Presentation on theme: "Problems with Voting Methods Voting System Criteria – Majority criterion – Condorcet Winner criterion – Independence of Irrelevant Alternatives – Pareto."— Presentation transcript:
Problems with Voting Methods Voting System Criteria – Majority criterion – Condorcet Winner criterion – Independence of Irrelevant Alternatives – Pareto criterion – Monotonicity Summary of Methods and Criteria Voting Methods Continued
Problems with the Plurality Method We’ve already seen an example of a problem with the plurality method: Remember the example with 3 candidates B, K and N. Suppose B gets 40%, and K and N each get 30% of the vote. A majority prefers someone other than B and yet B has a plurality of the votes and wins the election. We will now show that the Plurality method does not satisfy what is called “the Condorcet Winner Criterion.” To explain what this means, consider the following example… –Suppose a marching band has 100 members and there are 5 choices for which song they will play during the next show. Suppose the choices are Rico Suave (R), Hold My Hand (H), Strike it Up (S), Cantaloop (C) or Oops! I did it again (O). –Assume the band members decide to vote on which song to use and that they decide to use the plurality method to decide the winner.
Problems with plurality… Number of Voters (100 total) st RHC 2 nd HSH 3 rd COS 4 th OCO 5 th SRR Suppose the table below represents the preferences of the individual band members. By the plurality method Rico Suave (R) is a slight winner because R gets 49 votes – more than any other song. However, notice that not only does a majority prefer something besides R, that majority would consider R as their last choice among all choices.
Problems with plurality… Number of Voters (100 total) st RHC 2 nd HSH 3 rd COS 4 th OCO 5 th SRR Notice also that H is the first choice of 48 voters and also H is a second choice of 52 voters. Perhaps we might consider H as more representative of the preferences of the band as a whole. In fact, consider one-on-one comparisons with H and any other choice and a majority of the band will always pick H.
Problems with plurality… Number of Voters (100 total) st RHC 2 nd HSH 3 rd COS 4 th OCO 5 th SRR For example, let’s compare H and C head-to-head. We find that 97 voters (a clear majority) rank H over C. Every voter ranks H over choice O and also all voters rank H over S. How about comparing H and R ? We find that still a majority of voters (51 voters) would prefer candidate H over R. We have found that a majority of band members prefer H to any other choice and yet R will win by the plurality method.
Problems with the plurality method In our previous example, we found that H wins in one-on-one comparisons with each and every other candidate and yet H is not the winner by the plurality method. In the language of voting theory, we say that H is the Condorcet winner for this election because H beats every other candidate in one-on-one comparisons. We say that a voting procedure satisfies the Condorcet Winner Criterion (CWC) provided that for every possible set of preference lists either (1) there is no Condorcet winner or (2) the Condorcet winner is the unique winner of the election. The Plurality Method of Voting does not satisfy the Condorcet Winner Criterion as evidenced by the preceding example.
Voting Methods and Voting Method Criteria We will continue to consider other voting methods and other criteria. We can refer to these criteria as conditions that we would want an ideal voting system to satisfy. We may even consider them as basic principles of fairness with regards to voting. It is important to recognize one example is sufficient to show a particular method does not satisfy a given criterion – because we have shown that method doesn’t always satisfy the criterion. However, to show that a voting method satisfies a particular criterion, we must show it will satisfy that criterion in every possible example. Obviously, it’s impossible to try infinitely many possible examples, so proving a voting method satisfies a certain criterion requires an abstract theoretical approach that deductively proves that the criterion would be satisfied in any possible example.
Problems with the Borda Count Method Borda Count fails the majority criterion. Borda Count fails the independence of irrelevant alternatives criterion. The majority criterion (not in our book) states that if a candidate has a majority (more than half) of the first place votes, then that candidate should be the winner of the election. Independence of irrelevant alternatives (IIA) is a criterion that holds if there is a repeat election with the same voters and some non-winning candidates are removed, the previous winner should still win the election.
Straw Elections When studying the IIA criterion we sometimes refer to “straw elections”. A straw election is like a survey. The straw election is a first election to see what current preferences which is then followed later by another election. In the examples we study, we assume the same voters participate in both elections but in some cases their preferences might change from one election to the next.
Problems with Borda Count Consider the following example: By the Borda Count method, the winner of an election with this preference schedule would be… Number of Voters (11 total) st (3 points) ABC 2 nd (2 points) BCD 3 rd (1 point) CDB 4 th (0 points) DAA
Problems with Borda Count By the Borda Count method, the winner of an election with this preference schedule is … A gets 18 points, B gets 21 points, C gets 19 points, D gets 8 points so the winner is B. However, a majority (6 is more than half) of voters actually prefer A. This example shows how Borda Count can violate the majority criterion. Number of Voters (11 total) st (3 points) A (18) B (6) C (9) 2 nd (2 points) B (12) C (4) D (6) 3 rd (1 point) C (6) D (2) B (3) 4 th (0 points) D (0) A (0) A (0)
Problems with Borda Count To show how Borda Count fails the IIA criterion, suppose a straw election were held and the voters indicate the following preferences. Number of Voters (26 total) st (3pts) ABD 2 nd (2pts) BDA 3 rd (1pt) CCC 4 th (0pts) DAB Following the Borda count method, we calculate: A gets 50 points, B gets 52 points, C gets 26 points and D gets 28 points. So, by the Borda count method, B would win. Suppose that now a second election occurs with the same voters and preferences but that candidate C drops out. Who will win now?
Problems with Borda Count Number of Voters (26 total) st (2pts) ABD 2 nd (1pt) BDA 3 rd (0pts) DAB We remove C but we assume voters preference remain otherwise the same. Following the Borda count method, we calculate: A gets 32 points, B gets 30 points, and D gets 16 points. So, by the Borda count method, A would win. Wait a minute!! Look at that again. Before, when C was in the election, B would have won. Now that C dropped out, A is the winner! This illustrates how Borda Count can fail the Independence of Irrelevant Alternatives criterion.
Problems with sequential pairwise voting With sequential pairwise voting, in some cases, we can have a different winner of the election simply by changing the agenda. Sequential pairwise voting fails the Pareto criterion.
Problems with pairwise sequential voting The Pareto criterion holds that if every voter prefers some candidate (say A) over some other candidate (say B) then B should not be among the winners of the election. This criterion almost sounds like the opposite of the Condorcet criterion. However, notice that to show failure of the Pareto criterion, we must show there is at least one candidate (A) that everyone prefers over another (B) and yet B is a winner. To show failure of the Condorcet criterion, we must show there is a candidate (A) that everyone prefers head-to-head with all of the other candidates and yet A is not a winner.
Problems with sequential pairwise voting Consider the following preference schedule… Number of Voters (3 total) st ACB 2 nd BAD 3 rd DBC 4 th CDA Suppose we consider the candidates in alphabetical order: A, B, C, then D. Compare A with B: A wins 2 to 1. Compare A with C: C wins 2 to 1. Compare C with D: D wins 2 to 1. Thus, by sequential pairwise voting the winner is D. However, notice that every voter prefers B over D. In this example, there is a candidate (B) that every voter prefers over the winner. Thus, this example demonstrates how sequential pairwise voting can violate the Pareto criterion.
Problems with the Hare System Voting Method The monotonicity criterion holds that if a new election were held and the only change that is made is for some voter to move the winning candidate higher on their preference list, that candidate should still win the election. The Hare System does not satisfy the monotonicity criterion.
Problems with the Hare System To demonstrate how the Hare System fails the monotonicity criterion, consider the following example… Number of Voters (13 total) st ACBB 2 nd BBCA 3 rd CAAC Who will win using the Hare System of voting? B and C are tied for least first place votes (both have 4 first place votes which is less than A, which has 5). B and C are eliminated in the first round in the Hare System. With the current preference schedule, A is the winner of the election by the Hare System of voting. However, suppose that in the final minutes before the election, realizing his relative low standing among a majority of voters, candidate A makes a last minute appeal for support. Suppose this effort sways the opinion of the voter on the far-right column. Suppose that far-right column voter changes his true preference and ranks A higher…
Problems with the Hare System Suppose that, because of a change in the true preferences of a single voter, that voter moves candidate A higher on his preference list… Number of Voters (13 total) st ACBA 2 nd BBCB 3 rd CAAC Suppose the new preference schedule is as shown here. The voter on the far-right column has responded to candidate A’s last minute appeal for support and decided to rank A as first instead of second as he had previously. Now suppose the election is held with this preference schedule. Who actually wins by the Hare System? In the first round, candidate B is eliminated because initially B has the least first place votes…
Problems with the Hare System Number of Voters (13 total) st ACA 2 nd C 3 rd CAAC With B eliminated in round 1 with the Hare system, we have the table shown at left. We preserve relative preferences of voters with B eliminated and re-organize the table as shown below. Number of Voters (13 total) st ACCA 2 nd CAAC Now, by the Hare System, we declare C the winner of the election. This is because A has fewer first place votes than C and is eliminated leaving C the winner. This example has demonstrated that in spite of the fact that the only change made in the preferences among voters was a change that was favorable to A, that candidate did not remain as the winner of the election. We have shown the Hare System violates the monotonicity criterion.
Plurality fails IIA Consider the following preference table, Number of Voters (15 total) st ABC 2nd2nd BCB 3rd3rd CAA Initially, if the election were held with the current preference table, candidate A would win using plurality. Suppose, however that C dropped out. Number of Voters (15 total) st ABB 2 nd BAA 3 rd Without candidate C, assuming the voters’ preferences do not change, we find that now B has more votes and would win by plurality.
Summary of Voting Method Criteria Majority Criterion – If a candidate has a majority of first place votes then that candidate should be the winner of the election. Condorcet Winner Criterion – If a candidate beats every other candidate in one-on-one comparisons, then that candidate should be the winner of the election. Independence of Irrelevant Alternatives Criterion – If a second vote is held and a non-winning candidate drops out, then the previous winner should still win the election. Pareto Criterion – If there is some candidate A that everyone prefers to another candidate B, then B should not be the winner of the election. Monotonicity Criterion – If, after a second vote, the only changes in voter preferences are in favor of a certain candidate, that candidate should not lose the election.
Every Criterion is a Conditional Statement If a candidate has a majority of first place votes then that candidate should be the winner of the election. If a candidate beats every other candidate in one-on-one comparisons, then that candidate should be the winner of the election. If a second vote is held and a non-winning candidate drops out, then the previous winner should still win the election. If there is some candidate A that everyone prefers to another candidate B, then B should not be the winner of the election. If, after a second vote, the only changes in voter preferences are in favor of a certain candidate, then that candidate should not lose the election. This is an easy way to tell the difference between methods of voting and fairness criteria – the fairness criteria always can be written in the “If…then” form.
Summary of Voting Methods and Criteria Majority Criterion CWCIIAPareto criterion Monotonicity PluralityYesNo Yes Borda Count No Yes Sequential pairwise Yes No Yes Hare System YesNo YesNo ApprovalNo YesNoYes