Voting and Apportionment 14.3 Apportionment Methods 14.4 Apportionment One of the Most Precious Rights in Our Democracy Is the Right to Vote

Home , Apportionment (politics), Majority criterion

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CHAPTER 14 The concept of apportionment or fair division plays a vital role in the operation of corporations, politics, and educational institutions. For example, colleges and universities deal with large issues of apportionment such as the allocation of funds. In Section 14.3, you will encounter many different types of apportionment problems. David Butow/Corbis SABA 14.1 Voting Systems* 14.2 Voting Objections Voting and Apportionment 14.3 Apportionment Methods 14.4 Apportionment One of the most precious rights in our democracy is the right to vote. Objections We have elections to select the president of the United States, senators and representatives, members of the United Nations General Assembly, baseball players to be inducted into the Baseball Hall of Fame, and even “best” performers to receive Oscar and Grammy awards. There are many *Portions of this section were developed by ways of making the final decision in these elections, some simple, some Professor William Webb of Washington more complex. State University and funded by a National Electing senators and governors is simple: Have some primary elec- Science Foundation grant (DUE-9950436) tions and then a final election. The candidate with the most votes in the awarded to Professor V. S. Manoranjan. final election wins. Elections for president, as attested by the controver- sial 2000 presidential election, are complicated by our Electoral College system. Under this system, each state is allocated a number of electors selected by their political parties and equal to the number of its U.S. senators (always two), plus the number of its U.S. representatives (which may change each decade according to the size of each state’s population as determined in the census). These state electors cast their electoral votes (one for president and one for vice president) and send them to the president of the Senate who, on the following January 6, opens and reads them before both houses of Congress. The candidate for president with the most electoral votes, provided that it is an absolute majority (one over half the total), is declared president. Similarly, the vice presidential candidate with the absolute majority of electoral votes is declared vice president. At noon on January 20, the duly elected president and vice president are finally sworn into office. In this chapter we will look at several voting methods, the “fairness” Online Study Center of these methods, how votes are apportioned or divided among voters or For links to Internet sites related states, and the fairness of these apportionments. to Chapter 14, please access college.hmco.com/PIC/bello9e and click on the Online Study Center icon. V1 304470-ch14_pV1-V59 11/7/06 10:01 AM Page 2

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14.1 Voting Systems

Monsieur Butterfly and Pizza Too G S T A R N T I E When you vote in a presidential election, you are not directly voting for the pres- T D T ident! You are actually voting for electors, individuals who cast the electoral E votes on behalf of their party and states. They are the ones who elect the presi- G dent. Originally, electors were free to cast their votes as they pleased, but many of today’s electors are “bound” or “committed” by state law (25 states have such laws) to vote for the candidate who received the most popular votes in their state. HUMAN SIDE OF MATH In a typical U.S. election, voters vote for their first choices by using a ballot. A Marie-Jean so-called butterfly ballot used in Palm Beach County, Florida, during the 2000 Antoine presidential election is shown. There was some confusion about votes cast for Nicolas de Pat Buchanan (second hole) or Al Gore (third hole). Caritat, Marquis de Condorcet, was born (1743–1794) The Granger Collection September 17, 1743, in Ribemont, France. Condorcet distinguished himself as a writer, administrator, and politician. His most important work was the Essay on the Application of Analysis to the Probability of Majority Decisions (1785), in which he tried to combine mathematics and philosophy to apply to Robert Duyos/South Florida Sun-Sentinel social phenomena. One of A butterfly ballot used in Palm Beach County, Florida, during the 2000 presidential election. the major developments in this work is known as the About 460,000 votes were cast in Palm Beach County, and of those, 3400 Condorcet paradox, a topic were for Buchanan. Assuming that the remaining precincts in Florida would covered in this chapter. yield the same proportion of votes for Buchanan, how many of the approxi- mately 6 million votes cast in Florida would you project for Buchanan? Think Looking Ahead about it before you answer! In this chapter we shall study The proportion of votes for Buchanan in Palm Beach was different voting systems, the “flaws” or objections that can be 3400 34 raised about such systems, the 460,000 4600 methods used to fairly apportion resources among different If the same proportion applies to Florida, groups, and the objections to these apportionment methods. F 34 6,000,000 4600 or equivalently 4600F 34 6,000,000 F 44,348 304470-ch14_pV1-V59 11/7/06 10:01 AM Page 3

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Thus, you would expect about 44,347 Florida votes for Buchanan. (He actually got about 17,000 votes in Florida.) Moreover, the number of registered voters for Buchanan’s Reform Party in Palm Beach County was a mere 304 voters! What might be some of the reasons for this discrepancy?

There are two fundamentally different types of voting methods: preferential and nonpreferential. As the name suggests, a preferential voting system asks a voter to state a preference by ranking alternatives. This is usually done using a preference table or preference schedule. For example, suppose the Math Club wants to order some pizzas for the end-of-year party. The Pizza House offers a special: three different one-topping pizzas—one jumbo, one large, and one medium—for only $20. The question is, Which topping to order on which pizza? The club members decide that the most popular topping should go on the jumbo pizza, the second-choice topping on the large pizza, and the third choice on the medium pizza; the topping choices are pepperoni, sausage, mushrooms, or anchovies. Each club member could fill out a preference ballot, and the results for the ballots might be summarized as in Table 14.1. PhotoDisc/Getty Images TABLE 14.1

Choice Joan Richard Suzanne

First Sausage Sausage Pepperoni Second Pepperoni Pepperoni Mushrooms Third Mushrooms Anchovies Sausage Fourth Anchovies Mushrooms Anchovies

If you were only considering each person’s first choice, and Joan, Richard, and Suzanne were the only voters, sausage would win 2 votes to 1. We say that sausage received a majority (2 out of 3) of the first-place votes. A candidate with a majority of the votes is the one with more than half, or 50%, of the vote. Looking at the table, you might argue that pepperoni is a better choice because each voter has it listed as first or second choice. If all the Math Club members were voting, listing all the ballots would take a lot of space because with only four toppings, there would be 4! 24 different ballots to consider. If you had five toppings, there would be 5! 120 ballots. You can summarize the results of an election by showing how often a particular outcome was selected with a preference table. Do it in steps. 1. Replace the word sausage with the letter S, pepperoni with the letter P, and TABLE 14.2 so on.

5 2. If several people have exactly the same list of preferences, list them together. Suppose five people all vote S, P, M, A. This fact is shown by using a table of S votes like Table 14.2. The number 5 at the top indicates that five people had P the exact results S, P, M, A on their ballots. Note that the first choice S appears at the top, the second choice P is next, and so on. M 3. Assume that all the club members chose one of 3 or 4 different rankings. The A voting methods we will study will work the same way no matter how many of the 4! 24 possible rankings were chosen. 304470-ch14_pV1-V59 11/7/06 10:01 AM Page 4

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Now we are ready to analyze the results of elections using different voting systems: plurality, plurality with runoff, plurality with elimination, Borda count, and pairwise comparison.

A. The Plurality Method As you can see from the results of the 2000 presidential election in Table 14.3, if there are three or more candidates it is possible that no candidate receives a majority (more than 50%) of the votes. In this case, one method of selecting the winner is to select the candidate with the most votes. This method is called the plurality method. In a U.S. presidential election, the candidate with the most popular votes does not necessarily win!

TABLE 14.3 Results as of 6:00 P.M., EST, 11/17/2000

Candidates Votes Vote (%)

D Gore 49,921,267 49

R Bush 49,658,276 48

G Nader 2,756,008 3

RF Buchanan 447,927 0 No winner declared

Plurality Method Each voter votes for one candidate. The candidate with the most first-place votes is the winner.

Now let us go back to our pizza ballots.

EXAMPLE 1 Using the Plurality Method TABLE 14.4 The Math Club conducted an election, and the results were as shown in Table 14.4. 7542 (a) Did any of the rankings get the majority of the votes? ASPP (b) Which topping is the plurality winner? SPSM (c) Which topping comes in second? MMM S (d) Which topping comes in last? PAAA Solution Since plurality counts only first-place votes, we can see that A got 7 votes (see column 1, with 7 at the top), S got 5 votes (column 2), and P got 4 2 6 votes. Mushroom was never at the top, so it got no votes. (a) None of the rankings got a majority of the votes. Since there are 7 5 4 2 18 voters, more than 18/2 9 votes are needed for a majority. (b) A (anchovies) is the plurality winner with 7 votes. 304470-ch14_pV1-V59 11/7/06 10:01 AM Page 5

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(c) P (pepperoni) comes in second with 6 votes. (d) M (mushrooms) comes in last with no votes. As we saw in Example 1, a plurality is not necessarily a majority. There may be a situation with a large number of alternative choices where the winner might not get even 10% of the votes! Many political elections have only two candidates (or at least only two with a chance of winning). With only two choices, a plurality is necessarily a majority. However, there are also numerous instances with many candidates, including primary elections, electing members to the Baseball Hall of Fame, ranking football teams, and so on. In Example 1, the jumbo pizza ended up with anchovies (A) as the topping but you may have noticed that anchovies was the last choice of 5 4 2 11 voters. Since many people who don’t like anchovies really hate anchovies, it could well be the case that these 11 people—a clear majority—might not even want any of the jumbo pizza. Although this means more pizza for the 7 people who like anchovies, it doesn’t seem like the fairest way to choose. How can we overcome these difficulties? One way is to begin by eliminating all but the top two candidates and then make a head-to-head comparison between these two. Now the winner will have a majority! This variation of the plurality method is called plurality with runoff.

Plurality with Runoff Method Each voter votes for one candidate. If a candidate receives a majority of votes, that candidate is the winner. If no candidate receives a majority, eliminate all but the two top candidates and hold a runoff election. The candidate that receives a majority in the runoff election is the winner.

EXAMPLE 2 Using the Plurality with Runoff Method As you recall from Example 1, the election results were A, 7 votes; P, 6 votes; S, 5 votes; and M, 0 votes. Find the winner using the plurality with runoff method.

Solution TABLE 14.5 Since the top two vote getters were A and P, all others are eliminated, and we run an election between A and P. Look at Table 14.5 and mentally (or you can 7542 actually do it with a pencil) cross out all the S and M entries. Now look at ASPP the first column. There are 7 people who prefer A to P. The second, third, and fourth columns have 5 4 2 11 people who prefer P to A. (Note that SPSM in these columns we are only concerned with the fact that P is preferred MMM S over A, not the particular value of the preference.) Eleven is the majority of the 7 5 4 2 18 people voting in the election. Thus, A has 7 votes against PAAA P’s 11, and P is the new winner using the plurality with runoff method. The jumbo pizza will now have pepperoni! So far we have looked at the methods of plurality and plurality with runoff, two of the most widely used methods for political elections in many countries. Although they can be used to obtain a complete ranking of many alternatives, they are really designed to choose an overall winner. A major problem with both methods is that candidates who do not get either the most or second most first- place votes are immediately eliminated. Do we really want to place so much emphasis on first-place votes? 304470-ch14_pV1-V59 11/7/06 10:01 AM Page 6

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A fairly natural way to correct this emphasis on first-place votes is to use some kind of system that assigns a point value to each of the rankings and then counts points instead of votes. This kind of method is widely used in ranking sports teams such as in football polls, as well as in scoring track meets or selecting winners in music or television award shows. Historically, this method goes back to the eighteenth century and is named for Jean-Charles Borda (1733–1799), a French mathematician and nautical astronomer.

B. The Borda Count Method

The Borda Count Method Voters rank candidates from most to least favorable. Each last-place vote is awarded no point; each next-to-last-place vote is awarded one point, each third-from-last-place vote is awarded two points, and so on.* The candidate who receives the most points is the winner.

EXAMPLE 3 Using the Borda Count Method Find the winner of the election in Example 1 using the Borda count method.

Solution TABLE 14.6 Award 0, 1, 2, and 3 points to last, next to last, and so on. Counting the points for anchovies (A) in column 1 of Table 14.6, you get 7 first-place Points 7542 votes, worth 3 points each, a total of 3 7 21 points. Sausage (S) gets 3 ASPP 2 7 in column 1, 3 5 in column 2, 2 4 in column 3, and 1 2 in column 4 for a total of 14 15 8 2 39 points. Pepperoni (P) gets 2 SPSM 10 points in column 2, 12 in 3 and 6 in 4 for a total of 28 points. Finally, 1 MMM S mushrooms (M) get 7 points in column 1, 5 in 2, 4 in 3, and 4 in 4 for a total of 20 points. Thus, using the Borda count method, the rankings are S 0 PAAA (winner), P, A, and M with 39, 28, 21, and 20 points, respectively.

C. The Plurality with Elimination Method This method is a variation of the plurality method and may involve a series of elections.

Plurality with Elimination (The Hare Method) Each voter votes for one candidate. If a candidate receives a majority of votes, that candidate is 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 fewest votes, eliminate all candidates tied for fewest votes.) Repeat this process until a candidate receives a majority.

EXAMPLE 4 Using the Plurality with Elimination Method Consider the familiar pizza voting results. Which topping wins the election using the plurality with elimination method?

*Sometimes the last-place vote is awarded one point, next-to-last vote two points, and so on. 304470-ch14_pV1-V59 11/7/06 10:01 AM Page 7

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Solution First, let us count the number of first place votes in Table 14.7 to see if there is a majority.

TABLE 14.7 TABLE 14.8 TABLE 14.9

7542 7542 756

ASPP ASPP APP SPSM SPSS PAA MMM S PAAA PAAA

A has 7 votes (first column). S has 5 votes (second column). P has 4 2 6 votes (third and fourth columns). M has no votes. Since there are 7 5 4 2 18 voters, we need 10 votes for a majority. None of the toppings has a majority of the votes, but M received the fewest first-place votes, so M is eliminated, and all selections in each column below M move up one place, as shown in Table 14.8. Now A still has 7 votes, S has 5, and P has 4 2 6. Since S has the fewest votes, S is eliminated, and we are down to just P and A, as shown in Table 14.9. Now P is the clear majority winner with 5 6 11 votes. Thus, P is the winner of the election when we use the plurality with elimination method.

If we look at Examples 1–4, we can see that A is the winner using the plurality method, P is the winner using the plurality with runoff method, S is the winner using the Borda count method, and P is the winner using the plurality with elimination method. If a voting method is to indicate a group’s preference, the method used should not change the winner. This situation points out the importance of deciding on the voting system to be used before the election takes place. Of course, elections with only two candidates are easy because the winner will get at least half the votes—not only a plurality but also a majority. The difficulty arises when we have three or more candidates. If this is the case, we can compare candidates the easiest way we know: two at a time. This is the basis of the next voting method.

D. The Pairwise Comparison Method

Pairwise Comparison Method Voters rank candidates from most to least favorable. Each candidate is then compared with each of the other candidates. If candidate A is preferred to candidate B, then A receives one point. If candidate B is preferred to candidate A, then B receives one point. If there is a tie, each candidate receives one-half point. The candidate who receives the most overall points is the winner. *Sometimes the last-place vote is awarded one point, next-to-last two points, and so on. 304470-ch14_pV1-V59 11/7/06 10:01 AM Page 8

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For example, suppose we have three candidates: Alice, Bob, and Carol. We have to compare Alice versus Bob, Alice versus Carol, and Bob versus Carol. We could hold three separate elections, but it is possible to use the information in the preference tables we have used before. As the number of candidates grows, so do the number of head-to-head comparisons that need to be made. For n candidates, there are n(n 1) C(n, 2) 2 such comparisons. Thus, for n 10 candidates, we would need (10 9)/2 45 head-to-head comparisons. Let us use our preference tables to calculate the winner of all the possible head-to-head comparisons. The one clear-cut case is when one candidate beats all the others. This case even has a special name: A candidate who beats all the others in head-to-head comparisons is the Condorcet winner (named after the Marquis de Condorcet mentioned in the Human Side of Mathematics at the beginning of the chapter, who, like Borda, was an eighteenth-century French- man). As you might suspect, a big problem with using Condorcet winners is that often there is no such winner, as we shall see in the following example.

EXAMPLE 5 Using the Pairwise Comparison Method The results of an election involving three candidates, A, B, and C, are shown in TABLE 14.10 Table 14.10. Who wins the election using the pairwise comparison method?

234 Solution To determine the winner using the pairwise comparison method, we have to ABC compare A and B, A and C, and B and C. BCA Suppose the election is between just A and B (leave C out). CAB A: 2 votes from column 1 and 4 from column 3, a total of 6 votes B: 3 votes from column 2, a total of 3 votes Thus, A beats B 6 votes to 3, and A is awarded one point. Now, let us compare A and C (leave B out). A: 2 votes from column 1, a total of 2 votes C: 3 votes from column 2 and 4 votes from 3, a total of 7 votes Thus, C beats A 7 votes to 2, and C is awarded one point. Finally, let us compare B and C (leave A out). B: 2 votes from column 1 and 3 from column 2, a total of 5 votes C: 4 votes from column 3, a total of 4 votes Thus, B beats C 5 votes to 4, and B is awarded one point. What a dilemma! All the candidates have one point. There is no Condorcet winner in this election.

As we mentioned at the beginning of the section, there are two fundamentally different types of voting methods: preferential (those using a preference table) and nonpreferential. We will now discuss a nonpreferential voting method: approval voting. 304470-ch14_pV1-V59 11/7/06 10:01 AM Page 9

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E. Approval Voting Approval voting uses a different kind of preference table. The good news is that the table is much simpler in one respect: Each voter does not have to rank all the candidates first, second, third, and so on. Instead, each voter simply approves (A) or disapproves (D) each candidate. Thus, if you are a voter, you can vote for one candidate, two candidates, three candidates, and so on. Voting for two or more candidates doesn’t dilute your vote; each candidate that you approve of gets one full vote. When the votes are counted, the candidate with the most approval votes wins.

EXAMPLE 6 Using Approval Voting In Table 14.11, each row corresponds to a different candidate (W, X, Y, and Z), and each column corresponds to a different voter. An A means “approve” and a D means “disapprove.” Which of the candidates wins using approval voting?

TABLE 14.11

Candidate Voter 1 Voter 2 Voter 3 Voter 4 Voter 5 Voter 6 Voter 7 Voter 8

W ADAADDDD X AADDADAA Y DDADDAAA Z DDADAADA

Solution We examine each of the rows and count only the A’s. Row W has 3 A’s. Row X has 5 A’s. Row Y has 4 A’s. Row Z has 4 A’s. This means that candidate X (row 2) wins with 5 votes. Y and Z are tied with 4 votes each, and W is in last place with only 3 votes.

Like all voting methods, approval voting has its deficiencies, but it has a number of good features, too. It is simpler than the Borda count or plurality with elimination method, although not as simple as the plurality method. However, unlike the plurality method, it doesn’t rely only on first-place votes. It works well when voters can easily divide the candidates into “good” and “bad” categories. Approval voting is also good in situations where more than one winner is allowed. This occurs, for example, in electing players to the Baseball Hall of Fame. To be elected, an eligible player has to be named on 75% of the ballots. The voters are members of the Baseball Writers’ Association of America. They add one extra requirement: No one can vote for more than ten players. 304470-ch14_pV1-V59 11/7/06 10:01 AM Page 10

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EXERCISES 14.1

A The Plurality Method results are summarized in the preference schedule 1. Why did they have a vote recount in Florida below. during the 2000 presidential election? Because Florida law requires a recount when the winning Number of Voters margin in votes is less than 0.5% of the total Place 130 120 100 150 number of votes cast. First P T T S Candidate Votes Second RRRR R Bush 2,911,872 Third SSPP D Gore 2,910,942 Fourth T P S T G Nader 97,419 a. How many first-place votes are needed for a RF Buchanan 17,472 majority? b. Did any candidate receive a majority of first- a. What is the total number of votes shown in the place votes? table? c. Who is the winner by the plurality method? b. What is the difference between the number of votes obtained by Bush and by Gore? 5. Refer to the preference schedule in problem 4. c. What percent difference (to three decimal a. Which two candidates have the most first- places) is that? place votes? d. Does the difference require a recount? b. Which candidate is the winner using the plurality with runoff method? 2. Who is the winner in Florida under the plurality method? 6. The preference schedule below shows the rankings for four brands of auto tires, A, B, C, and D. 3. Four candidates, A, B, C, and D, are running for class president and receive the number of votes Number of Voters shown in the table. Place 13 12 10

6734 First A C D DCAB Second B B A CBDA Third D D B BDBC Fourth C A C AACD a. How many votes were cast in the election? a. How many votes were cast in the election? b. How many first-place votes are needed for a b. How many first-place votes are needed for a majority? majority? c. Did any brand receive a majority of first-place c. Did any candidate receive a majority of first- votes? place votes? d. Who is the winner by the plurality method? d. Who is the winner using the plurality method? 7. Refer to the preference schedule in problem 6. 4. Five hundred registered voters cast their prefer- a. Which two brands have the most first-place ence ballots for four candidates, P, T, R, and S. The votes? b. Which brand is the winner using the plurality with runoff method? 304470-ch14_pV1-V59 11/7/06 10:01 AM Page 11

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In problems 8–10, use the table below. A survey was Number of Voters conducted at Tampa International Airport to find the favorite vacation destination in Florida. The ranking Place 5 11 8 6 for four destinations, Busch Gardens (B), Disney First B O G J World (D), Epcot (E), and Sea World (S), are shown in the table. Second J J J G Third G G O O Number of Voters Fourth O B B B Place 20 15 10 First D E S 11. Which source wins using the plurality method? Second B B B 12. Which source wins using the plurality with runoff method? Third E D D 13. Which source wins using the Borda count Fourth S S E method?

B The Borda Count Method 14. Which source wins using the plurality with elimination method? 8. Find the winner and runner-up using the Borda count method. 15. Which source wins using the pairwise comparison method? C The Plurality with Elimination Method 9. Find the winner using the plurality with elimina- E Approval Voting tion method. 16. The results of a hypothetical election using approval voting are summarized in the table D The Pairwise Comparison Method below. An X indicates that the voter approves of 10. Find the winner using the pairwise comparison the candidate; a blank indicates no approval. Who method. is the winner using approval voting? 17. In problem 16, who is the winner using approval In problems 11–15, use the following information: A voting if Collins drops out of the race? group of patients suffering from a severe cold were informed that they needed at least 60 mg of vitamin C 18. Have you seen the new color choices for iMac daily. The possible sources of vitamin C were 1 orange computers? The iMac Club is sponsoring a week- (O), 2 green peppers (G), 1 cup of cooked broccoli (B), end event, and each participant will vote for his or 1 or 2 cup of fresh orange juice (J). The rankings for the her favorite iMac color using approval voting. The group are given in the table (above, right).

Table for Problem 16

Voters Candidates Richard Sally Thomas Uma Vera Walter Yvette Zoe

Adams X X X X X X Barnes X X X Collins X X X X 304470-ch14_pV1-V59 11/7/06 10:01 AM Page 12

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possible colors are strawberry, lime, grape, tan- 20. The Math Club uses approval voting to choose a gerine, and blueberry. Here is a summary of the faculty adviser for the upcoming year on the basis results. of the following responses: 12 participants voted for strawberry. Anne and Fran voted for Mr. Albertson. 7 participants voted for strawberry and Peter, Alex, and Jennifer voted for Ms. Baker blueberry. and Ms. Carr. 20 participants voted for grape and tangerine. William, Sam, Allison, and Betty voted for 18 participants voted for lime, grape, and Mr. Albertson, Ms. Baker, and Mr. Davis. tangerine. Joe, Katie, and Paul voted for Ms. Carr and 23 participants voted for blueberry and lime. Mr. Davis. 25 participants voted for tangerine. Jonathan voted for Mr. Davis. Use approval voting to determine the club’s a. How many total votes did Mr. Albertson favorite iMac color. receive? b. How many total votes did Ms. Baker receive? 19. A college class has decided to take a vote to c. How many total votes did Ms. Carr receive? determine which coffee flavors are to be served d. How many total votes did Mr. Davis receive? in the cafeteria. The choices are latte, cappuccino, e. Which teacher is selected as faculty adviser mocha, and Americano. The winning coffee flavor using approval voting? will be determined using approval voting on the basis of the following responses: In problems 21–30, use the following information: On September 23, 1993, 88 members of the International 12 students voted for latte and cappuccino. Olympics Committee (IOC) met in Monte Carlo to 5 students voted for cappuccino, mocha, and choose a site for the 2000 Summer Olympics. Five Americano. cities made bids: Beijing, (BC), Berlin (BG), Istan- 10 students voted for mocha and cappuccino. bul (I), Manchester (M), and Sydney (S). In the table 13 students voted for Americano and cappuccino. below is a summary of the site preferences of the The flavor with the most votes wins. committee members. a. How many total votes did latte receive? 21. Does any city have a majority of the first-place b. How many total votes did cappuccino receive? votes? If so, which city? c. How many total votes did mocha receive? d. How many total votes did Americano receive? 22. Which city has the most first-place votes? How e. Which coffee is selected by the class using many does it have? approval voting? 23. Which city is selected if the committee decides to use the plurality method?

Table for Problems 21–30

Number of Votes Choice 3 2 32 3 3 1 8 30 6

First I I BC M BG I M S BG Second BC BC I BC BC S S M S

Third M BG BG BG I BC BG BG M Fourth BG M M S S M I I BC Fifth SSSI MBGBCBCI 304470-ch14_pV1-V59 11/7/06 10:01 AM Page 13

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24. Which city is selected if the committee decides to 33. Did any city receive a majority of votes in the use the plurality with elimination method? second round? 25. Suppose the committee decides to give 5 points to 34. Why were four rounds of voting held? each city for every first-place selection it gets, 35. If the plurality with runoff method were used 4 points for every second-place selection, 3 points instead, which cities would have faced off in the for every third-place selection, 2 points for every runoff election? fourth-place selection, and 1 point for every fifth- place selection. If the winning city will be the city 36. If the plurality method were used, which city with the most points, which city will be selected? would have won the election? 26. Which city is selected if the committee decides to use the “regular” 4-3-2-1-0 Borda count method? Using Your Knowledge 27. Which city is selected if the committee decides to use the pairwise comparison method? Ace Cola has decided to begin a multimillion-dollar ad campaign to increase its lagging sales. The ads are to 28. Rank the cities from first to last using the “regu- be based on consumers’ preferring the taste of Ace lar” Borda count method. (Remember, you found Cola to its major competitors, Best Cola, Coala Cola, the Borda count winner in problem 26.) and Dkimjgo Cola. 29. Which city is selected if the committee decides to use approval voting? (Assume that each voter approves only his or her first two choices.) Change 30. Rank the cities from first to last using approval the voting. pace ... The following information will be used in problems 31–36. In July 2005, members of the International Ace Olympic Committee (IOC) met in Singapore to choose the site for the 2012 Summer Olympics. Five cities made bids: London (L), Paris (P), Madrid (MA), Moscow (MO), and New York (NY). The results of the with election, which used the plurality with elimination method, are shown below. (IOC members from countries with candidate cities were ineligible to vote while Ace their nation’s city was still in the running.)

LPMAMONY

First round 22 21 20 15 19 Second round 27 25 32 16 Third round 39 33 31 Fourth round 54 50

Source: http://news.bbc.co.uk/sport1/hi/other_sports/ olympics_2012/4656529.stm.

31. Which city won the election? Even koalas love 32. Did any city receive a majority of votes in the first round? 304470_ch14_pV1-V59 11/7/06 2:08 PM Page 14

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An independent testing agency conducted a carefully 37. Use the plurality method to find the preferred cola. controlled taste test on 50 randomly selected cola 38. Use the plurality with runoff method to find the drinkers. Their results are summarized in the table preferred cola. below. (In the table, A represents Ace, B represents Best, and so on.) 39. Use the Borda count method to find the preferred cola.

10 13 8 7 12 40. As an expert in the mathematics of voting, you are approached by Ace and offered a $25,000 consult- ABCCD ing fee if you can show that Ace is really the num- BADBA ber one cola. Find a point assignment for the Borda count method in which Ace comes in first. CDBAB (Hint: A gets a lot of second-place votes, so we DCADC want to make second place worth proportionally more. Remember, first place must still be worth more than second, so make the gap between second and third place larger.)

Research Questions

1. In the 2000 presidential election, there were more than two candidates. In how many other presidential elections have there been more than two candidates? 2. The 2000 presidential race was one of the closest in history. In what other years was the difference between the winner and runner-up less than 50 electoral votes? Who were the winners and runners-up of these elections? 3. Name five advantages of approval voting. References http://home.capecod.net/~pbaum/vote2.htm http://web.archive.org/collections/e2k.html www.infoplease.com/spot/closerace1.html www.multied.com/elections/ www.washingtonpost.com/wp-srv/onpolitics/elections/2000/results/ whitehouse/ www.archives.gov/federal_register/electoral_college/popular_vote_ 2000.html www.archives.gov/federal_register/electoral_college/votes_ 2000.html www.sa.ua.edu/ctl/math103/ 304470-ch14_pV1-V59 11/7/06 10:01 AM Page 15

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Web It Exercises

Have you heard of a cartogram? A cartogram is a map that relates regions based on their populations rather than their geographic sizes. After the 2004 presidential election, which pitted President George Bush against Senator John Kerry, much of the discussion centered on the “red states” versus the “blue states.” Go to www-personal.umich.edu/~mejn/election to see how a cartogram depicts the red state–blue state phenomenon. Which type of map represents the 2004 election results most accurately? Write a short essay about your conclusions.

14.2 Voting Objections

G S T A R Disaster 2000 N T I E T D In the preceding section we studied five preferential voting systems: plurality, T plurality with runoff, Borda count, plurality with elimination, and pairwise com- E

G parison. We also studied a nonpreferential voting system: approval voting. As we pointed out, all these systems have advantages and disadvantages and sometimes can produce different winners. Let us look at an actual example—the 2000 presidential election (Table 14.12).

TABLE 14.12 Results as of 5:58 P.M., EST, 11/17/2000

Candidates Votes Vote (%) States Won EV

D Gore 49,921,267 49 19 255

R Bush 49,658,276 48 29 246

G Nader 2,756,008 3 0 0

RF Buchanan 447,927 0 0 0 No winner declared Exit polls

As you can see from the results, Gore had more votes and Bush had won more states, but neither had won the electoral vote (EV) because it took 270 votes to win, and the 25 Florida electoral votes had not been decided as of November 17. If Gore got the 25 Florida votes, he would win. On the other hand, if Bush got them, he would be president. Of course, by now you know the rest of the story!

Is this fair? If we rely on the fact that Gore had the most votes, it would be fair to say that Gore was the winner. However, when we discussed the plurality method, we defined a majority as more than 50%. Should Gore win then? He should certainly beat Nader and Buchanan! But Bush also clearly beats Nader and Buchanan. Who is the winner then? Of course, you know the actual answer, but to make the discussion more precise, we will introduce four criteria that mathematicians and political scientists have agreed on as their fairness criteria for a voting system: the majority criterion, the head-to-head (Condorcet) criterion, the monotonicity criterion, and the irrelevant alternatives criterion. 304470-ch14_pV1-V59 11/7/06 10:01 AM Page 16

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A. The Majority Criterion It seems fair that if a candidate is the first choice of a majority of voters, then that candidate should be declared the winner. If this is not the case, then that voting method violates the majority criterion. Under this criterion (total number of votes), Gore should have been the winner. But as we know, Bush won the election.

Majority Criterion If a candidate receives a majority of first-place votes, then that candidate should be the winner.

EXAMPLE 1 Using the Majority Criterion La Cubanita Restaurant is conducting a survey to find out which is the most popular omelet among the western (W), bacon (B), and ham (H) omelets. The results of the survey are shown in Table 14.13.

TABLE 14.13

BREAKFAST 7AM TO 11AM Number of Votes CUBAN TOAST $ .99 CHEESE TOAST $1.45 Place 60 25 15 WESTERN OMELETTE $2.95 CHEESE OMELETTE $2.50 HAM OMELETTE $2.80 First W B B PLAIN OMELETTE $2.25 BACON OMELETTE $2.80 Second B H W CAFE CON LECHE SM.$1.40 LG.$1.70 HOT CHOCOLATE SM.$1.40 LG.$1.70 Third H W H *SERVED ON CUBAN BREAD* CHEESE 45¢

(a) Which omelet is the winner using the Borda count method? (b) Does the winner have a majority of votes?

Solution Using the Borda count method, W has 2(60) 1(15) 135 points B has 1(60) 2(25) 2(15) 140 points H has 1(25) 25 points (a) Using the Borda count method, the winner is B, the bacon omelette, with 140 points. (b) No. A majority of the people, 60 out of 100, chose the western omelette.

Note that although a majority of the people (60 out of 100) preferred the western omelet, under the Borda count method, the bacon omelet wins. Thus, in this example, the Borda count method violates the majority criterion; that is, a candidate with a majority of first-place votes can lose the election! 304470-ch14_pV1-V59 11/7/06 10:01 AM Page 17

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TABLE 14.14 EXAMPLE 2 Using the Majority Criterion An election to select their favorite airline, A, B, or C, is conducted among 32 stu- Number of Votes dents. The results are shown in Table 14.14. Which airline should be selected Place 8 6 18 under the specified method, and does the method satisfy the majority criterion? First A A B (a) The plurality method Second B C A (b) The Borda count method Third C B C (c) The plurality with elimination method (d) The pairwise comparison method

Solution (a) Using the plurality method, B is the winner with 18 out of 32 votes. Note that B received a majority of the votes, so the method of plurality does not violate the majority criterion. In general, a candidate who holds a majority of first-place votes also holds a plurality of first-place votes.

The plurality method never violates the majority criterion.

Note that the converse is not true: If you have a plurality of the votes, you do not necessarily have a majority of the votes. (b) Under the Borda count method we assign 0, 1, and 2 points to the third, second, and first places, respectively. The points for each airline are as follows: A: 2(8) 2(6) 1(18) 46 points B: 1(8) 2(18) 44 points C: 1(6) 6 points Thus, A is the winner under the Borda count method. Since airline B is the one holding the majority of first-place votes (18 out of 32), the Borda count method violates the majority criterion. Of course, the Borda count method does not always violate the majority criterion; it just has the potential to do so.

The Borda count method has the potential for violating the majority criterion.

(c) Since B has the majority of the votes (18 out of 32), B is the winner under plurality with elimination, so the majority criterion is not violated. In general, a candidate who holds a majority of first-place votes wins the election without having to hold a second election.

The plurality with elimination method never violates the majority criterion. 304470-ch14_pV1-V59 11/7/06 10:01 AM Page 18

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(d) Using the pairwise comparison involves the following cases and outcomes: A versus B (eliminate C) A: 8 6 14 B: 18 B wins 18 to 14. B is awarded 1 point. A versus C (eliminate B) A: 8 6 18 32 C: 0 A wins 32 to 0. A is awarded 1 point. B versus C (eliminate A) B: 8 18 26 C: 6 B wins 26 to 6. B is awarded 1 point. Since B has 2 points, B wins the election under the pairwise comparison method. In general, if a candidate holds a majority of first-place votes, this candidate always wins every pairwise (head-to-head) comparison.

The pairwise comparison method never violates the majority criterion.

Even though the Borda count method is the only method studied that violates the majority criterion, it does take into account the voters’ preferences by having all candidates ranked.

B. The Head-to-Head (Condorcet) Criterion Suppose four candidates, A, B, C, and D, are running for chair of the mathematics department. There are 20 voting members in the department, and the student newspaper performed a postelection survey of each of the 20 members in the department. Among other things, the survey asked the voters whom they preferred in a two-way race between candidate C (the one endorsed by students) and each of the other candidates. Here are the results. 11 voters preferred candidate C over candidate A. 11 voters preferred candidate C over candidate B. 17 voters preferred candidate C over candidate D. So, in head-to-head competition, candidate C won against each of the other candidates. Wouldn’t it seem unfair if candidate C was not declared the winner of the election? When the actual votes were tabulated, candidate A got 9 first- place votes, candidate B got no first-place votes, candidate C got 8 first-place votes, and candidate D got 3 first-place votes. If candidate C is not declared the winner, this would be a violation of the Condorcet criterion because C certainly wins when compared with every other candidate.

Head-to-Head (Condorcet) Criterion If a candidate is favored when compared head-to-head with every other candidate, then that candidate should be the winner. 304470-ch14_pV1-V59 11/7/06 10:01 AM Page 19

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EXAMPLE 3 Using the Head-to-Head Criterion Which sandwich is the most popular? La Cubanita restaurant conducted a survey among its customers to select the favorite sandwich from Cuban (C), pork (P), turkey (T), and vegetarian (V). The number of votes for each is shown in Table 14.15. Which sandwich should be selected under the specified method, and does the method satisfy the head-to-head criterion? (a) Head-to-head (b) Plurality (c) Borda count (d) Plurality with elimination (e) Pairwise comparison

TABLE 14.15

SANDWICHES Number of Voters CUBAN $3.49 SPECIAL $4.19 MEDIA NOCHE $3.29 Place 30 50 58 60 90 PORK $3.90 STEAK$3.99 BREADED $3.90 TURKEY $3.75 CLUB $3.95 First V V T P C HAM & CHEESE $3.35 CHICKEN$3.90 B.L.T. $3.50 Second T T V V P VEGETARIAN $3.50 TUNA $3.90 Third P C P T T *ADD LETTUCE & TOMATO* 30¢ Fourth C P C C V

Solution (a) We need a total of six head-to-head comparisons. A further look seems to indicate that P is the winner. Let us see why. P beats C in columns 1, 3, and 4 for 30 58 60 148 points, whereas C beats P in columns 2 and 5 for 50 90 140 points. Thus, P beats C. Comparing P and T, we see that P beats T in columns 4 and 5, obtaining 60 90 150 points, and T beats P in columns 1, 2, and 3, obtaining 30 50 58 138 points. Thus, P beats T 150 to 138. Comparing P and V, we see that P beats V in columns 4 and 5, and V beats P in columns 1, 2 and 3, so the score is the same as in the preceding comparison: P beats V 150 to 138. Thus, P is the favored candidate when compared head-to-head with every other candidate. (b) Using the plurality method, C wins with 90 votes.

The plurality method has the potential for violating the head-to-head criterion. 304470-ch14_pV1-V59 11/7/06 10:01 AM Page 20

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(c) Using the Borda count method, we assign 0, 1, 2, and 3 points to the fourth-, third-, second-, and first-place winners. The total points are C: 1(50) 3(90) 320 points P: 1(30) 1(58) 3(60) 2(90) 448 points T: 2(30) 2(50) 3(58) 1(60) 1(90) 484 points V: 3(30) 3(50) 2(58) 2(60) 476 points Using the Borda count method, T wins with 484 points.

The Borda count method has the potential for violating the head-to-head criterion.

(d) Using plurality with elimination, T is eliminated in the first round, P in the second round, and C in the third round. (Check this!) Thus, V is the winner 198 to 90 over C.

The plurality with elimination method has the potential for violating the head-to-head criterion.

(e) As you recall, in the pairwise comparison method each candidate is ranked and compared with each of the other candidates. Each time, the preferred candidate gets 1 point. Let us look at the comparisons. C and P C: 50 90 140 P: 30 58 60 148 P wins and gets 1 point. C and T C: 90 T: 30 50 58 60 198 T wins and gets 1 point. C and V C: 90 V: 30 50 58 60 198 V wins and gets 1 point. P and T P: 60 90 150 T: 30 50 58 138 P wins and gets 1 point. P and V P: 60 90 150 V: 30 50 58 138 P wins and gets 1 point. T and V T: 58 90 148 V: 30 50 60 140 T wins and gets 1 point. Thus, using the pairwise comparison method, P is the winner with 3 points.

The pairwise comparison method never violates the head-to-head criterion.

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14.2 Voting Objections V21

This example shows that the plurality, Borda count, and plurality with elimination methods may potentially violate the head-to-head criterion. Next, we shall introduce a third criterion called the monotonicity criterion that can be used to evaluate the fairness of an election and explore the possibility that the plurality with elimination method may have some further flaws.

C. Monotonicity Criterion When the outcome of a first election is not binding—for example, when a straw poll or survey is taken before the election—voters may change their preferences before the actual election. If a leading candidate gains votes at the expense of another candidate, the chances of winning for the leading candidate should increase because of the additional votes. However, this is not always the case! This strange result is a violation of the monotonicity criterion.

The Monotonicity Criterion If a candidate is the winner of a first nonbinding election and then gains additional support without losing any of the original support, then the candidate should be the winner of the second election.

We shall now discuss an example in which the winner of the first election (straw vote) gains additional votes before the actual election and still loses.

EXAMPLE 4 Using the Monotonicity Criterion Months before the actual vote to select the Heisman Trophy winner, it was claimed that one of the contenders was too old to win the trophy. A straw vote (first election) was conducted among 105 sportswriters, and the results were as shown in Table 14.16. After several weeks of heated discussion, five writers decided to change their ballots and award their first-choice votes to JH. The results of the new election are shown in Table 14.17.

TABLE 14.16 First Election TABLE 14.17 New Election

Number of Voters Number of Voters Place 42 30 23 10 Place 47 30 23 5

First JH CW DB DB First JH CW DB DB Second DB JH CW JH Second DB JH CW JH Third CW DB JH CW Third CW DB JH CW 304470-ch14_pV1-V59 11/7/06 10:01 AM Page 22

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Using the plurality with elimination method, (a) who is the winner of the first election? (b) who is the winner of the new election? (c) is the monotonicity criterion violated?

Solution (a) Using the plurality with elimination method, the first election results in the elimination of CW and then a win by JH over DB with a majority vote of 72 to 33. Thus, JH is the winner. (b) In the new election, using the plurality with elimination method, DB, with 23 5 28 points, is eliminated, and CW gets a majority of 30 23 53 votes over JH’s 47 5 52 votes. This time CW is the winner. (c) Although 5 voters changed from DB to JH in the new election, adding 5 votes to JH’s total, JH’s win in the first election was not repeated in the second election.

The plurality with elimination method has the potential for violating the monotonicity criterion.

By the way, a similar situation actually occurred in 2000 when selecting the Heisman winner among Chris Weinke (CW), Josh Heupel (JH), and Drew Brees (DB). The election, however, was actually conducted using the Borda count method, awarding each candidate 1, 2, and 3 points for third, second, and first place, respectively. Even though Weinke had reached the ripe old age of 28, he won by collecting 1628 points.

D. The Irrelevant Alternatives Criterion The fourth and last criterion we will study involves the removal (or introduction) of a candidate who has no chance of winning the election. For example, let us assume that we have an election among candidates A, B, and C with the results TABLE 14.18 shown in Table 14.18. It is easy to see that B is the winner using plurality, plurality with runoff, or Number pairwise comparison. But let us use the Borda count method. When we add up of Voters the points, we find Place 4 3 9 A: 23 points B: 22 points C: 3 points First A A B The race is close, but A wins out using the Borda count method. Second B C A It seems that in deciding between A and B, what people think of C shouldn’t matter; after all, C is completely out of the running. But look at the 3 voters rep- Third C B C resented in column 2. Suppose that after thinking it over a little more, they all decide that candidate C is even worse than they thought before and should be dropped to the bottom of their ballots or even dropped out of the election alto- gether! Note that the relative positions of A and B have not changed. With C at the bottom, the results are shown in Table 14.19 on page V23. 304470-ch14_pV1-V59 11/7/06 10:01 AM Page 23

14.2 Voting Objections V23

TABLE 14.19 Our Borda point count now becomes

Number A: 23 points B: 25 points C: 0 points of Voters B is now the winner using the Borda count method. Place 4 3 9 In other words, because some voters changed their minds about C, or because C dropped out of the race, the rankings of A and B were reversed. But the First A A B rankings of A and B should depend on how voters view A and B and not on what Second B B A they think of some other alternative. We call C an irrelevant alternative in ranking A and B. Third C C C

The Irrelevant Alternatives Criterion If a candidate is the winner of an election, and in a second election one or more of the losing candidates is removed, then the winner of the first election should be the winner of the second election.

All the methods we have studied have the potential to violate the irrelevant alternatives criterion.

EXAMPLE 5 Using the Four Fairness Criteria A group of 50 students ranked professors A, B, and C as shown in Table 14.20. If the plurality method is used to select the top professor, does the method satisfy the four fairness criteria we have studied?

TABLE 14.20

Number of Voters Place 28 12 10

First C B A Second B C B

Third A A C

Using the plurality method, C is the winner with 28 votes, which is a majority (56%) of the 50 votes cast. Thus, the majority criterion is satisfied. If we use the pairwise comparison method, we see that C beats A 40 to 10, C beats B 28 to 22, and B beats A 40 to 10, so C wins two points and the head-to-head criterion is satisfied. The monotonicity criterion is satisfied if we assume that a second election is held in which C picks up additional votes. C will certainly win the second election by plurality. Finally, if A or B drops out, C still wins by the plurality method, satisfying the irrelevant alternatives criterion. Thus, this particular election satisfies all four fairness criteria we have studied. Of course, each of the voting methods can be made to violate at least one of the fairness criteria. 304470-ch14_pV1-V59 11/7/06 10:01 AM Page 24

V24 14 Voting and Apportionment

Can we find a method that will satisfy all four criteria all the time? This question led to a long and futile search. In 1950, Kenneth Arrow, a U.S. econo- mist, made a very surprising discovery. He found that no voting method could ever satisfy these four conditions all the time. This idea it not restricted to the voting methods we know of now, but any voting method anybody might think of in the future as well. This fact is known as Arrow’s Impossibility Theorem. This discovery was a major factor in Arrow’s winning the Nobel Prize in economics.

THEOREM 14.1 Arrow’s Impossibility Theorem There is no possible voting method that will always simultaneously satisfy each of the four fairness criteria: 1. The majority criterion 2. The head-to-head criterion 3. The monotonicity criterion 4. The irrelevant alternatives criterion

In simple terms, Arrow’s discovery means that we can never find a voting method that does everything we want. Before you attempt the exercises, we summarize in Table 14.21 the four fairness criteria and indicate in Table 14.22 on page V25 which of the voting methods we have studied satisfies a particular criterion.

TABLE 14.21

Majority criterion If a candidate receives a majority of first-place votes, then that candidate should be the winner. Head-to-head If a candidate is favored when compared head-to-head (Condorcet) with every other candidate, then that candidate criterion should be the winner. Monotonicity criterion If a candidate is the winner of a first election and then gains additional support without losing any of the original support, then that candidate should be the winner of the second election. Irrelevant alternatives If a candidate is the winner of an election and in a criterion second election one or more of the losing candidates is removed, then the winner of the first election should be the winner of the second election. 304470-ch14_pV1-V59 11/7/06 10:01 AM Page 25

14.2 Voting Objections V25

TABLE 14.22

Plurality Borda with Pairwise Fairness Plurality Count Elimination Comparison Criterion Method Method Method Method

Majority Always May not Always Always criterion satisfies satisfy satisfies satisfies Head-to-head May not May not May not Always criterion satisfy satisfy satisfy satisfies Monotonicity Always Always May not Always criterion satisfies satisfies satisfy satisfies Irrelevant May not May not May not May not alternatives satisfy satisfy satisfy satisfy criterion

EXERCISES 14.2

A The Majority Criterion 2. Of course, you cannot rely solely on professional 1. Who makes the best Cuban sandwich in Tampa? judges, so the Tribune had readers vote for their According to a panel of Tampa Tribune judges favorite Cuban sandwich, and the outcome was who rated each sandwich anywhere from 1 (low) different! According to the people, the three best to 5 (high), the best three Cuban sandwiches are Cuban sandwiches are produced at La Septima produced at Wrights Gourmet (W), Puccetti’s (L), West Gate Bakery (G), and the Cuban Sand- Market (P), and La Segunda Central Bakery (S). wich Shop (C). The approximate number of votes Here is a table simulating the points in the voting. is shown in the table below.

Number Number of Voters of Points Place 600 300 200

Place 25 5 20 First L G C First W S P Second G C L Second P W S Third C L G Best Cuban Sandwich Third S P W a. Who is the winner using the plurality method? a. Who is the winner using the plurality method? b. Who is the winner using the Borda count b. Who is the winner using the Borda count method? method? c. Does the Borda count method violate the c. Does the Borda count method violate the majority criterion? majority criterion? d. Who is the winner using the pairwise compar- d. Who is the winner using the pairwise comparison method? ison method? 304470-ch14_pV1-V59 11/7/06 10:01 AM Page 26

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3. Do you drink coffee? Which kind do you prefer? Number of Voters Starbucks coffee offers latte (L), cappuccino (C), mocha (M), and Americano (A). The preferences Place 50 25 15 10 of 70 students surveyed at the University of South First S H B U Florida are shown in the table below. Second U S H B Number of Voters Third H U S S Place 10 20 30 10 Fourth B B U H First C C L L a. In a head-to-head comparison, is there a com- Second A M C C mercial preferred to all others? Third MAMA b. Is the head-to-head criterion satisfied if the plurality method is used to find the preferred Fourth L L A M commercial? Explain your answer. a. Which flavor is the winner using the Borda count method? C The Monotonicity Criterion b. Is the majority criterion satisfied? Explain 6. A company is planning to relocate to one of the your answer. larger counties in the United States: Los Angeles (L), Cook (C), or Harris (H). The Committee of 100 is to use the plurality with elimination method B The Head-to-Head (Condorcet) Criterion to select the county, and their preferences are 4. The Performing Arts Center Board is considering shown in the table below. After careful delibera- showing three different plays this season: Cats tion, the ten voters who voted C, L, H changed (C), A Chorus Line (L), and Les Miserables (M). their vote to L, C, H. Is the monotonicity criterion The ten members of the board rank the plays satisfied? Explain your answer. according to the preference table below. Number of Voters Number of Voters Place 25 30 10 35 Place 2 5 3 First C H C L First L C M Second H L L C Second C L C Third L C H H Third M M L

a. In a head-to-head comparison, is there a play D The Irrelevant Alternatives Criterion that is preferred to all others? 7. Which of the following has the highest cost of b. Is the head-to-head criterion satisfied if the living: Washington, D.C. (D), Alaska (A), or plurality method is used to determine the most Hawaii (H)? The table below shows the responses popular play? of 50 people who were asked that question. If Hawaii (H) is eliminated, is the irrelevant alterna- 5. Did you watch the Super Bowl this year? Which tives criterion satisfied? Explain your answer. commercials do you remember? Four of the all- time most memorable are Staples (S), 7 Up (U), Number of Voters Honda (H), and Budweiser (B). One hundred viewers were asked to watch and vote on the Place 20 18 12 commercial they preferred. The results are in the First D A H preference table (above, right). Second A H D Third H D A 304470-ch14_pV1-V59 11/7/06 10:01 AM Page 27

14.2 Voting Objections V27

8. Suppose that in problem 7 the responses of the 50 a. Who wins using pairwise comparisons? people are as shown in the table below. If Alaska b. Does any candidate beat every other candidate (A) is eliminated, is the irrelevant alternatives one on one, that is, in a head-to-head compari- criterion satisfied? Explain your answer. son? If so, which one? c. Who wins using the plurality method? Number of Voters d. Which fairness criteria, if any, are violated? Place 20 16 14 Explain. e. Suppose candidate C drops out, but the winner First D A H is still chosen using the plurality method. Is the Second A H D winner the same as in part (c)? If not, which candidate does win? Third H D A f. Which fairness criteria, if any, are violated? Explain. 9. The preference table below gives the results of an g. Who wins using the plurality with elimination election among three candidates, A, B, and C. method? (Assume candidate C is now back in.) h. Now suppose candidate A drops out, but the Number of Voters winner is still chosen using the plurality with Place 27 24 2 elimination method. Is the winner the same as in part (g)? If not, which candidate does win? First A B C i. Which fairness criteria, if any, are violated? Second C C B Explain. Third B A A 11. The preference schedule below gives the results of an election among four candidates, A, B, C, and D. a. Who wins using the plurality method? b. Does any candidate get a majority of the first- Number of Voters place votes? If so, which one? c. Who wins using the pairwise comparison Place 14 4 10 1 8 method? First A B C C D d. Does any candidate beat every other candidate one-on-one, that is, in a head-to-head compar- Second B D B D C ison? If so, which one? Third C C D B B e. Who wins using the Borda count method? f. Which fairness criteria, if any, are violated? Fourth DAAAA Explain. g. Suppose candidate B drops out, but the winner a. Who wins using the plurality with elimination is still chosen using the Borda count method. Is method? the winner the same as in part (e)? If not, which b. Who wins using the pairwise comparison candidate does win? method? h. Which fairness criteria, if any, are violated? c. Does any candidate beat every other candidate Explain. one on one, that is, in a head-to-head comparison? If so, which one? 10. The following preference table gives the results of d. Which fairness criteria, if any, are violated? an election among three candidates, A, B, and C. Explain.

Number of Voters Place 20 19 5

First A B C Second B C B Third C A A 304470-ch14_pV1-V59 11/7/06 10:01 AM Page 28

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12. The preference table below gives the results of an Number of Voters election among three candidates, A, B, and C. Place 33113311 Number of Voters First B A C C B A B E Place 7 8 10 4 Second A C B E ADAD First A B C A Third C D A B D E E A Second B C A C Fourth D E D D E C C C Third C A B B Fifth E B E A C B D B

a. Who wins using the plurality with elimination a. Who wins using pairwise comparisons? method? b. Suppose candidate C drops out, but the winner b. Suppose that the Florida Supreme Court inval- is still chosen using the pairwise comparison idates the results of the election, and everyone method? Is the winner the same as in part (a)? must revote. As it happens, everyone votes If not, which candidate does win? exactly as before except for the 4 voters in the c. Which fairness criteria, if any, are violated? last column of the table. These 4 voters, who Explain. originally voted A, C, B, decide to switch the order of their votes for A and C so that their 15. When using the pairwise comparison method, how new preference ballots are C, A, B. Who wins many comparisons need to be made if there are this new election using the plurality with elim- a. three candidates? b. four candidates? ination method? c. five candidates? d. n candidates? c. Which fairness criteria, if any, are violated? 16. When using the pairwise comparison method, Explain. how many comparisons must a candidate (say A) 13. The preference table below gives the results of an win to guarantee winning the election if there are election among three candidates, A, B, and C a. three candidates? b. four candidates? c. five candidates? d. n candidates? Number of Voters In Other Words Place 20 19 5

First A B C 17. Explain the majority criterion in your own words. Second B C B 18. Explain why the plurality method always satisfies the majority criterion. Third C A A 19. Explain why the pairwise comparison method a. Who wins using the plurality with elimination always satisfies the majority criterion. method? 20. Explain the monotonicity criterion in your own b. Suppose candidate A drops out, but the winner words. is still chosen using the plurality with elimination method. Is the winner the same as in part 21. Explain why the plurality with elimination method (a)? If not, which candidate does win? always satisfies the monotonicity criterion. c. Which fairness criteria, if any, are violated? 22. Explain the Condorcet criterion in your own words. Explain. 23. Explain why the pairwise comparison method 14. The following preference table (above, right) always satisfies the head-to-head (Condorcet) gives the results of an election among five candi- criterion. dates, A, B, C, D, and E. 24. Explain the irrelevant alternatives criterion in your own words. 25. Which of the five election techniques that we have studied is most likely to end in a tie? Explain. 304470-ch14_pV1-V59 11/7/06 10:01 AM Page 29

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Research Questions

1. There are several election procedures we have not discussed. Find out about Black’s election procedure and then write a short paragraph about it. 2. Write a short paragraph about the work and discoveries of Kenneth Arrow. 3. Can you reduce warfare by adopting advanced methods of voting? Write a short paragraph on this topic. For information, go to http://www.solutionscreative.com/voteadv.html. 4. Which election method is best? Write a research paper on the subject. You can get some background at http://electionmethods.org/. 5. How do Oscar nominees get chosen? Write a short essay on the subject. Go to http://electionmethods.org/. 6. You can’t have an effect on an election if you don’t vote. a. What was the voter turnout for the 2000 presidential election? b. What was the voter turnout for the 2004 presidential election? c. Which U.S. presidential election had the greatest turnout? What percent? d. Which state had the greatest turnout during the 2000 presidential election? What percent? e. Which state had the greatest turnout during the 2004 presidential election? What percent? To find out, go to www.fairvote.org/. 7. Which country had the best and worst average voter turnout in the 1990s? Go to www.fairvote.org/ to find out.

14.3 Apportionment Methods

In 1787, at the Constitutional Convention in Philadelphia, delegates of the 13 original states created a system of government with three branches: executive, legislative, and judicial. One of the most important issues was the representation of the states in the legislative branch. Smaller states wanted equal representation. In response, a Senate in which two senators represent each state was created. Larger states preferred proportional representation. Thus, a House of Representatives, in which each state receives a number of representatives proportional to its population, was created. Unfortunately, the founding fathers did not decide on the exact number of representatives for each state. In fact, Article 1, Section 2, of the Constitution states: Representatives shall be apportioned among the several states... according to their respective numbers. The number of representatives shall not exceed one for every thirty thousand, but each state shall have at least one representative. Historically, at least four apportionment methods have been implemented. 1792–1841: The Jefferson method 1842–1851, 1901–1941: The Webster method 1852–1900: The Hamilton (Vinton) method 1941–present: The Hill-Huntington method 304470-ch14_pV1-V59 11/7/06 10:01 AM Page 30

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We shall study the Hamilton, Jefferson, and Webster methods and omit the Hill- Huntington method because of its complexity. Instead, we will examine a similar method known as the John Quincy Adams method. The various methods will be presented in order of mathematical complexity rather than chronological order. Keep in mind that apportionment methods are not limited to governing bodies. Budget allocations, Super Bowl tickets, faculty and student senate seats, and many other items have to be fairly distributed, or apportioned.

Funding the Community College System G S T A R N T I E The community college system in the State of Florida consists of 28 colleges. T D T In 1999, the general budget allocation for the system amounted to more than E 1 billion dollars—$1,102,817,849 to be exact. How can we fairly distribute the G money among the 28 colleges? Here are some possibilities. 1. Divide the $1,102,817,849 equally among the 28 colleges. Each college gets 1,102,817,849 $39,386,351.75 28 Is this fair? Consider this: Miami-Dade had 98,924 students, whereas the Florida Keys had 4068. Under this equal allocation method, each will get the same amount, despite the disparity in their student populations. 2. We can also base funding on the number of students attending. In a recent year, the total student population in the Florida community college system was 737,864, so funding for each college would be proportional to the number of students attending that college. The amount for each college would be Number of students in the college 1,102,817,849 737,864 Is this method fair? Keep in mind that a college with, say, 1000 students each taking a 3-hour course will need to fund 3 1000 3000 student semester hours, whereas a college with 1000 full-time equivalent students (FTEs) will have to fund 40 1000 40,000 student semester hours. Note: One FTE is equivalent to 40 student semester hours. 3. Perhaps a fairer distribution would be to allocate the money on the basis of the number of FTEs in each college. The formula for the allocation to each college would then be Number of FTEs in the college $1,102,817,849 Total number of FTEs Can you think of any other way of fairly apportioning the $1,102,817,849 to the 28 colleges?

A. Apportionment Problems In Getting Started, we considered several formulas for apportioning money. To make these formulas more standard and the resulting allocation quotas more precise, we define the standard divisor (SD) and the standard quota (SQ) as follows: 304470-ch14_pV1-V59 11/7/06 10:01 AM Page 31

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Formula for Standard Divisor Formula for Standard Quota total population in the group population in the group SD SQ total number to be apportioned SD

EXAMPLE 1 Finding the Standard Divisor The five largest community colleges in Florida received about $324 million in total general education and general fund revenues. The number of FTEs (to the nearest 1000) in each of the five colleges is shown in Table 14.23.

TABLE 14.23

Miami Jacksonville Broward Valencia Daytona Total

30,000 17,000 13,000 12,000 9000 81,000

Find the standard divisor SD and the standard quota SQ for each college.

Solution total population in the group SD total number to be apportioned The total population in the group is the total number of FTEs, that is, 81,000. The total number to be apportioned is $324,000,000. Thus, 81,000 1 SD 0.00025 324,000,000 4000 1 means that 1 FTE gets $4000 a4000 b The standard quota SQ for each college is given in Table 14.24.

TABLE 14.24

population in the group 30,000 Miami SQ $120,000,000 SD 1 4000 > population in the group 17,000 Jacksonville SQ $68,000,000 SD 1 4000 > population in the group 13,000 Broward SQ $52,000,000 SD 1 4000 > population in the group 12,000 Valencia SQ $48,000,000 SD 1 4000 > population in the group 9,000 Daytona SQ $36,000,000 SD 1 4000 >

You may have noticed that since each FTE gets $4000, each college’s allocation will be ($4000 number of FTEs). 304470-ch14_pV1-V59 11/7/06 10:01 AM Page 32

V32 14 Voting and Apportionment

B. The Alexander Hamilton Method of Apportionment One of the earliest apportionment methods was the Hamilton method. Proposed to President George Washington in 1791, the method was promptly vetoed by the president—the first presidential veto in U.S. history! First, we give the procedure for apportioning a number of items into various groups using the Hamilton method and then discuss the presidential objections to the method.

Hamilton’s Method total population 1. Find SD total seats to be apportioned state population 2. Find SQ SD 3. Round SQ down to the nearest integer (lower quota). Each state should get at least that many seats but must get at least one seat. 4. Apportion additional seats one at a time to the states with the largest fractional part of the standard quotas.

EXAMPLE 2 Using the Hamilton Method Table 14.25 shows the population of the 15 states in the Union according to the 1790 census. Use the Hamilton method to apportion the 105 seats in the House of Representatives.

TABLE 14.25

State Population SQ Rounded Down Seats

Virginia 630,560 18.31 18 18 Massachusetts 475,327 13.80 13 14 Pennsylvania 432,879 12.57 12 13 North Carolina 353,523 10.27 10 10 New York 331,589 9.63 9 10 Maryland 278,514 8.09 8 8 Connecticut 236,841 6.88 6 7 South Carolina 206,236 5.99 5 6 New Jersey 179,570 5.21 5 5 New Hampshire 141,822 4.12 4 4 Vermont 85,533 2.48 2 2 Georgia 70,835 2.06 2 2 Kentucky 68,705 2.00 2 2 Rhode Island 68,446 1.99 1 2 Delaware 55,540 1.61 1 2 Total 3,615,920 98 105 304470-ch14_pV1-V59 11/7/06 10:01 AM Page 33

14.3 Apportionment Methods V33

Solution We use the four steps. 1. Since the total population is 3,615,920 and we have to apportion 105 seats, 3,615,920 SD 34,437.33 105 2. We first find SQs for Virginia, Massachusetts, and Delaware. 630,560 For Virginia, SQ 18.31 34,437.33 475,327 For Massachusetts, SQ 13.80 34,437.33 55,540 For Delaware, SQ 1.61 34,437.33 The SQ for all states, to two decimal places, is shown in column 3 of Table 14.25. 3. The SQs rounded down to the nearest integer are in column 4 of Table 14.25. Accordingly, Virginia, Massachusetts, and Delaware will get 18, 13, and 1 seat, respectively. Note that the total seats in column 4 add up to 98. What about the 105 98 7 seats that are left over? See step 4! 4. The additional seats are apportioned, one at a time, to the states with the largest fractional parts (South Carolina, Rhode Island, Connecticut, Massa- chusetts, New York, Delaware, and Pennsylvania). The actual number of seats apportioned is in column 5 of Table 14.25.

Note that the Hamilton method assigns 2 seats to Delaware, a state with a population of 55,540. However, it was stipulated in the Constitution that each seat in the House would represent a population of at least 30,000. Logically, 2 seats would have to represent 60,000 people, but Delaware only had a population of 55,540! Partially on the basis of this flaw, President Washington vetoed the use of the Hamilton plan to apportion the first House of Representatives. Instead, the Jefferson method, which assigned an extra seat to Virginia, Jeffer- son’s home state, was used. We shall use the Jefferson method to apportion the 105 seats in the original House of Representatives after we give one more example using the Hamilton method.

EXAMPLE 3 Using the Hamilton Method The second floor of Brandon Hospital houses five intensive care units: Medical (M), Surgical (S), Cardiac (C), Transitional (T), and Progressive (P). The maxi- mum number of patients that each unit can house is shown in Table 14.26. The total for all units is 90. The hospital has bought 50 recliners to be distributed among the five units. Use the Hamilton method to apportion the 50 recliners on the basis of the number of patients in each unit.

Solution The standard divisor SD 90/50 1.8, and the standard quota SQ is the number of patients in each unit divided by 1.8, as shown in column 3. Next, we 304470-ch14_pV1-V59 11/7/06 10:01 AM Page 34

V34 14 Voting and Apportionment

round down each SQ and enter the result in column 4. Note that the sum of all the rounded-down numbers in column 4 is 47. We apportion the 3 remaining recliners, one by one, to the units with the highest fractional parts: 0.89 (P), 0.67 (C), and 0.67 (S). Column 5 shows the actual number of recliners apportioned to each unit, with the bold numbers reflecting the extra recliner. Note that the sum of the numbers in column 5 is 50, the total number of recliners.

TABLE 14.26

Unit Patients SQ Rounded Down Actual Number Patient Rooms 231-276 Patient Rooms 211-220 Medical 15 15/1.8 8.330 88 M.I.C.U., S.I.C.U. & C.C.U. Surgical 30 30/1.8 16.67 16 16 1 17 P.C.U. & T.C.U. 1 7 North Elevators Cardiac 12 12/1.8 6.670 66 I.C.U. & C.C.U. Waiting Transitional 8 8/1.8 4.440 44 Progressive 25 25/1.8 13.89 13 13 1 14 Totals 90 50 47 50

Now you know how to apportion recliners as well as seats!

C. The Thomas Jefferson Method of Apportionment In Examples 2 and 3 some of the groups (Delaware and the Surgical Care Unit, for example) received additional items when applying step 4 in Hamilton’s method. Can we modify the standard quota SQ to overcome the possible inequity? Jefferson’s method attempts to do this by using a modified divisor MD that is slightly lower than the standard divisor SD to obtain a modified quota MQ that is slightly higher than the standard quota SQ. Does this sound confusing? Just remember that if you have the fraction population in the group SQ SD and you make the denominator SD slightly lower, the new modified quota MQ will be slightly higher. Here are the steps to apportion items using Jefferson’s method.

Jefferson’s Method 1. Find a modified divisor MD such that when each modified quota MQ is rounded down to the nearest integer, the sum of the resulting integers equals the number of items to be apportioned. 2. The apportionment for each group corresponds to the rounded down MQs found in step 1.

As we shall see, the challenge is to find that “magical” MD! 304470-ch14_pV1-V59 11/7/06 10:01 AM Page 35

14.3 Apportionment Methods V35

EXAMPLE 4 Using the Jefferson Method Use the Jefferson method to apportion the 50 recliners of Example 3.

Solution Table 14.27 shows the first four columns in Example 3. The standard divisor in Example 3 was 1.8, so let us make the modified divisor MD slightly lower—say, 1.7—to obtain the modified quotas shown in column 5 (rounded to two decimal places). We show the rounded down numbers corresponding to the actual number of recliners apportioned in the last column.

TABLE 14.27 Rounded Modified Rounded Unit Patients SQ Down Quota Down 15 15 Medical 15 1.8 8.33 8 1.7 8.82 8 30 30 Surgical 30 1.8 16.67 16 1.7 17.65 17 12 12 Cardiac 12 1.8 6.67 6 1.7 7.06 7 8 8 Transitional 8 1.8 4.44 4 1.7 4.71 4 25 25 Progressive 25 1.8 13.89 13 1.7 14.71 14 Total 90 50 47 50

Note that the sum of the rounded-down numbers in the last column adds up to 50 as required.

EXAMPLE 5 Using the Jefferson Method Use the Jefferson method to apportion the 1794 House of Representatives shown in Table 14.28. Note that the total U.S. population was 3,615,920, 105 seats were to be apportioned, and total population MQ MD

TABLE 14.28 State Population MQ Rounded Down

Virginia 630,560 19.11 19 Massachusetts 475,327 14.40 14 Pennsylvania 432,879 13.12 13 North Carolina 353,523 10.71 10 New York 331,589 10.05 10 Maryland 278,514 8.44 8 Connecticut 236,841 7.18 7 South Carolina 206,236 6.25 6 New Jersey 179,570 5.44 5 New Hampshire 141,822 4.30 4 Vermont 85,533 2.59 2 Georgia 70,835 2.15 2 Kentucky 68,705 2.08 2 Rhode Island 68,446 2.07 2 Delaware 55,540 1.68 1 Total 3,615,920 109.57 105 304470_ch14_pV1-V59 11/7/06 2:08 PM Page 36

V36 14 Voting and Apportionment

Solution The modified divisor MD 33,000 was mercifully supplied by Congress. We calculate some modified quotas and show the rest in the table. 630,560 For Virginia, 19.11 33,000 475,327 For Massachusetts, 14.40 33,000 55,540 For Delaware, 1.68 33,000 The rounded-down quotas corresponding to the number of seats under the Jefferson method are shown in the last column of Table 14.28.

D. The Daniel Webster Method of Apportionment The feasibility of the Jefferson method hinges on finding the “magic” modified divisor MD and was attacked on constitutional grounds. However, Jefferson pointed out that the Constitution only required that apportionment be based on population, and the modified divisor MD produced quotas that indeed reflected the population and, moreover, did so equally, since all states used the same divisor. The Jefferson method was used without incident until after the 1820 census, when a major flaw (to be discussed in the next section) was uncovered. When the same flaw appeared again following the 1830 census, the method was replaced by one proposed by Daniel Webster in 1832. By that time, the country had grown from 15 states with 3,615,920 people to 24 states with 12,860,702 people. The appeal of Webster’s method was its mathematical simplicity. Modified quotas are not rounded down to the nearest integer. Instead, they are rounded to the nearest integer using the mathematical rules we have studied: round up for fractions of 1 1 2 or more and down for fractions that are less than 2 . The bad news is that you still have to find that modified “magic” divisor MD. The good news is that the MD is usually supplied for us.

Webster’s Method 1. Find MD, the modified divisor. 2. Find MQ, the modified quota for each group. total population MQ MD 3. Round MQ in the usual manner for each group (up for 0.5 or more, down for less than 0.5). 4. The apportionment for each group corresponds to the values obtained in step 3, and the sum of the apportionments for all groups must equal the total number of items to be apportioned. 304470_ch14_pV1-V59 11/7/06 2:08 PM Page 37

14.3 Apportionment Methods V37

EXAMPLE 6 Using the Webster Method Use the Webster method to apportion the five states shown with their respective populations in the 1830 census in Table 14.29. The modified divisor MD selected by Webster was 49,800, and the total population was 12,860,702. (Source: www.census.gov/population/censusdata/table-16.pdf.)

Solution The modified quota MQ is found by dividing the state population by 49,800. For New York, MQ 1,918,608/49,800 38.53, or 39 New York gets 39 seats. For Pennsylvania, MQ 1,348,233/49,800 27.07, or 27 Pennsylvania gets 27 seats. For Kentucky, MQ 687,917/49,800 13.81, or 14 Kentucky gets 14 seats. For Vermont, MQ 280,652/49,800 5.64, or 6 Vermont gets 6 seats. For Louisiana, MQ 215,739/49,800 4.33, or 4 Louisiana gets 4 seats. The final results are shown in Table 14.29.

TABLE 14.29

Population State Population MQ Apportionment 49,800

New York 1,918,608 38.53 39 Pennsylvania 1,348,233 27.07 27 Kentucky 687,917 13.81 14 Vermont 280,652 5.64 6 Louisiana 215,739 4.33 4

E. The John Quincy Adams Method of Apportionment As we have mentioned, by 1830, politicians were once again struggling over the apportionment method to be used. The debate was so intense that former president John Quincy Adams, at the time a representative from Massachusetts, wrote in his memoirs: I passed an entirely sleepless night again. The iniquity of the Apportion- ment bill, and the disreputable means by which so partial and unjust a distribution of the representation had been effected, agitated me so that I could not close my eyes. Mr. Adams was referring to a proposal by James K. Polk of Tennessee, which used Jefferson’s method of apportionment with an increased divisor of 47,700. This increase favored the representation of some states but hurt the representation of some of the New England states. As a consequence, Adams proposed a new apportionment method that was similar to Jefferson’s but rounded up instead of down and, unfortunately, still used a “magic” divisor. Here is the procedure for Adams’s apportionment method. 304470-ch14_pV1-V59 11/7/06 10:01 AM Page 38

V38 14 Voting and Apportionment

Adams’s Method 1. Find a modified divisor MD such that when each group’s modified quota MQ is rounded up to the nearest integer, the sum of the resulting integers equals the number of items to be apportioned. 2. The apportionment for each group corresponds to the rounded up MQs found in step 1.

EXAMPLE 7 Using the Adams Method Use Adams’s method to apportion the recliners of Example 3.

Solution We have to find the modified “magic” divisor. Recall that in Example 4 the standard divisor 1.8 was slightly lowered to 1.7 to obtain the desired modified quotas. If the modified quotas are rounded up, the sum will be 9 18 8 5 15 55. To reduce this number, let us increase the divisor to 1.9 and round up the quota as shown in Table 14.30.

TABLE 14.30

Unit Patients Modified Quota Rounded Up 15 Medical 15 1.9 7.89 8 30 Surgical 30 1.9 15.79 16 12 Cardiac 12 1.9 6.32 7 8 Transitional 8 1.9 4.21 5 25 Progressive 25 1.9 13.16 14 Totals 90 50

Our modified “magic” divisor 1.9 did the trick; the rounded-up values, which correspond to the number of recliners each unit will get, add up to 50.

Before you attempt the exercises, Table 14.31 gives a summary of the apportionment methods we have studied and their important features. TABLE 14.31

Method Divisor Round the Quota Apportionment

Hamilton’s total population Down to the nearest Distribute left-over items to the groups with SD seats to be apportioned integer the largest fractional part until all items are distributed.

Jefferson’s MD is less than SD. Down to the nearest Apportion to each group its modified lower integer quota. Webster’s MD is less than, greater than, To the nearest Apportion to each group its modified or equal to SD. integer rounded quota. Adams’s MD is greater than SD. Up to the nearest Apportion to each group its modified upper integer quota. 304470-ch14_pV1-V59 11/7/06 10:01 AM Page 39

14.3 Apportionment Methods V39

As you can see from Table 14.31, Hamilton’s method rounded the standard quotas down to the nearest integer, Jefferson’s method rounded the modified quotas down to the nearest integer, Webster’s method rounded the modified quotas to the nearest integer, and Adams’s method rounded the modified quotas up to the nearest integer.

EXERCISES 14.3

A Apportionment Problems How much money will each of these two univer- In problems 1–7, use the following table: sities receive if the money is allocated according to the number of students in each institution? Answer to the nearest dollar. University Headcount FTE 3. The University of South Florida has 18,176 FTEs, University of Florida 41,652 29,646 whereas the University of West Florida has 4556. Florida State University 30,389 21,195 How much money will each of these two univer- Florida A&M University 11,324 8064 sities receive if the money is allocated according University of South to the number of FTEs in each institution? Answer Florida 31,555 18,176 to the nearest dollar. Florida Atlantic University 19,153 10,725 University of West Florida 7790 4556 University of Central B The Alexander Hamilton Method of Florida 30,009 18,312 Apportionment Florida International 4. Suppose that the state decides to apportion 200 new University 30,096 17,434 teaching positions on the basis of the number of University of North Florida 11,360 6697 students in each university. Florida Gulf Coast a. Use the Hamilton method to find the standard University 2893 1558 divisor SD. b. Use the Hamilton method to find the standard Subtotal E & G 216,221 136,363 quota SQ for Florida Atlantic and the Univer- sity of Central Florida. Answer to three deci- 1. The university system in the State of Florida con- mal places. sists of the ten universities listed in the table above. In a recent year, the general budget alloca- 5. Suppose the state decides to apportion 200 new tion for the system amounted to more than 2 bil- teaching positions on the basis of the number of lion dollars: $2,001,102,854. FTEs in each university. a. If the state decides to apportion the money a. Use the Hamilton method to find the standard equally among the ten universities, how much divisor SD. will each university get? Answer to the nearest b. Use the Hamilton method to find the standard dollar. quotient SQ for Florida Atlantic and the Uni- b. How much money will each student be allo- versity of Central Florida. Answer to three dec- cated if the money is apportioned equally imal places. among all the students? Answer to the nearest 6. Use the Hamilton method to apportion the 200 dollar. new teaching positions to each of the ten universi- c. How much money will each FTE (full-time ties on the basis of the number of students. equivalent) be allocated if the money is apportioned equally among all FTEs? Answer to the 7. Use the Hamilton method to apportion the 200 nearest dollar. new teaching positions to each of the 10 universities on the basis of the number of FTEs. Is the 2. The University of Florida has 41,652 students, number of positions for each university the same whereas Florida Gulf Coast University has 2893. as that obtained in problem 6? 304470_ch14_pV1-V59 11/7/06 2:08 PM Page 40

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8. Let us go back to the five intensive care units of Hillsborough 940,484 Example 3. Suppose the hospital buys 75 new intravenous (IV) pumps to be distributed among Manatee 243,531 the five units on the basis of the number of patients Orange 817,206 in each unit. Use the Hamilton method to fill in the blanks in the table below and apportion the 75 IV Pasco 330,704 units. Polk 457,347

Rounded Actual a. Find each county’s standard quota. Unit Patients SQ Down Number b. Using Hamilton’s method, find each county’s Medical 15 apportionment. Surgical 30 C The Thomas Jefferson Method of Cardiac 12 Apportionment Transitional 8 11. What leisure activities do you participate in? In the table below are five activities and the approx- Progressive 25 imate number of participants (in millions) in each. 90 Exercise 150 9. According to the Centers for Disease Control and Sports 90 Prevention, the states reporting the highest annual number of AIDS cases in a recent year are as Charity work 85 shown in the table below. In that same year, Home repair 130 federal spending on AIDS research amounted to $9,988 million. Computer hobbies 80

Sources: National Endowment for California 5637 the Arts; Statistical Abstract of the United States. Florida 5683 New York 7655 Suppose you wish to allocate $100 million to pro- mote leisure activities on the basis of the number Texas 3715 of participants. a. New Jersey 2061 Find the modified quota for each activity using the divisor 5.25. b. Find how much money should be apportioned Suppose the federal government wishes to allo- to each activity using Jefferson’s method. cate an additional $100 million for AIDS research on the basis of the number of cases in each of these 12. In the table below, the five U.S. charities receiving states. the highest donations (in millions) in a recent year a. Find each state’s standard quota. are shown. b. Find each state’s apportionment using Hamil- ton’s method Salvation Army $1230 10. Do you know where the 2012 Summer Olympics YMCA $ 630 will be held? One of the possible venues was Florida. Fidelity Investments $ 570 The populations of five counties where some American Cancer Society $ 560 of the events were to be held are as shown in the following table. Suppose $500 million is allocated American Red Cross $ 540 to these five counties on the basis of their respective Source: The Chronicle of Philanthropy. populations. 304470-ch14_pV1-V59 11/7/06 10:01 AM Page 41

14.3 Apportionment Methods V41

Suppose you are a philanthropist willing to donate Suppose the Immigration and Naturalization $150 million to these five charities on the basis of Service is planning on granting 700,000 visas next received donations. year. a. Find the modified quota for each charity using a. Find the modified quota using the divisor the divisor 23.3. 0.928. b. Find how much money should be apportioned b. How many visas should be allocated to each to each charity using Jefferson’s method. continent using Webster’s method? 15. The acreage of five county parks in Hillsborough D The Daniel Webster Method of County is shown in the table below. Suppose the Apportionment county wishes to distribute 75 new park rangers 13. How much do you spend on your pet annually? In among these five parks. the table below are the average annual costs (to the nearest $10) spent per household for several types Lake Park 600 of pets. E. G. Simmons 470 Dogs $190 Lettuce Lake 240 Cats $110 Lithia Springs 160 Birds $ 10 Eureka Springs 30 Horses $230 a. Find the modified quota for each park using the Source: U.S. Pet Ownership divisor 20.5. and Demographic Sourcebook. b. Find the number of rangers that should be allo- For every $500 spent on each of these four types cated to each park using Adams’s method. of pets, 16. Has your telephone area code been changed a. find the modified quota using the divisor 1.08. lately? With the popularity of cell phones increas- b. how much money should be apportioned to ing, more area codes are needed. The table below each pet category using Webster’s method? shows the number of existing area codes in five 14. What continents do immigrants to the United states. Unfortunately, there are a limited number States come from? The table below shows the of area codes that can be allocated to states. Sup- number of immigrants from each continent admit- pose we wish to allocate 25 new area codes to the ted to the United States in a recent year. five states listed on the basis of the number of area codes they already have. Europe 90,000 Texas 21 Asia 220,000 California 20 North America 255,000 Florida 12 South America 45,000 Ohio 10 Africa 40,000 Colorado 7 Source: U.S. Immigration and Naturalization Service. a. Find the modified quota for each state using the divisor 3.08. b. Find the number of area codes allocated to each state using Adams’s method. 304470-ch14_pV1-V59 11/7/06 10:01 AM Page 42

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In problems 17–20, use the headcount enrollment by county in the Florida State University System (to the nearest thousand) given in the following table:

County Dade Broward Hillsborough Orange Pinellas Headcount 33,000 21,000 17,000 11,000 11,000

Suppose the state decides to allocate $100 million to 22. a. Find each hospital’s modified quota using the these five counties. divisor 1.6745. b. Find each hospital’s apportionment using 17. a. Find the standard divisor. Jefferson’s method. b. Find each county’s standard quota. c. Find each county’s apportionment using Ham- 23. a. Find each hospital’s modified quota using the ilton’s method. divisor 1.7238. b. Find each hospital’s apportionment using 18. a. Find each county’s modified quota using the Adams’s method. divisor 916.05. b. Find each county’s apportionment using Jef- 24. Find each hospital’s apportionment using Web- ferson’s method. ster’s method. 19. a. Find each county’s modified quota using the divisor 948.60. Using Your Knowledge b. Find each county’s apportionment using Adams’s method. Students have suggested that many of the apportionment problems can be done using ratio and proportion. 20. Find each county’s apportionment using Web- Suppose you have four sports, A, B, C, and D, and the ster’s method with the standard quota. Student Government Association wishes to distribute $200,000 among the four on the basis of their average In problems 21–24, use the top-rated AIDS treatment attendance of 2000, 4000, 6000, and 8000 spectators, hospitals and their scores given in the table below. respectively.

Hospital Points 25. a. How much will each sport get if the $200,000 is distributed proportionately to its attendance? San Francisco General 100 b. How much will each sport get if the $200,000 Johns Hopkins Hospital 72 is apportioned using Hamilton’s, Jefferson’s, Adams’s, and Webster’s methods? Massachusetts General 62 c. Are the answers the same in parts (a) and (b)? Univ. of Calif. at San Francisco 56 Memorial Sloan-Kettering 50 In Other Words

Source: U.S. News and World Report. Describe in your own words how to calculate Suppose the best 200 AIDS specialists are to be 26. a standard divisor. 27. a standard quota. assigned to these 5 hospitals on the basis of the number of points in the survey. 28. Look at Table 14.31 on page V38. Name the method we have studied that rounds the modified 21. a. Find the standard divisor. quota b. Find each hospital’s standard quota. a. up. b. down. c. in the usual manner. c. Find each hospital’s apportionment using Ham- ilton’s method. 304470-ch14_pV1-V59 11/7/06 10:01 AM Page 43

14.4 Apportionment Objections V43

Research Questions

1. The 1824 election was marred by the so-called Corrupt Bargain. Describe the events that marred the election. What was claimed to be corrupt in the election? 2. The 2000 presidential election was disputed in court but was by no means the first disputed election in history. What was the first disputed presidential election in U.S. history? What was the dispute about? 3. Write a short paragraph describing the present method of apportionment for the House of Representatives. Make sure you mention the names and techniques involved with this method. 4. As we have discovered from this section, most of the apportionment methods used for the U.S. House of Representatives rely on the “magical” modified divisor MD. It is also apparent that a precise formula for computing this MD is not clearly prescribed (see Table 14.31 on page V38). See if you can uncover the mystery of how actual MDs have been obtained throughout history. 5. In December 2005, Iraq held an election to select members of its parliament. This was only the second proportional election ever held in Iraq (a proportional election is an electoral system in which all parties are represented in proportion to their voting strength). How were the members of parliament selected? Go to www.fairvote.org/blog/?p20 and find out. Write a short essay on your findings.

14.4 Apportionment Objections

As we saw in Section 14.2, there were several objections, or flaws, associated with the voting methods we studied. Similarly, there are several objections, or flaws, associated with Hamilton’s apportionment method. These objections are paradoxical and depend mainly on three factors: the number of items to be apportioned (objection: the Alabama paradox), the population in the group (objection: the population paradox), and the addition of one or more new groups that require apportionment (objection: the new-states paradox). The main objection to Jefferson’s, Webster’s, and Adams’s methods is that each has the potential of violating the quota rule.

The Quota Rule The apportionment for every group under consideration should always equal either the group’s upper quota (rounding up) or its lower quota (rounding down)

Even though Hamilton’s method will always satisfy the quota rule and Jefferson’s, Webster’s, and Adams’s methods will not, the method has equally serious flaws. The three main objections to Hamilton’s method are the Alabama paradox, the population paradox, and the new-states paradox. 304470-ch14_pV1-V59 11/7/06 10:01 AM Page 44

V44 14 Voting and Apportionment

Sweet Home Alabama G S T A R N T I E The Alabama paradox first surfaced after the 1870 census. With 270 members T D T in the House of Representatives, Rhode Island got 2 representatives, but E when the house size was increased to 280, Rhode Island lost a seat. After the G 1880 census, C. W. Seaton (chief clerk of the U.S. Census Office) computed apportionments for all house sizes between 275 and 350 members. He then wrote a letter to Congress pointing out that if the House of Representatives had 299 seats, Alabama would get 8 seats, but if the House of Representatives had 300 seats, Alabama would only get 7 seats! Again, this meant a loss of 1 seat for Alabama, even though the total number of house seats could be increased from 299 to 300. This objection or flaw has come to be known as the Alabama paradox. (Source: www.sa.ua.edu/ctl/math103/.)

A. The Alabama Paradox

The Alabama Paradox The Alabama paradox occurs when an increase in the total number of items to be apportioned results in a loss of items for a group.

EXAMPLE 1 Finding the Standard Divisor and Quota In the 1880 Census, the population of the United States was 50,189,209, and the population of Alabama was 1,262,505. Find the standard divisor and the standard quota when (a) 299 seats are to be apportioned. (b) 300 seats are to be apportioned.

Solution (a) The standard divisor is 50,189,209 167,856.89 299 and the standard quota is 1,262,505 7.52 167,856.89 (b) The standard divisor is 50,189,209 167,297.36 300 and the standard quota is 1,262,505 7.55 167,297.36

Does Example 1 prove that Alabama would get 8 seats when 299 seats were apportioned and only 7 when 300 seats were apportioned? The answer is no. To prove this, we would have to find the standard quotas for all the states and then 304470-ch14_pV1-V59 11/7/06 10:01 AM Page 45

14.3 Apportionment Objections V45

assign the leftover seats to the states with the largest fractional parts. Unfortu- nately, there were a total of 38 states in 1880, so the task would be monumental indeed. To learn more you can go to www.ctl.ua.edu/math103/ and look under “Apportionment.” For now, let us try a simple example.

EXAMPLE 2 The Alabama Paradox and the Hamilton Method Consider a country with a population of 50,000 and three states, A, B, and C, with the populations shown in Table 14.32. Show using Hamilton’s apportionment method that the Alabama paradox occurs if the number of seats is increased from 50 to 51.

Solution When the number of seats to be apportioned is 50, the standard divisor is 50,000 50 1000. The standard quotas SQs and the rounded-down values RDs are shown in columns 3 and 4 of Table 14.32. The state with the largest fractional part (0.5) is C. Thus, C gets the extra seat.

TABLE 14.32

State Population SQ RD Extra Final 25,200 A 25,200 1000 25.2 25 0 25 23,300 B 23,300 1000 23.3 23 0 23 1500 C 1500 1000 1.5 1 1 2 Total 50,000 49 50

If we repeat the apportionment in Table 14.33 using 51 seats, the standard 50,000 divisor is 51 980.39. Again, we show the standard quotas SQs and the rounded-down values in columns 3 and 4. The states with the largest fractional parts are B (0.77) and A (0.70), so each gets an extra seat.

TABLE 14.33

State Population SQ RD Extra Final 25,200 A 25,200 980.39 25.70 25 1 26 23,300 B 23,300 980.39 23.77 23 1 24 1500 C 1500 980.39 1.53 1 0 1 Total 50,000 49 51

Here we go again! Even though we increased the number of seats from 50 to 51, state C, formerly with 2 seats, ended up losing 1 seat—the Alabama paradox!

Note that this paradox can only occur when the number of objects to be apportioned increases. Thus, it seems reasonable to expect that if we hold the size of the House of Representatives to 435, as it has been for many years, no 304470_ch14_pV1-V59 11/7/06 2:08 PM Page 46

V46 14 Voting and Apportionment

objections or paradoxes should surface. Unfortunately, this is not the case. If the population of one or more states changes, one state could lose a seat to another state, even if its population is growing at a faster rate than the state that loses the seat. This paradox is known as the population paradox.

B. The Population Paradox

The Population Paradox The population paradox occurs when the population of group A is increasing faster than the population of group B, yet A loses items to group B.

The population paradox was discovered around 1900, when it was shown that a state could lose seats in the House of Representatives as a result of an increase in its population. (Virginia was growing much faster than Maine—about 60% faster—but Virginia lost a seat in the house, whereas Maine gained a seat.)

EXAMPLE 3 The Population Paradox and the Hamilton Method Table 14.34 shows the number of students taking mathematics, English, and science courses during the fall and spring semesters. If 100 full-time teaching positions are to be apportioned among the three departments on the basis of their respective course enrollments, (a) how many positions will each department get in the fall using Hamilton’s method? (b) how many positions will each department get in the spring using Hamilton’s method? (c) is the apportionment fair? Explain your reasoning (d) is this apportionment an example of the population paradox?

Solution Since there are 10,000 students and 100 positions, the standard divisor for the fall is 10,000/100 100. The standard quotas SQs and the number of positions for each department during the fall are shown in columns 3 and 4 of Table 14.34. The number of students for the spring is 10,030, so the standard divisor SQ is 10,030/100 100.3. The new standard quotas New SQs and the number of positions per department for the spring are shown in columns 6 and 7.

TABLE 14.34

Fall Semester Spring Semester Subject Number SQ Positions Number New SQ Positions

Math 951 9.51 10 961 961/100.3 9.58 9 English 1949 19.49 19 1969 1969/100.3 19.63 20 Science 7100 71.00 71 7100 7100/100.3 70.79 71 Total 10,000 100 10,030 100 304470-ch14_pV1-V59 11/7/06 10:01 AM Page 47

14.4 Apportionment Objections V47

(a) In the fall, mathematics gets 10 positions, English 19, and science 71. (b) In the spring, mathematics gets 9 positions, English 20, and science 71. 961 951 10 (c) No. The rate of growth for mathematics was 951 951 1.05%. 1969 1949 20 The rate of growth for English was 1949 1949 1.03%. Thus, mathematics was growing at a faster rate (1.05%) than English (1.03%), but despite this, mathematics lost 1 position (from 10 to 9), whereas English gained 1 position (from 19 to 20). (d) Yes. In this instance the mathematics population was increasing faster than the English population, yet mathematics lost one position to English.

C. The New-States Paradox As we have mentioned before, the objections to Hamilton’s apportionment method occur when the number of items to be apportioned changes (Alabama paradox) and when the population in the group changes (population paradox). We consider one more paradox that occurs when we add one or more groups that require apportionment (new-states paradox). The new-states paradox means that adding a new state with its fair share of seats can affect the number of seats due to other states. This has actually happened! The paradox was discovered in 1907 when Oklahoma became a state. Before Oklahoma became a state, the House of Representatives had 386 seats. Comparing Oklahoma’s population with that of other states, it was clear that Oklahoma should have 5 seats, so the house size was increased by 5 to 391 seats. The intent was to leave the number of seats unchanged for the other states. However, when the apportionment was mathematically recalculated, Maine gained a seat (from 3 to 4), and New York lost a seat (from 38 to 37).

The New-States Paradox The new-states paradox occurs when the addition of a new group changes the apportionment of another group.

EXAMPLE 4 The New-States Paradox and the Hamilton Method Suppose that in Example 3 the art department enrolls 600 students in the fall and that 5 additional positions are allocated, bringing the total number of students to 10,600 and the total number of positions to 105. (a) Use the Hamilton method to find the new fall apportionment for mathematics, English, science, and art. (b) How does the new fall apportionment compare with the original fall apportionment of Example 3? (c) Does the new apportionment seem fair? (d) Show that the new apportionment results in an occurrence of the new-states paradox.

Solution (a) To find the new fall semester apportionment, note that there are now 10,600 10,600 students and 105 positions, so the new standard divisor is 105 100.95. 304470-ch14_pV1-V59 11/7/06 10:01 AM Page 48

V48 14 Voting and Apportionment

The standard quotas and the final number of positions are shown in Table 14.35.

TABLE 14.35

New Fall Semester Original Fall Semester Subject Number SQ Positions Number SQ Positions

Math 951 9.42 10 951 9.51 10 English 1949 19.31 19 1949 19.49 19 Science 7100 70.33 70 7100 71.00 71 Art 600 5.94 6 Total 10,600 105 10,000 100

(b) The right side of Table 14.35 shows the original fall semester apportionments. As you can see, the science department lost 1 position (from 71 to 70) in the new apportionment, even though it did not lose any students. Presum- ably, the art department should have gotten the 5 new positions, but it got 6 instead. (c) The new apportionments do not seem fair because science lost 1 position to art. (d) The addition of one group (art) changed the apportionment of another group (science) and, by definition, is an occurrence of the new-states paradox.

Now we have seen that Hamilton’s method satisfies the quota rule but can produce the paradoxes we have studied in this section. Jefferson’s, Webster’s, and Adams’s methods can violate the quota rule because they are based on the philosophy that quotas can be conveniently modified. On the other hand, they do not produce the paradoxes we have studied. Can we find a perfect apportionment method that not only satisfies the quota rule but also avoids these paradoxes? Two mathematicians, Michael Balinski and H. Payton Young, proved in 1980 that there is no such method. Their result is called Balinski and Young’s Impossi- bility Theorem.

Balinski and Young’s Impossibility Theorem There is no apportionment method that satisfies the quota rule and avoids paradoxes.

Table 14.36 on page V49 shows which methods violate the quota rule or produce some of the paradoxes studied. 304470-ch14_pV1-V59 11/7/06 10:01 AM Page 49

14.4 Apportionment Objections V49

TABLE 14.36

Characteristic Hamilton’s Jefferson’s Adams’s Webster’s

May violate the quota rule No Yes Yes Yes May result in the Alabama paradox Yes No No No May result in the population paradox Yes No No No May result in the new-states paradox Yes No No No

So there we have it. Just as we could find no perfect voting method, there is also no perfect apportionment method!

EXERCISES 14.4

A The Alabama Paradox 3. Which are the four best theme parks in the United 1. Three Greek letter societies, , , and , have the States? According to Inside Track, a publication numbers of members as shown in the table below. that rates theme parks and attractions, the four If Hamilton’s method is used to apportion seats in best parks are Busch Gardens (B), King’s Island the Pan Hellenic Council, does the Alabama para- (K), Walt Disney World (W), and Six Flags Magic dox occur if the number of seats is increased Mountain (S). The number of votes received by a. from 30 to 31? b. from 60 to 61? each of the parks in a survey of 1020 persons is as shown in the table below. If Hamilton’s method is Society Members used to apportion 71 free tickets to visit the parks on the basis of the number of votes obtained in the 3220 survey, does the Alabama paradox occur if the 5000 number of tickets is increased to 72? If so, which park loses a ticket? 9780 Total 18,000 Park Votes B 405 2. A country consists of four states, A, B, C, and D, with the populations shown in the table below. If K 306 Hamilton’s method is used to apportion the legis- W 204 lature, does the Alabama paradox occur if the number of seats is increased S 105 a. from 104 to 105? b. from 114 to 115? Total 1020

City Population 4. The total number of minority students (in thousands) studying allopathic medicine in 1998 is A 1800 as shown in the following table. If Hamilton’s B 3720 method is used to allocate scholarships on the basis of enrollment, does the Alabama para- C 2330 dox occur when the number of scholarships is D 2150 increased from 24 to 25? Total 10,000 304470-ch14_pV1-V59 11/7/06 10:01 AM Page 50

V50 14 Voting and Apportionment

b. How many seats will each of the counties get Race Students in 10 years? White 44 c. Does the population paradox occur? Explain. Black 5 County Now 10 Years Hispanic 4 A 89,000 97,000 Asian 12 B 12,500 14,500 Total 65 C 22,500 24,700 Source: National Center for Health Statistics. Total 124,000 136,200

a. Which group(s) lose 1 scholarship? 7. The populations of three counties, A, B, and C, at b. Which group(s) gain 1 scholarship? present and in 10 years are shown in the table c. Which group(s) stay the same? below. “Lucky” Fulano, the state governor, suggested a 13-seat legislature apportioned according B The Population Paradox to population using Hamilton’s method. 5. The number of Medicare enrollees (in thousands) a. How many seats will each of the counties get for Delaware (D), Nebraska (N), and Kansas (K) in now? a recent year and the estimated projection for a b. How many seats will each of the counties get future year are as shown in the table below. Sup- in 10 years? pose the federal contribution is $11 million for each c. Does the population paradox occur? Explain. year and Hamilton’s method is used to apportion d. Which of the three counties will be unhappy the money on the basis of the number of enrollees. with reapportionment and why? a. How much will each state get in the most recent year? (Answer to the nearest million.) County Now 10 Years b. How much will each state get using the pro- A 89,000 97,000 jected population? (Answer to the nearest million.) B 125,000 145,000 c. Which state has the higher percent increase of C 225,000 247,000 enrollees, Delaware or Kansas? d. Does the population paradox occur when Total 439,000 489,000 Hamilton’s method is used to allocate the $11 million? Explain. 8. A nation consists of five states, A, B, C, D, and E, with populations 300, 156, 346, 408, and 590, State Number Projection respectively. Suppose 50 seats are to be apportioned on the basis of population using Hamilton’s Delaware 110 122 method. Nebraska 250 300 a. How many seats will each state receive? b. If the populations of states C and E were to Kansas 380 428 increase by 16 and 2, respectively, how many Total 740 850 seats would each state receive then? c. Does the population paradox occur? Explain. Source: Health Care Financing Administration. d. Which of the five states will be unhappy with 6. A state consists of three counties, A, B, and C, reapportionment and why? with the present populations and in 10 years shown in the following table. Suppose 100 seats C The New-States Paradox are to be apportioned on the basis of population 9. A company has two divisions: production (P) and using Hamilton’s method. sales (S). The number of employees in each is a. How many seats will each of the counties get now? 304470-ch14_pV1-V59 11/7/06 10:01 AM Page 51

14.4 Apportionment Objections V51

shown in the table below. There are 41 managers 11. A country has two states, A and B, with the popu- to be apportioned between the two divisions P lation (in hundreds) shown in the table below and and S. 100 seats in the legislature. a. Find each division apportionment using a. Find the apportionments for A and B using Hamilton’s method. Hamilton’s method. b. Suppose a new advertising division (A) with b. Suppose a third state C with a population of 114 employees and 8 new managers is to be 263 (hundred) is added with 6 additional seats. added. Does the new-states paradox occur Does the new-states paradox occur using using Hamilton’s method? Explain. Hamilton’s method? Explain.

Division Number State Population

P 402 A 4470 S 156 B 520 A114 C 263 Total 672 Total 5253

10. A country has two states, A and B, with the populations (in thousands) shown in the table below In Other Words and 30 seats in the legislature. a. Find the apportionments for A and B using 12. Describe in your own words the Alabama paradox. Hamilton’s method. b. Suppose a third state C with a population of 76 13. Describe in your own words the population (thousand) is added with 6 additional seats. paradox. Does the new-states paradox occur using 14. Describe in your own words the new-states Hamilton’s method? Explain. paradox.

State Population

A 268 B 104 C76 Total 448

Research Questions

1. When was the Alabama paradox discovered? Discuss the details of the discovery. 2. When was the population paradox discovered? Which states were involved? Discuss the details of the discovery. 3. When was the new-states paradox discovered? Which states were involved? Discuss the details of the discovery. 4. The results of the 2000 Census led to changes in the makeup of the U.S. House of Representatives. Some states gained delegates, and some states lost delegates. Which ones? Go to www.census.gov/population/www/censusdata/ apportionment.html to find out. 304470-ch14_pV1-V59 11/7/06 10:01 AM Page 52

V52 14 Voting and Apportionment

Chapter 14 Summary

Section Item Meaning 14.1 Plurality method Each voter votes for one candidate and the candidate with the most first-place votes is the winner. 14.1 Plurality with Each voter votes for one candidate. If the candidate receives a runoff method majority of votes, that candidate is the winner. Otherwise, eliminate all but the two top candidates and hold a runoff election. The candidate that receives a majority is the winner. 14.1 Borda count method Each voter ranks the candidates from most to least favorable, with each last-place vote awarded no point; each next to last place is awarded one point, each third from last place is awarded two points, and so on. The candidate that receives the most points is the winner. 14.1 Plurality with Each voter votes for one candidate. If a candidate receives a majority elimination method of votes, that candidate is 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 fewest votes, eliminate all candidates tied for fewest votes.) Repeat this process until a candidate receives a majority. 14.1 Pairwise comparison Each voter ranks candidates from most to least favorable. Each method candidate is then compared with each of the other candidates. If A is preferred to B, A gets 1 point. If B is preferred to A, B gets 1 point. If 1 there is a tie, each candidate receives 2 point. The candidate with the most overall points is the winner. 14.1 Approval voting Voters approve or disapprove each candidate. The candidate with the most approval votes wins. 14.2 Majority criterion If a candidate receives a majority of first-place votes, then that candidate should be the winner. 14.2 Head-to-head criterion If a candidate is favored when compared head-to-head with every other candidate, then that candidate should be the winner. 14.2 Monotonicity criterion If a candidate is the winner of a first election and then gains additional support without losing any of the original support, then the candidate should be the winner of the second election. 14.2 Irrelevant alternatives If a candidate is the winner of an election, and in a second election criterion one or more of the losing candidates are removed, then the winner of the first election should be the winner of the second election. 14.2 Arrow’s Impossibility There is no voting method that will always simultaneously satisfy all Theorem of the four fairness criteria. 14.3 Standard divisor (SD) Total population in a group/total number to be apportioned 14.3 Standard quota (SQ) Population in the group/standard divisor 14.3 Hamilton’s method A method in which the standard quota is rounded down and additional seats are apportioned to the states with the largest fractional part of the standard quotas. 304470_ch14_pV1-V59 11/7/06 2:08 PM Page 53

Chapter 14 Summary V53

Section Item Meaning 14.3 Jefferson’s method A method using a modified divisor so that when each group’s modified quota is rounded down to the nearest integer, the sum of the integers equals the number of items to be apportioned. 14.3 Webster’s method A method using a modified divisor so that when each group’s modified quota is rounded in the usual manner, the sum of the integers equals the number of items to be apportioned. 14.3 Adams’s method A method using a modified divisor so that when each group’s modified quota is rounded up to the nearest integer, the sum of the integers equals the number of items to be apportioned. 14.4 Quota rule The apportionment for every group under consideration should always equal either the group’s upper quota or its lower quota. 14.4 Alabama paradox Occurs when an increase in the total number of items to be apportioned results in a loss of items for a group. 14.4 Population paradox Occurs when the population of group A increases faster than the population of group B, yet A loses items to group B. 14.4 New-states paradox Occurs when the addition of a new group changes the apportionment of another group. 14.4 Balinski and Young’s There is no apportionment method that satisfies the quota rule and Impossibility Theorem avoids paradoxes.

Research Questions

1. What if you had an election and nobody came? The electoral commission in England is working on new election methods that would increase voter turnout. Research which methods are being used to do that and what the voter turnout was in last year’s local elections and in the 2005 general election in England. Is the turnout greater or less than that in U.S. local and general elections? 2. What election methods are used for events other than presidential elections? Find out! What are the procedures and voting methods used for a. the Grammy Awards? b. the Latin Grammy Awards? c. the Academy Awards? d. the Nobel Prize? e. the Heisman Trophy? 3. Write a research paper chronologically discussing all voting methods that have been used to apportion the U.S. House of Representatives. What method is currently used and what are the current objections to this method? 4. What impact did the 2000 census have on apportionment? Write a research paper on this question, paying particular attention to a. which states gained two seats as a result. b. which states lost two seats as a result. 304470-ch14_pV1-V59 11/7/06 10:01 AM Page 54

V54 14 Voting and Apportionment

References (for most of the preceding information) http://news.bbc.co.uk/1/hi/uk_politics/2988307.stm http://heismanmemorialtrophy.com/ www.oscars.org/index.html www.infoplease.com/ipa/A0150533.html www.almaz.com/nobel/ www.improb.com/ig/ig-pastwinners.html www.census.gov/prod/2001pubs/c2kbr01-7.pdf

Chapter 14 Practice Test

1. The results of an election involving four candidates, A, B, C, and D, are 10875 shown in the table to the left. ACDD a. Did any of the candidates receive a majority? b. Which candidate is the plurality winner? CDCB c. Which candidate comes in second? B BBC d. Which candidate comes in last? D AAA 2. Using the plurality with runoff method, who is the winner of the election in problem 1? 3. Using the Borda count method, who is the winner of the election in problem 1? 4. Using the plurality with elimination method, who is the winner of the election in problem 1? 5. 539 The results of an election involving three candidates, A, B, and C, are shown in the table to the left. Using the pairwise comparison method, who ABC is the winner a. between A and B? BCA b. between A and C? CAB c. between B and C? d. of the election? 6. The results of a hypothetical election using approval voting are summarized in the table below. An X indicates that the voter approves of the candidate; a blank indicates no approval.

VOTERS Candidates Thomas Uma Vera Walter Yvette Zoe

Adams X X X X X Barnes X X X Collins X X

a. Who is the winner using approval voting? b. Who is the winner if Adams drops out of the race? 304470-ch14_pV1-V59 11/7/06 10:01 AM Page 55

Chapter 14 Practice Test V55

7. An election to select their spring break destination, Aruba (A), Bahamas Number (B), or Cancun (C), is conducted among 64 students. The results are as of Voters shown in the table to the left. Which destination should be selected under Place 36 16 12 the specified method? Does the method satisfy the majority criterion? a. The plurality method First B A A b. The Borda count method Second A B C c. The plurality with elimination method d. The pairwise comparison method Third C C B 8. A restaurant conducted a survey among its customers to select their favorite entree from chicken (C), pork (P), turkey (T), and vegetarian (V). The number of votes for each was as shown in the table below. Which entree should be selected under the specified method, and does the method satisfy the head-to-head criterion? a. Head-to-head b. Plurality c. Borda count d. Plurality with elimination e. Pairwise comparison

Number of Voters Place 15 25 29 30 45

First V V T P C Second T T V V P Third P C P T T Fourth C P C C V

9. A group of 100 students ranked the three courses they liked best as shown Number in the table to the left. If the plurality method is used to select the top of Voters course, does the plurality method satisfy Place 56 24 20 a. the majority criterion? Explain. b. the head-to-head criterion? Explain. First C B A 10. As in problem 9, does the plurality method satisfy Second B C B a. the monotonicity criterion if we assume a second election is undertaken Third A A C and C gains additional support without losing any of the original support? Explain. b. the irrelevant alternatives criterion if we assume that either A or B drops out and a second election is undertaken? Explain. 11. Explain in your own words the meaning of Arrow’s Impossibility Theorem. 12. Five universities received $162 million in total general education and general fund revenues. The number of FTEs (to the nearest 1000) in each of the five universities is as shown in the table below.

A B C D E Total

15,000 8500 6500 6000 4500 40,500

Find the standard divisor SD and the standard quota SQ for each university. 304470-ch14_pV1-V59 11/7/06 10:01 AM Page 56

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13. A college is composed of five departments: mathematics (M), English (E), languages (L), art (A), and chemistry (C), with the number of faculty members shown in the table below. If 100 new positions are to be apportioned using Hamilton’s method and on the basis of the number of faculty in each department, what is the actual number of positions apportioned to each of the departments?

Department Faculty MQ Actual Number

Mathematics 30 English 60 Language 24 Art 16 Chemistry 50 Totals 180 90

14. Use the Jefferson method to apportion 90 (instead of 100) positions in problem 13. (Hint: The modified divisor must be slightly less than 2.) 15. In 1830, the population of Florida was 34,730. Use Webster’s method to find the number of seats for Florida if the modified divisor was 49,800. 16. Use Adams’s method to apportion 90 faculty positions in problem 13. (Hint: The modified divisor must be slightly more than 2.) 17. Consider a country with a population of 100,000 and three states in the legislature, A, B, and C, with the populations shown in the table below. Suppose 50 seats are apportioned using Hamilton’s method as shown in the first table. Fill in the blanks in the second table and determine if the Alabama paradox occurs when the number of seats is increased from 50 to 51 using Hamilton’s apportionment method. Explain your answer.

State Population SQ RD Extra Final

A 50,400 50,400/2000 25.2 25 0 25 B 46,600 46,600/2000 23.3 23 0 23 C 3000 3000/2000 1.5 1 1 2 Total 100,000 49 50

State Population SQ RD Extra Final

A 50,400 B 46,600 C 3000 Total 100,000 51

18. The table on page V57 shows the number of students taking mathematics, English, and science courses during the fall and spring semesters. Suppose that 100 full-time teaching positions were apportioned during the fall semester using Hamilton’s method with the results shown. 304470-ch14_pV1-V59 11/7/06 10:01 AM Page 57

Answers to Practice Test V57

a. How many positions will each department get in the spring using Hamilton’s method? b. Is the apportionment fair? Explain your answer. c. Is this apportionment an example of the population paradox? Explain.

Fall Semester Spring Semester Subject Number Fall SQ Positions Number Spring New SQ Positions

Math 476 9.5180 10 484 9.6338 9 English 975 19.4961 19 990 19.7054 20 Science 3550 70.9858 71 3550 70.6608 71 Total 5001 100 5024 100

19. A country has two states, A and B, with the population (in thousands) State Population shown in the table to the left and 41 seats in the legislature. A 804 a. Find the apportionments for A and B using Hamilton’s method. B 312 b. Suppose a third state C with a population of 228 (thousands) is added C 228 with 8 additional seats. Does the new-states paradox occur using Hamilton’s method? Explain. Total 1344 20. Explain the meaning of Balinski and Young’s Impossibility Theorem in your own words.

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1. a. No b. D 1 14.1 1 V4–V5 c. A d. B 2. D wins the runoff 20 to 10. 2 14.1 2 V5 3. C wins with 63 points. 3 14.1 3 V6 4. D 4 14.1 4 V6–V7 5. a. A wins 14 to 3. 5 14.1 5 V8 b. C wins 12 to 5. c. C wins 9 to 8. d. C is the winner. 6. a. Adams b. Barnes 6 14.1 6 V9 7. a. B. Yes 7 14.2 2 V17–V18 b. A. No. B has the majority but A wins under the Borda count method. c. B. Yes d. B. Yes 8. a. P. Yes b. C. No 8 14.2 3 V19–V20 c. T. No d. V. No e. P. Yes 9. a. Yes. C has the majority and wins under the 9 14.2 5 V23 plurality method. b. Yes. C has the majority and wins head-to- head against all other candidates. 304470-ch14_pV1-V59 11/7/06 10:01 AM Page 58

V58 14 Voting and Apportionment

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10. a. Yes. C will still win the second election 10 14.2 5 V23 under the plurality method. b. Yes. C will still win the second election under the plurality method when either A or B drops out. 11. There is no voting method that will always 11 14.2 V24 simultaneously satisfy each of the four fairness criteria. 12. SD 40,500/162,000,000 0.00025 12 14.3 1 V31 Standard Quotas (in millions) For A, 15,000/0.00025 60 For B, 8500/0.00025 34 For C, 6500/0.00025 26 For D, 6000/0.00025 24 For E, 4500/0.00025 18 Note that the total is $162 million.

13. 13 14.3 3 V32–V34 SQ RD Actual Number

30/1.8 16.67 16 16 1 17 60/1.8 33.33 33 33 24/1.8 13.33 13 13 16/1.8 8.89 8 8 1 9 50/1.8 27.78 27 27 1 28 100 97 100

14. Use 1.95 as the modified divisor, then round 14 14.3 4 V35 down.

MQ Actual

30/1.95 15.38 15 60/1.95 30.77 30 24/1.95 12.31 12 16/1.95 8.21 8 50/1.95 25.64 25

15. 34,730/49,800 0.6974, or 1 14 14.3 6 V37 16. Use 2.05 as the modified divisor, then round up. 16 14.3 7 V38

MQ Actual 30 2.05 14.63 15 60 2.05 29.27 30 24 2.05 11.71 12 16 2.05 7.80 8 50 2.05 24.39 25 304470-ch14_pV1-V59 11/7/06 10:01 AM Page 59

Answers to Practice Test V59

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17. 17 14.4 2 V45

SQ Extra Final 50,400 1960.78 25.70 25 1 26 46,600 1960.78 23.77 23 1 24 3000 1960.78 1.53 1 0 1

The Alabama paradox occurs because even though the number of seats was increased from 50 to 51, state C lost 1 seat, from 2 to 1. 18. a. 18 14.4 3 V46–V47

Position Number New SQ Positions

10 484 9.6338 9 19 990 19.7054 20 71 3550 70.6608 71

b. No. c. Yes. Mathematics lost 1 position and English gained 1 even though mathematics was 8 growing at a faster rate, 476 0.01681, than 15 English, 975 0.01538. 19. a. A gets 30 seats, and B 11. 19 14.4 4 V47–V48 b. Yes. The addition of the new state C caused A to lose 1 seat (from 30 to 29) and B to gain 1 (from 11 to 12). 20. There is no apportionment method that satisfies 20 14.4 V47–V48 the quota rule and avoids all paradoxes.