Slide 15-1 Copyright © 2005 Pearson Education, Inc.
SEVENTH EDITION and EXPANDED SEVENTH EDITION
Copyright © 2005 Pearson Education, Inc.
Chapter 15
Voting and Apportionment
Copyright © 2005 Pearson Education, Inc.
15.1
Voting Methods
Slide 15-4 Copyright © 2005 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?
Slide 15-5 Copyright © 2005 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.
TTTTBFourth
A
B
J
4
JBJAThird
BJAJSecond
AABTFirst
191214Number of Votes
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Example: Voting continued
b) How many students selected the candidates in this order: A, J, B, T?
3rd column of numbers, 9 people
c) How many students selected A as their first choice?
9 + 1 = 10
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Plurality Method
This is the most commonly used and 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.
Slide 15-8 Copyright © 2005 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
191214Number of Votes
From the table:Thomas received 14 votes Becky received 12 votes
Annette received 9 + 1 = 10 votes Jerry received 4 votes
Slide 15-9 Copyright © 2005 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.
Slide 15-10 Copyright © 2005 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.
Slide 15-11 Copyright © 2005 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.
Slide 15-12 Copyright © 2005 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
191214Number of Votes
Slide 15-13 Copyright © 2005 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
191214Number of Votes
Slide 15-14 Copyright © 2005 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.
Slide 15-15 Copyright © 2005 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.
Slide 15-16 Copyright © 2005 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
191214Number of Votes
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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
191214Number of Votes
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Example: Plurality with Elimination continued Still, no candidate received a majority. Annette
has the fewest number of first-place vote so she is eliminated.
New preference table Betty 26 Thomas 14 Betty is the winner. T
B
4
TTTBSecond
BBBTFirst
191214Number of Votes
Slide 15-19 Copyright © 2005 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 received 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.
Slide 15-20 Copyright © 2005 Pearson Education, Inc.
Example: Pairwise Comparison
Use the pairwise comparison method to determine the winner of the election for math club president.
Comparisons needed:
( 1) 4(3)6
2 2
n nc
Slide 15-21 Copyright © 2005 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
TTTTBFourth
A
B
J
4
JBJAThird
BJAJSecond
AABTFirst
191214Number of Votes
Slide 15-22 Copyright © 2005 Pearson Education, Inc.
Example: Pairwise Comparison continued 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 Annette would win the
election since she received 3 points, the most points from the pairwise comparison method.
TTTTBFourth
A
B
J
4
JBJAThird
BJAJSecond
AABTFirst
191214Number of Votes
Copyright © 2005 Pearson Education, Inc.
15.2
Flaws of Voting
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Majority Criterion
If a candidate receives a majority (more than 50%), of the first-place votes, that candidate should be declared the winner.
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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.
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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.
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Irrelevant Alternative 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.
Slide 15-28 Copyright © 2005 Pearson Education, Inc.
Summary of the Voting Methods and Whether They Satisfy the Fairness Criteria
May not satisfyMay 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 countPlurality
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15.3
Apportionment Methods
Slide 15-30 Copyright © 2005 Pearson Education, Inc.
Apportionment
The goal of apportionment is to determine a method to allocate the total number of items to be apportioned in a fair manner.
Four Methods Hamilton’s method Jefferson’s method Webster’s method Adam’s method
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Definitions
total populationStandard divisor =
number of items to be allocated
population for the particular groupStandard quota =
standard divisor
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Example
A Graduate school wishes to apportion 15 graduate assistantships among the colleges of education, business and chemistry based on their undergraduate enrollments. Find the standard quotas for the schools.
Slide 15-33 Copyright © 2005 Pearson Education, Inc.
Example continued
8020divisor = 534.67
15
14.999Standard quota
8020188029403200Population
TotalChemistryBusinessEducation
3200
534.675.985
2940
534.675.498
1880
534.673.516
Slide 15-34 Copyright © 2005 Pearson Education, Inc.
Hamilton’s Method
1. Calculate each group’s standard quota. 2. Round each standard quota down to the
nearest integer (the lower quota). Initially, each group receives its lower quota.
3. Distribute any leftover items to the groups with the largest fractional parts until all items are distributed.
Slide 15-35 Copyright © 2005 Pearson Education, Inc.
Example: Apportion the 15 graduate assistantships
15456Hamilton’s
13355Lower quota
14.999Standard quota
8020188029403200Population
TotalChemistryBusinessEducation
3200
534.675.985
2940
534.675.498
1880
534.673.516
Slide 15-36 Copyright © 2005 Pearson Education, Inc.
The Quota Rule
An apportionment for every group under consideration should always be either the upper quota or the lower quota.
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Jefferson’s Method
1. Determine a modified divisor, d, such that when each group’s modified quota is rounded down to the nearest integer, the total of the integers is the exact number of items to be apportioned. We will refer to the modified quotas that are rounded
down as modified lower quotas. 2. Apportion to each group its modified lower quota.
Slide 15-38 Copyright © 2005 Pearson Education, Inc.
Modified divisor = 480
15366Jefferson
3.91676.1256.67Modified quota
14.999Standard quota
8020188029403200Population
TotalChemistryBusinessEducation
3200
534.675.985
2940
534.675.498
1880
534.673.516
Slide 15-39 Copyright © 2005 Pearson Education, Inc.
Webster’s Method
1. Determine a modified divisor, d, such that when each group’s modified quota is rounded to the nearest integer, the total of the integers is the
exact number of items to be apportioned. We will refer to the modified quotas that are rounded down as
modified rounded quotas. 2. Apportion to each group its modified lower quota.
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Adams’s Method
1. Determine a modified divisor, d, such that when each group’s modified quota is rounded up to the nearest integer, the total of the integers is the
exact number ot items to be apportioned. We will refer to the modified quotas that are rounded down as
modified upper quotas. 2. Apportion to each group its modified lower quota.
Copyright © 2005 Pearson Education, Inc.
15.4
Flaws of the Apportionment Methods
Slide 15-42 Copyright © 2005 Pearson Education, Inc.
Three Flaws of Hamilton’s Method
The three flaws of Hamilton’s method are: the Alabama paradox, the population paradox, and the new-state paradox. These flaws apply only to Hamilton’s method and do not
apply to Jefferson’s method, Webster’s method, or Adam’s method.
In 1980 the Balinski and Young’s Impossibility Theorem stated that there is no perfect apportionment method that satisfies the quota rule and avoids any paradoxes.
Slide 15-43 Copyright © 2005 Pearson Education, Inc.
Alabama Paradox
The Alabama paradox occurs when an increase in the total number of items to be apportioned results in a loss of an item for the group.
Slide 15-44 Copyright © 2005 Pearson Education, Inc.
Example: Demonstrating the Alabama Paradox A large company with braches in three cities, must distribute
30 cell phones to the three offices. The cell phones will be apportioned based on the number of employees in each office shown in the table below.
900489250161Employees
Total321Office
Slide 15-45 Copyright © 2005 Pearson Education, Inc.
Example: Demonstrating the Alabama Paradox continued Apportion the cell phones using Hamilton’s
methods. Does the Alabama paradox occur using
Hamilton’s method if the number of new cell phones increased from 30 to 31. Explain.
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Example: Demonstrating the Alabama Paradox continued Based on 30 cell phones, the table is as follows:
900489250161Employees
291685Lower Quota
301686Hamilton’s apportionment
16.38.335.37Standard Quota
Total321Office
Slide 15-47 Copyright © 2005 Pearson Education, Inc.
Example: Demonstrating the Alabama Paradox continued Based on 31 cell phones, the table is as follows:
900489250161Employees
291685Lower Quota
311795Hamilton’s apportionment
16.848.615.55Standard Quota
Total321Office
Slide 15-48 Copyright © 2005 Pearson Education, Inc.
Example: Demonstrating the Alabama Paradox continued When the number of cell phones increased from
30 to 31, office one actually lost a cell phone, while the other two offices actually gained a cell phone under Hamilton’s apportionment.
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Population Paradox
The Population Paradox occurs when group A loses items to group B, although group A’s population grew at a faster rate than group B’s.
Slide 15-50 Copyright © 2005 Pearson Education, Inc.
Example: Demonstrating Population Paradox A school district with five elementary schools has funds for 54 scholarships. The student
population for each school is shown in the table below.
5400106311339331538733Population in 2003
5450111111339331540733Population in 2005
D E TotalCBASchool
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Example: Demonstrating Population Paradox continued Apportion the scholarships using Hamilton’s
method. If the school wishes to give the same number
of scholarships two years later, does a population paradox occur?
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Solution
Based on the population in 2003, the table is as follows:
16
15
15.38
1538
B
9
9
9.33
933
C
11
11
11.33
1133
D
54
52
5400
Total
11
10
10.63
1063
E
733Population in 2003
7Lower Quota
7Hamilton’s apportionment
7.33Standard Quota
ASchool
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Solution continued
Based on the population in 2005, the table is as follows:
15
15
15.258
1540
B
9
9
9.24
933
C
11
11
11.23
1133
D
54
53
5450
Total
11
11
11.01
1111
E
733Population in 2005
7Lower Quota
8Hamilton’s apportionment
7.262Standard Quota
ASchool
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Solution continued
In the school district in 2005, school B actually gives one of its scholarships to school A, even though the population in school B actually grew by 2 students.
Slide 15-55 Copyright © 2005 Pearson Education, Inc.
New State Paradox
The new-states paradox occurs when the addition of a new group changes the apportionment of another group.
Slide 15-56 Copyright © 2005 Pearson Education, Inc.
Summary
Small states
Small states
Large statesLarge statesAppointment method favors
NoNoNoYesMay produce the new-states paradox
NoNoNoYesMay produce the population paradox
NoNoNoYesMay produce the Alabama paradox
YesYesYesNoMay violate the quota rule
WebsterAdamsJeffersonHamilton
Apportionment Method
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