304470-ch14_pV1-V59 11/7/06 10:01 AM Page 1

C H A P T E R

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 can-
didate with the absolute majority of electoral votes is declared vice pres-
ident. At noon on January 20, the duly elected president and vice presi-
dent 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
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V2 14 Voting and Apportionment

14.1 Voting Systems

Monsieur Butterfly and Pizza Too
G START
IN E When you vote in a presidential election, you are not directly voting for the pres-
TT

D
ident! You are actually voting for electors, individuals who cast the electoral
GE

votes on behalf of their party and states. They are the ones who elect the presi-
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
so-called butterfly ballot used in Palm Beach County, Florida, during the 2000
The Granger Collection

Marie-Jean
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)
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
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14.1 Voting Systems V3

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 pref-
erence 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 sum-
marized as in Table 14.1.
PhotoDisc/Getty Images
TA B L E 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 major-
ity 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
TA B L E 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.
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V4 14 Voting and Apportionment

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!

TA B L E 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.

EX AM P LE 1
Using the Plurality Method

TA B L E 14 . 4 The Math Club conducted an election, and the results were as shown in
Table 14.4.
7 5 4 2
(a) Did any of the rankings get the majority of the votes?
A S P P
(b) Which topping is the plurality winner?
S P S M
(c) Which topping comes in second?
M M M S
(d) Which topping comes in last?
P A A A
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.
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14.1 Voting Systems V5

(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 plural-
ity 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.

EX AM P LE 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
TA B L E 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
7 5 4 2
actually do it with a pencil) cross out all the S and M entries. Now look at
A S P P 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
S P S M
in these columns we are only concerned with the fact that P is preferred
M M M 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
P A A A
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?
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V6 14 Voting and Apportionment

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 select-
ing 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.

EX AM P LE 3
Using the Borda Count Method

Find the winner of the election in Example 1 using the Borda count method.
Solution
TA B L E 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 7 5 4 2
votes, worth 3 points each, a total of 3  7
21 points. Sausage (S) gets
3 A S P P 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 S P S M
10 points in column 2, 12 in 3 and 6 in 4 for a total of 28 points. Finally,
1 M M M 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 P A A A
(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.

EX AM P LE 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.
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14.1 Voting Systems V7

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

TA B L E 14 . 7 TA B L E 14 . 8 TA B L E 14 . 9

7 5 4 2 7 5 4 2 7 5 6

A S P P A S P P A P P

S P S M S P S S P A A

M M M S P A A A

P A A A

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 win-
ner 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 plu-
rality 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 can-
didate 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.
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V8 14 Voting and Apportionment

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 candi-
dates, 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.

EX AM P LE 5
Using the Pairwise Comparison Method

The results of an election involving three candidates, A, B, and C, are shown in
TA B L E 14 .10 Table 14.10. Who wins the election using the pairwise comparison method?
2 3 4 Solution
To determine the winner using the pairwise comparison method, we have to
A B C compare A and B, A and C, and B and C.
B C A Suppose the election is between just A and B (leave C out).
C A B 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 fundamen-

tally different types of voting methods: preferential (those using a preference
table) and nonpreferential. We will now discuss a nonpreferential voting method:
approval voting.
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14.1 Voting Systems V9

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.

EX AM P LE 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?

TA B L E 14 .11

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

W A D A A D D D D

X A A D D A D A A

Y D D A D D A A A

Z D D A D A A D A

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.
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V10 14 Voting and Apportionment

E X E R C I S E S 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 R R R R
R Bush 2,911,872 Third S S P P
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 plu-
rality with runoff method?
2. Who is the winner in Florida under the plurality
method? 6. The preference schedule below shows the rank-
ings 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
6 7 3 4 First A C D
D C A B Second B B A
C B D A Third D D B
B D B C Fourth C A C
A A C D
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?
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14.1 Voting Systems V11

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 elimi-
nation 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
or 12 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
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V12 14 Voting and Apportionment

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 S S S I M BG BC BC I
304470-ch14_pV1-V59 11/7/06 10:01 AM Page 13

14.1 Voting Systems V13

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 coun-
tries with candidate cities were ineligible to vote while Ace
their nation’s city was still in the running.)

L P MA MO NY

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

V14 14 Voting and Apportionment

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-
A B C C D
ing fee if you can show that Ace is really the num-
B A D B A ber one cola. Find a point assignment for the
Borda count method in which Ace comes in first.
C D B A B
(Hint: A gets a lot of second-place votes, so we
D C A D C want to make second place worth proportionally
more. Remember, first place must still be worth
more than second, so make the gap between sec-
ond 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

14.2 Voting Objections V15

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 presiden-
tial 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 elec-
tion results most accurately? Write a short essay about your conclusions.

14.2 Voting Objections

G START Disaster 2000

IN In the preceding section we studied five preferential voting systems: plurality,
E
TT

plurality with runoff, Borda count, plurality with elimination, and pairwise com-
GE

parison. We also studied a nonpreferential voting system: approval voting. As

we pointed out, all these systems have advantages and disadvantages and some-
times can produce different winners. Let us look at an actual example—the 2000
presidential election (Table 14.12).

TA B L E 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) crite-
rion, the monotonicity criterion, and the irrelevant alternatives criterion.
304470-ch14_pV1-V59 11/7/06 10:01 AM Page 16

V16 14 Voting and Apportionment

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.

EX AM P LE 1
Using the Majority Criterion

La Cubanita Restaurant is conducting a survey to find out which is the most pop-
ular omelet among the western (W), bacon (B), and ham (H) omelets. The results
of the survey are shown in Table 14.13.

TA B L E 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

14.2 Voting Objections V17

TA B L E 14 .14 EX AM P LE 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 vio-
late 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, sec-
ond, 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 gen-
eral, 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

V18 14 Voting and Apportionment

(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 mathemat-
ics 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 pre-
ferred 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

14.2 Voting Objections V19

EX AM P LE 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

TA B L E 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

V20 14 Voting and Apportionment

(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.


304470-ch14_pV1-V59 11/7/06 10:01 AM Page 21

14.2 Voting Objections V21

This example shows that the plurality, Borda count, and plurality with elim-
ination 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 can-
didate 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.

EX AM P LE 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.

TA B L E 14 .16 First Election TA B L E 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

V22 14 Voting and Apportionment

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
TA B L E 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

TA B L E 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.

EX AM P LE 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?

TA B L E 14 . 2 0

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 major-
ity (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 elec-
tion 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.

TH EOR EM 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 fair-
ness criteria and indicate in Table 14.22 on page V25 which of the voting meth-
ods we have studied satisfies a particular criterion.

TA B L E 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

TA B L E 14 . 2 2

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

E X E R C I S E S 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 Cuba
n
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 compar- ison method?
ison method?
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V26 14 Voting and Apportionment

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 M A M A 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
C The Monotonicity Criterion
count method?
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
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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?
Place 14 4 10 1 8
c. Who wins using the pairwise comparison
method? First A B C C D
d. Does any candidate beat every other candidate
Second B D B D C
one-on-one, that is, in a head-to-head compar-
ison? If so, which one? Third C C D B B
e. Who wins using the Borda count method?
Fourth D A A A A
f. Which fairness criteria, if any, are violated?
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 compari-
son? 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
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V28 14 Voting and Apportionment

12. The preference table below gives the results of an

Number of Voters
election among three candidates, A, B, and C.
Place 3 3 1 1 3 3 1 1
Number of Voters
First B A C C B A B E
Place 7 8 10 4
Second A C B E A D A D
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 elimina-
tion 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.
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14.3 Apportionment Methods V29

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 representa-
tion of the states in the legislative branch. Smaller states wanted equal represen-
tation. 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 pro-
portional 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
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V30 14 Voting and Apportionment

We shall study the Hamilton, Jefferson, and Webster methods and omit the Hill-
Huntington method because of its complexity. Instead, we will examine a simi-
lar 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 START
IN The community college system in the State of Florida consists of 28 colleges.
E
TT

In 1999, the general budget allocation for the system amounted to more than
D
GE

1 billion dollars—$1,102,817,849 to be exact. How can we fairly distribute the

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 num-
ber 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 pre-
cise, we define the standard divisor (SD) and the standard quota (SQ) as follows:
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14.3 Apportionment Methods V31

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

EX AM P LE 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.

TA B L E 14 . 2 3

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

a means that 1 FTE gets $4000 b

1
4000
The standard quota SQ for each college is given in Table 14.24.

TA B L E 14 . 2 4

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). 
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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 proce-
dure 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.

EX AM P LE 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.

TA B L E 14 . 2 5

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

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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 popula-
tion 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.

EX AM P LE 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 num-
ber of patients in each unit divided by 1.8, as shown in column 3. Next, we
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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.

TA B L E 14 . 2 6

Unit Patients SQ Rounded Down Actual Number

Patient Rooms 231-276
Patient Rooms 211-220
Medical 15 15/1.8
8.330 8 8
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.
North Elevators
Cardiac 12 12/1.8
6.670 6 61
7
I.C.U. & C.C.U. Waiting Transitional 8 8/1.8
4.440 4 4

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!

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14.3 Apportionment Methods V35

EX AM P LE 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 num-
ber of recliners apportioned in the last column.

TA B L E 14 . 2 7
Rounded Modified Rounded
Unit Patients SQ Down Quota Down

Medical 15 15
1.8
8.33 8 15
1.7
8.82 8
Surgical 30 30
1.8
16.67 16 30
1.7
17.65 17
Cardiac 12 12
1.8
6.67 6 12
1.7
7.06 7
Transitional 8 8
1.8
4.44 4 8
1.7
4.71 4
Progressive 25 25
1.8
13.89 13 25
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. 

EX AM P LE 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

TA B L E 14 . 2 8
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 divi-
sor. 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.
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14.3 Apportionment Methods V37

EX AM P LE 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.
TA B L E 14 . 2 9

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 presi-
dent 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 representa-
tion 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 pro-
cedure for Adams’s apportionment method.
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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.

EX AM P LE 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.

TA B L E 14 . 3 0

Unit Patients Modified Quota Rounded Up

Medical 15 15
1.9
7.89 8
Surgical 30 30
1.9
15.79 16
Cardiac 12 12
1.9
6.32 7
Transitional 8 8
1.9
4.21 5
Progressive 25 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.
TA B L E 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 quo-
tas 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.

E X E R C I S E S 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 appor-
tioned 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 universi-
ties 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?
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V40 14 Voting and Apportionment

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.
New Jersey 2061
a. 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.
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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.
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V42 14 Voting and Apportionment

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 apportion-
ment 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.
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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/?p
20 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.
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V44 14 Voting and Apportionment

Sweet Home Alabama

G START
IN The Alabama paradox first surfaced after the 1870 census. With 270 members

E
TT

in the House of Representatives, Rhode Island got 2 representatives, but

D
GE

when the house size was increased to 280, Rhode Island lost a seat. After the
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.

EX AM P LE 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 stan-
dard 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
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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.

EX AM P LE 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 apportion-
ment 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
1000. The standard quotas SQs and the rounded-down values RDs are
50,000

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.

TA B L E 14 . 3 2

State Population SQ RD Extra Final

A 25,200 25,200
1000
25.2 25 0 25

B 23,300 23,300
1000
23.3 23 0 23

C 1500 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

divisor is 50,000
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.

TA B L E 14 . 3 3

State Population SQ RD Extra Final

A 25,200 25,200
980.39
25.70 25 1 26

B 23,300 23,300
980.39
23.77 23 1 24

C 1500 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
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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.)

EX AM P LE 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.

TA B L E 14 . 3 4

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

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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.
 951
(c) No. The rate of growth for mathematics was 961 951
951
10

1.05%.
 1949
The rate of growth for English was 19691949
1949
20

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 appor-
tionment (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.

EX AM P LE 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 mathemat-
ics, English, science, and art.
(b) How does the new fall apportionment compare with the original fall appor-
tionment 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
105
100.95.
students and 105 positions, so the new standard divisor is 10,600
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V48 14 Voting and Apportionment

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

TA B L E 14 . 3 5

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 apportion-
ments. 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

TA B L E 14 . 3 6

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!

E X E R C I S E S 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,
K 306
with the populations shown in the table below. If
Hamilton’s method is used to apportion the legis- W 204
lature, does the Alabama paradox occur if the
S 105
number of seats is increased
a. from 104 to 105? b. from 114 to 115? Total 1020

City Population 4. The total number of minority students (in thou-

A 1800
sands) studying allopathic medicine in 1998 is
as shown in the following table. If Hamilton’s
B 3720 method is used to allocate scholarships on the
C 2330
basis of enrollment, does the Alabama para-
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, sug-
gested 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 mil-
lion.) 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 appor-
tioned 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?
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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

A 114 C 263

Total 672 Total 5253

10. A country has two states, A and B, with the popu-

lations (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

C 76

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.
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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
there is a tie, each candidate receives 12 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
10 8 7 5
shown in the table to the left.
A C D D a. Did any of the candidates receive a majority?
b. Which candidate is the plurality winner?
C D C B
c. Which candidate comes in second?
B B B C d. Which candidate comes in last?
D A A A 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 3 9
5. 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
A B C is the winner
B C A
a. between A and B?
b. between A and C?
C A B 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 sup-
port? 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

V56 14 Voting and Apportionment

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

State Population
19. A country has two states, A and B, with the population (in thousands)
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
Total 1344
Hamilton’s method? Explain.
20. Explain the meaning of Balinski and Young’s Impossibility Theorem in
your own words.

Answers to Practice Test

IF YOU
ANSWER MISSED REVIEW

Question Section Example(s) Page(s)

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

IF YOU
ANSWER MISSED REVIEW

Question Section Example(s) Page(s)

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 81
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

IF YOU
ANSWER MISSED REVIEW

Question Section Example(s) Page(s)

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
English, 975
0.01538.
15

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.