12
Mathematics
and Politics
Copyright © 2015, 2011, 2008 Pearson Education, Inc.
Chapter 12, Unit A, Slide 1
Unit 12A
Voting: Does the
Majority Always Rule?
Copyright © 2015, 2011, 2008 Pearson Education, Inc.
Chapter 12, Unit A, Slide 2
Majority Rule
The simplest type of voting involves only two
choices. With majority rule, the choice receiving
more than 50% of the vote is the winner. Some
properties of majority rule are listed below.



Every vote has the same weight.
There is symmetry between the candidates.
If a vote for the loser were changed to a vote for
the winner, the outcome of the election would not
be changed.
Copyright © 2015, 2011, 2008 Pearson Education, Inc.
Chapter 12, Unit A, Slide 3
Example
The table shows the 2000 presidential election
official results. Describe the outcomes of the
popular and electoral votes.
Copyright © 2015, 2011, 2008 Pearson Education, Inc.
Chapter 12, Unit A, Slide 4
Example (cont)
Solution
The difference in popular vote totals between Gore
and Bush was
50,999,897 – 50,456,002 ≈ 543,895
Gore won the popular vote over Bush by a margin of
more than a half million votes. Neither Gore nor Bush
won a majority of the popular vote because a small
percentages of the total vote that went to Nader
and other candidates.
Copyright © 2015, 2011, 2008 Pearson Education, Inc.
Chapter 12, Unit A, Slide 5
Example (cont)
50,999,897
 0.4838  48.38%
Gore’s percentage:
105,405,100
50,999,897
Bush’s percentage:
 0.4838  48.38%
105,405,100
In the electoral vote, Bush won 271 out of the total 538
votes available, which gave him a slight majority of
271
 0.5037  50.37%
538
Because he won a majority of the electoral vote, Bush
became the President of the United States.
Copyright © 2015, 2011, 2008 Pearson Education, Inc.
Chapter 12, Unit A, Slide 6
Variations on Majority Rule
In some cases, a candidate or issue must receive
more than a majority of the vote to win ─ such as
60% of the vote, 75% of the vote, or a unanimous
vote. In these cases, a super majority is required.

A 60% super majority is required to end a
filibuster in the U.S. Senate.

A 2/3 super majority is required by both the House
and Senate to amend the U.S. Constitution.
Copyright © 2015, 2011, 2008 Pearson Education, Inc.
Chapter 12, Unit A, Slide 7
Example
Evaluate the outcome in each of the following
cases.
a. Of the 100 senators in the U.S. Senate, 59 favor
a new bill on campaign finance reform. The other 41
senators are adamantly opposed and start a
filibuster. Will the bill pass?
b. A criminal conviction in a particular state requires
a vote by 3/4 of the jury members. On a ninemember jury, seven jurors vote to convict. Is the
defendant convicted?
Copyright © 2015, 2011, 2008 Pearson Education, Inc.
Chapter 12, Unit A, Slide 8
Example (cont)
c. A proposed amendment to the U.S. Constitution
has passed both the House and the Senate with
more than the required 2/3 super majority. Each
state holds a vote on the amendment. The
amendment garners a majority vote in 36 of the 50
states. Is the Constitution amended?
d. A bill limiting the powers of the President has the
support of 73 out of 100 senators and 270 out of
435 members of the House of Representatives. But
the President promises to veto the bill if it is passed.
Will it become law?
Copyright © 2015, 2011, 2008 Pearson Education, Inc.
Chapter 12, Unit A, Slide 9
Example (cont)
Solution
a. The filibuster can be ended only by a vote of 3/5
of the Senate, or 60 out of the 100 senators. The 59
senators in favor of the bill cannot stop the filibuster.
The bill will not become law.
b. Seven out of nine jurors represents a super
majority of 7/9 = 77.8%. This percentage is more
than the required 3/4 (75%), so the defendant is
convicted.
Copyright © 2015, 2011, 2008 Pearson Education, Inc.
Chapter 12, Unit A, Slide 10
Example (cont)
Solution
c. The amendment must be approved by 3/4, or 75%, of
the states. But 36 out of 50 states represents only
36/50 = 72%, so the amendment does not become part
of the Constitution.
d. The bill has the support of 73/100 = 73% of the
Senate and 270/435 = 62% of the House. But overriding
a presidential veto requires a super majority vote of 2/3
= 66.7% in both the House and the Senate. The 62%
support in the House is not enough to override the veto,
so the bill will not become law.
Copyright © 2015, 2011, 2008 Pearson Education, Inc.
Chapter 12, Unit A, Slide 11
Example
The table shows the results for the three major
candidates in the 1992 U. S. presidential election.
Analyze the outcome. Could Bush have won if Perot
had not run?
Copyright © 2015, 2011, 2008 Pearson Education, Inc.
Chapter 12, Unit A, Slide 12
Example (cont)
Solution
Bill Clinton won a plurality, but there was no majority
winner of the popular vote. If Perot had not run, the
19% of the popular votes would have been divided
between Clinton and Bush. (Bush needed an
additional 12.3% of the popular vote to reach a 50%
total 37.7% + 12.3% = 50%). If 12.3/19 ≈ 65% of
the Perot voters preferred Bush to Clinton, Bush
could have won the popular vote. Note: winning the
popular vote does not guarantee winning the
electoral vote.
Copyright © 2015, 2011, 2008 Pearson Education, Inc.
Chapter 12, Unit A, Slide 13
Preference Schedule
Tabulating results from ballots requires a special
type of table, called a preference schedule, that tells
us how many voters chose each particular ranking
order among the candidates.
Copyright © 2015, 2011, 2008 Pearson Education, Inc.
Chapter 12, Unit A, Slide 14
Voting Methods with
Three or More Choices
Plurality method: The candidate with the most
first-place votes wins.
Single (top-two) runoff method: The two
candidates with the most first-place votes have a
runoff. The winner of the runoff is the winner of
the election.
Sequential runoff method: A series of runoffs is
held, eliminating the candidate with the fewest
first-place votes at each stage. Runoffs continue
until one candidate has a majority of the firstplace votes and is declared the winner.
Copyright © 2015, 2011, 2008 Pearson Education, Inc.
Chapter 12, Unit A, Slide 15
Voting Methods with
Three or More Choices
Point system (Borda count): Points are awarded
according to the rank of each candidate on each
ballot (first, second, third, …). The candidate with
the most points wins.
Method of pairwise comparisons (Condorcet
method): The candidate who wins the most
pairwise (one-on-one) contests is the winner of
the election.
Copyright © 2015, 2011, 2008 Pearson Education, Inc.
Chapter 12, Unit A, Slide 16
Example
Consider the following three-candidate preference
schedule for candidate A, B, and C. Can you find a
winner through pairwise comparisons? Explain.
Copyright © 2015, 2011, 2008 Pearson Education, Inc.
Chapter 12, Unit A, Slide 17
Example (cont)
Solution
Three pairwise comparisons are possible: A versus
B, B versus C, and A versus C. Columns 1 and 2
show A ranked ahead of B, but B is ahead of A in
column 3. Therefore, A beats B in a one-on-one
contest by 26 to 10. In the B versus C contest, B is
ranked ahead of C in columns 1 and 3, so B beats C
by 24 to 12. Similarly, A is ranked ahead of C only in
column 1, so C beats A by 22 to 14.
Copyright © 2015, 2011, 2008 Pearson Education, Inc.
Chapter 12, Unit A, Slide 18
Example (cont)
Summarizing, we have the following results for the
pairwise comparisons:
• A beats B
• B beats C
• C beats A
Because A beats B and B beats C, the first two
results suggest that A should also beat C. However,
the third result shows that A actually loses to C. The
results illustrate what is often called the Condorcet
paradox, in which pairwise comparisons do not
produce a clear winner.
Copyright © 2015, 2011, 2008 Pearson Education, Inc.
Chapter 12, Unit A, Slide 19
Example
Answer the following questions to be sure you
understand the preference schedule.
Copyright © 2015, 2011, 2008 Pearson Education, Inc.
Chapter 12, Unit A, Slide 20
Example (cont)
a. How many voters ranked candidates in the order E,
B, D, C, A?
b. How many voters had candidate E as their first
choice?
c. How many voters preferred candidate C over
candidate A?
Solution
a. The order E, B, D, C, A appears in the second-to-last
column of the table, and the last entry of that column
shows that 4 voters preferred this order.
Copyright © 2015, 2011, 2008 Pearson Education, Inc.
Chapter 12, Unit A, Slide 21
Example (cont)
Solution
b. Candidate E appears as first choice in the last two
columns, which represent a total of 4 + 2 = 6 voters.
c. The first column shows candidate A as first choice
and candidate C as fourth choice, which means the 18
voters who chose this order prefer candidate A to
candidate C. All the remaining columns show candidate
C ranked higher than candidate A; for example, the
second column shows 12 voters who put C in fourth
place and A in fifth place. Therefore, the total number of
voters who prefer C over A is 12 + 10 + 9 + 4 + 2 = 37.
Copyright © 2015, 2011, 2008 Pearson Education, Inc.
Chapter 12, Unit A, Slide 22