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 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 Votes = 40 Therefore, 40 students voted in the election. TTTTBFourth A B J 4 JBJAThird BJAJSecond AABTFirst Number of Votes

Slide 15-6 Copyright © 2005 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 c) How many students selected A as their first choice? = 10

Slide 15-7 Copyright © 2005 Pearson Education, Inc. 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 Number of Votes From the table: Thomas received 14 votes Becky received 12 votes Annette received = 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 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 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 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) (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 Number of Votes

Slide 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 Number of Votes

Slide 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 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 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 Number of Votes

Slide Copyright © 2005 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 Number of Votes

Slide Copyright © 2005 Pearson Education, Inc. 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 Number of Votes

Slide 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 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:

Slide Copyright © 2005 Pearson Education, Inc. Example: Pairwise Comparison continued Thomas versus Jerry  T = 14J = = 26  Jerry = 1 Thomas versus Annette  T = 14A = = 26  Annette = 1 Thomas versus Betty  T = 14B = = 26  Betty = 1 TTTTBFourth A B J 4 JBJAThird BJAJSecond AABTFirst Number of Votes

Slide Copyright © 2005 Pearson Education, Inc. Example: Pairwise Comparison continued continued Betty versus Annette  B = = 16A = = 24  Annette = 1 Betty versus Jerry  B = = 13J = = 27  Jerry = 1 Annette versus Jerry  A = = 22J = = 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 Number of Votes

Copyright © 2005 Pearson Education, Inc Flaws of Voting

Slide Copyright © 2005 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.

Slide Copyright © 2005 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.

Slide Copyright © 2005 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.

Slide Copyright © 2005 Pearson Education, Inc. 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 Copyright © 2005 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 countPlurality

Copyright © 2005 Pearson Education, Inc Apportionment Methods

Slide 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

Slide Copyright © 2005 Pearson Education, Inc. Definitions

Slide Copyright © 2005 Pearson Education, Inc. 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 Copyright © 2005 Pearson Education, Inc. Example continued Standard quota Population TotalChemistryBusinessEducation

Slide 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 Copyright © 2005 Pearson Education, Inc. Example: Apportion the 15 graduate assistantships 15456Hamilton’s 13355Lower quota Standard quota Population TotalChemistryBusinessEducation

Slide 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.

Slide Copyright © 2005 Pearson Education, Inc. 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 Copyright © 2005 Pearson Education, Inc. Modified divisor = Jefferson Modified quota Standard quota Population TotalChemistryBusinessEducation

Slide 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.

Slide Copyright © 2005 Pearson Education, Inc. 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 Flaws of the Apportionment Methods

Slide 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 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 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 Employees Total321Office

Slide 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.

Slide Copyright © 2005 Pearson Education, Inc. Example: Demonstrating the Alabama Paradox continued Based on 30 cell phones, the table is as follows: Employees Lower Quota Hamilton’s apportionment Standard Quota Total321Office

Slide Copyright © 2005 Pearson Education, Inc. Example: Demonstrating the Alabama Paradox continued Based on 31 cell phones, the table is as follows: Employees Lower Quota Hamilton’s apportionment Standard Quota Total321Office

Slide 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.

Slide Copyright © 2005 Pearson Education, Inc. 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 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 Population in Population in 2005 DETotalCBASchool

Slide Copyright © 2005 Pearson Education, Inc. 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?

Slide Copyright © 2005 Pearson Education, Inc. Solution Based on the population in 2003, the table is as follows: B C D Total E 733Population in Lower Quota 7Hamilton’s apportionment 7.33Standard Quota ASchool

Slide Copyright © 2005 Pearson Education, Inc. Solution continued Based on the population in 2005, the table is as follows: B C D Total E 733Population in Lower Quota 8Hamilton’s apportionment 7.262Standard Quota ASchool

Slide Copyright © 2005 Pearson Education, Inc. 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 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 Copyright © 2005 Pearson Education, Inc. Summary Small states Large states Appointment method favors No YesMay produce the new- states paradox No YesMay produce the population paradox No YesMay produce the Alabama paradox Yes NoMay violate the quota rule WebsterAdamsJeffersonHamilton Apportionment Method