Section 15.1 Voting Methods
What You Will Learn Plurality Method Borda Count Method Plurality with Elimination Pairwise Comparison Method Tie Breaking 15.1-2
Example 2: Voting for the Honor Society President Four students are running for president of the Honor Society: Antoine (A), Betty (B), Camille (C), and Don (D). The club members were asked to rank all candidates. The resulting preference table for this election is given in the table on the next slide. 15.1-3
Example 2: Voting for the Honor Society President 15.1-4
Example 2: Voting for the Honor Society President a) How many students voted in the election? b) How many students selected the candidates in this order: C, A, D, B? c) How many students selected A as their first choice? 15.1-5
Example 2: Voting for the Honor Society President Solution a) Add numbers in row labeled Number of Votes 19 + 15 + 11 + 7 + 2 = 54 15.1-6
Example 2: Voting for the Honor Society President Solution b) From the 2 nd column, 15 voted in the given order. 15.1-7
Example 2: Voting for the Honor Society President Solution c) Read across row that says First, find C, and read number above it. Add those: 15 + 2 = 17. 15.1-8
Plurality Method This is the most commonly used method, and it is the easiest method to use when there are more than two candidates. Each voter votes for one candidate. The candidate receiving the most votes is declared the winner. 15.1-9
Example 4: Electing the Honor Society President by the Plurality Method Consider the Honor Society election given in Example 2. Who is elected president using the plurality method? 15.1-10
Example 4: Electing the Honor Society President by the Plurality Method Solution Antoine: 7; Betty: 19; Don: 11; Camille: 15 + 2 = 17 Betty is elected president. She received 19/54 = 35% of the 1 st place votes; not a majority. 15.1-11
Borda Count Method Voters rank candidates from the most favorable to the least favorable. Each last-place vote is awarded one point, each next-to-last-place vote is awarded two points, each third-fromlast-place vote is awarded three points, and so forth. The candidate receiving the most points is the winner of the election. 15.1-12
Example 6: Electing the Honor Society President Using the Borda Count Method Use the Borda count method to determine the winner of the election for president of the Honor Society discussed in Example 2. Recall that the candidates are Antoine (A), Betty (B), Camille (C), and Don (D). For convenience, the preference table is reproduced on the next slide. 15.1-13
Example 6: Electing the Honor Society President Using the Borda Count Method 1 st place worth 4, 2 nd place worth 3, 3 rd place worth 2, 4 th place worth 1 15.1-14
Example 6: Electing the Honor Society President Using the Borda Count Method Solution Antoine: 7-1 st ; 7 4 = 28 34-2 nd ; 34 3 = 102 13-3 rd ; 13 2 = 26 Total 156 15.1-15
Example 6: Electing the Honor Society President Using the Borda Count Method Solution Betty: 19-1 st ; 19 4 = 76 35-4 th ; 35 1 = 35 Total 111 15.1-16
Example 6: Electing the Honor Society President Using the Borda Count Method Solution Camille: 17-1 st ; 17 4 = 68 11-2 nd ; 11 3 = 33 26-3 rd ; 26 2 = 52 Total 153 15.1-17
Example 6: Electing the Honor Society President Using the Borda Count Method Solution Don: 11-1 st ; 11 4 = 44 9-2 nd ; 9 3 = 27 15-3 rd ; 15 2 = 30 19-3 rd ; 19 1 = 19 Total 120 15.1-18
Example 6: Electing the Honor Society President Using the Borda Count Method Solution Antoine, with 156 points, receives the most points using the Borda count method and is declared the winner. 15.1-19
Plurality with Elimination Each voter votes for one candidate. If a candidate receives a majority of votes, that candidate is declared the winner. 15.1-20
Plurality with Elimination If no candidate receives a majority, eliminate the candidate with the fewest votes and hold another election. (If there is a tie for the fewest votes, eliminate all candidates tied for the fewest votes.) Repeat this process until a candidate receives a majority. 15.1-21
Example 8: Electing the Honor Society President Using the Plurality with Elimination Method Use the plurality with elimination method to determine the winner of the election for president of the Honor Society from Example 2. The preference table is shown on the next slide. Recall that A represents Antoine, B represents Betty, C represents Camille, and D represents Don. 15.1-22
Example 8: Electing the Honor Society President Using the Plurality with Elimination Method 15.1-23
Example 8: Electing the Honor Society President Using the Plurality with Elimination Method Solution 1 st place votes: Antoine: 7, Betty:19, Camille: 17, Don: 11 15.1-24
Example 8: Electing the Honor Society President Using the Plurality with Elimination Method Solution Total of 54 votes. No one has majority. Antoine is eliminated. We assume all voters rank their preferences the same. Column 1, 19 voters ranked the four as B, A, C, D; now it is B, C, D. Column 2 was C, A, D, B; now it is C, D, B. 15.1-25
Example 8: Electing the Honor Society President Using the Plurality with Elimination Method Solution Column 3 was D, C, A, B; now it is D, C, B. Column 4 was A, D, C, B; now it is D, C, B. Column 5 was C, D, A, B; now it is C, D, B. 15.1-26
Example 8: Electing the Honor Society President Using the Plurality with Elimination Method Solution The new preference table is: 15.1-27
Example 8: Electing the Honor Society President Using the Plurality with Elimination Method Solution 1 st place votes: Betty: 19, Camille: 17, Don: 18 15.1-28
Example 8: Electing the Honor Society President Using the Plurality with Elimination Method Solution No one has majority. Camille is eliminated. New preference table is: 15.1-29
Example 8: Electing the Honor Society President Using the Plurality with Elimination Method Solution 1 st place votes: Betty: 19, Don: 35 15.1-30
Example 8: Electing the Honor Society President Using the Plurality with Elimination Method Solution Don has a majority of first-place votes and is declared the winner using the plurality with elimination method. 15.1-31
Pairwise Comparison Method Voters rank the candidates. A series of comparisons in which each candidate is compared with each of the other candidates follows. If candidate A is preferred to candidate B, A receives one point. If candidate B is preferred to candidate A, B receives 1 point. If the candidates tie, each receives ½ point. 15.1-32
Pairwise Comparison Method After making all comparisons among the candidates, the candidate receiving the most points is declared the winner. 15.1-33
Number of Comparison The number of comparisons, c, needed when using the pairwise comparison method when there are n candidates is c n(n 1) 2 15.1-34
Example 10: Electing the Honor Society President Using the Pairwise Comparison Method Use the pairwise comparison method to determine the winner of the election for president of the Honor Society that was originally discussed in Example 2. 15.1-35
Example 10: Electing the Honor Society President Using the Pairwise Comparison Method Solution Antoine, Betty, Camille, Don 15.1-36
Example 10: Electing the Honor Society President Using the Pairwise Comparison Method Solution 4 candidates, n = 4, the number of comparisons needed is n n 1 c 4 3 6 2 2 The 6 comparisons are A versus B, A versus C, A versus D, B versus C, B versus D, and C versus D. 15.1-37
Example 10: Electing the Honor Society President Using the Pairwise Comparison Method Solution 1. The pairwise comparison of Antoine versus Betty is Antoine: 15 + 11 + 7 + 2 = 35 votes Betty: 19 votes Antoine wins this comparison and is awarded 1 point. 15.1-38
Example 10: Electing the Honor Society President Using the Pairwise Comparison Method Solution 2. The pairwise comparison of Antoine versus Camille is Antoine: 19 + 7 = 26 votes Camille: 15 + 11 + 2 = 28 votes Camille wins this comparison and is awarded 1 point. 15.1-39
Example 10: Electing the Honor Society President Using the Pairwise Comparison Method Solution 3. The pairwise comparison of Antoine versus Don is Antoine: 19 + 15 + 7 = 41 votes Don: 11 + 2 = 13 votes Antoine wins this comparison and is awarded a second point. 15.1-40
Example 10: Electing the Honor Society President Using the Pairwise Comparison Method Solution 4. The pairwise comparison of Betty versus Camille is Betty: 19 votes Camille: 15 + 11 + 7 + 2 = 35 votes Camille wins this comparison and is awarded a second point. 15.1-41
Example 10: Electing the Honor Society President Using the Pairwise Comparison Method Solution 5. The pairwise comparison of Betty versus Don is Betty: 19 votes Don: 15 + 11 + 7 + 2 = 35 votes Don wins this comparison and is awarded 1 point. 15.1-42
Example 10: Electing the Honor Society President Using the Pairwise Comparison Method Solution 6. The pairwise comparison of Camille versus Don is Camille: 19 + 15 + 2 = 36 votes Don: 11 + 7 = 18 votes Camille wins this comparison and is awarded a third point. 15.1-43
Example 10: Electing the Honor Society President Using the Pairwise Comparison Method Solution Antoine received 2 points, Betty received 0 points, Camille received 3 points, and Don received 1 point. Since Camille received 3 points, the most points from the pairwise comparison method, Camille wins the election. 15.1-44
Tie Breaking Breaking a tie can be achieved by either making an arbitrary choice, such as flipping a coin, or by bringing in an additional voter. Robert s Rule of Order: president of group votes only to break a tie or create a tie. Borda method: could choose person with most 1 st place votes. 15.1-45
Tie Breaking Pairwise comparison method: could choose the winner of a one-to-one comparison between the two candidates involved in the tie. Different tie-breaking methods could produce different winners. To remain fair, the method should be chosen in advance. 15.1-46
Section 15.2 Flaws of Voting
What You Will Learn Fairness Criteria Majority Criterion Head-to-Head Criterion Monotonicity Criterion Irrelevant Alternative Criterion 15.2-48
Fairness Criteria Mathematicians and political scientists have agreed that a voting method should meet the following four criteria in order for the voting method to be considered fair. Majority Criterion Head-to-head Criterion Monotonicity Criterion Irrelevant Alternatives Criterion 15.2-49
Majority Criterion If a candidate receives a majority (more than 50%) of the first-place votes, that candidate should be declared the winner. 15.2-50
Head-to-Head Criterion If a candidate is favored when compared head-to-head with every other candidate, that candidate should be declared the winner. 15.2-51
Monotonicity Criterion A candidate who wins a first election and then gains additional support without losing any of the original support should also win a second election. 15.2-52
Irrelevant Alternatives Criterion If a candidate is declared the winner of an election and in a second election one or more of the other candidates is removed, the previous winner should still be declared the winner. 15.2-53
Summary of the Voting Methods and Whether They Satisfy the Fairness Criteria Criteria Method Plurality Borda count Plurality with elimination Pairwise comparison Majority Always satisfies May not satisfy Always satisfies Always satisfies Head-tohead May not satisfy May not satisfy May not satisfy Always satisfies Monotonicity Always satisfies May not satisfy May not satisfy May not satisfy Irrelevant alternatives May not satisfy May not satisfy May not satisfy May not satisfy 15.2-54
Arrow s Impossibility Theorem It is mathematically impossible for any democratic voting method to simultaneously satisfy each of the fairness criteria: The majority criterion The head-to-head criterion The monotonicity criterion The irrevelant alternative criterion 15.2-55