One More Voting Method: Plurality with Elimination Our final voting method is an elimination method: we start with a large number of candidates, then successively remove weak candidates until very few candidates remain. The most common elimination method is the Plurality with Elimination Method: Step 1: If a candidate has a MAJORITY of first place votes, declare that candidate the winner. Step 2: If no candidate has a majority, then remove the candidate with the FEWEST number of first place votes. Repeat Step 1. January 30 and February 1 2012 1
Plurality with Elimination in Action Determine the winner of this election using Plurality with Elimination: 7 7 8 5 A C D B B B C A C A B C D D A D A candidate requires at more than 7 + 7 + 8 + 5 = 13.5 first place votes for a 2 majority. Since no candidate has a majority of first place votes, we proceed to Step 2. Who has the fewest first place votes? January 30 and February 1 2012 2
Plurality with Elimination in Action B has the fewest first place votes, so we remove B. 7 7 8 5 A C D //B //B //B C A C A //B C D D A D Pay attention! We are NOT removing the ballots that favored B. We are removing B from all ballots. Any candidate that was previously ranked below B will move up a spot. January 30 and February 1 2012 3
Plurality with Elimination in Action We rewrite the preference schedule with the three remaining candidates: 7 7 8 5 A C D A C A C C D D A D Notice that, upon removing B, the first and last columns involve the same ballot type. It will simplify things if we combine those columns: 12 7 8 A C D C A C D D A Now we repeat the Plurality with Elimination algorithm for this 3 candidate election. January 30 and February 1 2012 4
Plurality with Elimination in Action 12 7 8 A C D C A C D D A Step 1: Is there a Majority candidate? A majority requires at least 14 first place votes. There is no majority candidate, so go to step 2. Step 2: Eliminate candidate with fewest first place votes. In this case, C will be eliminated. January 30 and February 1 2012 5
Plurality with Elimination in Action Rewriting the table without C: Combining similar columns: 12 7 8 A //C D //C A //C D D A 12 7 8 A A D D D A 19 8 A D D A January 30 and February 1 2012 6
Plurality with Elimination in Action 19 8 A D D A Now repeat the Plurality with Elimination algorithm: Step 1: Is there a Majority candidate? Yes! A is the majority candidate, so we can finally declare a winner. A wins using Plurality with elimination. January 30 and February 1 2012 7
Re-examining the Election 7 7 8 5 A C D B B B C A C A B C D D A D We just showed that A is the Plurality with Elimination Winner of this election. On a recent worksheet you looked at this same election and found: B is the Borda Count Winner C is the Pairwise Comparison Winner D is the Plurality Winner. So, WHO SHOULD WIN? January 30 and February 1 2012 8
The Remainder of the Voting Theory Chapter We have now seen four different methods that can be used to determine the winner of an election. There are certainly many other methods used in practice, but these are four of the most common. Other common methods are combinations of these methods (For example, a candidate might be awarded 3 points for each first place, two points for each second place, 1 point for each third, and no points if below third, then the candidates with the lowest point totals are eliminated and the process is repeated until a majority candidate is found. This could be desribed as a Borda Count with Elimination ) Our focus for the remaining 4 class periods will be to understand the strengths and weaknesses of the four main methods. In particular, the previous example shows that it is possible for these four methods to all produce different winners for the same election. It s not the voting that s democracy; it s the counting. Tom Stoppard. Before determining which methods are actually fair voting methods, we look at a few more technical issues involving Plurality with Elimination. January 30 and February 1 2012 9
A Quick Plurality with Elimination Example An election involves 25 voters and 6 candidates. 2 7 6 1 1 4 4 A B B C C D E D F A B D A C F C D A A E F B D F D F C B E E C E E B A C A E F B F D Explain why you can determine the Plurality with Elimination winner in this election with almost no work. Solution: Step 1 of the elimination algorithm says that a Majority candidate should be selected as the winner. The elimination steps are only required IF the election does not already have a majority candidate. B is a majority candidate, so B wins using Plurality with Elimination. January 30 and February 1 2012 10
Plurality with Elimination: What to do in the case of a tie? Determine the winner, using Plurality with Elimination. 5 4 4 6 A B C D D C A A B A B C C D D B There is no Majority Candidate, so we should eliminate the candidate with the fewest first place votes. B and C are tied for least first place votes. There are several different logical tie breaking rules we could use. The most commonly used rule is that if there is a tie for who has the fewest first place votes, then both (or all) of the tie-ing candidates should be eliminated. In this case, we remove B and C. January 30 and February 1 2012 11
Plurality with Elimination: What to do in the case of a tie? 5 4 4 6 A //B //C D D //C A A //B A //B //C //C D D //B After removing B and C (moving A or D up if necessary) and combining similar ballots, we obtain 13 6 A D D A A now has a Majority, so A is the Plurality with Elimination Winner. January 30 and February 1 2012 12
Plurality with Elimination: Other tie breaking rules Return to the original election 5 4 4 6 A B C D D C A A B A B C C D D B B and C are tied for least first place votes. A different tie-breaking rule that is sometimes used in practice is to randomly choose one of the candidates with the fewest first place votes to be eliminated. Suppose we choose to eliminate B first. 5 4 4 6 A //B C D D C A A //B A //B C C D D //B January 30 and February 1 2012 13
Plurality with Elimination: Other tie breaking rules Remove B, move other candidates up where necessary, and combine similar columns: 5 8 6 A C D D A A C D C No candidate has a majority, so we eliminate the candidate with the fewest first place votes. In this case, we eliminate... A. Removing A and simplifying the schedule: 11 8 D C C D Thus, D is the Plurality with Elimination Winner. January 30 and February 1 2012 14
Plurality with Elimination: Other tie breaking rules This last example shows how delicate these voting methods can be. Not only does the winner of the election depend on which voting method is used, even a small change in a technical tie breaking rule can affect the winner of the election. We accept the first tie breaking rule as being the correct one. i.e., if multiple candidates receive the fewest number of first place votes, we eliminate all of these candidates. January 30 and February 1 2012 15
Measuring Fairness: Recall: A candidate is a Majority Candidate if this candidate receives more than half of the first place votes. A candidate is a Condorcet Candidate if this candidate wins all of the one-on-one comparisons involving this candidate. Typically, we think of Majority Candidates and Condorcet Candidates as being very strong candidates, and these types of candidates lead to our first two measures of fairness. January 30 and February 1 2012 16
Majority Criterion The Majority Criterion: If a candidate X receives a majority of first place votes, then candidate X should be declared the winner of the election. Please note that this is not a RULE. This is a notion of fairness. We believe that majority candidates are very strong candidates, and the Majority Criterion essentially says that we would think a voting method was unfair if a candidate had a majority of first place votes and did not win the election. A given voting method may satisfy or violate a given Fairness Criterion. Failing a fairness criterion simply acts as a strike against that particular voting method. January 30 and February 1 2012 17
Proofs that Some Methods Satisfy the Majority Criterion PLURALITY: Suppose candidate X is a Majority candidate. Since X receives more than half of the first place votes, X automatically receives more first place votes than any other candidate, so X will be declared the winner in the Plurality Method. This shows that the Plurality Method satisfies the Majority Criterion. PLURALITY WITH ELIMINATION: Suppose candidate X is a Majority candidate. By step 1 of the Plurality with Elimination algorithm, X is declared the winner using Plurality with Elimination. Therefore, the Plurality with Elimination Method satisfies the Majority Criterion. PAIRWISE COMPARISON: Suppose candidate X is a Majority candidate. Since X receives more than half of the first place votes, X will receive more votes than the other candidate in any one-on-one comparison. In other words, X will win all of her one-on-one comparisons, and so X wins more one-on-one comparisons than any other candidate, so X wins using Pairwise Comparisons. This shows that the Pairwise Comparison Method satisfies the Majority Criterion. January 30 and February 1 2012 18
Borda Count FAILS the Majority Criterion BORDA COUNT: Recall the example from the January 23 (Pairwise Comparison and Borda Count) set of slides. 21 26 3 A B C C C A B A B We saw that C won using Borda Count. Thus, B loses using Borda Count, despite the fact that B was a Majority Candidate. In fact, that set of slides contains several examples of elections in which the Majority Candidate LOSES using Borda. That set of slides even had an example of a candidate that received 79% of the first place votes and still lost using the Borda Count Method. January 30 and February 1 2012 19
Borda Count FAILS the Majority Criterion On an exam, you could be asked to construct an example which shows that a given criterion fails a given Fairness Criterion. If you have to construct an example on the fly, it is usually easiest to try to create a small example that does the trick. This is perhaps the smallest example of an election in which a Majority Candidate loses using Borda Count Method. 3 2 X Y Y Z Z X BORDA COUNT: Borda Count FAILS the Majority Criterion. January 30 and February 1 2012 20
What does it mean to SATISFY a Fairness Criterion? First, Fairness Criteria are applied to Voting Methods. They are not applied to the individual elections. To say that a voting method, VM, satisfies a given fairness criterion, FC, means that in EVERY ELECTION, whenever voting method VM is used to determine the winner of the election, AND the hypotheses of the fairness criterion are met, then the conclusion of the fairness criterion is guaranteed to hold. In order to show that a given voting method SATISFIES a given fairness criterion, you need to give a logical argument in which you assume the hypotheses of the Fairness Criterion, then, ONLY using those hypotheses and the rules for that voting method, demonstrate that the conclusion of the fairness criterion are true. Chances are that you have not had a lot of practice providing formal logical proofs and counterexamples, so we discuss a few ptifalls to avoid. January 30 and February 1 2012 21
Incorrect Arguments for Plurality Method and Majority Criterion Consider the election 21 26 3 A B C C C A B A B B receives more votes than any other candidate, so B is the winner using Plurality. Also, B is a Majority Candidate. Therefore, the Plurality Method satisfies the Majority Criterion. This is NOT a valid proof that the Plurality Method satisfies the Majority Criterion. All this shows is that, in this one example, the winner of the plurality method happened to be a majority candidate. A valid proof has to explain why, in ANY ELECTION WITH A MAJORITY CANDIDATE, said majority candidate MUST win using Plurality. To say that Plurality Method satisfies the Majority Criterion means that NO election, using Plurality, would ever produce a violation of the Majority Criterion. The above only shows that no violation occured in this particular example. January 30 and February 1 2012 22
Incorrect Arguments for Plurality Method and Majority Criterion 2 2 3 A B C C C A B A B C receives more votes than any other candidate, so C is the winner using the Plurality Method. C is not a Majority candidate (since a majority would require 4 or more first place votes.) The Plurality Method fails the Majority Criterion because a non-majority candidate won using Plurality. This is NOT a valid argument. The Majority Criterion does not say that the winner of the election must be a Majority Candidate. Rather, it says that IF the election has a Majority Candidate, then that candidate must win. Since this election did not have a Majority candidate, there is nothing to violate. If it helps, think of the Majority Criterion as saying A majority candidate should not lose instead of A majority candidate should win. January 30 and February 1 2012 23
What does it mean to FAIL a Fairness Criterion? To say that a voting method, VM, SATISFIES a given fairness criterion, FC, means that in EVERY ELECTION, whenever voting method VM is used to determine the winner of the election, AND the hypotheses of the fairness criterion are met, then the conclusion of the fairness criterion is guaranteed to hold. In order to show that a VOTING METHOD fails a fairness criterion, it is sufficient to find a single example of an election in which that fairness criterion is not met. For example, we showed that the Borda Count Method FAILS the Majority Criterion by finding a single example in which there was a Majority Candidate AND that Majority Candidate lost the election using the Borda Count Method. January 30 and February 1 2012 24
Incorrect Arguments for Borda Count Method and Majority Criterion Consider the election 2 1 A B C A B C A receives 2 3 + 1 2 = 8 Borda points, B receives 2 1 + 1 3 = 5 Borda points, C receives 2 2 + 1 1 = 5 Borda points. So A wins using Borda Method. Also, A is a Majority Candidate. Therefore, Borda Method satisfies the Majority Criterion. This is NOT a valid proof that the Plurality Method satisfies the Majority Criterion. All this shows is that, in this one example, the winner of the plurality method happened to be a majority candidate. A valid proof has to explain why, in ANY ELECTION WITH A MAJORITY CANDIDATE, said majority candidate MUST win using Borda Count. To say that Borda Cound satisfies the Majority Criterion means that NO election, using Borda Count, would ever produce a violation of the Majority Criterion. The above only shows that no violation occured in this particular example. January 30 and February 1 2012 25
Incorrect Arguments for Borda Count Method and Majority Criterion Suppose that candidate A is a Majority Candidate. Thus, A receives more than half of the first place votes. Since A has more votes than anyone else, A will have the highest Borda point total, so A will win using Borda Count Method. This shows that Borda Count satisfies the Majority Criterion. This is NOT a valid proof. In spirit, it is better than the previous invalid proof, since this TRIES to provide a logical argument that applies to all elections, instead of trying to draw general conclusions from one example. However, the implication Since A has more votes than anyone else, A will have the highest Borda point total is incorrect. (We have seen examples in which the candidate with the highest Borda point total had NO first place votes.) January 30 and February 1 2012 26
Recap To show that a voting method SATISFIES a fairness criterion, you need to provide a logical argument in which you assume the hypothesis of the voting method and, using ONLY those hypotheses and the rules for the given voting method, you show the conclusion of the voting method must be true. To show that a voting method FAILS a fairness criterion, it is enough to find a SINGLE election showing a violation. A violation consists of an election in which the hypotheses of that voting method are met, but the conclusion is not. January 30 and February 1 2012 27
Recap Plurality Borda Count Pairwise Comp Plur. with Elim Majority Criterion Satisfies Fails Satisfies Satisfies On an exam, you could be asked which methods satisfy the Majority Criterion and which fail it. You could also be asked to explain why a given method satisfies the Majority Criterion. In that case you would need to provide a proof, like we did a few slides ago. You could also be asked to construct an example showing that a given method fails the Majority Criterion. It is probably a good idea to commit the above proofs and countereamples to memory until you can reconstruct these proofs or counterexamples in your own words. January 30 and February 1 2012 28
Condorcet Criterion The Condorcet Criterion: If a candidate X is a Condorcet Candidate, then candidate X should be declared the winner of the election. Please note that this is not a RULE. This is a notion of fairness. We believe that Condorcet candidates are very strong candidates, and the Condorcet Criterion essentially says that we would think a voting method was unfair if a candidate was a Condorcet Candidate and did not win the election. January 30 and February 1 2012 29
Proof that Pairwise Comparisons Satisfies Condorcet Criterion PAIRWISE COMPARISONS: Suppose candidate X is a Condorcet candidate. X wins all of her one-on-one comparisons, and so X wins more one-on-one comparisons than any other candidate, so X wins using Pairwise Comparisons. This shows that the Pairwise Comparison Method satisfies the Condorcet Criterion. It turns out that the other three methods fail the Condorcet Criterion. January 30 and February 1 2012 30
The Other Methods Fail the Condorcet Criterion Recall the example from the January 25 Workhseet. C is a Condorcet Candidate. 7 7 8 5 A C D B B B C A C A B C D D A D We have seen that A is the Plurality with Elimination Winner of this election, B is the Borda Count Winner, and D is the Plurality Winner. Thus, the Condorcet Candidate C LOST in each of Plurality with Elimination, Borda Count, and Plurality, thus showing that each of these methods violates the Condorcet Criterion. January 30 and February 1 2012 31
Recap Majority Criterion Condorcet Criterion Plurality Satisfies Fails Borda Count Fails Fails Pairwise Comp Satisfies Satisfies Plur. with Elim Satisfies Fails On an exam, you could be asked which methods satisfy a given criterion and which methods fail it. You could also be asked to explain why a given method satisfies a given criterion. In that case you would need to provide a proof, like we did a few slides ago. You could also be asked to construct an example showing that a given method fails a given criterion. It is probably a good idea to commit the above proofs and countereamples to memory until you can reconstruct these proofs or counterexamples in your own words. January 30 and February 1 2012 32