SOLUTIONS MANUAL For EXCURSIONS IN MODERN MATHEMATICS 10TH EDITION Peter Tannenbaum by digitaldownload87

Table of Contents Solutions Chapter 1

Chapter 2

Chapter 3

Chapter 4

Chapter 5

Chapter 6

104

Chapter 7

120

Chapter 8

130

Chapter 9

148

Chapter 10

160

Chapter 11

173

Chapter 12

191

Chapter 13

211

Chapter 14

224

Chapter 15

233

Chapter 16

247

Chapter 17

261

iii

Chapter 1 WALKING 1.1. Ballots and Preference Schedules 1.

Number of voters 5 3 5 3 2 3 1st choice A A C D D B 2nd choice B D E C C E 3rd choice C B D B B A 4th choice D C A E A C 5th choice E E B A E D This schedule was constructed by noting, for example, that there were five ballots listing candidate C as the first preference, candidate E as the second preference, candidate D as the third preference, candidate A as the fourth preference, and candidate B as the last preference.

Number of voters 4 1st choice A 2nd choice D 3rd choice B 4th choice C 3. (a) 5 + 5 + 3 + 3 + 3 + 2 = 21

5 B C D A

6 C A D B

2 A C D B

(b) 11. There are 21 votes all together. A majority is more than half of the votes, or at least 11. (c) Chavez. Argand has 3 last-place votes, Brandt has 5 last-place votes, Chavez has no last-place votes, Dietz has 3 last-place votes, and Epstein has 5 + 3 + 2 = 10 last-place votes. 4. (a) 202 + 160 + 153 + 145 + 125 + 110 + 108 + 102 + 55 = 1160 (b) 581; There are 1160 votes all together. A majority is more than half of the votes, or at least 581. (c) Alicia. She has no last-place votes. Note that Brandy has 125 + 110 + 55 = 290 last-place votes, Cleo has 202 + 145 + 102 = 449 last-place votes, and Dionne has 160 + 153 + 108 = 421 last-place votes. 5.

Number of voters 37 36 24 13 5 1st choice B A B E C 2nd choice E B A B E 3rd choice A D D C A 4th choice C C E A D 5th choice D E C D B Here Brownstein was listed first by 37 voters. Those same 37 voters listed Easton as their second choice, Alvarez as their third choice, Clarkson as their fourth choice, and Dax as their last choice. Number of voters 1st choice 2nd choice 3rd choice 4th choice 5th choice

14 B A E C D

10 B D A E C

8 A B E D C

7 D C B E A

4 E B A C D

Chapter 1: The Mathematics of Elections 7.

Number of voters 14 10 8 7 4 A 2 3 1 5 3 B 1 1 2 3 2 C 5 5 5 2 4 D 4 2 4 1 5 E 3 4 3 4 1 Here 14 voters had the same preference ballot listing B as their first choice, A as their second choice, E as their third choice, D as their fourth choice, and C as their fifth and last choice.

Number of voters A B C D E

37 1 3 2 5 4

36 2 1 4 3 5

24 5 2 3 1 4

13 2 4 1 5 3

5 4 1 5 2 3

Number of voters 255 480 765 1st choice L C M 2nd choice M M L 3rd choice C L C (0.17)(1500) = 255; (0.32)(500) = 480; The remaining voters (51% of 1500 or 1500-255-480=765) prefer M the most, C the least, so that L is their second choice.

10.

Number of voters 450 900 225 675 1st choice A B C C 2nd choice C C B A 3rd choice B A A B 100% - 20% - 40% = 40% of the voters number 225 + 675 = 900. So, if N represents the total number of voters, then (0.40) N = 900 . This means there are N = 2250 total voters. 20% of 2250 is 450 (these voters have preference ballots A, C, B). 40% of 2250 is 900 (these voters have preference ballots B, C, A).

1.2. Plurality Method 11. (a) C. A has 15 first-place votes. B has 11 + 8 + 1 = 20 first-place votes. C has 27 first-place votes. D has 9 first-place votes. C has the most first-place votes with 27 and wins the election. (b) C, B, A, D. Candidates are ranked according to the number of first-place votes they received (27, 20, 15, and 9 for C, B, A, and D respectively). 12. (a) D. A has 21 first-place votes. B has 18 first-place votes. C has 10 + 1 = 11 first-place votes. D has 29 first-place votes. D has the most first-place votes with 29 and wins the election. (b) D, A, B, C. 13. (a) C. A has 5 first-place votes. B has 4 + 2 = 6 first-place votes. C has 6 + 2 + 2 + 2 = 12 first-place votes. D has no first-place votes. C has the most first-place votes with 12 and wins the election. (b) C, B, A, D. Candidates are ranked according to the number of first-place votes they received (12, 6, 5, and 0 for C, B, A, and D respectively). 14. (a) B. A has 6 + 3 = 9 first-place votes. B has 6 + 5 + 3 = 14 first-place votes. C has no first-place votes. D has 4 first-place votes. B has the most first-place votes with 14 and wins the election. Copyright © 2022 Pearson Education, Inc.

ISM: Excursions in Modern Mathematics, 10E

(b) B, A, D, C. Candidates are ranked according to the number of first-place votes they received (14, 9, 4 and 0 for B, A, D, and C respectively). 15. (a) D. A has 11% of the first-place votes. B has 14% of the first-place votes. C has 24% of the first-place votes. D has 23% + 19% + 9% = 51% of the first-place votes. E has no first-place votes. D has the largest percentage of first-place votes with 51% and wins the election. (b) D, C, B, A, E. Candidates are ranked according to the percentage of first-place votes they received (51%, 24%, 14%, 11% and 0% for D, C, B, A, and E respectively). 16. (a) C. A has 12% of the first-place votes. B has 15% of the first-place votes. C has 25% + 10% + 9% + 8% = 52% of the first-place votes. D has no first-place votes. E has 21% of the first-place votes. C has the largest percentage of first-place votes with 52% and wins the election. (b) C, E, B, A, D. Candidates are ranked according to the percentage of first-place votes they received (52%, 21%, 15%, 12%, and 0% for C, E, B, A and D respectively). 17. (a) A. A has 5 + 3 = 8 first-place votes. B has 3 first-place votes. C has 5 first-place votes. D has 3 + 2 = 5 first-place votes. E has no first-place votes. A has the most first-place votes with 8 and wins the election. (b) A, C, D, B, E. Candidates are ranked according to the number of first-place votes they received (8, 5, 5, 3, and 0 for A, C, D, B, and E respectively). Since both candidates C and D have 5 first-place votes, the tie in ranking is broken by looking at last-place votes. Since C has no last-place votes and D has 3 lastplace votes, candidate C is ranked above candidate D. 18. (a) A. A has 153 + 102 + 55 = 310 first-place votes. B has 202 + 108 = 310 first-place votes. C has 160 + 110 = 270 first-place votes. D has 145 + 125 = 270 first-place votes. Both A and B have the most firstplace votes with 310 so the tie is broken using last-place votes. A has no last-place votes. B has 125 + 110 + 55 = 290 last-place votes. So A wins the election. (b) A, B, D, C. Candidates are ranked according to the number of first-place votes they received (310, 310, 270 and 270 for A, B, D, and C respectively). In part (a), we saw that the tie between A and B was broken in favor of A. Since both candidates C and D have 270 first-place votes, the tie in ranking is broken by looking at last-place votes. Since C has 202 + 145 + 102 = 449 last-place votes and D has 160 + 153 + 108 = 421 last-place votes, candidate D is ranked above candidate C. 19. (a) A. A has 5 + 3 = 8 first-place votes. B has 3 first-place votes. C has 5 first-place votes. D has 3 + 2 = 5 first-place votes. E has no first-place votes. A has the most first-place votes with 8 and wins the election. (Note: This is exactly the same as in Exercise 17(a).) (b) A, C, D, B, E. Candidates are ranked according to the number of first-place votes they received (8, 5, 5, 3, and 0 for A, C, D, B, and E respectively). Since both candidates C and D have 5 first-place votes, the tie in ranking is broken by a head-to-head comparison between the two. But candidate C is ranked higher than D on 5 + 5 + 3 = 13 of the 21 ballots (a majority). Therefore, candidate C is ranked above candidate D. 20. (a) B. A has 153 + 102 + 55 = 310 first-place votes. B has 202 + 108 = 310 first-place votes. C has 160 + 110 = 270 first-place votes. D has 145 + 125 = 270 first-place votes. Both A and B have the most firstplace votes with 310 so the tie is broken by head-to-head comparison. But candidate A is ranked higher than B on 153 + 125 + 110 + 102 + 55 = 545 of the 1160 ballots (less than a majority). So B wins the tiebreaker and the election.

Chapter 1: The Mathematics of Elections

(b) B, A, D, C. Candidates are ranked according to the number of first-place votes they received (310, 310, 270 and 270 for B, A, D, and C respectively). In part (a), we saw that the tie between A and B was broken in favor of B. Since both candidates C and D have 270 first-place votes, the tie in ranking is broken by head-to-head comparison. Candidate C is ranked higher than D on 160 + 153 + 110 + 108 = 531 of the 1160 ballots (less than a majority). So in the final ranking, candidate D is ranked above candidate C.

1.3. Borda Count 21. (a) A has 4 × 15 + 3 × (9 + 8 + 1) + 2 × 11 + 1× 27 = 163 points. B has 4 × (11 + 8 + 1) + 3 × 15 + 2 × (27 + 9) + 1× 0 = 197 points. C has 4 × 27 + 3 × 0 + 2 × 8 + 1× (15 + 11 + 9 + 1) = 160 points. D has 4 × 9 + 3 × (27 + 11) + 2 × (15 + 1) + 1× 8 = 190 points. The winner is B. (b) B, D, A, C. Candidates are ranked according to the number of Borda points they received. 22. (a) A has 4 × 21 + 3 × 18 + 2 × (29 + 10) + 1× 1 = 217 points. B has 4 × 18 + 3 × (10 + 1) + 2 × 21 + 1× 29 = 176 points. C has 4 × (10 + 1) + 3 × (29 + 21) + 2 × 18 + 1× 0 = 230 points. D has 4 × 29 + 3 × 0 + 2 × 1 + 1× (21 + 18 + 10) = 167 points. The winner is C. (b) C, A, B, D. Candidates are ranked according to the number of Borda points they received. 23. (a) A has 4 × 5 + 3 × 2 + 2 × (6 + 2) + 1× (4 + 2 + 2) = 50 points. B has 4 × (4 + 2) + 3 × (2 + 2) + 2 × 2 + 1× (6 + 5) = 51 points. C has 4 × (6 + 2 + 2 + 2) + 3 × 0 + 2 × (5 + 4 + 2) + 1× 0 = 70 points. D has 4 × 0 + 3 × (6 + 5 + 4 + 2) + 2 × 2 + 1× (2 + 2) = 59 points. The winner is C. (b) C, D, B, A. Candidates are ranked according to the number of Borda points they received. 24. (a) A has 4 × (6 + 3) + 3 × (4 + 3) + 2 × 6 + 1× 5 = 74 points. B has 4 × (6 + 5 + 3) + 3 × 3 + 2 × 0 + 1× (6 + 4) = 75 points. C has 4 × 0 + 3 × (6 + 6 + 5) + 2 × (4 + 3 + 3) + 1× 0 = 71 points. D has 4 × 4 + 3 × 0 + 2 × (6 + 5) + 1× (6 + 3 + 3) = 50 points. The winner is B. (b) B, A, C, D. Candidates are ranked according to the number of Borda points they received. 25. Here we can use a total of 100 voters for simplicity. A has 5 × 11 + 4 × (24 + 23 + 19) + 3 × (14 + 9) + 2 × 0 + 1× 0 = 388 points. B has 5 × 14 + 4 × 0 + 3 × (24 + 11) + 2 × 23 + 1× (19 + 9) = 249 points. C has 5 × 24 + 4 × (14 + 11 + 9) + 3 × 23 + 2 × 19 + 1× 0 = 363 points. D has 5 × (23 + 19 + 9) + 4 × 0 + 3 × 0 + 2 × 14 + 1× (24 + 11) = 318 points. E has 5 × 0 + 4 × 0 + 3 × 19 + 2 × (24 + 11 + 9) + 1× (23 + 14) = 182 points. The ranking (according to Borda points) is A, C, D, B, E.

ISM: Excursions in Modern Mathematics, 10E

26. We use a total of 100 voters for simplicity. A has 5 × 12 + 4 × 0 + 3 × 9 + 2 × (25 + 21 + 10) + 1× (15 + 8) = 222 points. B has 5 × 15 + 4 × 9 + 3 × (21 + 12) + 2 × 8 + 1× (25 + 10) = 261 points. C has 5 × (25 + 10 + 9 + 8) + 4 × 0 + 3 × 0 + 2 × 15 + 1× (21 + 12) = 323 points. D has 5 × 0 + 4 × (21 + 15 + 12 + 10) + 3 × (25 + 8) + 2 × 0 + 1× 9 = 340 points. E has 5 × 21 + 4 × (25 + 8) + 3 × (15 + 10) + 2 × (12 + 9) + 1× 0 = 354 points. The ranking (according to Borda points) is E, D, C, B, A. 27. Haskins had 3 × 46 + 2 × 111 + 1× 423 = 138 + 222 + 423 = 783 points. Murray had 3 × 517 + 2 × 278 + 1× 60 = 1551 + 556 + 60 = 2167 points. Tagovailoa had 3 × 299 + 2 × 431 + 1× 112 = 897 + 862 + 112 = 1871 points. The ranking (according to Borda points) is Murray (2167), Tagovailoa (1871), and Haskins (783). 28. Bieber had 7 × 0 + 4 × 0 + 3 × 11 + 2 × 13 + 1× 5 = 64 points. Cole had 7 × 13 + 4 × 17 + 3 × 0 + 2 × 0 + 1× 0 = 159 points. Lynn had 7 × 0 + 4 × 0 + 3 × 0 + 2 × 3 + 1× 12 = 18 points. Morton had 7 × 0 + 4 × 0 + 3 × 18 + 2 × 10 + 1× 1 = 75 points. Verlander had 7 × 17 + 4 × 13 + 3 × 0 + 2 × 0 + 1× 0 = 171 points. As in Example 1.12, this uses a modified Borda count. In this case, first-place votes count 7 points rather than the usual 5. The ranking (according to modified Borda points) is Verlander (171), Cole (159), Morton (75), Bieber (64), and Lynn (18). 29. Each ballot has 4 × 1 + 3 × 1 + 2 × 1 + 1× 1 = 10 points that are awarded to candidates according to the Borda count. With 110 voters, there are a total of 110 × 10 = 1100 Borda points. So D has 1100 – 320 – 290 – 180 = 310 Borda points. The ranking is thus A (320), D (310), B (290), and C (180). 30. Each ballot has 7 × 1 + 4 × 1 + 3 × 1 + 2 × 1 + 1× 1 = 17 points that are awarded to candidates according to the Borda count. With 50 voters, there are a total of 50 × 17 = 850 Borda points. So E has 850 – 152 – 133 – 191 – 175 = 199 Borda points. The ranking is thus E (199), C (191), D (175), A (152), and B (133).

1.4. Plurality-with-Elimination 31. (a) A is the winner. Round 1: Candidate A B C D Number of first-place votes 15 20 27 9 D is eliminated. Round 2: The 9 first-place votes originally going to D now go to A. Candidate A B C D Number of first-place votes 24 20 27 B is eliminated. Round 3: There are 8 + 1 = 9 first-place votes originally going to B that now go to A. There are also 11 first-place votes going to B that would now go to D. But, since D is already eliminated, these 11 firstplace votes go to A. Candidate A B C D Number of first-place votes 44 27 Candidate A now has a majority of the first-place votes and is declared the winner. (b) A complete ranking of the candidates can be found by noting in part (a) when each candidate was eliminated. Since D was eliminated first, it is ranked last. Since B was eliminated next, it is ranked next to last. The final ranking is hence A, C, B, D.

Chapter 1: The Mathematics of Elections 32. (a) B is the winner. Round 1: Candidate A B C D Number of first-place votes 21 18 11 29 C is eliminated. Round 2: The 11 first-place votes originally going to C would next go to B. Candidate A B C D Number of first-place votes 21 29 29 Round 3: The 21 first-place votes going to A would next go to C. But C has been eliminated. So these 21 first-place votes go to B. Candidate A B C D Number of first-place votes 50 29 Candidate B now has a majority of the first-place votes and is declared the winner. (b) A complete ranking of the candidates can be found by noting in part (a) when each candidate was eliminated. Since C was eliminated first, it is ranked last. Since A was eliminated next, it is ranked next to last. The final ranking is hence B, D, A, C. 33. (a) C is the winner. Round 1: Candidate A B C D Number of first-place votes 5 6 12 0 Candidate C has a majority of the first-place votes and is declared the winner. (b) To determine a ranking, we ignore the fact that C wins and at the end of round 1, D is the first candidate eliminated. Round 2: No first-place votes are changed. Candidate A B C D Number of first-place votes 5 6 12 A is eliminated. Round 3: There are 5 first-place votes originally going to A that now go to C. Candidate Number of first-place votes The final ranking is C, B, A, D.

B 6

C 17

34. (a) B is the winner. Round 1: Candidate A B C D Number of first-place votes 9 14 0 4 Candidate B has a majority of the first-place votes and is declared the winner. (b) To determine a ranking, we ignore the fact that B wins and at the end of round 1, C is the first candidate eliminated. Round 2: No first-place votes are changed. Candidate A B C D Number of first-place votes 9 14 4 D is eliminated. Round 3: There are 4 first-place votes originally going to D that now go to A. Candidate Number of first-place votes The final ranking is B, A, D, C.

A 13

B 14

ISM: Excursions in Modern Mathematics, 10E

35. Round 1: Candidate A Number of first-place votes 8 E is eliminated. Round 2: No first-place votes change.

B 8

C 7

D 6

E 0

Candidate A B C D E Number of first-place votes 8 8 7 6 D is eliminated. Round 3: There are 4 first-place votes for D that move to candidate C and there are 2 first-place votes for D that move to candidate B. Candidate A B C D E Number of first-place votes 8 10 11 A is eliminated. Round 4: There are 5 first-place votes for A that move to candidate B and there are 3 first-place votes for A that move to candidate C (since D and E have both been eliminated). Candidate A B C D E Number of first-place votes 15 14 B now has a majority of the first-place votes and is declared the winner. The final ranking is B, C, A, D, E. 36. Round 1: Candidate A B C D E Number of first-place votes 8 4 11 12 5 B is eliminated. Round 2: There are 4 first-place votes for B that move to candidate E. Candidate A B C D E Number of first-place votes 8 11 12 9 A is eliminated. Round 3: There are 5 first-place votes for A that move to candidate E (since B is eliminated), 2 first-place votes for A that move to candidate D (since B is eliminated), and 1 first-place vote for A that moves to C. Candidate A B C D E Number of first-place votes 12 14 14 C is eliminated. Round 4: There are 6 first-place votes for C that move to candidate E (since A is eliminated), 5 first-place votes for C that move to candidate D (since both A and B are eliminated), and there is 1 first-place vote for C that moves to candidate E (C earned this vote earlier when candidate A was eliminated). Candidate A B C D E Number of first-place votes 19 21 E now has a majority of the first-place votes and is declared the winner. The final ranking is E, D, C, A, B. 37. (a) D is the winner. Round 1: Candidate A B C D E Percentage of first-place votes 11 14 24 51 0 Candidate D has a majority of the first-place votes and is declared the winner. (b) To determine a ranking, we ignore the fact that D wins and at the end of round 1, E is the first candidate eliminated. Round 2: No first-place votes are changed.

Chapter 1: The Mathematics of Elections Candidate A B C D E Percentage of first-place votes 11 14 24 51 A is eliminated. Round 3: The 11% of the first-place votes that went to A now go to C. Candidate A B C D E Percentage of first-place votes 14 35 51 B is eliminated. Round 4: The 14% of the first-place votes that went to B now go to C. Candidate A B C D E Percentage of first-place votes 49 51 A complete ranking of the candidates can be found by noting when each candidate was eliminated. The final ranking is hence D, C, B, A, E. 38. (a) C is the winner. Round 1: Candidate A B C D E Percentage of first-place votes 12 15 52 0 21 Candidate C has a majority of the first-place votes and is declared the winner. (b) To determine a ranking, we ignore the fact that C wins and at the end of round 1, D is the first candidate eliminated. Round 2: No first-place votes are changed. Candidate A B C D E Percentage of first-place votes 12 15 52 21 A is eliminated. Round 3: The 12% of the first-place votes that went to A now go to B (since D has already been eliminated). Candidate A B C D E Percentage of first-place votes 27 52 21 E is eliminated. Round 4: The 14% of the first-place votes that went to E now go to B (since D has already been eliminated). Candidate A B C D E Percentage of first-place votes 48 52 A complete ranking of the candidates can be found by noting when each candidate was eliminated. The final ranking is hence C, B, E, A, D. 39. Round 1: Candidate A B C D E Number of first-place votes 8 8 7 6 0 Candidates E, D, and C are all eliminated. Round 2: There are 4 first-place votes for D that go to B (since C has been eliminated). There are 2 firstplace votes for D that go to B. There are 7 first-place votes for C that go to A (since both D and E are eliminated). Candidate A B C D E Number of first-place votes 15 14 A now has a majority of the first-place votes and is declared the winner.

ISM: Excursions in Modern Mathematics, 10E

40. Round 1: Candidate A B C D E Number of first-place votes 8 4 11 12 5 Candidates B, E, and A are all eliminated. Round 2: There are 4 first-place votes for B that go to C (since E has been eliminated). There are 5 first-place votes for E that go to D (since A has been eliminated). There are 5 first-place votes for A that go to C (since both B and E are eliminated), there are 2 first-place votes for A that go to D (since B is eliminated), and 1 first-place vote for A that goes to C. Candidate A B C D E Number of first-place votes 21 19 C now has a majority of the first-place votes and is declared the winner.

1.5. Pairwise Comparisons 41. (a) Candidate D is the winner. A versus B: 15 + 9 = 24 votes to 27 + 11 + 8 + 1 = 47 votes (B wins). B gets 1 point. A versus C: 15 + 11 + 9 + 8 + 1 = 44 votes to 27 votes (A wins). A gets 1 point. A versus D: 15 + 8 + 1 = 24 votes to 27 + 11 + 9 = 47 votes (D wins). D gets 1 point. B versus C: 15 + 11 + 9 + 8 + 1 = 44 votes to 27 votes (B wins). B gets 1 point. B versus D: 15 + 11 + 8 + 1 = 35 votes to 27 + 9 = 36 votes (D wins). D gets 1 point. C versus D: 27 + 8 = 35 votes to 15 + 11 + 9 + 1 = 36 votes. (D wins). D gets 1 point. The final tally is 1 point for A, 2 points for B, 0 points for C, and 3 points for D. (b) A complete ranking for the candidates is found by tallying points. In this case, the final ranking is D (3 points), B (2 points), A (1 point), and C (0 points). 42. (a) Candidate C is the winner. A versus B: 29 + 21 = 50 votes to 18 + 10 + 1 = 29 votes (A wins). A gets 1 point. A versus C: 21 + 18 = 39 votes to 29 + 10 + 1 = 40 votes (C wins). C gets 1 point. A versus D: 21 + 18 + 10 = 49 votes to 29 + 1 = 30 votes (A wins). A gets 1 point. B versus C: 18 votes to 29 + 21 + 10 + 1 = 61 votes (C wins). C gets 1 point. B versus D: 21 + 18 + 10 + 1 = 50 votes to 29 votes (B wins). B gets 1 point. C versus D: 21 + 18 + 10 + 1 = 50 votes to 29 votes. (C wins). C gets 1 point. The final tally is 2 points for A, 1 point for B, 3 points for C, and 0 points for D. (b) A complete ranking for the candidates is found by tallying points. In this case, the final ranking is C (3 points), A (2 points), B (1 point), and D (0 points). 43. (a) Candidate C is the winner. A versus B: 6 + 5 = 11 votes to 4 + 2 + 2 + 2 + 2 = 12 votes (B wins). B gets 1 point. A versus C: 5 + 2 = 7 votes to 6 + 4 + 2 + 2 + 2 = 16 votes (C wins). C gets 1 point. A versus D: 5 + 2 + 2 = 9 votes to 6 + 4 + 2 + 2 = 14 votes (D wins). D gets 1 point. B versus C: 4 + 2 = 6 votes to 6 + 5 + 2 + 2 + 2 = 17 votes (C wins). C gets 1 point. B versus D: 4 + 2 + 2 + 2 = 10 votes to 6 + 5 + 2 = 13 votes (D wins). D gets 1 point. C versus D: 6 + 2 + 2 + 2 + 2 = 10 votes to 5 + 4 = 9 votes. (C wins). C gets 1 point. The final tally is 0 points for A, 1 point for B, 3 points for C, and 2 points for D. (b) A complete ranking for the candidates is found by tallying points. In this case, the final ranking is C (3 points), D (2 points), B (1 point), and A (0 points).

Chapter 1: The Mathematics of Elections

44. (a) Candidate B is the winner. A versus B: 6 + 4 + 3 = 13 votes to 6 + 5 + 3 = 14 votes (B wins). B gets 1 point. A versus C: 6 + 4 + 3 + 3 = 16 votes to 6 + 5 = 11 votes (A wins). A gets 1 point. A versus D: 6 + 6 + 3 + 3 = 18 votes to 5 + 4 = 9 votes (A wins). A gets 1 point. B versus C: 6 + 5 + 3 = 14 votes to 6 + 4 + 3 = 13 votes (B wins). B gets 1 point. B versus D: 6 + 5 + 3 = 14 votes to 6 + 4 + 3 = 13 votes (B wins). B gets 1 point. C versus D: 6 + 6 + 5 + 3 + 3 = 23 votes to 4 votes. (C wins). C gets 1 point. The final tally is 2 points for A, 3 points for B, 1 point for C, and 0 points for D. (b) The final ranking is B (3 points), A (2 points), C (1 point), and D (0 points). 45. Candidate D is the winner. A versus B: 24% + 23% + 19% + 11% + 9% = 86% of the votes to 14% of the votes (A wins). A versus C: 23% + 19% + 11% = 53% of the votes to 24% + 14% + 9% = 47% of the votes (A wins). A versus D: 24% + 14% + 11% = 49% of the votes to 23% + 19% + 9% = 51% of the votes (D wins). A versus E: 24% + 23% + 19% + 14% + 11% + 9% = 100% of the votes to 0% of the votes (A wins). B versus C: 14% of the votes to 24% + 23% + 19% + 11% + 9% = 86% of the votes (C wins). B versus D: 24% + 14% + 11% = 49% of the votes to 23% + 19% + 9% = 51% of the votes (D wins). B versus E: 24% + 23% + 14% + 11% = 72% of the votes to 19% + 9% = 28% of the votes (B wins). C versus D: 24% + 14% + 11% = 49% of the votes to 23% + 19% + 9% = 51% of the votes (D wins). C versus E: 24% + 23% + 14% + 11% + 9% = 81% of the votes to 19% of the votes (C wins). D versus E: 23% + 19% + 14% + 9% = 65% of the votes to 24% + 11% = 35% of the votes (D wins). The final tally is 3 points for A, 1 point for B, 2 points for C, 4 points for D, and 0 points for E. 46. Candidate C is the winner. A versus B: 25% + 12% + 10% = 47% of the votes to 21% + 15% + 9% + 8% = 53% of the votes (B wins). A versus C: 21% + 12% = 33% of the votes to 25% + 15% + 10% + 9% + 8% = 67% of the votes (C wins). A versus D: 12% + 9% = 21% of the votes to 25% + 21% + 15% + 10% + 8% = 79% of the votes (D wins). A versus E: 12% + 9% = 21% of the votes to 25% + 21% + 15% + 10% + 8% = 79% of the votes (E wins). B versus C: 21% + 15% + 12% = 48% of the votes to 25% + 10% + 9% + 8% = 52% of the votes (C wins). B versus D: 15% + 9% = 24% of the votes to 25% + 21% + 12% + 10% + 8% = 76% of the votes (D wins). B versus E: 15% + 12% + 9% = 36% of the votes to 25% + 21% + 10% + 8% = 64% of the votes (E wins). C versus D: 25% + 10% + 9% + 8% = 52% of the votes to 21% + 15% + 12% = 48% of the votes (C wins). C versus E: 25% + 10% + 9% + 8% = 52% of the votes to 21% + 15% + 12% = 48% of the votes (C wins). D versus E: 15% + 12% + 10% = 37% of the votes to 25% + 21% + 9% + 8% = 63% of the votes (E wins). The final tally is 0 points for A, 1 point for B, 4 points for C, 2 points for D, and 3 points for E. 47. A versus B: 7 + 5 + 3 = 15 votes to 8 + 4 + 2 = 14 votes (A wins). A gets 1 point. A versus C: 8 + 5 + 3 = 16 votes to 7 + 4 + 2 = 13 votes (A wins). A gets 1 point. A versus D: 8 + 5 + 3 = 16 votes to 7 + 4 + 2 = 13 votes (A wins). A gets 1 point. A versus E: 5 + 3 + 2 = 10 votes to 8 + 7 + 4 = 19 votes (E wins). E gets 1 point. B versus C: 8 + 5 + 2 = 15 votes to 7 + 4 + 3 = 14 votes (B wins). B gets 1 point. B versus D: 8 + 5 = 13 votes to 7 + 4 + 3 + 2 = 16 votes (D wins). D gets 1 point. B versus E: 8 + 5 + 4 + 2 = 19 votes to 7 + 3 = 10 votes (B wins). B gets 1 point. C versus D: 8 + 7 + 5 = 20 votes to 4 + 3 + 2 = 9 votes (C wins). C gets 1 point. C versus E: 7 + 5 + 4 + 2 = 18 votes to 8 + 3 = 11 votes (C wins). C gets 1 point. D versus E: 5 + 4 + 3 + 2 = 14 votes to 8 + 7 = 15 votes (E wins). E gets 1 point. The final tally is 3 points for A, 2 points for B, 2 points for C, 1 point for D, and 2 points for E. Now B, C, and E each have 2 points. In head-to-head comparisons, B beats C and B beats E so that B is ranked higher than C and E. Also, C beats E so C is ranked higher than E as well. The final ranking is thus A, B, C, E, D.

ISM: Excursions in Modern Mathematics, 10E

48. A versus B: 6 + 5 + 5 + 5 + 5 + 2 + 1 = 29 votes to 11 votes (A wins). A gets 1 point. A versus C: 7 + 5 + 5 + 2 + 1 = 20 votes to 20 votes (tie). A and C each get ½ point. A versus D: 6 + 5 + 5 + 5 + 2 + 1 = 24 votes to 16 votes (A wins). A gets 1 point. A versus E: 7 + 6 + 5 + 5 + 5 + 2 + 1 = 31 votes to 9 votes (A wins). A gets 1 point. B versus C: 7 + 5 + 5 + 4 + 2 = 23 votes to 17 votes (B wins). B gets 1 point. B versus D: 6 + 5 + 5 + 4 + 2 + 1 = 23 votes to 17 votes (B wins). B gets 1 point. B versus E: 7 + 5 + 5 + 4 + 2 = 23 votes to 17 votes (B wins). B gets 1 point. C versus D: 6 + 5 + 5 + 4 + 1 = 21 votes to 19 votes (C wins). C gets 1 point. C versus E: 7 + 6 + 5 + 5 + 1 = 24 votes to 16 votes (C wins). C gets 1 point. D versus E: 7 + 5 + 5 + 2 = 19 votes to 21 votes (E wins). E gets 1 point. The final tally is 3.5 points for A, 3 points for B, 2.5 points for C, 0 points for D, and 1 point for E. The final ranking is thus A, B, C, E, D. 49. (a) With five candidates, there are a total of 4 + 3 + 2 + 1 = 10 pairwise comparisons. Each candidate is part of 4 of these (one against each other candidate). So, to find the number of points each candidate earns, 1 we simply subtract the losses from 4. The 10 points are distributed as follows: E wins 1 points, D 2 1 1 1 wins 2 points, C gets 3 points, B gets 2 points, and A gets the remaining 10 − 1 − 2 − 3 − 2 = 1 point. 2 2 2 So A loses 3 pairwise comparisons. (b) Candidate C, with 3 points, is the winner. [The complete ranking is C, D, B, E, A.] 1 50. (a) Since there are a total of (6 × 5) / 2 = 15 pairwise comparisons, F must have won 15 – 1 – 2 – 2 - 3 2 1 1 1 2 = 4 of them (A earned 1 point, B and C each earned 2 points, D earned 3 , and E earned 2 2 2 2 points). This means F lost 1 pairwise comparison.

(b) Candidate F, with 4 points, is the winner.

1.6. Fairness Criteria 51. First, we determine the winner using the Borda count. A has 4 × 6 + 3 × 0 + 2 × 0 + 1 × (2 + 3) = 29 points. B has 4 × 2 + 3 × 6 + 2 × 3 + 1 × 0 = 32 points. C has 4 × 3 + 3 × 2 + 2 × 6 + 1 × 0 = 30 points. D has 4 × 0 + 3 × 3 + 2 × 2 + 1 × 6 = 19 points. So candidate B is the winner using Borda count. However, candidate A has 6 of the 11 votes (a majority) and beats all three other candidates (B, C, and D) in head-to-head comparisons. That is, candidate A is a Condorcet candidate. Since this candidate did not win using the Borda count, this is a violation of the Condorcet criterion. 52. First, we determine the winner is candidate B using the plurality-with-elimination method (see Exercise 32(a)). In Exercise 42, however, we saw that candidate C was a Condorcet candidate (beating each other candidate in head-to-head comparisons and earning 3 points in the process). Since candidate C did not win using plurality-with-elimination, this is a violation of the Condorcet criterion.

Chapter 1: The Mathematics of Elections

53. The winner of this election is candidate R using the plurality method. Now F is clearly a nonwinning candidate. Removing F as a candidate leaves the following preference table. Number of voters 49 48 3 1st choice R H H 2nd choice H S S 3rd choice O O O 4th choice S R R In a recount, candidate H would be the winner using the plurality method. This is a violation of the IIA criterion. 54. First, we determine the winner using the Borda count. A has 4 × 14 + 3 × 0 + 2 × 0 + 1 × (10 + 8 + 4 + 1) = 79 points. B has 4 × 4 + 3 × (14 + 10) + 2 × (8 + 1) + 1 × 0 = 106 points. C has 4 × (10 + 1) + 3 × 8 + 2 × (14 + 4) + 1 × 0 = 104 points. D has 4 × 8 + 3 × (4 + 1) + 2 × 10 + 1 × 14 = 81 points. So candidate B (Boris) is the winner using Borda count. Now D (Dave) is clearly a nonwinning (irrelevant) candidate. Removing D as a candidate leaves the following preference table. Number of voters 14 10 8 4 1 1st choice A C C B C 2nd choice B B B C B 3rd choice C A A A A We now recount using the Borda count. A has 3 × 14 + 2 × 0 + 1 × (10 + 8 + 4 + 1) = 65 points. B has 3 × 4 + 2 × (14 + 10 + 8 + 1) + 1 × 0 = 78 points. C has 3 × (10 + 8 + 1) + 2 × 4 + 1 × 14 = 79 points. In a recount, candidate C (Carmen!) would be the winner using Borda count, a violation of the IIA criterion. 55. First, we determine the winner using plurality-with-elimination. Round 1: Candidate A B C D Number of first-place votes 8 3 5 5 Candidate E is eliminated. Round 2: No votes are shifted.

E 0

Candidate A B C D E Number of first-place votes 8 3 5 5 Candidate B is eliminated. Round 3: The 3 first-place votes for B now go to A (since E has been eliminated). Candidate A B C D E Number of first-place votes 11 5 5 Candidate A now has a majority (11 of the 21 votes) and is declared the winner. Now C is clearly a nonwinning (irrelevant) candidate. Removing C as a candidate leaves the following preference table. Number of voters 5 5 1st choice A E 2nd choice B D 3rd choice D B 4th choice E A We now recount using the plurality-with-elimination.

3 A D B E

3 D B E A

3 B E A D

2 D B A E

ISM: Excursions in Modern Mathematics, 10E

Round 1: Candidate A B D E Number of first-place votes 8 3 5 5 Candidate B is eliminated. Round 2: The 3 first-place votes that went to B now shift to E. Candidate A B D E Number of first-place votes 8 5 8 Candidate D is eliminated. Round 3: D had 5 first-place votes. Of these, 3 go to E and 2 go to A (since B was eliminated). Candidate A B D E Number of first-place votes 10 11 Candidate E now has a majority (11 of the 21 votes) and is declared the winner. Remember that candidate A was the winner before candidate C was removed. This is a violation of the IIA criterion. 56. If X has a majority of the first-place votes, then X will win every pairwise comparison (it is ranked above all other candidates on more than half the ballots) and is, therefore, the winner under the method of pairwise comparisons. 57. If X is the Condorcet candidate, then by definition X wins every pairwise comparison and is, therefore, the winner under the method of pairwise comparisons. 58. When a voter moves a candidate up in his or her ballot the number of first place votes for that candidate either increases or stays the same. It follows that if X had a plurality of the first place votes and a voter changes his or her ballot to rank X higher, then X still has a plurality. 59. When a voter moves a candidate up in his or her ballot, that candidate’s Borda points increase. It follows that if X had the most Borda points and a voter changes his or her ballot to rank X higher, then X still has the most Borda points. 60. When a voter moves a candidate up in his or her ballot it can’t hurt the candidate in a pairwise comparison— the candidate wins the same pairwise comparisons as before and possibly a few more. It follows that if X won the most pairwise comparisons and a voter changes his or her ballot to rank X higher, then X still wins the most pairwise comparisons.

JOGGING 61. Suppose the two candidates are A and B and that A gets a first-place votes and B gets b first-place votes and suppose that a > b. Then A has a majority of the votes and the preference schedule is Number of voters a b 1st choice A B 2nd choice B A It is clear that candidate A wins the election under the plurality method, the plurality-with-elimination method, and the method of pairwise comparisons. Under the Borda count method, A gets 2a + b points while B gets 2b + a points. Since a > b, 2a + b > 2b + a and so again A wins the election. 62. In this variation of the Borda count each candidate gets 1 less point per ballot. It follows that if N is the number of voters, each candidate gets N fewer points than he or she would under the standard Borda count method. Since each candidate’s total points gets decreased by the same number N, the ranking of the candidates remains the same. 63. The number of points under this variation is complementary to the number of points under the standard Borda count method: a first place is worth 1 point instead of N, a second place is worth 2 points instead of N – 1,…, a last place is worth N points instead of 1. It follows that having the fewest points here is equivalent to having the most points under the standard Borda count method, having the second fewest is equivalent to having the second most, and so on. Copyright © 2022 Pearson Education, Inc.

Chapter 1: The Mathematics of Elections

As another way to see this, suppose candidates C1 and C2 receive p1 and p2 points respectively using the Borda count as originally described in the chapter and r1 and r2 points under the variation described in this exercise. Then, for an election with k voters, we have p1 + r1 = p2 + r2 = k ( N + 1) . So if p1 < p2, we have –p1 > –p2 and so k(N + 1) – p1 > k(N + 1) – p2 which implies r1 > r2. Consequently the relative ranking of the candidates is not changed. 64. Use the Reverse Borda count method described in Exercise 63. In this variation the ranking of a candidate on a ballot equals the number of Borda points the candidate gets from that ballot. It follows that the average ranking of a candidate equals the total Borda points for that candidate divided by N (the number of voters). The candidate with the lowest average ranking is the candidate with the least total points and thus the winner of the election. 65. (a) Each of the voters gave Ohio State 25 points, so we compute 1625/25 to find 65 voters in the poll. (b) Let x = the number of Florida’s second-place votes. Then 1529 = 24 ⋅ x + 23 ⋅ ( 65 − x ) and so x = 34. This leaves 65 – 34 = 31 third-place votes for Florida. (c) Let y = the number of Michigan’s second-place votes. Then 1526 = 24 ⋅ y + 23 ⋅ ( 65 − y ) and so y = 31. This leaves 65 – 31 = 34 third-place votes for Michigan. 66. (a) Under the Borda count, each voter gives X more points than Y. Hence, the Borda count for X will be greater than the Borda count for Y and so X will rank above Y. (b) Since every voter prefers candidate X to candidate Y, a pairwise comparison between X and Y results in a win for X. If Y wins a pairwise comparison against Z, then X must also win against Z (by transitivity – since X is above Y on each ballot, when Y is above Z it must be that X is also above Z). Therefore X will have at least one more point than Y (from the head-to-head between X and Y). Thus, X will rank above Y under pairwise comparisons. 67. By looking at Deandre Ayton’s vote totals, it is clear that 1 point is awarded for each third-place vote (0 points is too few and 2 points is too many for each third-place vote). Let x = points awarded for each secondplace vote. Then, 1x + 63 = 66 gives x = 3 points for each second-place vote. Next, let y = points awarded for each first-place vote. Looking at Luka Doncic’s votes, we see that 98 y + 2 × 3 = 496 . Solving this equation gives y = 5 points for each first-place vote. 68. (a) Recall that using plurality-with-elimination in the Math Club election resulted in B being eliminated in the first round, C being eliminated second, A being eliminated third, and D winning (see Example 1.13). If top-two IRV is used, candidates B and D are both eliminated in round one. The 4 first-place votes for B would transfer to candidate C. The 8 first-place votes for D would also transfer to candidate C. The new preference table is shown below. Number of voters

1st choice

2nd choice C A We then have C winning the election -- clearly different than the plurality-with-elimination outcome. (b) Many examples are possible. (Example 1.21 shows that plurality-with-elimination violates the monotonicity criterion and in that example, since there are only 3 candidates from the start, top-two IRV and plurality-withelimination are identical methods and give the same result.)

ISM: Excursions in Modern Mathematics, 10E

1st choice

2nd choice

3rd choice C A A In this election B is a Condorcet candidate (since B beats A by a ‘score’ of14 to 10 and B beats C by a score of 16 to 8) and yet B is eliminated first using top-two IRV making C the winner. 69. (a) Round 1: Number of voters

1st choice

2nd choice

3rd choice

4th choice D A A A Candidate A with 23 last-place votes is eliminated. Round 2:

Number of voters

1st choice

2nd choice

3rd choice D D B C B Candidate B has 8 + 1 = 9 last-place votes, C has 4 last-place votes, and D has 14 + 10 = 24 last-place votes. So candidate D is eliminated. Round 3: Number of voters

1st choice

2nd choice C B B C B B has more last-place votes (meaning C has more first-place votes) and C is declared the winner. (b) Consider the election given by the preference schedule below. Here A is a Condorcet candidate but, having the most last-place votes, is eliminated in the first round. Number of voters

1st choice

2nd choice

3rd choice

4 choice A C D B D (c) Consider the election given by the preference schedule below. Here B wins under the Coombs method (C is eliminated first and B is preferred to A by 19 voters). However, if 8 voters move B from their 3rd choice to their 2nd choice, then C wins (since A would be eliminated first and C is preferred head-to-head over B). Number of voters

1st choice

2nd choice

3rd choice

Chapter 1: The Mathematics of Elections

70. (a) In round 1, no candidate has a majority of the 37 votes. Number of voters

1st choice

In round 2, counting 1st and 2nd place votes, A totals 14 votes, B totals 14 + 10 + 4 = 28 votes, C totals 10 + 8 + 1 = 19 votes, and D total 8 + 4 + 1 = 13 votes. Number of voters

1st choice

2nd choice

Both B and C have a majority of the 37 votes. However, candidate B has more and is declared the winner. (b) C is a Condorcet candidate in the Math Club election. However, as shown in part (a), B wins the election using the Bucklin method. (c) Say a ballot has candidate X ranked in the kth position. If and when the election gets to round k that vote contributes to X’s total under the Bucklin method. It follows that if X wins in round k and a voter moves X up in his or her ballot then X would still win (in round k or possibly in an earlier round).

RUNNING 71. One reasonable approach would be to use variables xk , 1 ≤ k ≤ 5 , to represent the number of points handed out for kth place. With so many variables, there is some flexibility in determining the values of these variables. To start, focus attention on Antetonkoumpo and Harden and variables x1 and x2 . The following two equations must hold: 78 x1 + 23 x2 = 941 23 x1 + 78 x2 = 776 Multiplying the first equation by 78 and the second equation by -23 produces the two equations: 6084 x1 + 1794 x2 = 73398 −529 x1 − 1794 x2 = −17848 Adding these equations eliminates the variable x2 and gives 5555 x1 = 55550 which has solution x1 = 10 . The first equation then gives 780 + 23 x2 = 941 so that x2 = 7 . We then turn our attention to Stephen Curry. His votes yield equation 16 x3 + 25 x4 + 20 x5 = 175 in which three of the four terms are multiples of 5. This observation means that variable x3 must also be a multiple of 5. Since the value of x3 is a positive integer strictly less than 7, we deduce that x3 = 5 . From that, we have 16(5) + 25 x4 + 20 x5 = 175 or 25 x4 + 20 x5 = 95 . Since x4 and x5 are both positive integers strictly less than 5, we can easily guess and check that x4 = 3 and x5 = 1 . The point values on each ballot are thus determined to be 10 points for 1st, 7 points for 2nd, 5 points for 3rd, 3 points for 4th, and 1 point for 5th. 72. (a) In this election, C is the winner under the plurality method yet a majority of the voters (4 out of 7) prefer both A and B over C. Number of voters

1st choice

2nd choice

3rd choice

ISM: Excursions in Modern Mathematics, 10E

(b) We may use the same example as in part (a). Both A and B are eliminated in the first round and so C is the winner under the plurality-with-elimination method. (c) Suppose there are k voters and N candidates. Under the Borda count method, the total number of points for all the candidates is kN ( N + 1) / 2 (each voter contributes 1 + 2 +  + N = N ( N + 1) / 2 points to the total), so the average number of points per candidate is k ( N + 1) / 2 . Now suppose that X is a candidate that loses to every other candidate in a one-to-one comparison. The claim is that under the Borda count method, X will receive less than the average k ( N + 1) points and therefore cannot be the winner. (One way to see this is as follows: Suppose that the N candidates are presented to the voters one at a time, with each voter keeping an updated partial ranking of the candidates as they are being presented. Suppose, moreover, that X is the first candidate to be presented. When this happens, X starts with kN points (X is the only candidate in every voter’s ballot and therefore has all the first-place votes.) As soon as the next candidate (say Y) is presented, X’s points will drop by more than k/2 (X loses to Y in a one-to-one comparison means that more than k/2 of the voters have ranked Y above X.) By the same argument each time one of the subsequent candidates is presented X’s points drop by more than k/2. By the time we are done, X must have less than kN − k2 ( N − 1) = k2 ( N + 1) points.) 73. (a) The Math Club election serves as an example. Candidate A has a majority of last-place votes (23 of the 37) and yet wins the election when the plurality method is used. Number of voters

1st choice

2nd choice

3rd choice

4 choice D A A A A (b) In this election, C is the winner under the plurality method (in the case of the tie in round 1, eliminate all candidates with the fewest number of votes). However, a majority of the voters (4 out of 7) prefer both A and B over C. Number of voters

1st choice

2nd choice

3rd choice C C B (c) Under the method of pairwise comparisons, if a majority of voters have candidate X ranked last on their ballot, then candidate X will never win a head-to-head comparison (since any other candidate Y is preferred to X by a majority of voters). Thus, X will end up with no points under the method and cannot win the election. (d) Suppose there are N candidates, x is the number of voters placing X last, and y is the remaining number of voters. Since a majority of the voters place X last, x > y. The maximum number of points that X can receive using a Borda count is x + Ny. (One point for each of the x last place votes and, assuming all other voters place X in first-place, N points for each of the other y voters.) The total number of points given out by each voter is 1 + 2 + 3 + ... + N = N(N + 1)/2 and so the total number of points given out by all x + y voters is N(N + 1)(x + y)/2. Since some candidate must receive at least 1/N of the total points, some candidate must receive at least (N + 1)(x + y)/2 points. Since x > y, we have (N + 1)(x + y)/2 = (Nx + x + Ny + y)/2 > (Ny + x + Ny + y)/2 > x + Ny, and consequently, X cannot be the winner of the election using the Borda count method.

Chapter 1: The Mathematics of Elections

74. In this election, D is preferred over A by a majority of the voters (12 to 10) and yet all four complete rankings rank A above D. Number of voters

1st choice

2nd choice

3rd choice

4th choice

Method

Winner

2nd place

3rd place

Last place

Plurality

Plurality with elimination

D D

Borda count

C,B (tied)

Pairwise comparison

A,B (tied)

C,D (tied)

75. Suppose there are k voters and N candidates. Case 1: k is odd, say k = 2t + 1. Suppose the candidate with a majority of the first-place votes is X. The fewest possible Borda points X can have is F(t + 1) + t [when there are (t + 1) votes that place X first and the remaining votes place X last]. The most Borda points that any other candidate can have is (N – 1)(t + 1) + Ft [when there are (t + 1) voters that place the candidate second and the remaining voters place that candidate first]. Thus, the majority criterion will be satisfied when F(t+ 1) + t > (N – 1)(t + 1) + Ft, which after simplification implies F > N(t + 1) – (2t + 1), or F > N ( k 2+1 ) − k . Case 2: k is even, say k = 2t. An argument similar to the one given in case 1 gives the inequality F(t + 1) + (t – 1) > (N – 1)(t + 1) + F(t – 1) which after simplification implies F > [N(t + 1) – 2t]/2, or F >  N ( k2 + 1) − k  / 2 .

APPLET BYTES 76. (a) Plurality method: A has 89 first-place votes, B has 91 first-place votes, and C has 73 first-place votes. So, using the plurality method, B wins.

Borda count method: A has 3 × 49 + 3 × 40 + 2 × 40 + 1× 51 + 2 × 43 + 1× 30 = 514 points. B has 2 × 49 + 1× 40 + 3 × 40 + 3 × 51 + 1× 43 + 2 × 30 = 514 points. C has 1× 49 + 2 × 40 + 1× 40 + 2 × 51 + 3 × 43 + 3 × 30 = 490 points. So, using the Borda count method, there is a tie between candidates A and B (each with 514 points). Plurality-with-elimination method: In round 1, B has 91 votes, A has 89 votes, and C has 73 votes. So candidate C is eliminated. In an election between A and B, A has 49 + 40 + 43 = 132 votes and B has 40 + 51 + 30 = 121 votes so B is eliminated and A is declared the winner. Pairwise comparisons method: A vs. B – A wins 132-121 and gets 1 point. A vs. C – A wins 129-124 and gets 1 point. B vs. C – B wins 140-113 and gets 1 point.

ISM: Excursions in Modern Mathematics, 10E

The final tally is A – 2 points, B – 1 point, C – 0 points. So, using the pairwise comparisons method, candidate A wins. (b) One possible manipulation is to change two votes from B>C>A to A>B>C. This makes A the winner under all four voting methods studied. Number of voters

1st choice

2nd choice

3rd choice

77. Many preference schedules are possible. Giving candidate C more strength relative to A and B is one approach to take. One such schedule moves 9 votes away from A>B>C (4 go to A>C>B and 5 go to C>B>A) and 5 votes away from B>C>A to C>B>A. In this new schedule, candidate B wins using the plurality method, candidate C wins using the Borda count method, candidate A wins using the plurality-with-elimination method, and candidate C wins using pairwise comparisons. Number of voters

1st choice

2nd choice

3rd choice

78. (a) Plurality method: A has 93 first-place votes, E has 81 first-place votes, B has 44 first-place votes, D has 42 first-place votes, and C has 40 first-place votes. So, using the plurality method, A wins.

Borda count method: A has 5 × 93 + 1× 44 + 4 × 10 + 2 × 30 + 2 × 42 + 1× 81 = 774 points. B has 4 × 93 + 5 × 44 + 2 × 10 + 1× 30 + 1× 42 + 2 × 81 = 846 points. C has 3 × 93 + 2 × 44 + 5 × 10 + 5 × 30 + 4 × 42 + 3 × 81 = 978 points. D has 2 × 93 + 4 × 44 + 1× 10 + 3 × 30 + 5 × 42 + 4 × 81 = 996 points. E has 1× 93 + 3 × 44 + 3 × 10 + 4 × 30 + 3 × 42 + 5 × 81 = 906 points. So, using the Borda count method, candidate D wins with 996 points. Plurality-with-elimination method: Candidate E wins (C is eliminated first, followed by D, B, and then A). Pairwise comparisons: A vs. B – A wins 175-125 and gets 1 point. A vs. C – C wins 93-207 and gets 1 point. A vs. D – D wins 103-197 and gets 1 point. A vs. E – E wins 103-197 and gets 1 point. B vs. C – C wins 137-163 and gets 1 point. B vs. D – D wins 147-153 and gets 1 point. B vs. E – E wins 137-163 and gets 1 point. C vs. D – D wins 133-167 and gets 1 point. C vs. E – C wins 175-125 and gets 1 point. D vs. E – D wins 179-121 and gets 1 point. So, using the pairwise comparisons method, candidate D wins (4 points for D, 3 points for C, 2 points for E, 1 point for A, 0 points for B). Since candidate D is a Condorcet candidate, we conclude that plurality method and the plurality-withelimination method both violate the Condorcet criterion in this election. Copyright © 2022 Pearson Education, Inc.

Chapter 1: The Mathematics of Elections (b) We aim to make D a winning candidate when using the plurality method. One way to do this is to count all first place votes for candidate A, D, and E and divide that number by 3. We then make sure candidate D has at least that many first-place votes and candidates A and E have less than that many. Since 93 + 42 + 81 = 216, we will take 21 votes away from the 93 voters ranking A>B>C>D>E and take 10 votes away from the voters ranking E>D>C>B>A giving these votes to the 42 voters ranking D>C>E>A>B. This makes D the winner under all four voting methods studied in this chapter. Number of voters

1 choice 2 choice 3 choice 4 choice 5 choice

79. D is preferred over A by a majority of the voters (11 to 10) but all four complete rankings rank A above D. Number of voters

1st choice

2nd choice

3rd choice

4th choice

Method

Winner

2nd place

3rd place

Last place

Plurality

Borda count

Plurality with elimination

Pairwise comparison

A,B (tied)

C,D (tied)

Chapter 2 WALKING 2.1. Weighted Voting 1. (a) A generic weighted voting system with N = 5 players is described using notation [q : w1 , w2 , w3 , w4 , w5 ] where q represents the value of the quota and wi represents the weight of player Pi . In this case, the players are the partners, w1 = 15 , w2 = 12 , w3 = w4 = 10 , and w5 = 3 . Since the total number of votes is 15 + 12 + 10 + 10 + 3 = 50 and the quota is determined by a simple majority (more than 50% of the total number of votes), q = 26. That is, the partnership can be described by [26 :15,12,10,10,3] . 1 2 (b) Since   50 = 33 , we choose the quota q as the smallest integer larger than this value which is 34. 3 3 The partnership can thus be described by [34 :15,12,10,10,3] .

2. (a) There are 100 votes (each one representing 1% ownership). A simple majority requires more than half of that or q = 51. That is, the partnership can be described by [51: 30, 25, 20,16,9] . 2 2 (b) Since  100 = 66 , we choose the quota q as the smallest integer larger than this value which is 67. 3 3 The partnership can thus be described by [67 : 30, 25, 20,16,9] .

3. (a) The quota must be more than half of the total number of votes. This system has 6 + 4 + 3 + 3 + 2 + 2 = 20 total votes. Since 50% of 20 is 10, the smallest possible quota would be 11. Note: q = 10 is not sufficient. If 10 votes are cast in favor and 10 cast against a motion, that motion should not pass. (b) The largest value q can take is 20, the total number of votes. (c)

3 × 20 = 15, so the value of the quota q would be 15. 4

(d) The value of the quota q would be strictly larger than 15. That is, 16. 4. (a) The quota must be more than half of the total votes. This system has 10 + 6 + 5 + 4 + 2 = 27 total votes. 1 × 27 = 13.5, so the smallest value q can take is 14. 2 (b) The largest value q can take is 27, the total number of votes. (c)

2 × 27 = 18 3

(d) 19 5. (a) P1 , the player with the most votes, does not have veto power since the other players combined have 3 + 3 + 2 = 8 votes and can successfully pass a motion without him. The other players can’t have veto power either then as they have fewer votes and hence less (or equal) power. So no players have veto power in this system.

Chapter 2: The Mathematics of Power (b) P1 does have veto power here since the other players combined have only 3 + 3 + 2 = 8 votes and cannot successfully pass a motion without him. P2 , on the other hand, does not have veto power since the other players combined have the 4 + 3 + 2 = 9 votes necessary to meet the quota. Since P3 and P4 have the same or fewer votes than P2 , it follows that only P1 has veto power. (c) P4 , the player with the fewest votes, does not have veto power here since the other players combined have 4 + 3 + 3 = 10 votes and can pass a motion without him. P3 , on the other hand, does have veto power since the other players combined have only 4 + 3 + 2 = 9 votes. Since P1 and P2 have the same or more votes than P3 , it follows that P1 , P2 , and P3 all have veto power. (d) P4 , the player with the fewest votes, has veto power since the other players combined have only 4 + 3 + 3 = 10 votes and cannot pass a motion without him. Since the other players all have the same or more votes than P4 , it follows that all players have veto power in this system. 6. (a) P1 does have veto power here since the other players combined have only 4 + 2 + 1 = 7 votes and cannot pass a motion without him. P2 , on the other hand, does not have veto power since the other players combined have the 8 + 2 + 1 = 11 votes necessary to meet the quota. Since P3 and P4 have the same or fewer votes than P2 , it follows that only P1 has veto power. (b) Based on (a), we know that P1 still has veto power since only the quota has changed (increased). It would now be even more difficult for the other players to pass a motion without P1 . P2 also has veto power here since the other players combined have 8 + 2 + 1 = 11 votes and cannot pass a motion without him. P3 , on the other hand, does not have veto power since the other players combined have 8 + 4 + 1 = 13 votes. Since P4 has fewer votes than P3 , it follows that only P1 and P2 have veto power. (c) Based on (a) and (b), we know that P1 and P2 still have veto power since only the quota has changed (increased). P3 now also has veto power here since the other players combined have only 8 + 4 + 1 = 13 votes and cannot pass a motion without him. P4 , on the other hand, does not have veto power since the other players combined have 8 + 4 + 2 = 14 votes. (d) P4 , the player with the fewest votes, has veto power since the other players combined have only 8 + 4 + 2 = 14 votes and cannot pass a motion without him. Since the other players all have the same or more votes than P4 , it follows that all four players have veto power in this system. 7. (a) In order for all three players to have veto power, the player having the fewest votes (the weakest) must have veto power. In order for that to happen, the quota q must be strictly larger than 7 + 5 = 12. The smallest value of q for which this is true is q = 13. (b) In order for P2 to have veto power, the quota q must be strictly larger than 7 + 3 = 10. The smallest value of q for which this is true is q = 11. [Note: When q = 11, we note that P3 does not have veto power since the other two players have 7 + 5 = 12 votes.] 8. (a) In order for all five players to have veto power, the player having the fewest votes (the weakest) must have veto power. In order for that to happen, the quota q must be strictly larger than 10 + 8 + 6 + 4 = 28. The smallest value of q for which this is true is q = 29. (b) In order for P3 to have veto power, the quota q must be strictly larger than 10 + 8 + 4 + 2 = 24. The smallest value of q for which this is true is q = 25. [Note: When q = 25, we note that P4 does not have veto power since the other four players have 10 + 8 + 6 + 2 = 26 votes.] Copyright © 2022 Pearson Education, Inc.

ISM: Excursions in Modern Mathematics, 10E

9. To determine the number of votes each player has, write the weighted voting system as [49: 4x, 2x, x, x]. (a) If the quota is defined as a simple majority of the votes, then x is the largest integer satisfying 4x + 2x + x + x 49 > which means that 8 x < 98 or x < 12.25 . So, x = 12 and the system can be 2 described as [49: 48, 24, 12, 12]. (b) If the quota is defined as more than two-thirds of the votes, then x is the largest integer satisfying 2 ( 4x + 2x + x + x) 49 > which means that 16 x < 147 or x < 9.1875 . So, x = 9 and the system can be 3 described as [49: 36, 18, 9, 9]. (c) If the quota is defined as more than three-fourths of the votes, then x is the largest integer satisfying 3(4x + 2x + x + x) which means that 24 x < 196 or x < 8.167 . So, x = 8 and the system can be 49 > 4 described as [49: 32, 16, 8, 8]. 10. (a) [121: 96, 48, 48, 24, 12, 12]; if the quota is defined as a simple majority of the votes, then x is the largest 8x + 4 x + 4 x + 2 x + x + x integer satisfying 121 > . 2 (b) [121: 72, 36, 36, 18, 9, 9]; if the quota is defined as more than two-thirds of the votes, then x is the largest 2 (8 x + 4 x + 4 x + 2 x + x + x) . integer satisfying 121 > 3 (c) [121: 64, 32, 32, 16, 8, 8]; if the quota is defined as more than three-fourths of the votes, then x is the 3(8x + 4 x + 4 x + 2 x + x + x) . largest integer satisfying 121 > 4

2.2. Banzhaf Power 11. (a) w1 + w3 = 7 + 3 = 10 (b) Since (7 + 5 + 3) / 2 = 7.5, the smallest allowed value of the quota q in this system is 8. For {P1 , P3 } to be a winning coalition, the quota q could be at most 10. So, the values of the quota for which {P1 , P3 } is winning are 8, 9, and 10. (c) Since 7 + 5 + 3 = 15, the largest allowed value of the quota q in this system is 15. For {P1 , P3 } to be a losing coalition, the quota q must be strictly greater than its weight, 10. So, the values of the quota for which {P1 , P3 } is losing are 11, 12, 13, 14, and 15. 12. (a) 8 + 6 + 4 = 18 (b) Since (10 + 8 + 6 + 4 + 2) / 2 = 15, the smallest allowed value of the quota q in this system is 16. For {P2 , P3 , P4 } to be a winning coalition, the quota q could be at most 18. So, the values of the quota for which {P2 , P3 , P4 } is winning are 16, 17, and 18. (c) Since 10 + 8 + 6 + 4 + 2 = 30, the largest allowed value of the quota q in this system is 30. For {P2 , P3 , P4 } to be a losing coalition, the quota q must be strictly greater than its weight, 18. So, the values of the quota for which {P2 , P3 , P4 } is losing are integer values from 19 to 30. Copyright © 2022 Pearson Education, Inc.

Chapter 2: The Mathematics of Power

13. P1 is critical (underlined) three times; P2 is critical three times; P3 is critical once ; P4 is critical once. The total number of times the players are critical is 8 (number of underlines). The Banzhaf power distribution is β1 = 3 / 8; β 2 = 3 / 8; β 3 = 1/ 8; β 4 = 1/ 8 . 14. P1 is critical (underlined) seven times; P2 is critical five times; P3 is critical three times; P4 is critical three times; P5 is critical one time. The total number of times the players are critical (all underlines) is 19. The Banzhaf power distribution is β1 = 7 /19; β 2 = 5 /19; β3 = 3 /19; β4 = 3 /19; β5 = 1/19. 15. (a) P1 is critical since the other players only have 5 + 2 = 7 votes. P2 is also critical since the other players only have 6 + 2 = 8 votes. However, P4 is not critical since the other two players have 6 + 5 = 11 (more than q = 10). (b) The winning coalitions are those whose weights are 10 or more. These are: {P1 , P2 }, {P1 , P3 }, {P1 , P2 , P3 }, {P1 , P2 , P4 }, {P1 , P3 , P4 }, {P2 , P3 , P4 }, {P1 , P2 , P3 , P4 }. (c) We underline the critical players in each winning coalition: {P1 , P2 }, {P1 , P3 }, {P1 , P2 , P3 }, {P1 , P2 , P4 },

{P1 , P3 , P4 }, {P2 , P3 , P4 }, {P1 , P2 , P3 , P4 }. Then, it follows that β1 =

5 3 1 ; β 2 = β3 = ; β 4 = . 12 12 12

16. (a) All the players are critical in this coalition since the total weight of the coalition (3 + 1 + 1 = 5) is exactly the same as the quota. If any one player were to leave the coalition, the remaining players would not have enough votes to meet the quota. (b) The winning coalitions are those whose weights are 5 or more. These are: {P1 , P2 }, {P1 , P2 , P3 }, {P1 , P2 , P4 }, {P1 , P3 , P4 }, {P1 , P2 , P3 , P4 }. (c) We underline the critical players in each winning coalition: {P1 , P2 }, {P1 , P2 , P3 }, {P1 , P2 , P4 },

{P1 , P3 , P4 }, {P1 , P2 , P3 , P4 }. Then, it follows that β1 =

5 3 1 1 ; β2 = ; β3 = ; β 4 = . 10 10 10 10

17. (a) The winning coalitions (with critical players underlined) are: {P1 , P2 } , {P1 , P3 } , and {P1 , P2 , P3 } . P1 is critical three times; P2 is critical one time; P3 is critical one time. The total number of times the players are critical is 5. The Banzhaf power distribution is β1 = 3 / 5; β2 = 1/ 5; β3 = 1/ 5. (b) The winning coalitions (with critical players underlined) are: {P1 , P2 } , {P1 , P3 } , and {P1 , P2 , P3 } . P1 is critical three times; P2 is critical one time; P3 is critical one time. The total number of times the players are critical is 5. The Banzhaf power distribution is β1 = 3/ 5; β2 = 1/ 5; β3 = 1/ 5. The distributions in (a) and (b) are the same. 18. (a) The winning coalitions (with critical players underlined) are: {P1 , P2 } and {P1 , P2 , P3 } . P1 is critical twice; P2 is critical twice; P3 is never critical. The total number of times the players are critical is 4. The Banzhaf power distribution is β1 = 2 / 4; β 2 = 2 / 4; β 3 = 0 / 4.

ISM: Excursions in Modern Mathematics, 10E

(b) The winning coalitions (with critical players underlined) are: {P1 , P2 } and {P1 , P2 , P3 } . P1 is critical twice; P2 is critical twice; P3 is never critical. The total number of times the players are critical is 4. The Banzhaf power distribution is β1 = 2 / 4; β 2 = 2 / 4; β 3 = 0 / 4. The distributions in (a) and (b) are the same. 19. (a) The winning coalitions (with critical players underlined) are: {P1 , P2 , P3 } , {P1 , P2 , P4 } , {P1 , P2 , P5 } , {P1 , P3 , P4 } , {P1 , P2 , P3 , P4 } , {P1 , P2 , P3 , P5 } , {P1 , P2 , P4 , P5 } , {P1 , P3 , P4 , P5 } , {P2 , P3 , P4 , P5 } ,

{P1 , P2 , P3 , P4 , P5 } . The Banzhaf power distribution is β1 = 8 / 24; β 2 = 6 / 24; β 3 = 4 / 24; β 4 = 4 / 24;

β 5 = 2 / 24. (b) The situation is like (a) except that {P1 , P2 , P5 } , {P1 , P3 , P4 } and {P2 , P3 , P4 , P5 } are now losing coalitions. In addition, P2 is now critical in {P1 , P2 , P3 , P4 } , P3 is now critical in {P1 , P2 , P3 , P5 } , P4 is now critical in {P1 , P2 , P4 , P5 } , P5 is now critical in {P1 , P3 , P4 , P5 } , and P1 is now critical in the grand coalition {P1 , P2 , P3 , P4 , P5 } . The winning coalitions (with critical players underlined) are now: {P1 , P2 , P3 } , {P1 , P2 , P4 } , {P1 , P2 , P3 , P4 } , {P1 , P2 , P3 , P5 } , {P1 , P2 , P4 , P5 } , {P1 , P3 , P4 , P5 } , {P1 , P2 , P3 , P4 , P5 } .

The Banzhaf power distribution is β1 = 7 /19; β 2 = 5 /19; β 3 = 3 /19; β 4 = 3 /19; β 5 = 1/19. (c) This situation is like (b) with the following exceptions: {P1 , P2 , P4 } and {P1 , P3 , P4 , P5 } are now losing coalitions; P3 is critical in {P1 , P2 , P3 , P4 } ; P5 is critical in {P1 , P2 , P4 , P5 } ; and P2 is critical in {P1 , P2 , P3 , P4 , P5 } . The winning coalitions (with critical players underlined) are now: {P1 , P2 , P3 } , {P1 , P2 , P3 , P4 } , {P1 , P2 , P3 , P5 } , {P1 , P2 , P4 , P5 } , {P1 , P2 , P3 , P4 , P5 } . The Banzhaf power distribution

is β1 = 5 /15; β 2 = 5 /15; β 3 = 3 /15; β 4 =1/15; β 5 = 1/15. (d) Since the quota equals the total number of votes in the system, the only winning coalition is the grand coalition and every player is critical in that coalition. The Banzhaf power distribution is easy to calculate in the case since all players share power equally. It is β1 = 1/ 5; β 2 =1/ 5; β 3 =1/ 5; β 4 =1/ 5; β 5 =1/ 5. 20. (a) P1 is a dictator and the other players are dummies. Thus P1 is the only critical player in each winning coalition. The Banzhaf power distribution is β1 = 1; β 2 = β 3 = β 4 = 0. (b) The winning coalitions (with critical players underlined) are: {P1 , P2 } , {P1 , P3 } , {P1 , P4 } , {P1 , P2 , P3 } , {P1 , P2 , P4 } , {P1 , P3 , P4 } , {P1 , P2 , P3 , P4 } . The Banzhaf power distribution is β1 =

β2 =

7 = 70%; 10

1 1 1 = 10%; β 3 = = 10%; β 4 = = 10%. 10 10 10

(c) This situation is like (b) with the following exceptions: {P1 , P4 } is now a losing coalition; P1 and P2 are both critical in {P1 , P2 , P4 }; P1 and P3 are both critical in {P1 , P3 , P4 } . The winning coalitions (with critical players underlined) are now: {P1 , P2 } , {P1 , P3 } , {P1 , P2 , P3 } , {P1 , P2 , P4 } , {P1 , P3 , P4 } , {P1 , P2 , P3 , P4 } . The Banzhaf power distribution is β1 =

6 2 2 = 60%; β 2 = = 20%; β 3 = = 20%; 10 10 10

β 4 = 0.

Chapter 2: The Mathematics of Power (d) The winning coalitions (with critical players underlined) are now: {P1 , P2 , P3 } , {P1 , P2 , P3 , P4 } . The

Banzhaf power distribution is β1 =

2 2 2 ; β 2 = ; β 3 = ; β 4 = 0. 6 6 6

21. (a) A player is critical in a coalition if that coalition without the player is not on the list of winning coalitions. So, in this case, the critical players are underlined below. {P1 , P2 } , {P1 , P3 } , {P1 , P2 , P3 } (b) P1 is critical three times; P2 is critical one time; P3 is critical one time. The total number of times all players are critical is 3 + 1 + 1 = 5. The Banzhaf power distribution is β1 = 3 / 5; β 2 = 1/ 5; β3 = 1/ 5. 22. (a) A player is critical in a coalition if that coalition without the player is not on the list of winning coalitions. {P1 , P2 }, {P1 , P2 , P3 }, {P1 , P2 , P4 }, {P1 , P2 , P3 , P4 } (b) β1 = 4 / 8 , β 2 = 4 / 8 , β 3 = β 4 = 0 23. (a) The winning coalitions (with critical players underlined) are {P1 , P2 } , {P1 , P3 } , {P2 , P3 } , {P1 , P2 , P3 } ,

{P1 , P2 , P4 } , {P1 , P2 , P5 } , {P1 , P2 , P6 } , {P1 , P3 , P4 } , {P1 , P3 , P5 } , {P1 , P3 , P6 } , {P2 , P3 , P4 } , {P2 , P3 , P5 } , {P2 , P3 , P6 } . (b) These winning coalitions (with critical players underlined) are {P1 , P2 , P4 } , {P1 , P3 , P4 } , {P2 , P3 , P4 } , {P1 , P2 , P3 , P4 } , {P1 , P2 , P4 , P5 } , {P1 , P2 , P4 , P6 } , {P1 , P3 , P4 , P5 } , {P1 , P3 , P4 , P6 } , {P2 , P3 , P4 , P5 } ,

{P2 , P3 , P4 , P6 } , {P1 , P2 , P3 , P4 , P5 } , {P1 , P2 , P3 , P4 , P6 } , {P1 , P2 , P4 , P5 , P6 } , {P1 , P3 , P4 , P5 , P6 } , {P2 , P3 , P4 , P5 , P6 } , {P1 , P2 , P3 , P4 , P5 , P6 } . (c) P4 is never a critical player since every time it is part of a winning coalition, that coalition is a winning coalition without P4 as well. So, β 4 = 0. (d) A similar argument to that used in part (c) shows that P5 and P6 are also dummies. One could also argue that any player with fewer votes than P4 , a dummy, will also be a dummy. So, P4 , P5 , and P6 will never be critical -- they all have zero power.

The only winning coalitions with only two players are {P1 , P2 }, {P1 , P3 }, and {P2 , P3 }; and both players are critical in each of those coalitions. All other winning coalitions consist of one of these coalitions plus additional players, and the only critical players will be the ones from the two-player coalition. So P1 , P2 , and P3 will be critical in every winning coalition they are in, and they will all be in the same number of winning coalitions, so they all have the same power. Thus, the Banzhaf power distribution is β1 = 1/ 3; β 2 = 1/ 3; β3 = 1/ 3; β 4 = 0; β 5 = 0; β 6 = 0. 24. (a) Three-player winning coalitions (with critical players underlined): {P1 , P2 , P3 }, {P1 , P2 , P4 }, {P1 , P2 , P5 },

{P1 , P3 , P4 }, {P2 , P3 , P4 }. (b) Four-player winning coalitions (with critical players underlined): {P1 , P2 , P3 , P4 }, {P1 , P2 , P3 , P5 },

{P1 , P2 , P3 , P6 }, {P1 , P2 , P4 , P5 }, {P1 , P2 , P4 , P6 }, {P1 , P2 , P5 , P6 }, {P1 , P3 , P4 , P5 }, {P1 , P3 , P4 , P6 }, {P1 , P3 , P5 , P6 }, {P2 , P3 , P4 , P5 }, {P2 , P3 , P4 , P6 }. Copyright © 2022 Pearson Education, Inc.

ISM: Excursions in Modern Mathematics, 10E

(c) Five-player winning coalitions (with critical players underlined): {P1 , P2 , P3 , P4 , P5 }, {P1 , P2 , P3 , P4 , P6 }, {P1 , P2 , P3 , P5 , P6 }, {P1 , P2 , P4 , P5 , P6 }, {P1 , P3 , P4 , P5 , P6 }, {P2 , P3 , P4 , P5 , P6 }. 15 13 11 9 3 1 (d) β1 = 52 ; β2 = 52 ; β3 = 52 ; β4 = 52 ; β5 = 52 ; β6 = 52 . Note: The grand coalition is also a

winning coalition. However, it has no critical players. 25. (a) { A, B} , { A, C} , {B, C} , { A, B, C} , { A, B, D} , { A, C , D} , {B, C , D} , { A, B, C , D} (b) A, B, and C have Banzhaf power index of 4/12 each; D is a dummy. (D is never a critical player, and the other three clearly have equal power.) 26. (a) { A, B} , { A, C} , { A, D} , { A, B, C} , { A, B, D} , { A, C , D} , {B, C , D} , { A, B, C , D} (b) A has Banzhaf power index of 6/12; B, C, and D have Banzhaf power index of 2/12 each.

2.3. Shapley-Shubik Power 27. P1 is pivotal (underlined) ten times; P2 is pivotal (again, underlined) ten times; P3 is pivotal twice ; P4 is pivotal twice. The total number of times the players are pivotal is 4! = 24 (number of underlines). The Shapley-Shubik power distribution is σ 1 = 10 / 24; σ 2 = 10 / 24; σ 3 = 2 / 24; σ 4 = 2 / 24 . 28. P1 is pivotal (underlined) ten times; P2 is pivotal six times; P3 is pivotal six times ; P4 is pivotal twice. The total number of times the players are pivotal is 4! = 24. The Shapley-Shubik power distribution is σ 1 = 10 / 24; σ 2 = 6 / 24; σ 3 = 6 / 24; σ 4 = 2 / 24 . 29. (a) There are 3! = 6 sequential coalitions of the three players. Each pivotal player is underlined. < P1 , P2 , P3 >, < P1 , P3 , P2 >, < P2 , P1 , P3 >, < P2 , P3 , P1 >, < P3 , P1 , P2 >, < P3 , P2 , P1 > (b) P1 is pivotal four times; P2 is pivotal one time; P3 is pivotal one time. The Shapley-Shubik power distribution is σ 1 = 4 / 6; σ 2 = 1/ 6; σ 3 = 1/ 6. 30. (a) There are 3! = 6 sequential coalitions of the three players. Each pivotal player is underlined. < P1 , P2 , P3 >, < P1 , P3 , P2 >, < P2 , P1 , P3 >, < P2 , P3 , P1 >, < P3 , P1 , P2 >, < P3 , P2 , P1 > (b) σ 1 = 2 / 6; σ 2 = 2 / 6; σ 3 = 2 / 6. 31. (a) Since P1 is a dictator, σ 1 = 1, σ 2 = 0, σ 3 = 0, and σ 4 = 0 . (b) There are 4! = 24 sequential coalitions of the four players. Each pivotal player is underlined. < P1 , P2 , P3 , P4 >, < P1 , P2 , P4 , P3 >, < P1 , P3 , P2 , P4 >, < P1 , P3 , P4 , P2 >,

< P1 , P4 , P2 , P3 >, < P1 , P4 , P3 , P2 >, < P2 , P1 , P3 , P4 >, < P2 , P1 , P4 , P3 >, < P2 , P3 , P1 , P4 >, < P2 , P3 , P4 , P1 >, < P2 , P4 , P1 , P3 >, < P2 , P4 , P3 , P1 >, < P3 , P1 , P2 , P4 >, < P3 , P1 , P4 , P2 >, < P3 , P2 , P1 , P4 >, < P3 , P2 , P4 , P1 >, < P3 , P4 , P1 , P2 >, < P3 , P4 , P2 , P1 >, < P4 , P1 , P2 , P3 >, < P4 , P1 , P3 , P2 >, < P4 , P2 , P1 , P3 >, < P4 , P2 , P3 , P1 >, < P4 , P3 , P1 , P2 >, < P4 , P3 , P2 , P1 > P1 is pivotal 16 times; P2 is pivotal 4 times; P3 is pivotal 4 times; P4 is pivotal 0 times. The ShapleyShubik power distribution is σ 1 = 16 / 24; σ 2 = 4 / 24; σ 3 = 4 / 24; σ 4 = 0. Copyright © 2022 Pearson Education, Inc.