Chapter : The Mathematics of Voting (c) 5 + 4 + 3 + 2 + = 5 (d) 5 20 = 300 (e) 300 69 70 64 48 = 49 25. C wins with 500 5 20 25 = 40 points; the points sum to 4 50 + 3 50 + 2 50 + 50 = 500. 26. E. There are a total of 40 5 = 600 points given out. Therefore, candidate E has 600 (39 + 2 + 80 + 3) = 47 points. D. Plurality-with-Elimination Method 27. (a) Professor Argand Round : E Number of first-place votes 8 3 5 5 0 E is eliminated. Round 2: Eliminating E does not affect the first-place votes of the candidates. Using the table from Round, we see that B now has the fewest number of first-place votes, so B is eliminated. The three votes originally going to B would next go to E, but E has also been eliminated. So B s three votes go to A. E Number of first-place votes 5 5 Candidate A now has a majority of the first-place votes and is declared the winner. (b) Since Professor Chavez is not under consideration, the resulting preference schedule would contain only four candidates. Number of voters 5 3 5 3 2 3 st choice A A E D D B 2 nd choice B D D B B E 3 rd choice D B A E A A 4 th choice E E B A E D (c) Professor Epstein. Round : Number of first-place votes 8 3 5 5 B is eliminated. Round 2: The three votes originally going to B go to E. Number of first-place votes 8 5 8 The 5 votes that D had in round 2 would go to B except that B has been eliminated. Instead, three of these votes go to E and two of these five votes go to A. Number of first-place votes 0 Candidate E now has a majority of the first-place votes and is declared the winner. 6 Copyright 200 Pearson Education, Inc. Publishing as Prentice Hall.
Chapter : The Mathematics of Voting (d) A was the winner of the original election, and, in the reelection without candidate C (an irrelevant alternative) it turned out that candidate E was declared the winner. This violates the Independence-of- Irrelevant-Alternatives criterion. 28. B is the winner. Round : Number of first-place votes 30 330 270 330 C is eliminated. Round 2: Number of first-place votes 420 490 330 Number of first-place votes 595 645 B now has a majority of the first-place votes and is declared the winner. 29. (a) Dante is the winner. Round : E Number of first-place votes 8 0 2 3 Candidate D has a majority of the first-place votes and is declared the winner. (b) Since Dante is a majority winner, in this case the winner is determined in the first round. (c) If there is a choice that has a majority of the first-place votes, then that candidate will be the winner under the plurality-with-elimination method. So, the plurality-with-elimination method satisfies the majority criterion. 30. (a) C is the winner. Round : Number of first-place votes 2 4 0 (b) C has a majority (4 of 27) of the first-place votes. (c) If there is a choice that has a majority of the first-place votes, then that candidate will be the winner under the plurality-with-elimination method in the first-round. 3. B is the winner. Round : Percent of first-place votes 40 35 25 0 Round, we see that candidate C now has the smallest percentage of first-place votes, so C is eliminated. The 25% of the votes originally going to C now go to B. Percent of first-place votes 40 60 Candidate B now has a majority of the first-place votes and is declared the winner. 32. Candidate B is the winner. Round : Number of first-place votes 48% 28% 24% 0% 7 Copyright 200 Pearson Education, Inc. Publishing as Prentice Hall.
Chapter : The Mathematics of Voting Round, we see that candidate C now has the fewest number of first-place votes, so C is eliminated. Number of first-place votes 48% 52% Candidate B now has a majority of the first-place votes and is declared the winner. 33. (a) Dungeons (South Africa) is the winner. Round : Number of first-place votes 0 9 5 7 Candidate C is eliminated. Round 2: The five votes originally going to C now go to D. Number of first-place votes 0 9 2 Candidate B is eliminated. Of the 9 votes going to B in Round 2, four will go to D and five will go to A. Number of first-place votes 5 6 Candidate D has a majority of the first-place votes and is declared the winner. (b) B is the Condorcet candidate. In a head-to-head contest, B beats A, 6 votes to 5 votes. In a head-to-head contest, B beats C, 6 votes to 5 votes. In a head-to-head contest, B beats D, 9 votes to 8 votes. (c) Banzai Pipeline (B), which is the Condorcet candidate, fails to win the election under the plurality-withelimination method. So, the plurality-with-elimination method fails to satisfy the Condorcet criterion. 34. (a) Clinton is the winner. Round : Number of first-place votes 6 5 6 0 Candidate Round, we see that candidate B now has the fewest number of first-place votes, so B is eliminated. Number of first-place votes 6 Candidate A is eliminated. Clinton is the winner. (b) Buford is the winner. Round : Number of first-place votes 4 5 8 0 Candidate Round, we see that candidate A now has the fewest number of first-place votes, so A is eliminated. Number of first-place votes 9 8 Candidate C is eliminated. Buford is the winner. 8 Copyright 200 Pearson Education, Inc. Publishing as Prentice Hall.
Chapter : The Mathematics of Voting (c) The monotonicity criterion. E. Pairwise Comparisons Method 35. (a) C is the winner. A versus B: 3 votes to 3 votes (tie). A gets ½ point, B gets ½ point. A versus C: 8 votes to 8 votes (C wins). C gets point. A versus D: 4 votes to 2 votes (A wins). A gets point. A versus E: 2 votes to 5 votes (A wins). A gets point. B versus C: 8 votes to 8 votes (C wins). C gets point. B versus D: 8 votes to 8 votes (D wins). D gets point. B versus E: 8 votes to 8 votes (E wins). E gets point. C versus D: 3 votes to 3 votes (tie). C gets ½ point, D gets ½ point. C versus E: 3 votes to 3 votes (tie). C gets ½ point, E gets ½ point. D versus E: 5 votes to votes. D gets point. The final tally, is 2 points for A, point for B, 3 points for C, 2 points for D, and points for E. Candidate C is the winner. (b) The preference schedule without Alberto is shown below. Number of voters 8 6 5 5 2 st choice C E E D D 2 nd choice B D C C E 3 rd choice D B D E B 4 th choice E C B B C (c) Without candidate A, the method of pairwise comparisons produces the following results. B versus C: 8 votes to 8 votes (C wins). C gets point. B versus D: 8 votes to 8 votes (D wins). D gets point. B versus E: 8 votes to 8 votes (E wins). E gets point. C versus D: 3 votes to 3 votes (tie). C gets 2 point, D gets 2 point. C versus E: 3 votes to 3 votes (tie). C gets 2 point, E gets 2 point. D versus E: 5 votes to votes. D gets point. The final tally, is 0 points for B, 2 points for C, 2 points for D, and points for E. Candidate D, Dora, is now the winner. (d) C was the winner of the original election, and, in the re-election without candidate A (an irrelevant alternative) it turned out that candidate D was declared the winner. This violates the Independence-of- Irrelevant-Alternatives criterion. 36. Candidate C is the winner. A versus B: 3 votes to 24 votes (A wins). A gets point. A versus C: 8 votes to 37 votes (C wins). C gets point. A versus D: 26 votes to 29 votes (D wins). D gets point. B versus C: 23 votes to 32 votes (C wins). C gets point. B versus D: 32 votes to 23 votes (B wins). B gets point. C versus D: 39 votes to 6 votes. (C wins). C gets point. The final tally is point for A, point for B, 3 points for C, and point for D. 37. A versus B: 3 votes to 8 votes (A wins). A gets point. A versus C: votes to 0 votes (A wins). A gets point. A versus D: votes to 0 votes (A wins). A gets point. 9 Copyright 200 Pearson Education, Inc. Publishing as Prentice Hall.
Chapter : The Mathematics of Voting A versus E: 0 votes to votes (E wins). E gets point. B versus C: votes to 0 votes (B wins). B gets point. B versus D: 8 votes to 3 votes (D wins). D gets point. B versus E: 6 votes to 5 votes (B wins). B gets point. C versus D: 3 votes to 8 votes (C wins). C gets point. C versus E: 8 votes to 3 votes (C wins). C gets point. D versus E: 3 votes to 8 votes (D wins). D gets point. The final tally is 3 points for A, 2 points for B, 2 points for C, 2 points for D, and point for E. Candidate A, Professor Argand, is the winner. 38. Candidate D is the winner. A versus B: votes to 3 votes (B wins). B gets point. A versus C: 6 votes to 8 votes (A wins). A gets point. A versus D: votes to 3 votes (D wins). D gets point. A versus E: 7 votes to 7 votes (A wins). A gets point. B versus C: 22 votes to 2 votes (B wins). B gets point. B versus D: 0 votes to 4 votes (D wins). D gets point. B versus E: 23 votes to vote (B wins). B gets point. C versus D: 0 votes to 4 votes (D wins). D gets point. C versus E: 7 votes to 7 votes (C wins). C gets point. D versus E: 23 votes to vote (D wins). D gets point. The final tally is 2 points for A, 3 points for B, point for C, 4 points for D, and 0 points for E. 39. (a) With five candidates, there are a total of 4 + 3 + 2 + = 0 pairwise comparisons. Each candidate is part of four of these. So, to find the number of points each candidate earns, we simply subtract the losses from 4. The 0 points are distributed as follows: E wins points, D wins 2 points, C gets 3 points, B gets 2 points, and A gets the remaining 0 2 3 2 = point. (b) Candidate C, with 3 points, is the winner. 40. (a) Since there are a total of (6 5) / 2 = 5 pairwise comparisons, F must have won 5 2 3 3 - = 2 of them (A earned 2 points, B and C each earned 3 points, and D and E each earned - points). (b) B and C tie for first place. F. Ranking Methods 4. (a) A has 0 first-place votes. B has 5 + 4 = 9 first-place votes. C has 5 first-place votes. D has 7 first-place votes. Winner A: Second place: B. Third place: D. Last place: C. (b) A has 4 0 + 3 0 + 2 (7 + 5 + 5) + 4 = 78 points. B has 4 (5 + 4) + 3 7 + 2 0 + 5 = 82 points. C has 4 5 + 3 (0 + 5 + 4) + 2 0 + 7 = 84 points. D has 4 7 + 3 5 + 2 4 + (0 + 5) = 66 points. Winner: C. Second place: B. Third place: A. Last place: D. 0 Copyright 200 Pearson Education, Inc. Publishing as Prentice Hall.