Thinking Mathematically (6th Edition)

Published by Pearson
ISBN 10: 0321867327
ISBN 13: 978-0-32186-732-2

Chapter 13 - Voting and Apportionment - 13.2 Flaws of Voting Methods - Exercise Set 13.2 - Page 862: 16

Answer

(a) Candidate A is declared the winner using the plurality-with-elimination method. (b) After six voters change their votes, we can see the new preference table below. Candidate C is declared the winner using the plurality-with-elimination method. The monotonicity criterion is not satisfied. Initially, Candidate A was selected as the winner. After some votes were changed in favor of Candidate A, Candidate A was not selected as the winner.
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Work Step by Step

(a) With the plurality-with-elimination method, the candidate with the fewest number of first-place votes is eliminated in each round. After that candidate is eliminated, the other candidates that were ranked below that candidate on each ballot move up one spot on that ballot. The rounds continue in this way until only one candidate remains, and this candidate is declared the winner. In round 1, we can count the number of first-place votes for each candidate. Candidate A: 14 Candidate B: 12 Candidate C: 10 Candidate D: 6 In round 1, Candidate D has the fewest number of first-place votes, so Candidate D is eliminated. After Candidate D is eliminated, the other candidates that were ranked below Candidate D on each ballot move up one spot on that ballot. In round 2, we can count the number of first-place votes for each candidate. Candidate A: 14 Candidate B: 12 Candidate C: 10 + 6 = 16 In round 2, Candidate B has the fewest number of first-place votes, so Candidate B is eliminated. After Candidate B is eliminated, the other candidates that were ranked below Candidate B on each ballot move up one spot on that ballot. In round 3, we can count the number of first-place votes for each candidate. Candidate A: 14 + 12 = 26 Candidate C: 10 + 6 = 16 In round 3, Candidate C has the fewest number of first-place votes, so Candidate C is eliminated. After Candidate C is eliminated, Candidate A is the only candidate remaining, so Candidate A is declared the winner. Candidate A is declared the winner using the plurality-with-elimination method. (b) After six voters change their ballot, we can see the new preference table below. We can go through the plurality-with-elimination method again to select the winner. In round 1, we can count the number of first-place votes for each candidate. Candidate A: 14 + 6 = 20 Candidate B: 12 Candidate C: 10 Candidate D: 0 In round 1, Candidate D has the fewest number of first-place votes, so Candidate D is eliminated. After Candidate D is eliminated, the other candidates that were ranked below Candidate D on each ballot move up one spot on that ballot. In round 2, we can count the number of first-place votes for each candidate. Candidate A: 14 + 6 = 20 Candidate B: 12 Candidate C: 10 In round 2, Candidate C has the fewest number of first-place votes, so Candidate C is eliminated. After Candidate C is eliminated, the other candidates that were ranked below Candidate C on each ballot move up one spot on that ballot. In round 3, we can count the number of first-place votes for each candidate. Candidate A: 14 + 6 = 20 Candidate C: 12 + 10 = 22 In round 3, Candidate A has the fewest number of first-place votes, so Candidate A is eliminated. After Candidate A is eliminated, Candidate C is the only candidate remaining, so Candidate C is declared the winner. Candidate C is declared the winner using the plurality-with-elimination method. The monotonicity criterion is not satisfied. Initially, Candidate A was selected as the winner. After some votes were changed in favor of Candidate A, Candidate A was not selected as the winner. Therefore, the monotonicity criterion is not satisfied.
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