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Chapter 7: Voting Systems
Section 7.1: Voting Methods
Every couple of years or so, voters go to the polls to cast ballots for their choices for mayor, governor, senator, president, etc. Then the election officials count the ballots and declare a winner. But how do the election officials determine who the winner is. If there are only two candidates, then there is no problem figuring out the winner. The candidate with more than 50% of the votes wins. This is known as the majority. So the candidate with the majority of the votes is the winner.
Majority Rule: This concept means that the candidate (choice) receiving more than 50% of the vote is the winner.
But what happens if there are three candidates, and no one receives the majority? That depends on where you live. Some places decide that the person with the most votes wins, even if they don’t have a majority. There are problems with this, in that someone could be liked by 35% of the people, but is disliked by 65% of the people. So you have a winner that the majority doesn’t like. Other places conduct runoff elections where the top two candidates have to run again, and then the winner is chosen from the runoff election. There are some problems with this method. First, it is very costly for the candidates and the election office to hold a second election. Second, you don’t know if you will have the same voters voting in the second election, and so the preferences of the voters in the first election may not be taken into account.
So what can be done to have a better election that has someone liked by more voters yet doesn't require a runoff election? A ballot method that can fix this problem is known as a preference ballot.
Preference Ballots: Ballots in which voters choose not only their favorite candidate, but they actually order all of the candidates from their most favorite down to their least favorite.
Note: Preference Ballots are transitive: If a voter prefers choice A to choice B and also prefers choice B to choice C, then the voter must prefer choice A to choice C.
To understand how a preference ballot works and how to determine the winner, we will look at an example.
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Example 7.1.1: Preference Ballot for the Candy Election Suppose an election is held to determine which bag of candy will be opened. The choices (candidates) are Hershey’s Miniatures (M), Nestle Crunch (C), and Mars’ Snickers (S). Each voter is asked to fill in the following ballot, by marking their first, second, and third place choices.
Figure 7.1.1: Preference Ballot for the Candy Election
Each voter fills out the above ballot with their preferences, and what follows is the results of the election.
Table 7.1.2: Ballots Cast for the Candy Election Voter Anne Bob Chloe Dylan Eli Fred 1 st^ choice C M C M S S 2 nd^ choice S S M C M M 3 rd^ choice M C S S C C
Voter George Hiza Isha Jacy Kalb Lan 1 st^ choice S^ S^ S^ M^ C^ M 2 nd^ choice M M M C M C 3 rd^ choice C C C S S S
Voter Makya Nadira Ochen Paki Quinn Riley 1 st^ choice S^ S^ C^ C^ S^ S 2 nd^ choice M M M M M M 3 rd^ choice C C S S C C
Now we must count the ballots. It isn’t as simple as just counting how many voters like each candidate. You have to look at how many liked the candidate in first-place, second place, and third place. So there needs to be a better way to organize the results. This is known as a preference schedule.
Preference Schedule: A table used to organize the results of all the preference ballots in an election.
Candy Preference Crunch _________ Miniatures _________ Snickers _________
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The Borda Count Method (Point System) : Each place on a preference ballot is assigned points. Last place receives one point, next to last place receives two points, and so on. Thus, if there are N candidates, then first-place receives N points. Now, multiply the point value for each place by the number of voters at the top of the column to find the points each candidate wins in a column. Lastly, total up all the points for each candidate. The candidate with the most points wins.
Example 7.1.4: The Winner of the Candy Election—Borda Count Method Using the preference schedule in Table 7.1.3, find the winner using the Borda Count Method.
Table 7.1.3: Preference Schedule for the Candy Election Number of voters 3 1 4 1 9 1 st^ choice M M C C S 2 nd^ choice C S M S M 3 rd^ choice S C S M C
The third choice receives one point, second choice receives two points, and first choice receives three points. There were three voters who chose the order M, C, S. So M receives 33 = 9 points for the first-place, C receives 32 = 6 points, and S receives 3*1 = 3 points for those ballots. The same process is conducted for the other columns. The table below summarizes the points that each candy received.
Table 7.1.4: Preference Schedule of the Candy Election with Borda Count Points Number of voters 3 1 4 1 9 1 st^ choice M 9
M
C
C
S
2 nd^ choice C 6
S
M
S
M
3 rd^ choice S 3
C
S
M
C
Adding up these points gives, M = 9 + 3 + 8 + 1 + 18 = 39 C = 6 + 1 + 12 + 3 + 9 = 31 S = 3 + 2 + 4 + 2 + 27 = 38
Thus, Hershey’s Miniatures wins using the Borda Count Method.
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So who is the winner? With one method Snicker’s wins and with another method Hershey’s Miniatures wins. The problem is that it all depends on which method you use. Therefore, you need to decide which method to use before you run the election.
The Plurality with Elimination Method (Sequential Runoffs) : Eliminate the candidate with the least amount of 1st place votes and re-distribute their votes amongst the other candidates. Repeat this process until you find a winner. Note: At any time during this process if a candidate has a majority of first-place votes, then that candidate is the winner.
Example 7.1.5: The Winner of the Candy Election—Plurality with Elimination Method Using the preference schedule in Table 7.1.3, find the winner using the Plurality with Elimination Method.
Table 7.1.3: Preference Schedule for the Candy Election Number of voters 3 1 4 1 9 1 st^ choice M M C C S 2 nd^ choice C^ S^ M^ S^ M 3 rd^ choice S C S M C
This isn’t the most exciting example, since there are only three candidates, but the process is the same whether there are three or many more. So look at how many first-place votes there are. M has , C has , and S has 9. So M is eliminated from the preference schedule.
Table 7.1.5: Preference Schedule for the Candy Election with M Eliminated Number of voters 3 1 4 1 9 1 st^ choice M^ M^ C^ C^ S 2 nd^ choice C S M S M 3 rd^ choice S C S M C
So the preference schedule becomes:
Table 7.1.6: Preference Schedule for the Candy Election with M Eliminated Number of voters 3 1 4 1 9 1 st^ choice C S C C S 2 nd^ choice S^ C^ S^ S^ C
And then we can condense it down to:
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Table 7.1.8: Summary of One-on-One Match-Ups for the Candy Election Match-Up 1 Match-Up 2 Match-Up 3 M vs. C M vs. S S vs. C 13 to 5 8 to 10 10 to 8 Winner of Match-Up 1 : M Winner of Match-Up 2 : S Winner of Match-Up 3 : S
M: 1 S: 2 C: 0 Thus, S wins the election. Note: If any one given match-up ends in a tie, then both candidates receive ½ point each for that match-up.
The problem with this method is that many overall elections (not just the one-on-one match-ups) will end in a tie, so you need to have a tie-breaker method designated before beginning the tabulation of the ballots. Another problem is that if there are more than three candidates, the number of pairwise comparisons that need to be analyzed becomes unwieldy. So, how many pairwise comparisons are there?
In Example 7.1.6, there were three one-on-one comparisons when there were three candidates. You may think that means the number of pairwise comparisons is the same as the number of candidates, but that is not correct. Let’s see if we can come up with a formula for the number of candidates. Suppose you have four candidates called A, B, C, and D. A is to be matched up with B, C, and D (three comparisons). B is to be compared with C and D, but has already been compared with A (two comparisons). C needs to be compared with D, but has already been compared with A and B (one more comparison). Therefore, the total number of one-on-one match-ups is 3 + 2 + 1 = 6 comparisons that need to be made with four candidates. What about five or six or more candidates? Looking at five candidates, the first candidate needs to be matched-up with four other candidates, the second candidate needs to be matched-up with three other candidates, the third candidate needs to be matched-up with two other candidates, and the fourth candidate needs to only be matched-up with the last candidate for one more match-up. Thus, the total is 4 + 3 + 2 + 1 = 10 pairwise comparisons when there are five candidates.
Now, for six candidates, you would have (^5) + 4 + 3 + 2 + 1 = 15 pairwise comparisons to do. Continuing this pattern, if you have N candidates then there are
( N^ −^1 )^ +^ ( N −^2 )+^ ^ +^3 +^2 +^1 pairwise comparisons. For small numbers of candidates,
it isn’t hard to add these numbers up, but for large numbers of candidates there is a shortcut for adding the numbers together. It turns out that the following formula is true:
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N N
N N
− + − + + + + =. Thus, for 10 candidates, there are
9 8 7 6 5 4 3 2 1 10 10^ (^1 )^ 10 9(^ ) 45
- = = = pairwise comparisons. So you
can see that in this method, the number of pairwise comparisons to do can get large quite quickly.
Now that we have reviewed four different voting methods, how do you decide which method to use? One question to ask is which method is the fairest? Unfortunately, there is no completely fair method. This is based on Arrow’s Impossibility Theorem.
Arrow's Impossibility Theorem: No voting system can satisfy all four fairness criteria in all cases.
This brings up the question, what are the four fairness criteria? They are guidelines that people use to help decide which voting method would be best to use under certain circumstances. They are the Majority Criterion, Condorcet Criterion, Monotonicity Criterion, and Independence of Irrelevant Alternatives Criterion.
Fairness Criteria: The Majority Criterion (Criterion 1): If a candidate receives a majority of the 1st-place votes in an election, then that candidate should be the winner of the election. The Condorcet Criterion (Criterion 2): If there is a candidate that in a head-to-head comparison is preferred by the voters over every other candidate, then that candidate should be the winner of the election. This candidate is known as the Condorcet candidate. The Monotonicity Criterion (Criterion 3): If candidate X is a winner of an election and, in a re-election, the only changes in the ballots are changes that favor X, then X should remain a winner of the election. The Independence of Irrelevant Alternatives Criterion (Criterion 4): If candidate X is a winner of an election and one (or more) of the other candidates is removed and the ballots recounted, then X should still be a winner of the election.
Example 7.1.7: Condorcet Criterion Violated Suppose you have a vacation club trying to figure out where it wants to spend next year’s vacation. The choices are Hawaii (H), Anaheim (A), or Orlando (O). The preference schedule for this election is shown below in Table 7.1.9.
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Now using the Plurality with Elimination Method, Adams has 47 first-place votes, Brown has 24, and Carter has 29. This time, Brown is eliminated first instead of Carter. Two of Brown’s votes go to Adams and 22 of Brown’s votes go to Carter. Now, Adams has 47 + 2 = 49 votes and Carter has 29 + 22 = 51 votes. Carter wins the election. This doesn’t make sense since Adams had won the election before, and the only changes that were made to the ballots were in favor of Adams. However, Adams doesn’t win the re-election. The reason that this happened is that there was a difference in who was eliminated first, and that caused a difference in how the votes are re-distributed. In this example, the Plurality with Elimination Method violates the Monotonicity Criterion.
Example 7.1.9: Majority Criterion Violated Suppose a group is planning to have a conference in one of four Arizona cities: Flagstaff, Phoenix, Tucson, or Yuma. The votes for where to hold the conference are summarized in the preference schedule shown below in Table 7.1.12.
Table 7.1.12: Preference Schedule for Conference City Number of voters 51 25 10 14 1 st^ choice Flagstaff Phoenix Yuma Tucson 2 nd^ choice Phoenix Yuma Phoenix Phoenix 3 rd^ choice Tucson Tucson Tucson Yuma 4 th^ choice Yuma Flagstaff Flagstaff Flagstaff
If we use the Borda Count Method to determine the winner then the number of Borda points that each candidate receives are shown in Table 7.1.13.
Table 7.1.13: Preference Schedule for Conference City with Borda Points Number of voters 51 25 10 14 1 st^ choice 4 points
Flagstaff 204
Phoenix 100
Yuma 40
Tucson 56 2 nd^ choice 3 points
Phoenix 153
Yuma 75
Phoenix 30
Phoenix 42 3 rd^ choice 2 points
Tucson 102
Tucson 50
Tucson 20
Yuma 28 4 th^ choice 1 point
Yuma 51
Flagstaff 25
Flagstaff 10
Flagstaff 14
The totals of all the Borda points for each city are:
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Flagstaff: 204 + 25 + 10 + 14 =253 points Phoenix: 153 + 100 + 30 + 42 =325 points Yuma: 51 + 75 + 40 + 28 =194 points Tucson: 102 + 50 + 20 + 56 =228 points
Phoenix wins using the Borda Count Method. However, notice that Flagstaff actually has the majority of first-place votes. There are 100 voters total and 51 voters voted for Flagstaff in first place (51/100 = 51% or a majority of the first- place votes). So, Flagstaff should have won based on the Majority Criterion. This shows how the Borda Count Method can violate the Majority Criterion.
Example 7.1.10: Independence of Irrelevant Alternatives Criterion Violated A committee is trying to award a scholarship to one of four students: Anna (A), Brian (B), Carlos (C), and Dmitri (D). The votes are shown below.
Table 7.1.14: Preference Schedule for Scholarship Number of voters 5 5 6 4 1 st^ choice D A C B 2 nd^ choice A^ C^ B^ D 3 rd^ choice C B D A 4 th^ choice B D A C
Using the Method of Pairwise Comparisons: A vs B: 10 votes to 10 votes, A gets ½ point and B gets ½ point A vs C: 14 votes to 6 votes, A gets 1 point A vs D: 5 votes to 15 votes, D gets 1 point B vs C: 4 votes to 16 votes, C gets 1 point B vs D: 15 votes to 5 votes, B gets 1 point C vs D: 11 votes to 9 votes, C gets 1 point
So A has 1½ points, B has 1 point, C has 2 points, and D has 1 point. So Carlos is awarded the scholarship.
Now suppose it turns out that Dmitri didn’t qualify for the scholarship after all. Though it should make no difference, the committee decides to recount the vote. The preference schedule without Dmitri is below.
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candidate running. As an example, if a Democrat, a Republican, and a Libertarian are all running in the same race, and you happen to prefer the Libertarian candidate. However, you are afraid that the Democratic candidate will win if you vote for the Libertarian candidate, so instead you vote for the Republican candidate. You have voted insincerely to your true preference.
Approval Voting: Since there is no completely fair voting method, people have been trying to come up with new methods over the years. One idea is to have the voters decide whether they approve or disapprove of candidates in an election. This way, the voter can decide that they would be happy with some of the candidates, but would not be happy with the other ones. A possible ballot in this situation is shown in Table 7.1.17:
Table 7.1.17: Approval Voting Ballot Candidate Approve Disapprove Smith X Baker X James X Paulsen X
This voter would approve of Smith or Paulsen, but would not approve of Baker or James. In this type of election, the candidate with the most approval votes wins the election.
One issue with approval voting is that it tends to elect the least disliked candidate instead of the best candidate. Another issue is that it can result in insincere voting as described above. As a reminder, there is no perfect voting method. Arrow proved that there never will be one. So make sure that you determine the method of voting that you will use before you conduct an election.
Section 7.2: Weighted Voting
Voting Power: There are some types of elections where the voters do not all have the same amount of power. This happens often in the business world where the power that a voter possesses may be based on how many shares of stock he/she owns. In this situation, one voter may control the equivalent of 100 votes where other voters only control 15 or 10 or fewer votes. Therefore, the amount of power that each voter possesses is different. Another example is in how the President of the United States is elected. When a person goes to the
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polls and casts a vote for President, he or she is actually electing who will go to the Electoral College and represent that state by casting the actual vote for President. Each state has a certain number of Electoral College votes, which is determined by the number of Senators and number of Representatives in Congress. Some states have more Electoral College votes than others, so some states have more power than others. How do we determine the power that each state possesses? To figure out power, we need to first define some concepts of a weighted voting system. The individuals or entities that vote are called players. The notation for the players is P P 1 , 2 (^) , P 3 , , PN , where N is the number of players. Each player controls a certain number
of votes, which are called the weight of that player. The notation for the weights is w 1 (^) , w 2 (^) , w 3 , , wN , where w 1 is the weight of P 1 , w 2 is the weight of P 2 , etc. In order for a
motion to pass, it must have a minimum number of votes. This minimum is known as the quota. The notation for quota is q. The quota must be over half the total weights and cannot be more than total weight. In other words,
1 2 3 2 1 2 3
N N
w w w w q w w w w
The way to denote a weighted voting system is [ q : w 1 , w 2 , w 3 , , wN ].
Example 7.2.1: Weighted Voting System A company has 5 shareholders. Ms. Lee has 30% ownership, Ms. Miller has 25%, Mr. Matic has 22% ownership, Ms. Pierce has 14%, and Mr. Hamilton has 9%. There is a motion to decide where best to invest their savings. The company’s by- laws define the quota as 58%. What does this voting system look like?
Treating the percentages of ownership as the votes, the system looks like:
Example 7.2.2: Valid Weighted Voting System Which of the following are valid weighted voting systems?
a. [ 8 : 5, 4, 4,3, 2]
The quota is 8 in this example. The total weight is 5 + 4 + 4 + 3 + 2 = 18. Half of 18 is 9, so the quota must be 9 < q ≤ 18. Since the quota is 8, and 8 is not more than 9, this system is not valid.
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are needed to pass a motion. That also means that any player can stop a motion from passing. A player that can stop a motion from passing is said to have veto power. In the
system [ 8 : 5, 4, 2] , player three has a weight of two. Players one and two can join
together and pass any motion without player three, and player three doesn’t have enough weight to join with either player one or player two to pass a motion. So player three has no power. A player who has no power is called a dummy.
Example 7.2.3: Dictator, Veto Power, or Dummy?
In the weighted voting system [ 57 : 23, 21,16,12] , are any of the players a dictator
or a dummy or do any have veto power.
Since no player has a weight higher than or the same as the quota, then there is no dictator. If players one and two join together, they can’t pass a motion without player three, so player three has veto power. Under the same logic, players one and two also have veto power. Player four cannot join with any players to pass a motion, so player four’s votes do not matter. Thus, player four is a dummy.
Now that we have an understanding of some of the basic concepts, how do we quantify how much power each player has? There are two different methods. One is called the Banzhaf Power Index and the other is the Shapely-Shubik Power Index. We will look at each of these indices separately.
Banzhaf Power Index:
A coalition is a set of players that join forces to vote together. If there are three players P P 1 , 2 and P 3 then the coalitions would be:
{ P 1^ } { ,^ P 2^ } { ,^ P 3^ } {,^ P P 1 ,^2^ } { ,^ P P 1 ,^3^ } { ,^ P 2^ ,^ P 3^ } {,^ P P 1 ,^2^ , P 3^ }.
Not all of these coalitions are winning coalitions. To find out if a coalition is winning or not look at the sum of the weights in each coalition and then compare that sum to the quota. If the sum is the quota or more, then the coalition is a winning coalition.
Example 7.2.4: Coalitions with Weights
In the weighted voting system[17 :12,7,3 ] , the weight of each coalition and
whether it wins or loses is in the table below.
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Table 7.2.1: Coalition Listing Coalition Weight Win or Lose?
{ P 1^ }^12 Lose
{ P 2^ }^7 Lose
{ P 3^ }^3 Lose
{ P P 1 , 2^ }^19 Win
{ P P 1 , 3^ }^15 Lose
{ P 2^ , P 3^ }^10 Lose
{ P P 1 ,^2^ , P 3^ }^22 Win
In each of the winning coalitions you will notice that there may be a player or players that if they were to leave the coalition, the coalition would become a losing coalition. If there is such a player or players, they are known as the critical player(s) in that coalition.
Example 7.2.5: Critical Players
In the weighted voting system[17 :12,7,3 ] , determine which player(s) are critical
player(s). Note that we have already determined which coalitions are winning coalitions for this weighted voting system in Example 7.2.4. Thus, when we continue on to determine the critical player(s), we only need to list the winning coalitions.
Table 7.2.2: Winning Coalitions and Critical Players Coalition Weight Win or Lose? Critical Player
{ P P 1 , 2^ }^19 Win^ P^1 , P^2
{ P P 1 , 2 , P 3 }^22 Win^ P^1 , P^2
Notice, player one and player two are both critical players two times and player three is never a critical player.
Banzhaf Power Index:
The Banzhaf power index is one measure of the power of the players in a weighted voting system. In this index, a player’s power is determined by the ratio of the number of times that player is critical to the total number of times any and all players are critical.
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Example 7.2.7: Banzhaf Power Index
Example 7.2.1 had the weighted voting system of [ 58 : 30, 25, 22,14,9]. Find the
Banzhaf power index for each player.
Since there are five players, there are 31 coalitions. This is too many to write out, but if we are careful, we can just write out the winning coalitions. No player can win alone, so we can ignore all of the coalitions with one player. Also, no two- player coalition can win either. So we can start with the three player coalitions.
Table 7.2.4: Winning Coalitions and Critical Players Winning Coalition Critical Player
{ P P 1 ,^2^ , P 3^ } P P 1 ,^2^ , P 3
{ P P 1 ,^2^ , P 4^ } P P 1 ,^2^ , P 4
{ P P 1 ,^2^ , P 5^ } P P 1 ,^2^ , P 5
{ P P P 1 ,^3 , 4^ } P P P 1 ,^3 , 4
{ P P P 1 ,^3 , 5^ } P P P 1 ,^3 , 5
{ P 2^ ,^ P P 3 , 4^ } P 2^ ,^ P P 3 , 4
{ P P 1 ,^2^ ,^ P P 3 , 4 }
{ P P 1 ,^2^ ,^ P P 3 , 5^ } P 1
{ P P 1 ,^2^ ,^ P 4^ , P 5^ } P P 1 , 2
{ P P P 1 ,^3 ,^4^ , P 5^ } P P 1 , 3
{ P 2^ ,^ P P 3 ,^4^ , P 5^ } P 2^ ,^ P P 3 , 4
{ P P 1 , 2 , P P 3 , 4 , P 5 }
So player one is critical eight times, player two is critical six times, player three is critical six times, player four is critical four times, and player five is critical two times. Thus, the total number of times any player is critical is T = 26.
Banzhaf power index for P 1 =
Banzhaf power index for P 2 = 6 3 26 13
Banzhaf power index for P 3 =
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Banzhaf power index for P 4 = 4 2 26 13
Banzhaf power index for P 5 =
Every player has some power. Player one has the most power with 30.8% of the power. No one has veto power, since no player is in every winning coalition.
Shapely-Shubik Power Index: Shapely-Shubik takes a different approach to calculating the power. Instead of just looking at which players can form coalitions, Shapely-Shubik decided that all players form a coalition together, but the order that players join a coalition is important. This is called a sequential coalition. Instead of looking at a player leaving a coalition, this method examines what happens when a player joins a coalition. If when a player joins the coalition, the coalition changes from a losing to a winning coalition, then that player is known as a pivotal player. Now we count up how many times each player is pivotal, and then divide by the number of sequential coalitions. Note, that in reality when coalitions are formed for passing a motion, not all players will join the coalition. The sequential coalition is used only to figure out the power each player possess.
As an example, suppose you have the weighted voting system of [17 :12,7,3 ]. One of the
sequential coalitions is P P 1 , 2 (^) , P 3 which means that P 1 joins the coalition first, followed
by P 2 joining the coalition, and finally, P 3 joins the coalition. When player one joins the coalition, the coalition is a losing coalition with only 12 votes. Then, when player two joins, the coalition now has enough votes to win (12 + 7 = 19 votes). Player three joining doesn’t change the coalition’s winning status so it is irrelevant. Thus, player two is the
pivotal player for this coalition. Another sequential coalition is P P P 1 , 3 , 2. When player
one joins the coalition, the coalition is a losing coalition with only 12 votes. Then player three joins but the coalition is still a losing coalition with only 15 votes. Then player two joins and the coalition is now a winning coalition with 22 votes. So player two is the pivotal player for this coalition as well.
How many sequential coalitions are there for N players? Let’s look at three players first. The sequential coalitions for three players ( P 1 , P 2 , P 3 ) are: P P 1 , 2 (^) , P 3 (^) , P P P 1 , 3 , 2 (^) , P 2 (^) , P P 1 , 3 (^) , P 2 (^) , P P 3 , 1 (^) , P P P 3 , 1 , 2 (^) , P P 3 , 2 (^) , P 1.
Note: The difference in notation: We use { } for coalitions and sequential
coalitions.
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