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A Mathematical View of Our World. 1 st ed. Parks, Musser, Trimpe, Maurer, and Maurer. Chapter 3. Voting and Elections. Section 3.1 Voting Systems. Goals Study voting systems Plurality method Borda count method Plurality with elimination method Pairwise comparison method

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### A Mathematical View of Our World

1st ed.

Parks, Musser, Trimpe, Maurer, and Maurer

### Chapter 3

Voting and Elections

Section 3.1Voting Systems
• Goals
• Study voting systems
• Plurality method
• Borda count method
• Plurality with elimination method
• Pairwise comparison method
• Discuss tie-breaking methods
3.1 Initial Problem
• The city council must select among 3 locations for a new sewage treatment plant.
• A majority of city councilors say they prefer site A to site B.
• A majority of city councilors say they prefer site A to site C.
• In the vote site B is selected.
• Did the councilors necessarily lie about their preferences before the election?
• The solution will be given at the end of the section.
Voting Systems
• The following voting methods will be discussed:
• Plurality method
• Borda count method
• Plurality with elimination method
• Pairwise comparison method
Plurality Method
• When a candidate receives more than half of the votes in an election, we say the candidate has received a majority of the votes.
• When a candidate receives the greatest number of votes in an election, but not more than half, we say the candidate has received a plurality of the votes.
Question:

Suppose in an election, the vote totals are as follows. Andy gets 4526 first-place votes. Lacy gets 1901 first-place votes. Peter gets 2265 first-place votes.

Choose the correct statement.

a. Andy has a majority.

b. Andy has a plurality only.

Plurality Method, cont’d
• In the plurality method:
• Voters vote for one candidate.
• The candidate receiving the most votes wins.
• This method has a couple advantages:
• The voter chooses only one candidate.
• The winner is easily determined.
Plurality Method, cont’d
• The plurality method is used:
• In the United States to elect senators, representatives, governors, judges, and mayors.
• In the United Kingdom and Canada to elect members of parliament.
Example 1
• Four persons are running for student body president. The vote totals are as follows:
• Under the plurality method, who won the election?
Example 1, cont’d
• Solution: With 2554 votes, Charles has a plurality and wins the election.
• Note that there were a total of 9658 votes cast.
Example 2
• Three candidates ran for Attorney General in Delaware in 2002. The vote totals were as follows:
• Vivian Houghton: 13,860
• What percent of the votes did each candidate receive and who won the election?
Example 2, cont’d
• Solution: A total of 228,557 votes were cast.
Borda Count Method
• In the Borda count method:
• Voters rank all of the m candidates.
• Votes are counted as follows:
• A voter’s last choice gets 1 point.
• A voter’s next-to-last choice gets 2 points.
• A voter’s first choice gets m points.
• The candidate with the most points wins.
Borda Count Method, cont’d
• The main advantage of the Borda count method is that it uses more information from the voters.
• A variation of the Borda count method is used to select the winner of the Heisman trophy.
Example 3
• Four persons are running for student body president. Voters rank the candidates as shown in the table below.
• Under the Borda count method, who is elected?
Example 3, cont’d
• Solution: Convert the votes to points.
Example 3, cont’d
• Solution: Total the points for each person:
• Aaron: 9436 + 4104 + 5572 + 3145 = 22,257
• Bonnie: 9828 + 10,497 + 4948 + 1228 = 26,501
• Charles: 10,216 + 7101 + 3468 + 3003 = 23,788
• Dion: 9152 + 7272 + 5328 + 2282 = 24,034
• Bonnie has the most points and is the winner.
Example 3, cont’d
• Note that in this same election:
• Charles won using the plurality method because he had more first place votes than any other candidate.
• Bonnie won using the Borda count method because her point total was highest, due to having many second-place votes.
Plurality with Elimination Method
• In the plurality with elimination method:
• Voters choose one candidate.
• If one candidate receives a majority of the votes, that candidate is selected.
• If no candidate receives a majority, eliminate the candidate who received the fewest votes and do another round of voting.
Plurality with Elimination, cont’d
• Cont’d:
• This process is repeated until someone receives a majority of the votes and is declared the winner.
• The plurality with elimination method is used:
• To select the location of the Olympic games.
• In France to elect the president.
Plurality with Elimination, cont’d
• Rather than needing to potentially conduct multiple votes, the voters can be asked to rank all candidates during the first election.
• A preference table is used to display these rankings.
Example 4
• Four persons are running for department chairperson. The 17 voters ranked the candidates 1st through 4th.
• Under plurality with elimination, who is the winner?
Example 4, cont’d
• Solution: Some voters had the same preference ranking. Identical rating have been grouped to form the preference table below.
• The number at the top of each column indicates the number of voters who shared that ranking.
Example 4, cont’d
• Solution, cont’d: The first-place votes for each candidate are totaled:
• Alice: 6; Bob: 4; Carlos: 4; Donna: 3
• Donna, who has the fewest first-place votes, is eliminated.
Example 4, cont’d
• Solution, cont’d: A new preference table, without Donna, must be created.
• Donna is eliminated from each column.
• Any candidates ranked below Donna move up.
Example 4, cont’d
• Solution, cont’d: The first-place votes for each candidate are totaled:
• Alice: 7; Bob: 4; Carlos: 6
• No candidate received a majority.
• Bob, who has the fewest first-place votes, is eliminated.
Example 4, cont’d
• Solution, cont’d: A new preference table, without Bob, must be created.
• Bob is eliminated from each column.
• Any candidates ranked below Bob move up.
Example 4, cont’d
• Solution, cont’d: The first-place votes for each candidate are totaled:
• Alice: 9; Carlos: 8
• Alice received a majority and is the winner.
Pairwise Comparison Method
• In the pairwise comparison method:
• Voters rank all of the candidates.
• For each pair of candidates X and Y, determine how many voters prefer X to Y and vice versa.
• If X is preferred to Y more often, X gets 1 point.
• If Y is preferred to X more often, Y gets 1 point.
• If the candidates tie, each gets ½ a point.
• The candidate with the most points wins.
Pairwise Comparison, cont’d
• The pairwise comparison method is also called the Condorcet method.
Example 5
• Three persons are running for department chair. The 17 voters rank all the candidates, as shown in the preference table below.
• Under the pairwise comparison method, who wins the election?
Example 5, cont’d
• Solution: There are 3 pairs of candidates to compare:
• Alice vs. Bob
• Alice vs. Carlos
• Bob vs. Carlos
• For each pair of candidates, delete the third candidate from the preference table and consider only the two candidates in question.
Example 5, cont’d
• Solution, cont’d:
• We say Alice is preferred to Bob 10 to 7.
Example 5, cont’d
• Solution, cont’d:
• Alice is preferred to Carlos 9 to 8, so Alice receives another point.
Example 5, cont’d
• Solution, cont’d:
• Carlos is preferred to Bob 10 to 7, so Carlos receives one point.
Example 5, cont’d
• Solution, cont’d: The final point totals are:
• Alice: 2 points
• Bob: 0 points
• Carlos: 1 point
• Alice wins the election.
Question:

Candidate B is the winner of an election with the following preference table .

What voting method could have been used to determine the winner?

a. Plurality method

b. Borda count method

c. Plurality with elimination method

d. Pairwise comparison method

Voting Methods, cont’d
• The four voting systems studied here can produce different winners even when the same voter preference table is used.
• Any of the four methods can also produce a tie between two or more candidates, which must be broken somehow.
Tie Breaking
• A tie-breaking method should be chosen before the election.
• To break a tie caused by perfectly balanced voter support, election officials may:
• Make an arbitrary choice.
• Flipping a coin
• Drawing straws
• Bring in another voter.
• The Vice President votes when the U. S. Senate is tied.
3.1 Initial Problem Solution
• A majority of city councilors said they preferred site A to site B and also site A to site C. If B won the election, did they necessarily lie?
• Solution:
• The councilors would not have to lie in order for this to happen. This situation can occur with some voting methods.
Initial Problem Solution, cont’d
• For example, this situation could occur if the voting method used was plurality with elimination.
• Suppose 11 councilors ranked the sites as shown in the table below.
Initial Problem Solution, cont’d
• Notice that in this scenario:
• Site A is preferred to site B 7 to 4.
• Site A is preferred to site C 7 to 4.
• However, in the vote count:
• Site A, with the fewest first-place votes, is eliminated.
• In the second round of voting, site B wins.
Section 3.2Flaws of the Voting Systems
• Goals
• Study fairness criteria
• The majority criterion
• Montonicity criterion
• Irrelevant alternatives criterion
• Study fairness of voting methods
• Arrow impossibility theorem
• Approval voting
3.2 Initial Problem
• The Compromise of 1850 averted civil war in the U.S. for 10 years.
• Henry Clay proposed the bill, but it was defeated in July 1850.
• A short time later, Stephen Douglas was able to get essentially the same proposals passed.
• How is this possible?
• The solution will be given at the end of the section.
Flaws of Voting Systems
• We have seen that the choice of voting method can affect the outcome of an election.
• Each voting method studied can fail to satisfy certain criteria that make a voting method “fair”.
Fairness Criteria
• The fairness criteria are properties that we expect a good voting system to satisfy.
• Four fairness criteria will be studied:
• The majority criterion
• The monotonicity criterion
• The irrelevant alternatives criterion
The Majority Criterion
• If a candidate is the first choice of a majority of voters, then that candidate should be selected.
Question:

Candidate A won an election with 3000 of the 8500 votes. Was the majority criterion necessarily violated?

a. yes

b. no

The Majority Criterion, cont’d
• For the majority criterion to be violated:
• A candidate must have more than half of the votes.
• This same candidate must not win the election.
• Note:
• This criterion does not say what should happen if no candidate receives a majority.
• This criterion does not say that the winner of an election must win by a majority.
The Majority Criterion, cont’d
• If a candidate is the first choice of a majority of voters, then that candidate will always win using:
• The plurality method.
• The plurality with elimination method.
• The pairwise comparison method.
• In both of these methods any candidate with more than half the vote will always win.
The Majority Criterion, cont’d
• If a candidate is the first choice of a majority of voters, then that candidate might not win using:
• The Borda count method.
• The candidate with the most points may not be the candidate with the most first-place votes.
Example 1
• Four cities are being considered for an annual trade show. The preferences of the organizers are given in the table.
Example 1, cont’d
• Which site has a majority of first-place votes?
• Which site wins using the Borda count method?
Example 1, cont’d
• Solution: There are 9 votes, so a majority would be 5 or more votes.
• The first place vote totals are:
• Chicago: 5; Seattle: 3; Phoenix: 1; Boston: 0
• Chicago has a majority of first-place votes.
Example 1, cont’d
• Solution, cont’d: Find the point totals for Borda count.
• The points are calculated as follows:
• Chicago: 5(4) + 0(3) + 2(2) + 2(1) = 26
• Seattle: 3(4) + 4(3) + 2(2) + 0(1) = 28
• Phoenix: 1(4) + 2(3) + 2(2) + 4(1) = 18
• Boston: 0(4) + 3(3) + 3(2) + 3(1) = 18
• Under the Borda count method, Seattle is the winner.
Example 1, cont’d
• Note that Chicago had a majority of first-place votes, but under the Borda count method Seattle was the winner.
• This is an example of the Borda count method failing the majority criterion. In this case, we would say that the Borda count method was unfair.
• If a candidate is favored when compared separately with each of the other candidates, then the favored candidate should win the election.
• This is also called the Condorcet criterion.
• A candidate must be preferred pairwise to every other candidate.
• This same candidate must not win the election.
• Note:
• This criterion does not say what should happen if no candidate is preferred pairwise to every other candidate.
• If a candidate is favored pairwise to every other candidate, then that candidate will always win using:
• The pairwise comparison method.
• This candidate will earn the most points from the pairwise comparisons.
• If a candidate is favored pairwise to every other candidate, then that candidate might not win using:
• The plurality method.
• The plurality with elimination method.
• The Borda count method.
Example 2
• Seven people are choosing an option for a retirement party: catering, picnic, or restaurant. The preferences are shown in the table below.
• Which site is selected using the plurality method?
Example 2, cont’d
• Solution:
• The picnic has the most votes, 3, so it is the winning option under the plurality method.
Example 2, cont’d
• Solution, cont’d:
• The pairwise comparisons are made:
• R is preferred to P 4 to 3.
• R is preferred to C 5 to 2.
• R is preferred separately to every other candidate, but R did not win the election. This is a violation of the head-to-head criterion.
The Monotonicity Criterion
• Suppose a particular candidate, X, wins an election.
• If, hypothetically, this election were redone and the only changes were that some voters switched X with the candidate they had ranked one higher, then X should still win.
• This criterion is only used in special cases.
Monotonicity Criterion, cont’d
• The monotonicity criterion is always satisfied by:
• The plurality method.
• The Borda count method.
• The pairwise comparison method.
Monotonicity Criterion, cont’d
• The monotonicity criterion is not always satisfied by:
• The plurality with elimination method.
Example 3
• Teachers are voting for a union president from the candidates Akst, Bailey, and Chung. The preferences are shown in the table below.
Example 3, cont’d
• Who will win using the plurality with elimination method?
• Solution: The first-place vote totals are:
• Akst: 14; Bailey: 12; Chung: 15.
• No candidate received a majority of at least 21 votes, so Bailey is eliminated.
Example 3, cont’d
• Solution, cont’d:
• After Bailey’s elimination, a new preference table is created.
• Akst now has 26 first-place votes, and is the winner.
Example 3, cont’d
• If 4 of the 5 teachers who ranked the candidates CAB changed to a ranking of ACB, would this affect the outcome of the election?
• Solution: Akst won the first election and the only changes now are that 4 teachers moved him from 2nd to 1st place.
Example 3, cont’d
• Solution, cont’d: The new preference table is shown below.
Example 3, cont’d
• Solution, cont’d: The first-place vote totals are:
• Akst: 18; Bailey: 12; Chung: 11.
• No candidate received a majority of at least 21 votes, so Chung is eliminated.
Example 3, cont’d
• Solution, cont’d:
• After Chung’s elimination, a new preference table is created.
• Bailey now has 22 votes, and is the winner.
Example 3, cont’d
• Solution, cont’d:
• The only changes in the preference table were ones that favored Akst, who won the first election.
• However, Akst ended up losing the modified vote to Bailey.
• This is a violation of the monotonicity criterion.
The Irrelevant Alternatives Criterion
• Suppose a candidate, X, is selected in an election.
• If, hypothetically, this election were redone with one or more of the unselected candidates removed from the vote, then X should still win.
Irrelevant Alternatives Criterion, cont’d
• The irrelevant alternatives criterion is not always satisfied by any of the 4 voting methods studied.
Example 4
• The 5 members of a book club are voting on what book to read next: a mystery, a historical novel, or a science fiction fantasy. The preference table is shown below.
Example 4, cont’d
• Which of the books is selected using the plurality with elimination method?
• Solution: The first-place vote totals are:
• M: 2; H: 1; S: 2.
• No book has a majority, so H is eliminated.
Example 4, cont’d
• Solution, cont’d: After H is eliminated a new preference table is created.
• Book M receives 3 first-place votes, and is the winner. The book club will read the mystery.
Example 4, cont’d
• If the science fiction book is removed from the list, is the irrelevant alternatives criterion violated in the new election?
• Solution: A new preference table, without S, is created.
Example 4, cont’d
• Solution, cont’d:
• In the new table, M has 2 votes and H has 3.
• Book H is the new winner, violating the irrelevant alternatives criterion.
Arrow Impossibility Theorem
• The Arrow Impossibility Theorem states that no system of voting will always satisfy all of the 4 fairness criteria.
• This fact was proved by Kenneth Arrow in 1951.
Question:

The results of an election using the plurality method were analyzed. It was found that the election did not violate any of the four fairness criteria. Does this contradict the Arrow Impossibility Theorem?

a. yes

b. no

Approval Voting
• No voting system is always fair, but we can explore systems that are unfair less often than others. One such system is called approval voting.
• In approval voting:
• Each voter votes for all candidates he/she considers acceptable.
• The candidate with the most votes is selected.
Example 5
• Three candidates are running for two positions. There are 9 voters and the votes are shown in the table below.
• Who is the winner under approval voting?
Example 5, cont’d
• Solution: The vote totals are:
• Ammee: 6
• Bonnie: 7
• Celeste: 5
• Bonnie and Ammee are selected for the two positions.
3.2 Initial Problem Solution
• Henry Clay presented the Compromise of 1850 as one bill containing all the proposals.
• Of the 60 senators, a majority would not approve the bill because they disagreed on individual issues within the bill.
Initial Problem Solution, cont’d
• Stephen Douglas presented each proposal of the Compromise in a separate bill.
• A (different) majority of the senators passed each proposal and the Compromise of 1850 went into effect, although a majority never approved the measures as a whole.
Section 3.3Weighted Voting Systems
• Goals
• Study weighted voting systems
• Coalitions
• Dummies and dictators
• Veto power
• Study the Banzhaf power index
3.3 Initial Problem
• A stockholder owns 17% of the shares of a company.
• Among the other 3 stockholders, no one owns more than 32% of the shares.
• Why will no one listen to the stockholder with 17%?
• The solution will be given at the end of the section.
Weighted Voting Systems
• In a weighted voting system, an individual voter may have more than one vote.
• The number of votes that a voter controls is called the weight of the voter.
• An example of a weighted voting system is the election of the U.S. President by the Electoral College.
Weighted Voting Systems, cont’d
• The weights of the voters are usually listed as a sequence of numbers between square brackets.
• For example, the voting system in which Angie has a weight of 9, Roberta has a weight of 12, Carlos has a weight of 8, and Darrell has a weight of 11 is represented as [12, 11, 9, 8].
Weighted Voting Systems, cont’d
• The voter with the largest weight is called the “first voter”, written P1.
• The weight of the first voter is represented by W1.
• The remaining voters and their weights are represented similarly, in order of decreasing weights.
Example 1
• The voting system in which Angie has a weight of 9, Roberta has a weight of 12, Carlos has a weight of 8, and Darrell has a weight of 11 was represented as [12, 11, 9, 8].
• In this case, P1 = Roberta, P2 = Darrell, P3 = Angie, and P4 = Carlos.
• Also, W1 = 12, W2 = 11, W3 = 9, and W4 = 8.
Weighted Voting Systems, cont’d
• Yes or no questions are commonly called motions.
• A final decision of ‘No’ defeats the motion and leaves the status quo unchanged.
• A final decision of ‘Yes” passes the motion and changes the status quo.
Weighted Voting Systems, cont’d
• A simple majority requirement means that a motion must receive more than half of the votes to pass.
• A supermajority requirement means that the minimum number of votes required to pass a motion is set higher than half of the total weight.
• A common supermajority is two-thirds of the total weight.
Weighted Voting Systems, cont’d
• The weight required to pass a motion is called the quota.
• Example: A simple majority quota for the weighted voting system [12, 11, 9, 8] would be 21.
• Half of the total weight is (12 + 11 + 9 + 8)/2 = 40/2 = 20. More than half of the weight would be at least 21 ‘Yes’ votes.
Question:

Given the weighted voting system [10, 9, 8, 8, 5], find the quota for a supermajority requirement of two-thirds of the total weight.

a. 27

b. 21

c. 26

d. 20

Weighted Voting Systems, cont’d
• The quota for a weighted voting system is usually added to the list of weights.
• Example: For the weighted voting system [12, 11, 9, 8] with a quota of 21 the complete notation is [21 : 12, 11, 9, 8].
Example 2
• Given the weighted voting system

[21 : 10, 8, 7, 7, 4, 4], suppose P1, P3, and P5 vote ‘Yes’ on a motion.

• Is the motion passed or defeated?
Example 2, cont’d
• Solution:
• The given voters have a combined weight of 10 + 7 + 4 = 21.
• The quota is met, so the motion passes.
Example 3
• Given the weighted voting system

[21 : 10, 8, 7, 7, 4, 4], suppose P1, P5, and P6 vote ‘Yes’ on a motion.

• Is the motion passed or defeated?
Example 3, cont’d
• Solution:
• The given voters have a combined weight of 10 + 4 + 4 = 18.
• The quota is not met, so the motion is defeated.
Coalitions
• Any nonempty subset of the voters in a weighted voting system is called a coalition.
• If the total weight of the voters in a coalition is greater than or equal to the quota, it is called a winning coalition.
• If the total weight of the voters in a coalition is less than the quota, it is called a losing coalition.
Question:

Given the weighted voting system [27: 10, 9, 8, 8, 5], is the coalition {P1, P4, P5} a winning coalition or a losing coalition?

a. winning

b. losing

Example 4
• For the weighted voting system

[8: 6, 5, 4], list all possible coalitions and determine whether each is a winning or losing coalition.

Example 4, cont’d
• Solution: Each coalition and its status is listed in the table below.
Coalitions, cont’d
• In a weighted voting system with n voters, exactly 2n- 1 coalitions are possible.
• Example:
• How many coalitions are possible in a weighted voting system with 7 voters?
• Solution: The formula tells us there are

27 - 1 = 128 – 1 = 127 coalitions.

Example 5
• The voting weights of EU members in a council in 2003 are shown in the table.
Example 5, cont’d
• If resolutions must receive 71% of the votes to pass, what is the quota?
• How many coalitions are possible?
Example 5, cont’d
• Solution:
• There are 87 votes total. So the quota is 71% of 87, or approximately 62 votes.
• There are n = 15 members, so there are 215 – 1 = 32,767 coalitions possible.
Dictators and Dummies
• A voter whose presence or absence in any coalition makes no difference in the outcome is called a dummy.
• A voter whose presence or absence in any coalition completely determines the outcome is called a dictator.
• When a weighted voting system has a dictator, the other voters in the system are automatically dummies.
Veto Power
• In between the complete power of a dictator and the zero power of a dummy is a level of power called veto power.
• A voter with veto power can defeat a motion by voting ‘No’ but cannot necessarily pass a motion by voting ‘Yes’.
• Any dictator has veto power, but a voter with veto power is not necessarily a dictator.
Example 6
• Consider the weighted voting system

[12: 7, 6, 4].

• List all the coalitions and determine whether each is a winning or losing coalition.
• Are there any dummies or dictators?
• Are there any voters with veto power?
Example 6, cont’d
• Solution:
• Each coalition and its status is listed in the table below.
Example 6, cont’d
• Solution, cont’d:
• Removing the third voter from any coalition does not change the status of the coalition. P3 is a dummy.
Example 6, cont’d
• Solution, cont’d:
• No voter has complete power to pass or defeat a motion. There is no dictator.
Example 6, cont’d
• Solution, cont’d:
• If P1 is not in a coalition, then it is a losing coalition. P1 has veto power.
Question:

In the weighted voting system

[27: 10, 9, 8, 8, 5], is P1 a:

a. dictator

b. dummy

c. voter with veto power

d. none of the above

Example 7
• Consider the weighted voting system [10: 10, 5, 4].
• Are there any dummies, dictators, or voters with veto power?
Example 7, cont’d
• Solution:
• P1 has enough weight to pass a motion by voting ‘Yes’ no matter how anyone else votes.
• If P1 votes ‘No’, the motion will not pass no matter how anyone else votes.
• P1 is a dictator and thus all other voters are dummies.
Critical Voters
• If a voter’s weight is large enough so that the voter can change a particular winning coalition to a losing coalition by leaving the coalition, then that voter is called a critical voter in that winning coalition.
Question:

Given the weighted voting system [27: 10, 9, 8, 8, 5], is the voter P4 a critical voter in the winning coalition {P1, P2, P4, P5}?

a. yes

b. no

Example 8
• Consider the weighted voting system [21 : 10, 8, 7, 7, 4, 4].
• Which voters in the coalition {P2, P3, P4, P5 } are critical voters in that coalition?
Example 8, cont’d
• Solution: The weight in the winning coalition is 26.
• If P2 leaves, the weight goes down to

26 – 8 = 18 < quota.

• If P3 leaves, the weight goes down to

26 – 7 = 19 < quota.

Example 8, cont’d
• Solution cont’d:
• If P4 leaves, the weight goes down to

26 – 7 = 19 < quota.

• If P5 leaves, the weight goes down to

26 – 4 = 22 > quota.

• The critical voters in this coalition are P2, P3, and P4.
The Banzhaf Power Index
• The more times a voter is a critical voter in a coalition, the more power that voter has in the system.
• The Banzhaf power of a voter is the number of winning coalitions in which that voter is critical.
Banzhaf Power Index, cont’d
• The sum of the Banzhaf powers of all voters is called the total Banzhaf power in the weighted voting system.
• An individual voter’s Banzhaf power index is the ratio of the voter’s Banzhaf power to the total Banzhaf power in the system.
• The sum of the Banzhaf power indices of all voters is 100%.
Banzhaf Power Index, cont’d
• An individual voter’s Banzhaf power index is calculated using the following process:
• Find all winning coalitions for the system.
• Determine the critical voters for each winning coalition.
• Calculate each voter’s Banzhaf power.
• Find the total Banzhaf power in the system.
• Divide each voter’s Banzhaf power by the total Banzhaf power.
Example 9
• For the weighted voting system [18 : 12, 7, 6, 5], determine:
• The total Banzhaf power in the system.
• The Banzhaf power index of each voter.
Example 9, cont’d
• Solution Step 1: Find all the winning coalitions.
Example 9, cont’d
• Solution Step 2: Determine the critical voters for each winning coalition.
• Remove each voter one at a time and check to see whether the resulting coalition is still a winning coalition.
• This work is shown in the next slides.
Example 9, cont’d
• Solution Step 3: Count the number of times each voter is a critical voter:
• P1: 5 times
• P2: 3 times
• P3: 3 times
• P4: 1 time
• Step 4: The total Banzhaf power in the system is 5 + 3 + 3 + 1 = 12
Example 9, cont’d
• Solution Step 5: Divide each voter’s Banzhaf power by the total Banzhaf power to find the Banzhaf power indices.
3.3 Initial Problem Solution
• One of the 4 stockholders owns 17% of the shares of a company. Among the other 3, no one owns more than 32%. Why will no one listen to the stockholder with 17%?
Initial Problem Solution, cont’d
• We know one stockholder owns 32% and one owns 17%, so the other 51% is split between the remaining two stockholders.
• Suppose the other percents are 26% and 25%.
Initial Problem Solution, cont’d
• The winning coalitions in this case are:
• {32%, 26%, 25%, 17%}
• {32%, 26%, 25%}
• {32%, 26%, 17%}
• {32%, 25%, 17%}
• {26%, 25%, 17%}
• {32%, 26%}
• {32%, 25%}
• {26%, 25%}
Initial Problem Solution, cont’d
• The voter with 17% is in several winning coalitions, but removing that voter does not cause any of them to become losing coalitions.
• The voter with 17% is not a critical voter in any winning coalition.
• The reason no one will listen to the voter is that he or she is a dummy.