Chapter 12 a slide show summer 2014

Chapter 12AVoting Theory

(Mathematics and Politics)

Majority Rule

• Involves only two choices (no more than that)

• Every vote has the same weight

• Whoever has the most votes (over 50%) wins

Super Majority Rule

• Takes more than 50% (more than half) to win or for the vote to pass successfully

Super Majority: Comments

• Examples:• Senate: Requires a 60% vote for a law to pass

• Trial by jury: Might require 75% to 100% vote

• Amendment to U.S. Constitution: • Requires a 2/3 vote of the Senate

• Then requires a 2/3 vote of the House of Reps.

• Then requires a 3 / 4 vote of the 50 states

U.S. Presidential ElectionsThe Electoral College votes equals the same number of members in the U.S. Congress (538).

North Carolina has 2 Senators and 13 Representatives = 15 Electoral Votes

The winner of our vote ( “the popular vote”) in N.C. gets all of the electoral votes (no matter what margin of victory.

Only Nebraska and Maine are exceptions to this “winner-take-all” system.It takes a majority (270 electoral votes) to win the election.

Voting with Three or More Choices

• Plurality (most votes gets the win)

• Preference voting (various methods or rules)• Plurality (again)

• Single runoff

• Sequential runoffs

• Point system (Borda count)

• Pairwise comparisons (Condorcet count)

• Sometimes there is no winner with these methods

Example: Math Club Election

• Rather than select one person, this ballot allows people to vote their preferences in order from “1st choice” to “whatever”

• There are 120 possible permutations (arrangements of your voting choices) for this ballot

• We’ll simplify the situation for this lecture

Ballot: From these students:AlonzoBonnieCarolDavidEugene

Select your choices in order:

1st Choice ______2nd Choice ______3rd Choice ______4th Choice ______5th Choice ______

The 55 ballots are collected, grouped, and we make…

Ballot1st A2nd D3rd E4th C5th B


















Ballot1st B2nd E3rd D4th C5th A












Ballot1st C2nd B3rd E4th D5th A










Ballot1st D2nd C3rd E4th B5th A









Ballot1st E2nd B3rd D4th C5th A




Ballot1st E2nd C3rd D4th B5th A


… a preference schedule For example, 4 people used their ballots to select Eugene as their 1st choice, Bonnie as their 2nd choice, David as their 3rd choice, Carol as their 4th

choice, and Alonzo as their 5th

choice.

Preference Schedule for the Club Election

1st Choice A B C D E E

2nd Choice D E B C B C

3rd Choice E D E E D D

4th Choice C C D B C B

5th Choice B A A A A A

Number of Voters

18 12 10 9 4 2







Believe it or not, there are five general voting methods that are used to choose a winner.• The plurality method

• The single runoff

• The sequential runoff

• The point system (Borda count)

• The pairwise comparisons (Condorcet count)

• You can actually get a different winner (depending on which voting method is used)

Plurality Method

• Count the 1st place votes only

• The person with the most 1st place votes is declared the winner

• Voters don’t really have to “rank” or “prefer” the candidates – only the voter’s “first choice” is needed

Plurality MethodPreference Schedule for the Club Election






Number of Voters

18 12 10 9 4 2

In our example, Alonzo has 18 first place votes, Bonnie has 12 first place votes, Carol has 10 first place votes, David has 9 first place votes, and Eugene has 6 first place votes (4 + 2).

Alonzo is the winner in the plurality method of voting. Note that 37 voters (12 + 10 + 9 + 4 + 2) did not select Alonzo as a first choice, so this is not a “majority” decision.

Plurality Method: Comments

• Applications / Advantages

• Most elections for political office in the United States are decided using the plurality method when there are three or more choices involved – an extension of the majority rule style

• “Costwise” this is the least expensive and the least complicated kind of election –voters don’t usually like complex ballots

• Drawbacks

• The winner might not be the “first choice” that most people prefer

• With lots of candidates, the percentage of votes needed to win using the plurality method can be a really low number

Single Runoff

• STEP 1: Determine the top two candidates (choices) with the most 1st

place votes

• STEP 2: Then look at the “other places” where the top two candidates (choices) were ranked

• Get the winner from there

Single RunoffPreference Schedule for the Club Election






Number of Voters

18 12 10 9 4 2

STEP 1: In our example, Alonzo has 18 first place votes and Bonnie has 12 first place votes. So Alonzo

and Bonnie will be in the runoff.

STEP 2: Now look at the other 25 voters. All of them (10 + 9 + 4 + 2) prefer Bonnie over Alonzo.So Bonnie wins: Bonnie 37

Alonzo 18

Single Runoff: Comments

• Sometimes works well when the “initial winner” does not have the majority of the votes

• Can still, on occasion, be unfair

Sequential Runoffs

• STEP 1: The candidate (choice) with the least number of 1st place votes overall is eliminated.

• STEP 2: Use the remaining candidates (choices) and eliminate the one with the least number of 1st place votes from that group.

• STEP 3: Continue this process until a winner is determined.

Sequential RunoffsPreference Schedule for the Club Election






Number of Voters

18 12 10 9 4 2

STEP 1: In the first runoff, Eugene has the least number of first place votes (4 + 2). Therefore, Eugeneis eliminated.

We are now down to Alonzo, Bonnie, Carol, and David for the second runoff.


1st Choice A B C D

2nd Choice D B C B C

3rd Choice D D D



Number of Voters

18 12 10 9 4 2

STEP 2: With Eugene eliminated, the other choices move up. This is the rule for sequential runoffs. Note that the number of voters (bottom row) did not change.

Sequential Runoffs

Preference Schedule for the Club Election

1st Choice A B C D B C

2nd Choice D D B C D D

3rd Choice C C D B C B


5th Choice

Number of Voters

18 12 10 9 4 2

STEP 2: With Eugene eliminated, the other choices move up. This is the rule for sequential runoffs. We count the number of first choices here:

Alonzo 18, Bonnie 16 (12 + 4), Carol 12 (10 + 2), David 9David has the least number of first place votes, so David is now eliminated.


1st Choice A B C D B C

2nd Choice D D B C D D

3rd Choice C C D B C B


5th Choice

Number of Voters

18 12 10 9 4 2

STEP 2: With David eliminated, the other choices move up.


1st Choice A B C B C

2nd Choice B C

3rd Choice C C B C B


5th Choice

Number of Voters

18 12 10 9 4 2


1st Choice A B C C B C

2nd Choice C C B B C B

3rd Choice B A A A A A

4th Choice

5th Choice

Number of Voters

18 12 10 9 4 2

STEP 3: We are down to three candidates. Count the number of first place votes.Alonzo 18, Bonnie 16 (12 + 4), Carol 21(10 + 4 + 2)

Bonnie has the least number of first place votes, so Bonnie is now eliminated.


1st Choice A B C C B C

2nd Choice C C B B C B

3rd Choice B A A A A A

4th Choice

5th Choice

Number of Voters

18 12 10 9 4 2

STEP 3: Bonnie is eliminated and the other candidates move up the list.


1st Choice A C C C C C

2nd Choice C A A A A A

3rd Choice

4th Choice

5th Choice

Number of Voters

18 12 10 9 4 2

STEP 4: Alonzo now has 18 first place votes, but Carol has 37 first place votes (12 + 10 + 9 + 4 + 2). Alonzo is eliminated – and Carol win the election.

Sequential Runoffs: Comments

• Applications “in real life”• International Olympic Committee to choose host cities

• Australia to elect members of the House of Representatives

Point System (called a Borda Count)

• STEP 1: Assign point values • “Last place” is awarded 1 point

• “Next to last place” is award 2 points

• Continue the pattern

• It is important to assign points “from the bottom and then ‘up’ “

• STEP 2:• Tally the points

• The winner is the one with the most overall points

Point System (Borda Count)Preference Schedule for the Club Election

1st Choice (5) A B C D E E

2nd Choice (4) D E B C B C

3rd Choice (3) E D E E D D

4th Choice (2) C C D B C B

5th Choice (1) B A A A A A

Number of Voters

18 12 10 9 4 2

Alonzo has these point values:18 x 5 points (first place) = 9012 x 1 point (last place) = 1210 x 1 point (last place) = 109 x 1 point (last place) = 94 x 1 point (last place) = 42 x 1 point (last place) = 2

TOTAL POINTS: 127 points


1st Choice (5) A B C D E E

2nd Choice (4) D E B C B C

3rd Choice (3) E D E E D D

4th Choice (2) C C D B C B

5th Choice (1) B A A A A A

Number of Voters

18 12 10 9 4 2

Bonnie has these point values:18 x 1 points (last place) = 1812 x 5 points (first place) = 6010 x 4 points (second place) = 409 x 2 points (fourth place) = 184 x 4 points (second place) = 162 x 2 points (fourth place) = 4

TOTAL POINTS: 156 points







Number of Voters

18 12 10 9 4 2

Hopefully you have the algorithm (method of tallying the points) in your head now.The final tallies are:

Alonzo: (18 x 5) + (12 X 1) + (10 X 1) + (9 X 1) + (4 x 1) + (2 x 1) = 127 points Bonnie: (18 x 1) + (12 X 5) + (10 X 4) + (9 X 2) + (4 x 4) + (2 x 2) = 156 points Carol: (18 x 2) + (12 X 2) + (10 X 5) + (9 X 4) + (4 x 2) + (2 x 4) = 162 points David: (18 x 4) + (12 X 3) + (10 X 2) + (9 X 5) + (4 x 3) + (2 x 3) = 191 points *** WinnerEugene: (18 x 3) + (12 X 4) + (10 X 3) + (9 X 3) + (4 x 5) + (2 x 5) = 189 points

Point System (Borda Count): Comments

• Borda Count “in real life”• Iindividual sports awards (Heisman Trophy winner, NBA Rookie of the Year,

NFL MVP, etc.)

• College football polls

• Music industry awards

• Hiring of school employees and corporate executives

• Despite its flaws, experts in voting theory consider the Borda count method one of the best, if not the very best, method for deciding elections with many candidates.

Pairwise Comparisons (Condorcet Count)

• STEP 1: Take all possible “head-to-head pairings” and select the winner in each pairing.

• STEP 2: The candidate with the most “wins” in considered the overall winner

Pairwise Comparisons (Condorcet Count)Preference Schedule for the Club Election






Number of Voters

18 12 10 9 4 2

Here are the possible “pairings” in our example: ***

A versus B A versus C A versus D A versus E

B versus C B versus D B versus E

C versus D C versus E

D versus E

***Think in terms of “five football teams playing against each other in a one-to-one contest.”

“The ‘team’ with the best overall win-loss record is the winner” ***







Number of Voters

18 12 10 9 4 2





D versus E

Example:A versus BA gets 18 preference votesB gets 12+10+9+4+2 = 37

So B wins the head-to-head contest over A.







Number of Voters

18 12 10 9 4 2





D versus E

Example:B versus EB gets 12+ 10=22 preference votesE gets 18+9+4+2=33 votes

So E wins the head-to-head contest over B.







Number of Voters

18 12 10 9 4 2


A versus B A versus C A versus D A versus E18 – 37 18 – 37 18 - 37 18 – 37 B versus C B versus D B versus E16 – 39 26 – 29 22 - 33C versus D C versus E12 – 43 19 - 36D versus E27 - 28

The winners:

B C D E

C D E

D E

E

A 0 winsB 1 winC 2 winsD 3 winsE 4 wins

Eugene is the overall winner in this voting method.

Pairwise Comparisons(Condorcet Count): Comments

•Applications “in real life”

•“Round-robin” tournaments (every team plays every other team

•NFL Drafts (coaches & executives selecting players)

•The principal weakness is that this method fails to take into

consideration a voter’s other preferences beyond first choice and in so

doing can lead to some very bad election results.

•It does not always produce a clear winner

Insincere (Strategic) Voting

• The candidate (choice) we really want has no change of “winning”

• Therefore, we will vote on a lesser candidate (choice) with a better chance of winning

• In closely contested elections a few insincere voters can completely change the outcome of an election.

Voting Theory

• Essential ingredients of every election: voters, candidates (choices),

and ballots

• First half is voting;

• Second half is counting.

• Arrow’s Impossible Theorem:• For elections involving three or more candidates (choices), a method for

determining election results that is democratic and always fair is mathematically impossible.

Documents

Chapter 12 a slide show summer 2014