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Chapter 15 Section 1 - Slide 1Copyright © 2009 Pearson Education, Inc.
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Copyright © 2009 Pearson Education, Inc. Chapter 15 Section 1 - Slide 2
Chapter 15
Voting and Apportionment
Chapter 15 Section 1 - Slide 3Copyright © 2009 Pearson Education, Inc.
WHAT YOU WILL LEARN
• Preference tables• Voting methods• Flaws of voting methods
Copyright © 2009 Pearson Education, Inc. Chapter 15 Section 1 - Slide 4
Section 1
Voting Methods
Chapter 15 Section 1 - Slide 5Copyright © 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?
Chapter 15 Section 1 - Slide 6Copyright © 2009 Pearson Education, Inc.
Example: Voting (continued)
a) How many students voted in the election?Add the row labeled Number of Votes14 + 12 + 9 + 4 + 1 = 40Therefore, 40 students voted in the election.
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ABJ
4
JBJAThirdBJAJSecondAABTFirst
191214# of Votes
Chapter 15 Section 1 - Slide 7Copyright © 2009 Pearson Education, Inc.
Example: Voting (continued)
b) How many students selected the candidates in this order: A, J, B, T?
3rd column of numbers, 9 people
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ABJ
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JBJAThirdBJAJSecondAABTFirst
191214# of Votes
Chapter 15 Section 1 - Slide 8Copyright © 2009 Pearson Education, Inc.
Example: Voting (continued)
c) How many students selected A as their first choice?
9 + 1 = 10
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JBJAThirdBJAJSecondAABTFirst
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Chapter 15 Section 1 - Slide 9Copyright © 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.
Chapter 15 Section 1 - Slide 10Copyright © 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.
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JBJAThirdBJAJSecondAABTFirst
191214# of Votes
Chapter 15 Section 1 - Slide 11Copyright © 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
Chapter 15 Section 1 - Slide 12Copyright © 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.
Chapter 15 Section 1 - Slide 13Copyright © 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.
Chapter 15 Section 1 - Slide 14Copyright © 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
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ABJ
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JBJAThirdBJAJSecondAABTFirst
191214# of Votes
Chapter 15 Section 1 - Slide 15Copyright © 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
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JBJAThirdBJAJSecondAABTFirst
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Chapter 15 Section 1 - Slide 16Copyright © 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.
Chapter 15 Section 1 - Slide 17Copyright © 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.
Chapter 15 Section 1 - Slide 18Copyright © 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
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Chapter 15 Section 1 - Slide 19Copyright © 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
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BBAASecond
AABTFirst
191214# of Votes
Chapter 15 Section 1 - Slide 20Copyright © 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
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BBBTFirst
191214# of Votes
Chapter 15 Section 1 - Slide 21Copyright © 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.
Chapter 15 Section 1 - Slide 22Copyright © 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:
c
n(n 1)
2
4(3)
26
Chapter 15 Section 1 - Slide 23Copyright © 2009 Pearson Education, Inc.
Example: Pairwise Comparison (continued) Thomas versus Jerry
T = 14 J = 12 + 9 + 4 + 1 = 26 Jerry = 1 Thomas versus Annette
T = 14 A = 12 + 9 + 4 + 1 = 26 Annette = 1 Thomas versus Betty
T = 14 B = 12 + 9 + 4 + 1 = 26 Betty = 1
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JBJAThirdBJAJSecondAABTFirst
191214# of Votes
Chapter 15 Section 1 - Slide 24Copyright © 2009 Pearson Education, Inc.
Example: Pairwise Comparison (continued) Betty versus Annette
B = 12 + 4 = 16 A = 14 + 9 + 1 = 24 Annette = 1 Betty versus Jerry
B = 12 + 1 = 13 J = 14 + 9 + 4 = 27 Jerry = 1 Annette versus Jerry
A = 12 + 9 + 1 = 22 J = 14 + 4 = 18 Annette = 1
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191214# of VotesAnnette would win
with 3 total points, the most from the pairwise comparison method.
Copyright © 2009 Pearson Education, Inc. Chapter 15 Section 1 - Slide 25
Section 2
Flaws of Voting
Chapter 15 Section 1 - Slide 26Copyright © 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
Chapter 15 Section 1 - Slide 27Copyright © 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.
Chapter 15 Section 1 - Slide 28Copyright © 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.
Chapter 15 Section 1 - Slide 29Copyright © 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.
Chapter 15 Section 1 - Slide 30Copyright © 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.
Chapter 15 Section 1 - Slide 31Copyright © 2009 Pearson Education, Inc.
Summary of the Voting Methods and Whether They Satisfy the Fairness Criteria
May not satisfy
May not satisfy
May not satisfy
May not satisfy
Irrelevant alternatives
Always satisfies
May not satisfy
Always satisfies
Always satisfies
Monotonicity
Always satisfies
May not satisfy
May not satisfy
May not satisfy
Head-to-head
Always satisfies
Always satisfies
May not satisfy
Always satisfies
Majority
Pairwise comparison
Plurality with elimination
Borda count
PluralityMethod
Criteria
Chapter 15 Section 1 - Slide 32Copyright © 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