divisor. 2. The procedures of Adams and Jefferson are the greatest quota violators, followed by Dean’s procedure then the Hill–Huntington procedure. Webster’s procedure rarely violates quota. The following data [2] was obtained using the Monte Carlo simulation for a US-like nation of 50 states and 435-seat house. It shows the expected number of quota violations per 1000 apportionment problems.

230

Chapter 3 • Apportionment

Procedure

Adams

Dean

H–H

Webster

Jefferson

Expected number of quota violations

1000

15.40

2.86

0.61

1000

3 Another point in favor of Webster’s procedure is that it shows no tendency to favor larger or smaller states. The three procedures to the left of Webster’s in Note 3.5.23 tend to favor smaller states (increasingly from right to left). Jefferson’s procedure, on the opposite end, have a strong tendency to favor larger states. (For statistics about bias estimates toward larger or smaller states, see [2]).   We further the heuristic argument begun in Note 3.5.23, by trying to understand why divisor procedures can violate the quota. The use of a divisor smaller than the natural divisor scales up all the quotas by the same factor. Obviously the incremental increase in the larger quotas will be larger than the incremental increase in the smaller ones. Therefore, the larger states are likely to add more seats than the smaller ones do; and this could involve exceeding the upper natural quota by the larger states. On the other hand, the use of a divisor larger than the natural divisor scales down all the quotas by the same factor. Obviously, the incremental decrease in the larger quotas will be larger than the incremental decrease in the smaller quotas, causing the larger states to drop more seats than the smaller states do, which could make the larger states go below the lower natural quotas. The above observations are only statistical and do not always hold. In Example 3.4.29 we detected a violation of quota rule by Webster’s procedure. The student is asked to process the same data in that example using the other standard procedures. The results are shown in the following table:

Procedure

Interval of proper divisors

Adams

1345 < d < 1351

5

2

2

1

Dean

1013.25 < d < 1021.5

6

2

1

1

H–H

955.302 < d < 963.07

6

2

1

1

Webster

908 < d < 914.15

(7)

1

1

1

Jefferson

742.75 < d < 848.85

(7)

1

1

1

A

B

C

D

Note that, in this example, the procedure of Webster violates the quota rule (along with Jefferson’s procedure) while the procedures of Adams, Dean, and Hill–Huntington do not. The following is another interesting example. Example 3.6.9 Apportion 20 seats among the following eight states using the five standard divisor procedures and detect all violations of the quota rule.  Solution The results are shown in the following table. Computations are left to the student.

3

231 3.6 · Apportionment Paradoxes

State

Population

NQ

Adams 1202

Dean 918

H–H 915

Webster 913

Jefferson 850

A

6846

6.846

6

7

7

7

(8)

B

5942

5.942

5

(7)

(7)

(7)

6

C

1205

1.205

2

1

1

1

1

D

1204

1.204

2

1

1

1

1

E

1203

1.203

2

1

1

1

1

F

1201

1.201

1

1

1

1

1

G

1200

1.200

1

1

1

1

1

H

1199

1.199

1

1

1

1

1

The total population is 20,000, the natural divisor is 20,000 = 1000 person/seat; and the 20 divisor used for each procedure is shown next to the procedure’s name. Notice the quota violations (marked by parenthesis). The only procedure that conformed with the quota rule in this example is Adams’ procedure! Note 3.6.10 Now that we have found Webster’s procedure to be unbiased and to violate quota with less frequency than the procedure of Hill–Huntington, it is natural to ask: Why was Webster’s procedure replaced with the procedure of Hill–Huntington? Here are some good reasons: 1. The Hill–Huntington procedure automatically satisfies the constitutional requirement that each state should be represented by at least one seat. Webster’s procedure, on the other hand, can give zero seats to a state which would require an intervention from outside the procedure, or an increase in the house size, in order to satisfy the constitutional requirement. 2. The difference in per capita representations and the difference in their reciprocals (the district sizes) provide two different measures of inequity, minimized by two different procedures, namely, Webster’s procedure and Dean’s procedure, respectively. On the other hand, the ratio of per capita representations and the ratio of their reciprocals (the larger district size divided by the smaller one) provide one and the same measure of inequity, which is minimized by the Hill–Huntington procedure. The measure of inequity that the Hill–Huntington procedure minimizes is therefore more “robust” in some sense. Of course, this is a point in favor of the Hill–Huntington procedure. 3. Measuring inequity by a ratio makes more sense than measuring it by difference. Consider, for example, three states: A with 41 seats per 100 citizens, B with 21 seats per 100 citizens, and C with just one seat per 100 citizens. Measuring inequity by the difference in per capita representation would lead to the conclusion that there is as much inequity between A and B as there is between B and C: 21 21 1 20 41 − = − = · 100 100 100 100 100

232

Chapter 3 • Apportionment

Measuring inequity, on the other hand, by the ratio of per capita representations, would lead to the conclusion that the inequities between A, B and between B, C are

3

41 = 1.952 · · · 21

and

21 = 21, 1

respectively, which is more expressive of the considerable difference in inequities. For a more intuitive example, think of three people A, B, and C with bank account balances of $2,000,001, $1,000,001 and $1, respectively. If we adopt the difference as a measure of inequity, we are forced to say that the wealth gap between A and B is same as the wealth gap between B and C. Ratio makes more sense as a measure of inequity because it leads to the conclusion that A is approximately twice as rich as B while B is about one million times as rich as C.   Apart from the standard paradoxes like the ones presented in this chapter, and others that will be encountered in the exercises, one may come across some anomalies that are not known with special names in the literature. The following is an example. Example 3.6.11 The following two tables show the population data of a nation consisting of three states A, B, and C based on the censuses of the years 2000 and 2010. The only change that occurred in the 10-year period from 2000 to 2010 is that 20 people migrated from state B to state C. No change in the population of A or the total population occurred.

Year 2000 data

State A B C Total Population 1414 3465 5121 10,000

Year 2010 data

State A B C Total Population 1414 3445 5141 10,000

The Hill–Huntington apportionment for the year 2000 is 1 seat for A, 4 seats for B, and 5 seats for C. In the year 2010, the same procedure apportions 2 seats to A, 3 seats to B, and 5 seats to C. (We leave it to the student to do the computations.) Note that although the population movement was from B to C, the seat movement was from B to A which did not receive any additional population. The following diagram illustrates this rather unexpected result. A ← seat ← B → population → C To understand why the Hill–Huntington computations led to this anomaly, notice that in the year 2000, the natural quota of A was 1.414 which misses the cut-off point of rounding √ ( 1 × 2) by about 0.0002 only and rounds down to 1. B was very lucky to get 4 seats because its natural quota of 3.465 was just about 0.001 higher than the cut-off point of rounding

233 3.7 · Applications of Priority Formulas

√ ( 3 × 4). C’s natural quota of 5.121 comfortably rounds down to 5. The natural divisor works with 1 seat for A, 4 seats for B and 5 seats for C. The group of 20 people who left B for C after the 2000 census was large enough to bring the natural quota of B below the cut-off point of rounding in its interval, but was too small to qualify the natural quota of C for rounding up instead of down. The sum of rounded natural quotas became only 9 and a lower divisor had to be used in order to absorb the surplus seat. This enabled A to pick up that surplus seats since its natural quota was much closer to the cut-off point in its interval than the natural quota of B was after the group of 20 people had left it.  Similar behavior by the other standard divisor procedures will occur in the exercises.

3.7

Applications of Priority Formulas

The term priority formulas in this section’s title refers to the formulas proven in Propositions 3.4.6, 3.4.14, 3.4.25, 3.4.37, and 3.5.25. Our first application of those formulas is a simple algorithm that apportions seats, using a divisor procedure, one seat at time. In this section we shall drop the assumption that the number of seats ≥ the number of states.

3.7.1 One-By-One Seat Apportionment Example 3.7.1 We use Webster’s procedure to apportion 10 seats among the states A, B, and C with populations 135, 333, and 532 people, respectively. At any point of the process, let us represent the present apportionment by the apportionment vector (i, j, k) where i, j , and k are the numbers of seats apportioned to A, B and C, respectively. We start with 0 seats for each of the three states, so the initial apportionment vector is (0, 0, 0). By Axiom 3 in 3.2.1, the first seat should certainly go to C since it has the largest population. The new vector is (0, 0, 1). Which of the three states gets the second seat? With each seat added, the divisor goes down and from the priority formula proven in Proposition 3.4.25 we see that the state with the highest value of p 1 gets that added seat. At present, the values of 

135 0+

, 1 2

333 0+

, 1 2

p n+ 12

532 1+



1 2

n+ 2

for the three states are given by the priority vector   = 270, 666, 354.6 ·

Therefore, the second seat goes to B, leading to the apportionment vector (0, 1, 1) and the corresponding priority vector 

135 0+

1 2

,

333 1+

1 2

,

532 1+

1 2



  = 270, 222, 354.6 ·

3

234

3

Chapter 3 • Apportionment

Therefore, the third seat goes to C, leading the apportionment vector (0, 1, 2) and the priority vector  135 333 532 , , = (270, 222, 212.8) · 0 + 12 1 + 12 2 + 12 The fourth seat then goes to A, leading to the apportionment vector (1, 1, 2) and the priority vector  135 333 532 , , = (90, 222, 212.8) , 1 + 12 1 + 12 2 + 12 and the fifth seat goes to B. The following table shows all the steps necessary to apportion 10 seats. Seats

Apportionment vector

Priority vector for next seat

Next seat

1

(0, 0, 1)

(270, 666, 354.666)

B

2

(0, 1, 1)

(270, 222, 354.666)

C

3

(0, 1, 2)

(270, 222, 212.8)

A

4

(1, 1, 2)

(90, 222, 212.8)

B

5

(1, 2, 2)

(90, 133.2, 212.8)

C

6

(1, 2, 3)

(90, 133.3, 152)

C

7

(1, 2, 4)

(90, 133.3, 118.222)

B

8

(1, 3, 4)

(90, 95.142, 118.222)

C

9

(1, 3, 5)

(90, 95.142, 96.727)

C

10

(1, 3, 6)

(90, 95.142, 81.846)

B

Note that the state with the highest priority in each line gets the new seat in the next line. For instance, the highest priority in the ninth line is that of state C and therefore state C goes from 5 seats in a nine-seat house (ninth line) to 6 seats in a ten-seat house (tenth line).  A similar algorithm can be used with each divisor procedure.

3.7.2 The Quota-Divisor Procedures In their landmark result of 1983 (Theorem 3.3.18), Balinski & Young proved that populationmonotonicity and the quota rule are incompatible. The same authors, however, were able to reconcile house-monotonicity with the quota rule by means of a modified version of the one-by-one seat apportionment [2], that creates a house-monotone quota procedure out of each divisor procedure. We call these procedures, the quota-Jefferson procedure, the quotaWebster procedure, and so on. House-monotonicity is a built-in quality of the process of one-by-one seat apportionment since the “next” seat goes to some state without taking any seat away from the “present” share

235 3.7 · Applications of Priority Formulas

of any state. (Incidentally, this provides an immediate proof to Theorem 3.6.1 since divisor procedures can be implemented using one-by-one seat apportionment as in Sect. 3.7.1). The quota rule can then be enforced by checking whether or not assigning the next seat according to the established priority rule of the underlying divisor procedure violates the quota rule. If this violates the quota rule, we go downward on the priority list till we find a state that can get that seat without causing a violation of the quota rule. The following example explains the process. Example 3.7.2 We use the quota-Jefferson procedure and the quota-Adams to apportion up to 21 seats among the following three states

State

A

B

C

Total

Population

112

132

756

1000

According to Jefferson’s priority, the state with the highest (Proposition 3.4.6.) Step 1:

Step 2:

Step 3:

Step 4:

Step 5:

Step 6:

p n+1

gets the next seat

We start with the initial apportionment vector (0, 0, 0). By Axiom 3 in 3.2.1, the 1st seat goes to C and we move to the apportionment vector (0, 0, 1) with the 112 132 756 corresponding priority vector ( 0+1 , 0+1 , 1+1 ) = (112, 132, 378). According to Jefferson’s priority, the 2nd seat should go to C (since it has the highest priority in Step 1), but we must check on the quota rule before implementing Jefferson’s “recommendation.” With two seats, The natural divisor is 1000 2 = 132 756 , , ) = (0.224, 0.264, 1.512). 500 and the natural quotas vector is ( 112 500 500 500 Therefore, the apportionment vector (0, 0, 2) satisfies quota and is adopted. The 112 132 756 corresponding priority vector is ( 0+1 , 0+1 , 2+1 ) = (112, 132, 252). Before giving the 3rd seat to C, as suggested by the priority vector in the Step 2, we must check the natural quota vector (0.336, 0.396, 2.268) in a 3-seat house (calculations are left to the student). Clearly, the apportionment vector (0, 0, 3) is within quota and is therefore adopted. The corresponding priority vector is 112 132 756 ( 0+1 , 0+1 , 3+1 ) = (112, 132, 189). Before giving the 4th seat to C, as suggested by the priority vector in Step 3, we must check the natural quota vector (0.448, 0.528, 3.024) in a 4-seat house. Clearly, the apportionment vector (0, 0, 4) is within quota and is therefore adopted. 112 132 756 , 0+1 , 4+1 ) = (112, 132, 151.2). The corresponding priority vector is ( 0+1 Before giving the 5th seat to C, as suggested by the priority vector in Step 4, we must check the natural quota vector (0.560, 0.660, 3.780) in a 5-seat house. The apportionment vector (0, 0, 5) violates the quota rule since the natural quota of C is only 3.780, so we must pass over C in favor of B who has the next highest Jefferson priority and the apportionment vector (0, 1, 4) is within quota. The corresponding 112 132 756 priority vector is ( 0+1 , 1+1 , 4+1 ) = (112, 66, 151.2). Before giving the 6th seat to C, as suggested by the priority vector in Step 5, we must check the natural quota vector (0.672, 0.792, 4.536) in a 6-seat house.

3

236

3

Step 7:

Step 8:

Chapter 3 • Apportionment

Clearly, the apportionment vector (0, 1, 5) is within quota and is therefore adopted. 112 132 756 The corresponding priority vector is ( 0+1 , 1+1 , 5+1 ) = (112, 66, 126). Before giving the 7th seat to C, as suggested by the priority vector in Step 6, we must check the natural quota vector (0.784, 0.924, 5.292) in a 7-seat house. Clearly, the apportionment vector (0, 1, 6) is within quota and is therefore adopted. 112 132 756 The corresponding priority vector is ( 0+1 , 1+1 , 6+1 ) = (112, 66, 108). Before giving the 8th seat to A, as suggested by the priority vector in Step 7, we must check the natural quota vector (0.896, 1.056, 6.048) in an 8-seat house. Clearly, the apportionment vector (1, 1, 6) is within quota and is therefore adopted. 112 132 756 The corresponding priority vector is ( 1+1 , 1+1 , 7+1 ) = (56, 66, 108).

The following two tables show the apportionments of up to 21 seats using the quota-Jefferson and the quota-Adams procedures. When the priority rule of the underlying divisor procedure (Jefferson or Adams) has to be overridden in order to stay within quota, an ∗ is entered.  Quota-Jefferson Procedure

Seats

Apportionment vector

Priority vector for next seat

Next NQ vector

Next seat

1

(0, 0, 1)

(112.00, 132.00, 378.00)

(0.224, 0.264, 1.512)

C

2

(0, 0, 2)

(112.00, 132.00, 252.00)

(0.336, 0.396, 2.268)

C

3

(0, 0, 3)

(112.00, 132.00, 189.00)

(0.448, 0.528, 3.024)

C

4

(0, 0, 4)

(112.00, 132.00, 151.20)

(0.560, 0.660, 3.780)

B∗

5

(0, 1, 4)

(112.00, 66.00, 151.20)

(0.672, 0.792, 4.536)

C

6

(0, 1, 5)

(112.00, 66.00, 126.00)

(0.784, 0.924, 5.292)

C

7

(0, 1, 6)

(112.00, 66.00, 108.00)

(0.896, 1.056, 6.048)

A

8

(1, 1, 6)

(56.00, 66.00, 108.00)

(1.008, 1.188, 6.804)

C

9

(1, 1, 7)

(56.00, 66.00, 94.50)

(1.120, 1.320, 7.560)

A

10

(1, 1, 8)

(56.00, 66.00, 84.00)

(1.232, 1.452, 8.316)

C

11

(1, 1, 9)

(56.00, 66.00, 75.60)

(1.344, 1.584, 9.072)

C

12

(1, 1, 10)

(56.00, 66.00, 68.72)

(1.456, 1.716, 9.828)

B∗

13

(1, 2, 10)

(56.00, 44.00, 68.72)

(1.568, 1.848, 10.584)

C

14

(1, 2, 11)

(56.00, 44.00, 63.00)

(1.680, 1.980, 11.340)

C

15

(1, 2, 12)

(56.00, 44.00, 58.15)

(1.792, 2.112, 12.096)

C

16

(1, 2, 13)

(56.00, 44.00, 54.00)

(1.904, 2.244, 12.852)

A

17

(2, 2, 13)

(37.33, 44.00, 54.00)

(2.016, 2.376, 13.608)

C

18

(2, 2, 14)

(37.33, 44.00, 50.40)

(2.128, 2.508, 14.364)

C

19

(2, 2, 15)

(37.33, 44.00, 47.25)

(2.240, 2.640, 15.120)

C

20

(2, 2, 16)

(37.33, 44.00, 44.47)

(2.352, 2.772, 15.876)

B∗

21

(2, 3, 16)

(37.33, 33.00, 44.47)

(2.464, 2.904, 16.632)

C

3

237 3.7 · Applications of Priority Formulas

Quota-Adams Procedure

Seats

Apportionment vector

Priority vector for next seat

Next NQ vector

1

(0, 0, 1)

(∞ , ∞, 756.00)

(0.224, 0.264, 1.512)

Next seat B

2

(0, 1, 1)

(∞, 132.00, 756.00)

(0.336, 0.396, 2.268)

C∗

3

(0, 1, 2)

(∞, 132.00, 378.00)

(0.448, 0.528, 3.024)

C∗

4

(0, 1, 3)

(∞, 132.00, 252.00)

(0.560, 0.660, 3.780)

A

5

(1, 1, 3)

(112.00, 132.00, 252.00)

(0.672, 0.792, 4.536)

C

6

(1, 1, 4)

(112.00, 132.00, 189.00)

(0.784, 0.924, 5.292)

C

7

(1, 1, 5)

(112.00, 132.00, 151.20)

(0.896, 1.056, 6.048)

C

8

(1, 1, 6)

(112.00, 132.00, 126.00)

(1.008, 1.188, 6.804)

B

9

(1, 2, 6)

(112.00, 66.00, 126.00)

(1.120, 1.320, 7.560)

C

10

(1, 2, 7)

(112.00, 66.00, 108.00)

(1.232, 1.452, 8.316)

C∗

11

(1, 2, 8)

(112.00, 66.00, 94.50)

(1.344, 1.584, 9.072)

C∗

12

(1, 2, 9)

(112.00, 66.00, 84.00)

(1.456, 1.716, 9.828)

A

13

(2, 2, 9)

(56.00, 66.00, 84.00)

(1.568, 1.848, 10.584)

C

14

(2, 2, 10)

(56.00, 66.00, 75.60)

(1.680, 1.980, 11.340)

C

15

(2, 2, 11)

(56.00, 66.00, 68.72)

(1.792, 2.112, 12.096)

C

16

(2, 2, 12)

(56.00, 66.00, 63.00)

(1.904, 2.244, 12.852)

B

17

(2, 3, 12)

(56.00, 44.00, 63.00)

(2.016, 2.376, 13.608)

C

18

(2, 3, 13)

(56.00, 44.00, 58.15)

(2.128, 2.508, 14.364)

C

19

(2, 3, 14)

(56.00, 44.00, 54.00)

(2.240, 2.640, 15.120)

C∗

20

(2, 3, 15)

(56.00, 44.00, 50.40)

(2.352, 2.772, 15.876)

A

21

(3, 3, 15)

(37.33, 44.00, 50.40)

(2.464, 2.904, 16.632)

C

The following are the apportionments of 20 seats and 21 seats by the procedures of Jefferson, quota-Jefferson, Adams, and quota-Adams. The ∗ indicates a quota violation.

20 seats:

State A B C

Jefferson 2 2 16

q-Jefferson 2 2 16

Adams 3 3 14 ∗

q-Adams 2 3 15

21 seats:

State A B C

Jefferson 2 2 17 ∗

q-Jefferson 2 3 16

Adams 3 3 15

q-Adams 3 3 15

We refer the interested reader to the monograph by Balinski and Young [2] for a proof of the existence, at each step of the construction of the quota-divisor procedures, of a state that can receive the next seat without causing a violation of the quota rule. In Theorem 3.7.7 below, we address the case of only three states. A few simple results are needed first.

238

3

Chapter 3 • Apportionment

Proposition 3.7.3 If the total population of all states in a nation is p citizens and the house size is h seats, then a state Aj with pj citizens and aj seats satisfies upper quota if and only if aj − 1
0 receives k seats where k is the unique non-negative integer such that s(k − 1, k) ≤ q < s(k, k + 1). 2. If two states X and Y are apportioned m seats and n seats, respectively, we define the θ -inequity (denoted θ (X : m, Y : n)) between the two states by θ (X : m, Y : n) = |(pcr(X : m))2 − (pcr(Y : n))2 |. Prove that the mean square apportionment procedure minimizes the pairwise θ -inequities and that it is the only divisor procedure with this minimization property.

3

250

3

Chapter 3 • Apportionment

48. For each of Tables 3.1 through 3.7 in the beginning of this exercise set, use the mean square procedure to apportion 10 seats and detect all quota violations. You must determine the interval of proper divisors in each case. 49. For each of Tables 3.1 through 3.7, 1. Find the largest house size that allows the smallest state to get just one seat and the corresponding number of seats apportioned to each state, when the mean square procedure is used. 2. Find the smallest house size that allows the smallest state to get just one seat and the corresponding number of seats apportioned to each state, when the mean square procedure is used. 50. Prove that 1. limn→∞ {s(n, n + 1) − m(n, n + 1)} = 0. 2. For n = 0, 1, 2, · · · s(n, n + 1) − m(n, n + 1) > s(n + 1, n + 2) − m(n + 1, n + 2)· 51. For the population data in Table 3.2 above, the following table shows Webster’s apportionment of 10 seats.

State

Population

Seats

Divisor interval

A

3580

4

795.555 < d < 1022.857

B

2550

3

728.571 < d < 1020.000

C

2515

2

1006.00 < d < 1676.666

D

1355

1

903.333 < d < 2710.000

In the fourth column of the above table we use Proposition 3.4.25 to compute the interval of divisors that allow a state to receive the indicated number of seats. Compared to the natural divisor of 1000 person per seat, the population difference of only 35 between B and C does not justify the one seat difference between these two states. Can you resolve this issue by changing the house size? Exercises 52 and 53 you are asked to prove theorems for the procedures of Jefferson and Adams similar to Theorems 3.5.12, 3.5.14 and 3.5.21 for the procedures of Webster, Hill– Huntington, and Dean, respectively. It is assumed that state A (with m seats and a population of x people) is the poor state; and state B (with n seats and a population of y people) is the rich state. 52. Prove that the inequity measure associated with Jefferson’s procedure is n xy − m. 53. Prove that the inequity measure associated with Adams’ procedure is n − m yx .

251

Answers to Selected Exercises Chapter 1 Answers to Exercises 1, 2, 3, 4, 5, 6, 29, 30 are presented in the following table. Scenario

CW

CL

Plurality

P with RO

Hare

Coombs

Borda

Hare–Borda

1

–

–

c

a

a

a

a, c

a

2

–

–

a

a

a

a

b

a

3

d

c

a

b

b

d

d

d

4

a

b

a, b

a

a, b, c

a

a

a

5

c

d

a

b

c

c

c

c

6

–

d

a

a

a

a

b

a

07. e is a Condorcet winner that loses in all of the procedures named. 08. 1. b is a Condorcet winner. 2. b is the plurality winner. 3. b is the Hare winner. 4. b is a Coombs non-winner. 09. 1. e is a Condorcet winner, b is a Condorcet loser. 2. a is the winner. 4. No. (From 1&2.) 10. d 11. e 12. Yes. 13. Yes. 18. d wins regardless of the agenda, since it is a Condorcet winner. 19. a loses regardless of the agenda since it is a Condorcet loser; b wins with agenda acdb; c wins with agenda adbc; d wins with agenda abcd. 20. a wins with agenda cdba; b wins with agenda acdb; c wins with agenda dbac; d wins with agenda bacd. 22. a, b, c can win with agendas edcba, edacb, edbac, respectively. d is a nonCondorcet loser that loses regardless of the agenda. 26. e is the only Pareto dominated alternative. It wins with agenda cdabe. 27. b and e are Pareto dominated. No agenda can save b since it is also a Condorcet loser, while e wins with agenda bcdae. 31. 1. a a Condorcet winner and is second to last in Borda count. 2. a a Condorcet loser and is second in Borda count. 32. e is the sole winner of Borda count. It is also the sole winner of the Hare–Borda procedure since it is a Condorcet winner.

© Springer Nature Switzerland AG 2019 Sherif El-Helaly, The Mathematics of Voting and Apportionment, Compact Textbooks in Mathematics, https://doi.org/10.1007/978-3-030-14768-6

252

Answers to Selected Exercises

33. The Condorcet tournament ranking is d first, b second, a third, c fourth. Borda count ranking is d first, a second, b third, c fourth. Note also that b (who is placed second by the Condorcet tournament) is Pareto dominated. 35. 2. a is a Condorcet winner. b is the sole winner of the top-bottom procedure, c is the sole winner of Hare’s procedure, and a is the sole winner of Coombs’ procedure. The top-bottom procedure does not satisfy the Condorcet winner criterion. 36. μ = {a} while μs = {a, b}. Voter s prefers {a, b} to {a} since, in his sincere ranking, he prefers b to a. 39. No, because none of the voters could be the Arrow’s dictator. 40. 1. {a, c} and {a, d}. 3. (λ)/{a, c} = (λ)/{a, c}; (λ)/{a, d} = (λ)/{a, d}. 41. 2. d ↔μ a and a ↔μ b but b →μ d; c →μ b and b ↔μ a but a →μ c; d ↔μ a and a →μ c but c →μ d. 42. 1. This scenario was created from a complete directed graph as in Example 1.3.23. The graph can be easily recovered from the scenario. a is a weak Condorcet winner, c is a strong Condorcet loser, and e is a Condorcet neutralizer. 2. a and e. 3. {a, e}. 43. 1. a is a weak Condorcet winner, d is a Condorcet neutralizer, and c is a strong Condorcet loser. 2. a and d. 3. {a, c, d} (The strong Condorcet loser c made it to the Hare–Borda social choice set.) 44. Assume we have n candidates. The total number of Copeland points is (n−1)n ;a 2 weak Condorcet winner receives ≥ n2 points and a Condorcet neutralizer receives points. To show that a strong Condorcet loser can join the social choice exactly n−1 2 set of the Hare–Copeland procedure, construct a complete directed graph in which a loses to b and is tied with everyone else, and b loses to everyone except a. McGarvey’s theorem guarantees the existence of a scenario representable by this graph. 45. Exercise 42: Hare–Copeland social choice set is {a, e} (same as the Hare–Borda social choice set.) Exercise 43: Hare–Copeland social choice set is {a, d} (different from the Hare– Borda social choice set.)

Chapter 2 01. 02. 03. 04. 06.

8, 6, 4, 2, 0 7, 5, 3, 1 8, 10 9,11 In System 2, the most you get, if you miss a or b, is 11 points. Therefore, to meet the quota of 12, you need to include both a and b; and you need to include at least two more of c, d, e, f, g, and h. 6 6 2 2 07. 1. bi(a) = 10 , bi(b) = 26 , bi(c) = 26 , bi(d) = 26 , bi(e) = 26 . No dummies or 26 veto-powered voters.

253 Answers to Selected Exercises

08. 09. 10. 11. 12. 13.

14. 15. 16. 17.

9 7 5 3 1 2. bi(a) = 25 , bi(b) = 25 , bi(c) = 25 , bi(d) = 25 , bi(e) = 25 . No dummies or veto-powered voters. 8 6 4 4 2 , bi(b) = 24 , bi(c) = 24 , bi(d) = 24 , bi(e) = 24 . No dummies or 3. bi(a) = 24 veto-powered voters. 1 2 4 bi(a) = bi(f ) = 14 , bi(b) = bi(e) = 14 , bi(c) = bi(d) = 14 . No dummies; each of c and d has veto power. 8 8 2 2 2 , bi(b) = 22 , bi(c) = 22 , bi(d) = 22 , bi(e) = 22 . No dummies or bi(a) = 22 veto-powered voters. 7 7 3 3 3 , bi(b) = 23 , bi(c) = 23 , bi(d) = 23 , bi(e) = 23 . No dummies or bi(a) = 23 veto-powered voters. 6 2 2 0 bi(a) = 10 , bi(b) = 20 , bi(c) = 20 , bi(d) = 20 , bi(e) = 20 . e is a dummy and a 20 has veto power. 8 6 2 2 2 , bi(b) = 20 , bi(c) = 20 , bi(d) = 20 , bi(e) = 20 . There are no bi(a) = 20 dummies and a has veto power. 9 5 3 3 1 bi(a) = 21 , bi(b) = 21 , bi(c) = 21 , bi(d) = 21 , bi(e) = 21 . There are no dummies and a has veto power. Both changes benefit e but the change in Exercise 12 benefits e more. acbf d eg acgf deb gcaf bed ecgf dba acbf d eg acgf deb gcaf bed ecgf dba Yes Yes

18.

q 6 7 8

a 6/12 5/10 4/8

19.

Index bi si

b 2/12 3/10 4/8

a 2/12 4/24

c 2/12 1/10 0/8

b 4/12 8/24

Dummmies None None a, b

d 2/12 1/10 0/8

c 4/12 8/24

d 2/12 4/24

Veto-powered None a c, d

No dummies or veto-powered voters.

20. No. 21. Weights: a:3, b:3, c:1, d:1, e:1; q=7. Index

a

b

c

d

e

bi

7/17

7/17

1/17

1/17

1/17

si

54/120

54/120

4/120

4/120

4/120

22. Weights: a:3, b:3, c:2, d:2, e:2; q=7. Index

a

b

c

d

e

bi

7/29

7/29

5/29

5/29

5/29

si

30/120

30/120

20/120

20/120

20/120

254

25.

Answers to Selected Exercises

Index bi si

f 10/42 168/720

g 10/42 168/720

i 10/42 168/720

b 6/42 108/720

n 6/42 108/720

l 0/42 0/720

l

is

a

dummy. No veto-powered voters. 29. Weights: a:2, b:2, c:1, d:1, e:1; q=5 30. 6 × 1 × 4 × 4 × 4 × 4 × 6 × 6 = 55, 296 31. Each of the five permanent members has a weight of 7, each of the ten nonpermanent members has a weight of 1; and q = 39 Index

Permanent members

Non-permanent members

bi

212/1270  0.1669291

21/1270  0.0165354

si

421/2145  0.1962703

4/2145  0.0018648

Chapter 3 Answers to Exercises 14, 17, 20, 23, 26, 48 are presented in Tables 1, 2, 3, 4, 5, 6, and 7. Quota violations are circled. Table 1

Dean 1020.84– 1061.25

H-H 1000.51– 1000.55

Webster 980– 996.36

Mean square 926.31– 960.96

Jeff 782.85– 816.66

State

Pop

Adams 1225–1370

A

5480

5

5

5

6

6

6

B

2450

2

2

2

2

3

3

C

1415

2

2

2

1

1

1

D

0655

1

1

1

1

0

0

State

Pop

Adams 1257.5– 1275

Dean 1044.17– 1047.91

H-H 1026.75– 1033.45

A

3580

3

3

B

2550

3

3

C

2515

2

D

1355

2

Table 2

Webster 1006–1020

Mean square (natural divisor)

Jeff 838.34–850

4

4

4

4

3

3

3

3

3

2

2

2

2

1

1

1

1

1

255 Answers to Selected Exercises

Table 3

State Pop

Adams 1345.5–1797

Dean (natural divisor)

Webster H-H (natural (natural divisor) divisor)

Mean square (natural divisor)

Jeff 895.5–897

A

5382 5

5

5

5

5

6

B

1796 2

2

2

2

2

2

C

1791 2

2

2

2

2

1

D

1031 1

1

1

1

1

1

Table 4

Webster Adams Dean H-H (natural State Pop 1277.8–1594 1171.32–1192.5 1124.30–1127.12 divisor) B

6389  5 1594 2

 5 2

C

1590 2

2

D

0427 1

1

1

A

Mean square (natural divisor)

Jeff 797–798.62

6

6

2

2

2

 8 1

1

2

2

1

0

0

0

6

Table 5

Dean Adams 1004.60– State Pop 1257–1500.2 1160.86

H-H 1002.37– 1157.42

Webster (natural Mean square divisor) 880.95–997.91

Jeff 833.45–937.62

7501  6 1257 2

7

7

8

8

8

B

1

1

1

1

1

C

1012 1

1

1

1

1

1

D

0230 1

1

1

0

0

0

Dean 1050.87– 1092.75

H-H 1030.26– 1046.51

Webster Jeff (natural Mean square divisor) (natural divisor) 818.86–857.91

A

Table 6

Adams 1146.4– State Pop 1433 A

5732 5

5

6

6

6

B

1826 2

2

2

2

2

6 2

C

1457 2

2

1

1

1

1

D

0985 1

1

1

1

1

1

256

Answers to Selected Exercises

Table 7

State Pop

Dean Adams 1116.61– 1202.5–1443 1163.25

H-H 1096.73– 1113.29

Webster 1034–1046

Mean square Jeff 959.87– 775.5–801.6 980.93

 6 2

7

7

7

B

7215  6 1551 2

1

1

2

 9 1

C

0711 1

1

1

1

1

0

D

0523 1

1

1

1

0

0

A

01. Table 4: Four quota solutions exist: A:7, B:2, C:1, D:0 A:7, B:1, C:1, D:1 A:6, B:2, C:2, D:0 A:6, B:2, C:1, D:1 02. A:6, B:2, C:2, D:0 03. A:6, B:2, C:1, D:1 06. 10 seats: A:2, B:2, C:3, D:3; 11 seats: A:1, B:2, C:4, D:4. A is the Alabama state. 07. 10 seats: A:1, B:1, C:3, D:5; 11 seats: A:0, B:1, C:4, D:6. A is the Alabama state. 08. 5 seats: A:1, B:1, C:1, D:1, E:1; 6 seats: A:2, B:2, C:2, D:0, E:0. There are two Alabama states in this case: D and E. 09. 1. E and F are excluded. 2. Votes of E + Votes of F = 1,000,000 votes (thrown away.) 3. A:34, B:25, C:23, D:8, E:0, F :0. 10. 1. A:7, B:2, C:1, D:0; 2. A:6, B:2, C:2, D:0; 3. A:6, B:2, C:1, D:1; 4. A:6, B:2, C:1, D:1. p 11. In Remark 3.3.8, the larger qδ the smaller qi the larger qi , which is equal to di where p is the population of the state and d is the natural divisor. 13. The initial allocation, which is the integer part of the natural quota can only increase (or stay the same) when the house size increases from h to h + 1. 15. For the data in Table 4 1. Largest house size is 44. A:29, B:7, C:7, D:1; 2. Smallest house size is 21. A:14, B:3, C:3, D:1 16. B: 3 or 4; C: 14, 15 or 16; D: 0 18. For the data in Table 4 Largest house size is 24. A:15, B:4, C:4, D:1 19. B: 4; C: 13, 14 or 15; D: 1 21. For the data in Table 4 1. Largest house size is 35. A:22, B:6, C:6, D:1; 2. Smallest house size is 12. A:7, B:2, C:2, D:1 22. B: 4; C: 14, 15 or 16; D: 1 24. For the data in Table 4 Largest house size is 32. A:21, B:5, C:5, D:1

257 Answers to Selected Exercises

25. B: 4; C: 14, 15 or 16; D: 1 27. For the data in Table 4 Largest house size is 31. A:20, B:5, C:5, D:1 28. B: 4; C: 14, 15 or 16; D: 1 33. A ← seat ← B → population → C 39. (0, 0, 0, 0) → (1, 0, 0, 0) → (2, 0, 0, 0) → (2, 1, 0, 0) → (2, 1, 1, 0) → (2, 1, 1, 1) → (3, 1, 1, 1) → (4, 1, 1, 1) → (5, 1, 1, 1) → (6, 1, 1, 1) → (6, 2, 1, 1). In the last step we had to override Webster’s priority in order to stay within quota. 43. The one with the highest population among the three states gets that surplus seat. 44. See the first note in Notes 3.6.08. 45. No. Upper quota and lower quota are both violated. See the first note in Notes 3.6.08. 49. For the data in Table 4 1. Largest house size is 35. A:24, B:5, C:5, D:1; 2. Smallest house size is 18. A:11, B:3, C:3, D:1

259

References 1. Arrow, K.J.: Social Choice and Individual Values, 2nd edn. Yale University, New Haven (1963) 2. Balinski, M.L., Young, H.P.: Fair Representation: Meeting the Ideal of One Man One Vote, 2nd edn. Brookings Institution, Washington, D.C. (2001) 3. Black, D.: Theory of Committees and Elections. Cambridge University Press, Cambridge (1958) 4. Felsenthal, D.S., Machover, M.: The Measurement of Voting Power. Edward Elgar, Cheltenham (1998) 5. Fishburn, P.C.: The Theory of Social Choice. Princeton University, Princeton (1973) 6. Gaertner, W.: A Primer in Social Choice Theory. Oxford University, New York (2006) 7. Gärdenfors, P.: Manipulation of social choice functions. J. Econ. Theory 13, 217–228 (1976) 8. Gibbard, A.: Social Choice and the Arrow Conditions. Unpublished Manuscript (1969) 9. Gura, E.-Y., Maschler, M.B.: Insights into Game Theory. Cambridge University, New York (2008) 10. Heckelman, J.C., Miller, N.R. (ed.): Handbook of Social Choice and Voting. Edward Elgar, Cheltenham (2015) 11. Kelly, J.S.: Social Choice Theory. Springer, Berlin (1988) 12. Laruelle, A., Valenciano, F.: Voting and Collective Decision-Making. Cambridge University, New York (2008) 13. Laslier, J.-F.: Tournament Solutions and Majority Voting. Springer, Berlin (1997) 14. McGarvey, D.C.: A Theorem on the Construction of Voting Paradoxes. Econometrica 21 (4), 608–610 (1953) 15. Moon, J.W.: Topics on Tournaments in Graph Theory. Holt, Rinehart and Winston, New York (1968) 16. Nitzan, S.: Collective Preference and Choice. Cambridge University, New York (2010) 17. Nurmi, H.: Comparing Voting Systems. D. Reidel, Boston (1987) 18. Pukelsheim, F.: Proportional Representation: Apportionment Methods and their Applications. Springer, New York (2014) 19. Riker, W.H.: Liberalism Against Populism. Waveland, Long Grove (1982) 20. Schwartz, T.: The Logic of Collective Choice. Columbia University, New York (1986) 21. Sen, A.: A possibility theorem on majority decision. Econometrica 34, 491–496 (1966) 22. Straffin Jr., P.D.: Topics in the Theory of Voting. Birkhäuser, Boston (1980) 23. Suzumura, K.: Rational Choice, Collective Decisions, and Social Welfare. Cambridge University, London (1983) 24. Taylor, A.: Mathematics and Politics. Springer, New York (1995) 25. Taylor, A., Zwicker, W.: A Characterization of weighted voting. Proc. Am. Math. Soc. 115, 1089–1094 (1992) 26. Taylor, A., Zwicker, W.: Simple games and magic squares. J. Comb. Theory Series A 71, 67–88 (1995) 27. Taylor, A., Zwicker, W.: Simple Games. Princeton University, Princeton (1999) 28. Young, H.P.: Condorcet’s theory of voting. Am. Polit. Sci. Rev. 82, 1231–1244 (1988b) 29. Young, H.P.: Equality in Theory and Practice. Princeton University, Princeton (1994)

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Index A Adams, J.Q., 175, 192 Adams’ procedure, 175, 185–193, 203, 219, 221, 225, 229, 231, 245, 250 – rounding rule of, 185 – threshold divisors, 187–189, 191, 223, 244 Agenda, 28–33, 104, 108–110, 251 Agenda voting, 28–33, 104, 109, 110 Alabama paradox, 169–172, 243–245 Anonymity (of a two-alternative rule), 65–67 Anonymous social choice function, 97, 100, 102 Arithmetic mean, 199–201, 216, 217, 249 Arrow, K., 67, 68 Arrow’s axioms, 68–70, 72–74, 78, 112 Arrow’s (impossibility) theorem, 67–74, 78, 111 Axioms of apportionment, 162–166 B Balinski, M., 172, 174, 234, 237 Balinski-Young theorem, 172–175, 241 Banzhaf, J.F. III., 117 Banzhaf distribution of power, 123–125, 127, 131, 132, 153–155 Banzhaf index, 117, 123–128, 131–133, 138, 140–142, 153 Banzhaf score, 117–123, 125–128, 138 Black’s theorem, 61–65, 74 Bush, G.W., 4–6 C Cartesian product of sets, 136–137 Coalition, 116–129, 132, 133, 137–157 Collective quasi-rationality, 74–76, 112 Collective rationality, 68, 71–75, 113 Combination, 133–136 Complete graph, 22, 37–40, 85, 252 Condorcet loser, 23–24, 26, 27, 29–31, 35, 36, 43, 46–48, 81, 84–91, 105, 107–109, 111, 113, 251, 252 Condorcet loser criterion, 26, 29, 31, 35, 43, 108 Condorcet neutralizer, 84–92, 113, 252

Condorcet paradox, 25, 32, 36, 54, 97, 99, 102, 104 Condorcet tournament, 21–24, 29, 31, 34, 35, 37–39, 52–55, 58, 60, 61, 63–65, 74, 81, 84–88, 105, 107–109, 111, 113, 252 Condorcet-type classifications of alternatives, 84–88, 92 Condorcet winner, 23–27, 29, 30, 35, 36, 43, 44, 46–48, 50, 51, 53, 81, 84–91, 98–102, 104–107, 111–113, 251, 252 Condorcet winner criterion, 26, 29, 35, 43, 44, 50, 98–102, 104, 107, 112, 252 Consistency of apportionment procedures, 227–233 Conventional rounding, 161, 162, 166, 167, 192, 193 D Dean’s procedure, 217–220, 222–225, 229, 231, 246 – rounding rule of, 218 – threshold divisors, 222–225 Dictator, 65, 68, 73–75, 77, 78, 102, 139–144, 252 Directed graph, 22, 37–40, 85, 252 District size, 160, 168, 171, 172, 215, 216, 231 Divisor procedures, 162, 174–227, 229, 230, 233–237, 241, 247–249 Dominant voter, 139–144 Dummy voter, 101 E Elimination procedures, 6–21 Equity criteria, 207–225 Equivalent relation, 147 Equivalent yes-no voting system, 116, 128, 146–149, 152, 153, 156, 168, 181, 194, 244, 248 Exact apportionment, 160–163 F Felsenthal–Machover example, 126–128

© Springer Nature Switzerland AG 2019 Sherif El-Helaly, The Mathematics of Voting and Apportionment, Compact Textbooks in Mathematics, https://doi.org/10.1007/978-3-030-14768-6

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Index

G Gärdenfors, P., 93, 94 Gärdenfors preference on sets of alternatives, 92, 94–95 Gärdenfors theorem, 93, 98–102 Generalizations of standard social choice procedures, 79–83, 107 Geometric mean, 199–201, 217, 218 Gibbard, A., 74, 92 Gibbard–Satterthwaite theorem, 92, 93, 102 Gibbard’s axioms, 75–77 Gibbard’s theorem (on oligarchy), 74–78 – proof, 74–778 Gore, A., 4, 5 Graph representation, 22, 23, 29–32, 34, 36, 37, 39, 40, 51, 53, 59, 81–84, 89–91, 105–109, 111, 113 Graph theory, 22

K Killer amendment, 32–35, 110 L λ-blocking set of voters, 75 λ-decisive set of voters, 69, 73, 74, 78 Loser, 1, 18, 20, 22–24, 26, 27, 29–31, 35, 36, 43, 46–48, 81, 84–91, 105, 107–109, 111, 113, 201, 233, 251, 252 Losing coalition, 116, 156 Lower natural quota, 163–166, 174, 178, 186, 192, 230, 248 Lowndes’ procedure, 168–169, 171, 175, 228, 243, 244, 248 Lowndes, W., 166, 168, 169, 173, 226

I Independence of irrelevant alternatives (IIA), 55–59, 64, 65, 68, 70, 72, 74, 75, 111, 113 Indifference, 8, 41, 52, 60–62, 65–68, 79–114 Inferior, 31, 60, 61 Initial allocation, 164–172, 174, 175, 244, 245, 248, 256 Intensity of preference, 56 Irrelevant alternative, 44, 46, 56, 75

M Magic square, 150–152 Magic square voting system, 150–157 Magic sum, 150 Majority, 2–22, 24–28, 31, 33, 35, 41, 43, 47, 52, 62, 64–67, 81–83, 86, 102, 107, 111 Manipulable social choice functions, 93, 98, 102–114 May’s theorem, 65–67 McGarvey’s theorem, 37–40, 79, 85, 252 Mean (arithmetic), 199–201, 216, 217, 249 – geometric, 199–201, 216, 217 – harmonic, 216–219 Measures of inequity, 208–211, 231 – δ-inequity, 209–213 – ϕ-inequity, 215–217 – ρ-inequity, 209–211, 213–215 Measures of inequity (monotone, symmetric), 208 Membership index of a voter, 147 Modified quota, 176, 177, 180, 183, 185, 190, 208, 226 Modified divisor, 176–178, 180, 183, 184, 186, 189, 190, 192, 193, 195–198, 208, 223 Monotone (rounding rule), 176, 226 Monotonicity (of a two-alternative rule), 66, 67 Monotonicity social choice procedure, 20

J Jason–Linusson example, 244 Jefferson, T., 21, 175 Jefferson’s procedure, 175, 177–186, 188, 190, 191, 193, 219, 221, 229, 230, 245, 250 – rounding rule of, 177 – threshold divisors, 181–184, 188

N Nader, R., 5 Natural divisor, 159, 160, 162, 165, 169–171, 173, 176, 177, 179, 180, 182–184, 186, 188–191, 193–199, 203–206, 208, 222–227, 229–231, 233, 235, 250, 254–256 Natural (exact) quota, 159

H Hamilton, A., 166, 168 Hamilton’s procedure, 166–167, 169, 171, 174, 227, 243, 244, 248 Harmonic mean, 217–219 Hill, J., 175 Hill–Huntington procedure, 199–208, 213–215, 220, 224, 229, 231, 241, 246 – rounding rule of, 200–202, 214 – threshold divisors, 205, 206, 223, 224, 244 House-monotone, 170, 171, 226, 234, 241 Huntington, E.V., 175

263 Index

Nearly λ-blocking set of voters, 75 Nearly λ-decisive set of voters, 69, 73, 74, 78 Neutrality (of a two-alternative rule), 66, 67 Neutral social choice function, 101 n-trade-robust, 148 O Oligarchy, 74–78, 112 One-by-one seat apportionment, 233–235 P Pareto dominated, 31, 110, 251, 252 Pareto inferior/superior, 31 Pareto, V., 30 Parity of Banzhaf scores, 122–123 Per capita representation, 207, 209, 215, 216, 231, 232 Permutation, 66, 67, 95–97, 128–131, 133–138, 140–142, 144, 154, 155 Pivotal voter, 129, 130, 133, 140, 154, 155 Plurality, 2–4, 6–9, 13, 14, 17, 19–21, 24, 26, 37, 42, 52, 79, 81, 103, 104, 106, 164, 251 Plurality procedure, 3–6, 8, 9, 13, 14, 20, 25, 26, 31, 42, 48, 79, 81, 92, 105, 107, 108 +/− table, 118–122, 125, 127, 131, 138 Population monotone, 172, 174, 226, 241 Population paradox, 172 Proof of Arrow’s (impossibility) theorem, 68, 69 Proper divisors, 176–184, 186, 188–192, 195, 196, 198, 203–208, 213, 214, 221, 223–230, 245, 246, 250 Proper subcoalition, 116, 156 Proper supercoalition, 116, 156 Proportional representation, 163, 164 Q Quasi ranking, 74 Quasi social welfare function, 74 Quasi transitive, 74 Quota – apportionment, 164, 165, 173, 243 – lower natural, 163–166, 174, 178, 186, 192, 230, 248 – modified, 176, 177, 180, 183, 185, 190, 208, 226 – natural, 160–172, 174, 176–180, 185, 186, 191–193, 198, 199, 206–208, 224, 225, 230, 233, 235, 236, 238, 244, 248, 256

– procedures, 162, 166–175, 206, 207, 225, 226, 234, 241, 245, 248 – rule, 163, 179, 180, 185, 186, 192, 197, 198, 206, 224, 230, 231, 234, 235, 237, 239–241 – solution, 163–166, 173, 174, 256 – upper natural, 163, 174, 178, 179, 185, 192, 230, 248 – in yes-no voting, 116, 119–128, 131–133, 140, 144, 145, 148–150, 153–157 Quota-divisor procedures, 234–250 R Ranking, 34, 45, 46, 50, 52–69, 74, 79, 80, 84, 89, 90, 93–104, 111, 252 Restricting the domain, 61 S Satisfies lower quota, 238 Satisfies upper quota, 238 Scoring procedure, 41–52, 79 Semi ranking, 52–55, 57, 59–62, 65, 74, 111 Semi social welfare function, 52, 57–60, 64, 68–70, 74–77, 112 Sen, A., 61 Shapley–Shubik distribution of power, 130, 131, 155 Shapley–Shubik index of power, 128–132 Shapley–Shubik score, 130, 133, 138, 141, 142, 144 Singly monotone, 20, 112 Social choice, 1–115, 252 – procedure (function), 1, 3, 17, 18, 20, 21, 24, 26, 28, 31, 41, 79–115 – set, 1, 20, 26, 27, 31, 52, 79, 83, 88–93, 97, 100, 104, 107, 108, 111, 113, 114, 252 Stable social choice function, 93 State (poor, rich), 207–212, 214, 216, 217, 250 Stays within quota, 163 Strategic voting, 45 Strategy-proof social choice function, 93, 102 Strong Condorcet loser, 85, 86, 88–91, 113, 252 Strong dictator, 65, 68 Subcoalition, 116, 121, 128, 156 Supercoalition, 116, 121, 156 Superior, 31, 60, 61, 133 Surplus seats, 164–175, 182–184, 197, 198, 205, 207, 227, 228, 233, 244, 245, 248, 257

264

Index

T Taylor, A.D., 118, 144, 148–150, 152 Threshold (critical) divisors, 183 Tied winners, 1, 6, 11, 24 Trade-equivalence class, 147, 157 Trade-equivalent, 147–149 Trade-robust, 147–152 Trades among coalitions, 146–147 Transitive, 60–62, 64, 65, 68, 74 Two-alternative majority rule, 21, 28, 35, 41, 52, 62, 64–67, 81, 86, 107 Two-alternative rule, 65–67 U United Nations Security Council (UNSC), 115, 157 2000 Unites States Presidential Election, 4–6 Unrestricted domain, 68, 71–77 Upper natural quota, 163, 174, 178, 179, 185, 192, 230, 248 V Veto power, 115, 121, 125–128, 131, 140–142, 153–155, 157, 253 Veto-powered voter, 139–144, 252–254 Violates lower quota, 163, 229, 240, 241, 2339 Violates upper quota, 229

W Washington, G., 175 Weak Condorcet winner, 84–91, 113, 252 Weakly Pareto efficient social choice function, 31, 108 Weak Pareto condition, 68 Weak preference, 60 Webster, D., 175, 192, 223, 225, 229–231, 246, 250 Webster’s procedure, 175, 192–200, 203, 208, 211–213, 215, 218, 219, 229–231, 233, 245 – rounding rule of, 192, 193, 213, 218 – threshold divisors, 195–197, 244 Weighted/weightable yes-no voting system, 116, 126, 140, 144–157 Weight of a coalition, 139, 147 Winner, 1, 3–30, 34–36, 42–54, 62, 67, 79, 81, 84–91, 98–102, 104–109, 111–113, 251, 252 Winning coalition, 116–126, 128, 129, 132, 137–146, 149–151, 153–156 Winning set of alternatives, 79 Y Young, H.P., 172, 174, 227, 234, 237 Z Zwicker, W.S., 144, 149, 150, 152