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Chapter 15 Section 1 - Slide 1 Copyright © 2009 Pearson Education, Inc. AND.

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Presentation on theme: "Chapter 15 Section 1 - Slide 1 Copyright © 2009 Pearson Education, Inc. AND."— Presentation transcript:

1 Chapter 15 Section 1 - Slide 1 Copyright © 2009 Pearson Education, Inc. AND

2 Copyright © 2009 Pearson Education, Inc. Chapter 15 Section 1 - Slide 2 Chapter 15 Voting and Apportionment

3 Chapter 15 Section 1 - Slide 3 Copyright © 2009 Pearson Education, Inc. WHAT YOU WILL LEARN Preference tables Voting methods Flaws of voting methods

4 Copyright © 2009 Pearson Education, Inc. Chapter 15 Section 1 - Slide 4 Section 1 Voting Methods

5 Chapter 15 Section 1 - Slide 5 Copyright © 2009 Pearson Education, Inc. Example: Voting Voting for Math Club President: Four students are running for president of the Math Club: Jerry, Thomas, Annette and Becky. The club members were asked to rank all candidates. The resulting preference table for this election is shown on the next slide. a) How many students voted in the election? b) How many students selected the candidates in this order: A, J, B, T? c) How many students selected A as their first choice?

6 Chapter 15 Section 1 - Slide 6 Copyright © 2009 Pearson Education, Inc. Example: Voting (continued) a) How many students voted in the election? Add the row labeled Number of Votes 14 + 12 + 9 + 4 + 1 = 40 Therefore, 40 students voted in the election. TTTTBFourth A B J 4 JBJAThird BJAJSecond AABTFirst 191214 # of Votes

7 Chapter 15 Section 1 - Slide 7 Copyright © 2009 Pearson Education, Inc. Example: Voting (continued) b) How many students selected the candidates in this order: A, J, B, T? 3 rd column of numbers, 9 people TTTTBFourth A B J 4 JBJAThird BJAJSecond AABTFirst 191214 # of Votes

8 Chapter 15 Section 1 - Slide 8 Copyright © 2009 Pearson Education, Inc. Example: Voting (continued) c) How many students selected A as their first choice? 9 + 1 = 10 TTTTBFourth A B J 4 JBJAThird BJAJSecond AABTFirst 191214 # of Votes

9 Chapter 15 Section 1 - Slide 9 Copyright © 2009 Pearson Education, Inc. 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.

10 Chapter 15 Section 1 - Slide 10 Copyright © 2009 Pearson Education, Inc. Example: Plurality Method Who is elected math club president using the plurality method? We will assume that each member would vote for the person he or she listed in first place. TTTTBFourth A B J 4 JBJAThird BJAJSecond AABTFirst 191214 # of Votes

11 Chapter 15 Section 1 - Slide 11 Copyright © 2009 Pearson Education, Inc. Example: Plurality Method (continued) Thomas would be elected since he received the most votes. Note that Thomas received 14/40, or 35%, of the first-place votes, which is less than a majority. Thomas received 14 votes Becky received 12 votes Annette received 10 votes Jerry received 4 votes

12 Chapter 15 Section 1 - Slide 12 Copyright © 2009 Pearson Education, Inc. 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-from-last- place vote is awarded three points, and so forth. The candidate receiving the most points is the winner of the election.

13 Chapter 15 Section 1 - Slide 13 Copyright © 2009 Pearson Education, Inc. Example: Borda Count Use the Borda count method to determine the winner of the election for math club president. Since there are four candidates, a first-place vote is worth 4 points, a second-place vote is worth 3 points, a third-place vote is worth 2 points, and a fourth-place vote is worth 1 point.

14 Chapter 15 Section 1 - Slide 14 Copyright © 2009 Pearson Education, Inc. Example: Borda Count (continued) Thomas 14 first place votes 0 second place 0 third place 26 fourth place 14(4) + 0 + 0 + 26(1) = 82 Annette 10 first place votes 12 second place 18 third place 0 fourth place 10(4) + 12(3) + 18(2) + 0 = 112 TTTTBFourth A B J 4 JBJAThird BJAJSecond AABTFirst 191214 # of Votes

15 Chapter 15 Section 1 - Slide 15 Copyright © 2009 Pearson Education, Inc. Example: Borda Count (continued) Betty 12 first place votes 5 second place 9 third place 14 fourth place 12(4) + 5(3) + 9(2) + 14 = 95 Jerry 4 first place votes 23 second place 13 third place 0 fourth place 4(4) + 23(3) + 13(2) + 0 = 111 TTTTBFourth A B J 4 JBJAThird BJAJSecond AABTFirst 191214 # of Votes

16 Chapter 15 Section 1 - Slide 16 Copyright © 2009 Pearson Education, Inc. Example: Borda Count (continued) Thomas - 82 Annette - 112 Betty - 95 Jerry - 111 Annette, with 112 points, receives the most points and is declared the winner.

17 Chapter 15 Section 1 - Slide 17 Copyright © 2009 Pearson Education, Inc. Plurality with Elimination Each voter votes for one candidate. If a candidate receives a majority of votes, that candidate is declared the winner. 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.

18 Chapter 15 Section 1 - Slide 18 Copyright © 2009 Pearson Education, Inc. Example: Plurality with Elimination Use the plurality with elimination method to determine the winner of the election for president of the math club. Count the number of first place votes Annette 10 Betty 12 Thomas 14 Jerry 4 TTTTBFourth A B J 4 JBJAThird BJAJSecond AABTFirst 191214 # of Votes

19 Chapter 15 Section 1 - Slide 19 Copyright © 2009 Pearson Education, Inc. Example: Plurality with Elimination (continued) Since 40 votes were cast, a candidate must have 20 first place votes to receive a majority. Jerry had the fewest number of first place votes, so he is eliminated. Redo the table. Thomas 14 Annette 10 Betty 16 T A B 4 TTTBThird BBAASecond AABTFirst 191214# of Votes

20 Chapter 15 Section 1 - Slide 20 Copyright © 2009 Pearson Education, Inc. Example: Plurality with Elimination (continued) Still, no candidate received a majority. Annette has the fewest number of first-place votes, so she is eliminated. New preference table Betty 26 Thomas 14 Betty is the winner. T B 4 TTTBSecond BBBTFirst 191214# of Votes

21 Chapter 15 Section 1 - Slide 21 Copyright © 2009 Pearson Education, Inc. 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. After making all comparisons among the candidates, the candidate receiving the most points is declared the winner.

22 Chapter 15 Section 1 - Slide 22 Copyright © 2009 Pearson Education, Inc. Example: Pairwise Comparison Use the pairwise comparison method to determine the winner of the election for math club president. Number of comparisons needed:

23 Chapter 15 Section 1 - Slide 23 Copyright © 2009 Pearson Education, Inc. Example: Pairwise Comparison (continued) Thomas versus Jerry  T = 14J = 12 + 9 + 4 + 1 = 26 Jerry = 1 Thomas versus Annette  T = 14A = 12 + 9 + 4 + 1 = 26Annette = 1 Thomas versus Betty  T = 14B = 12 + 9 + 4 + 1 = 26Betty = 1 TTTTBFourth A B J 4 JBJAThird BJAJSecond AABTFirst 191214 # of Votes

24 Chapter 15 Section 1 - Slide 24 Copyright © 2009 Pearson Education, Inc. Example: Pairwise Comparison (continued) Betty versus Annette  B = 12 + 4 = 16A = 14 + 9 + 1 = 24Annette = 1 Betty versus Jerry  B = 12 + 1 = 13J = 14 + 9 + 4 = 27Jerry = 1 Annette versus Jerry  A = 12 + 9 + 1 = 22J = 14 + 4 = 18Annette = 1 TTTTBFourth A B J 4 JBJAThird BJAJSecond AABTFirst 191214 # of Votes Annette would win with 3 total points, the most from the pairwise comparison method.

25 Copyright © 2009 Pearson Education, Inc. Chapter 15 Section 1 - Slide 25 Section 2 Flaws of Voting

26 Chapter 15 Section 1 - Slide 26 Copyright © 2009 Pearson Education, Inc. 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

27 Chapter 15 Section 1 - Slide 27 Copyright © 2009 Pearson Education, Inc. Majority Criterion If a candidate receives a majority (more than 50%) of the first-place votes, that candidate should be declared the winner.

28 Chapter 15 Section 1 - Slide 28 Copyright © 2009 Pearson Education, Inc. 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.

29 Chapter 15 Section 1 - Slide 29 Copyright © 2009 Pearson Education, Inc. 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.

30 Chapter 15 Section 1 - Slide 30 Copyright © 2009 Pearson Education, Inc. 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.

31 Chapter 15 Section 1 - Slide 31 Copyright © 2009 Pearson Education, Inc. Summary of the Voting Methods and Whether They Satisfy the Fairness Criteria May not satisfy Irrelevant alternatives Always satisfies May not satisfy Always satisfies Monotonicity Always satisfies May not satisfy Head-to- head Always satisfies May not satisfy Always satisfies Majority Pairwise comparison Plurality with elimination Borda count PluralityMethod Criteria

32 Chapter 15 Section 1 - Slide 32 Copyright © 2009 Pearson Education, Inc. 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


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