Copyright 2013, 2010, 2007, Pearson, Education, Inc. Section 15.1 Voting Methods

Copyright 2013, 2010, 2007, Pearson, Education, Inc.

Section 15.1

Voting Methods


What You Will Learn

Plurality MethodBorda Count MethodPlurality with EliminationPairwise Comparison MethodTie Breaking

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Example 2: Voting for the Honor Society PresidentFour 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.

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Example 2: Voting for the Honor Society President

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Example 2: Voting for the Honor Society Presidenta) 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?

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Example 2: Voting for the Honor Society PresidentSolutiona) Add numbers in row labeled

Number of Votes19 + 15 + 11 + 7 + 2 = 54

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Example 2: Voting for the Honor Society PresidentSolutionb) From the 2nd column, 15 voted

in the given order.

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Example 2: Voting for the Honor Society PresidentSolutionc) Read across row that says First,

find C, and read number above it. Add those: 15 + 2 = 17.

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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.

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Example 4: Electing the Honor Society President by the Plurality MethodConsider the Honor Society election given in Example 2. Who is elected president using the plurality method?

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Example 4: Electing the Honor Society President by the Plurality MethodSolutionAntoine: 7; Betty: 19; Don: 11;Camille: 15 + 2 = 17Betty is elected president.She received 19/54 = 35% of the 1st place votes; not a majority.

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Borda Count MethodVoters 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-from-last-place vote is awarded three points, and so forth.The candidate receiving the most points is the winner of the election.

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Example 6: Electing the Honor Society President Using the Borda Count MethodUse 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.

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Example 6: Electing the Honor Society President Using the Borda Count Method1st place worth 4, 2nd place worth 3, 3rd place worth 2, 4th place worth 1

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Example 6: Electing the Honor Society President Using the Borda Count MethodSolutionAntoine: 7 - 1st; 7 × 4 = 2834 - 2nd; 34 × 3 = 10213 - 3rd;13 × 2 = 26Total 156

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Example 6: Electing the Honor Society President Using the Borda Count MethodSolutionBetty: 19 - 1st; 19 × 4 = 7635 - 4th; 35 × 1 = 35Total 111

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Example 6: Electing the Honor Society President Using the Borda Count MethodSolutionCamille: 17 - 1st; 17 × 4 = 6811 - 2nd; 11 × 3 = 3326 - 3rd;26 × 2 = 52Total 153

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Example 6: Electing the Honor Society President Using the Borda Count MethodSolutionDon: 11 - 1st; 11 × 4 = 449 - 2nd; 9 × 3 = 2715 - 3rd;15 × 2 = 3019 - 3rd;19 × 1 = 19Total 120

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Example 6: Electing the Honor Society President Using the Borda Count MethodSolutionAntoine, with 156 points,

receives the most points using the Borda count method and is declared the winner.

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Plurality with Elimination

Each voter votes for one candidate.

If a candidate receives a majority of votes, that candidate is declared the winner.

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Plurality with EliminationIf 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.

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Example 8: Electing the Honor Society President Using the Plurality with Elimination MethodUse 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.

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Example 8: Electing the Honor Society President Using the Plurality with Elimination Method

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Example 8: Electing the Honor Society President Using the Plurality with Elimination MethodSolution1st place votes: Antoine: 7, Betty:19, Camille: 17, Don: 11

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Example 8: Electing the Honor Society President Using the Plurality with Elimination MethodSolutionTotal 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.

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Example 8: Electing the Honor Society President Using the Plurality with Elimination MethodSolutionColumn 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.

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Example 8: Electing the Honor Society President Using the Plurality with Elimination MethodSolutionThe new preference table is:

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Example 8: Electing the Honor Society President Using the Plurality with Elimination MethodSolution1st place votes:Betty: 19, Camille: 17, Don: 18

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Example 8: Electing the Honor Society President Using the Plurality with Elimination MethodSolutionNo one has majority. Camille is eliminated. New preference table is:

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Example 8: Electing the Honor Society President Using the Plurality with Elimination MethodSolution1st place votes:Betty: 19, Don: 35

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Example 8: Electing the Honor Society President Using the Plurality with Elimination MethodSolution

Don has a majority of first-place votes and is declared the winner using the plurality with elimination method.

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Pairwise Comparison MethodVoters 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.

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Pairwise Comparison MethodAfter making all comparisons among the candidates, the candidate receiving the most points is declared the winner.

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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

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Example 10: Electing the Honor Society President Using the Pairwise Comparison MethodUse the pairwise comparison method to determine the winner of the election for president of the Honor Society that was originally discussed in Example 2.

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Example 10: Electing the Honor Society President Using the Pairwise Comparison MethodSolutionAntoine, Betty, Camille, Don

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Example 10: Electing the Honor Society President Using the Pairwise Comparison MethodSolution4 candidates, n = 4, the number of comparisons needed is

The 6 comparisons are A versus B,A versus C, A versus D, B versus C,B versus D, and C versus D.

c

n n 1 2

4 3

26

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Example 10: Electing the Honor Society President Using the Pairwise Comparison MethodSolution1. The pairwise comparison of Antoine versus Betty isAntoine: 15 + 11 + 7 + 2 = 35 votesBetty: 19 votesAntoine wins this comparison and is awarded 1 point.

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Example 10: Electing the Honor Society President Using the Pairwise Comparison MethodSolution2. The pairwise comparison of Antoine versus Camille isAntoine: 19 + 7 = 26 votesCamille: 15 + 11 + 2 = 28 votesCamille wins this comparison and is awarded 1 point.

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Example 10: Electing the Honor Society President Using the Pairwise Comparison MethodSolution3. The pairwise comparison of Antoine versus Don isAntoine: 19 + 15 + 7 = 41 votesDon: 11 + 2 = 13 votesAntoine wins this comparison and is awarded a second point.

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Example 10: Electing the Honor Society President Using the Pairwise Comparison MethodSolution4. The pairwise comparison of Betty versus Camille isBetty: 19 votesCamille: 15 + 11 + 7 + 2 = 35 votesCamille wins this comparison and is awarded a second point.

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Example 10: Electing the Honor Society President Using the Pairwise Comparison MethodSolution5. The pairwise comparison of Betty versus Don isBetty: 19 votesDon: 15 + 11 + 7 + 2 = 35 votesDon wins this comparison and is awarded 1 point.

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Example 10: Electing the Honor Society President Using the Pairwise Comparison MethodSolution6. The pairwise comparison of Camille versus Don isCamille: 19 + 15 + 2 = 36 votesDon: 11 + 7 = 18 votesCamille wins this comparison and is awarded a third point.

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Example 10: Electing the Honor Society President Using the Pairwise Comparison MethodSolutionAntoine 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.

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Tie BreakingBreaking 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 1st place votes.

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Tie BreakingPairwise 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.

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Documents

Copyright 2013, 2010, 2007, Pearson, Education, Inc. Section 15.1 Voting Methods