§ 1.5 The Method of Pairwise Comparisons (Copeland’s Method)

Pairwise Comparisons The Idea: Hold a ‘round-robin’ tournament in which every candidate is matched one-on-one with every other candidate. Each one-on-one matchup is called a pairwise comparison. In a pairwise comparison between candidates X and Y each vote is assigned to either X or Y, the vote going to whichever of the two candidates is higher on the ballot. The winner of the comparison is the one with the most votes, and each win is worth 1 point. Each tie is worth 1/2 point. (Losses are worth nothing!)

Example: Let’s look at the Muppet example again--this time under a pairwise comparison system. Number of voters 21 15 12 7 1st Choice Piggy Gonzo Fozzie Kermit 2nd Choice 3rd Choice 4th Choice

Example: Let’s look at the Muppet example again--this time under a pairwise comparison system. Piggy versus Kermit: 21 votes to 34 votes (Kermit wins). Kermit gets 1 point. Number of voters 21 15 12 7 1st Choice Piggy Gonzo Fozzie Kermit 2nd Choice 3rd Choice 4th Choice

Example: Let’s look at the Muppet example again--this time under a pairwise comparison system. Gonzo versus Fozzie: 36 votes to 19 votes (Gonzo wins). Gonzo gets 1 point. Number of voters 21 15 12 7 1st Choice Piggy Gonzo Fozzie Kermit 2nd Choice 3rd Choice 4th Choice

If we continued these comparisons we would get a ‘scorecard’ like the one that follows: Piggy versus Kermit: 21 votes to 34 votes (Kermit wins). Kermit gets 1 point. Piggy versus Gonzo: 21 votes to 34 votes (Gonzo wins). Gonzo gets 1 point. Piggy versus Fozzie: 21votes to 34 votes (Fozzie wins). Fozzie gets 1 point. Kermit versus Gonzo: 28 votes to 27 votes (Kermit wins). Kermit gets 1 point. Kermit versus Fozzie: 43 votes to 12 votes (Kermit wins). Kermit gets 1 point. Gonzo versus Fozzie: 36 votes to 19 votes (Gonzo wins). Gonzo gets 1 point. Example: Let’s look at the Muppet example again--this time under a pairwise comparison system. Number of voters 21 15 12 7 1st Choice Piggy Gonzo Fozzie Kermit 2nd Choice 3rd Choice 4th Choice

Next we tally each candidates points. Piggy -. 0 Kermit -. 3 Gonzo - Next we tally each candidates points. . . Piggy - 0 Kermit - 3 Gonzo - 2 Fozzie - 1 . . .and find that the winner is Kermit. Example: Let’s look at the Muppet example again--this time under a pairwise comparison system. Number of voters 21 15 12 7 1st Choice Piggy Gonzo Fozzie Kermit 2nd Choice 3rd Choice 4th Choice

Aside: What is the point of all these examples? Asking, “Who is the winner of the election?” is incredibly ambiguous. Determining the winner of an election depends on the method of counting as much as it does on the actual votes cast.

‘Pros’ of Pairwise Comparison The Pairwise Comparison method satisfies all of the following: The Majority Criterion The Condorcet Criterion The Monotonicity Criterion

The Pairwise Comparison Method What’s wrong with this method?

Example: Suppose we are holding an election for a committee post and the candidates are A, B, C, D and E. If we hold an election using the Pairwise Comparison method we get the following results: A versus B: 14 votes to 30 votes. B gets 1 point. A versus C: 32 votes to 12 votes. A gets 1 point. A versus D: 26 votes to 18 votes. A gets 1 point. A versus E: 36 votes to 8 votes. A gets 1 point. B versus C: 20 votes to 24 votes. C gets 1 point. B versus D: 22 votes to 22 votes. B and D get 1/2 point. B versus E: 28 votes to 16 votes. B gets 1 point. C versus D: 24 votes to 20 votes. C gets 1 point. C versus E: 20 votes to 24 votes. E gets 1 point. D versus E: 36 votes to 8 votes. D gets 1 point. Number of votes 4 12 8 2 1st Choice A B C D E 2nd Choice 3rd Choice 4th Choice 5th Choice

Example: Suppose we are holding a election for a committee post and the candidates are A, B, C, D and E. The points then tally as. . A - 3 points B - 2 1/2 points C - 2 points D - 1 1/2 points E - 1 point . . .and A is the winner of the election. Number of votes 4 12 8 2 1st Choice A B C D E 2nd Choice 3rd Choice 4th Choice 5th Choice

Example: Suppose we are holding a election for a committee post and the candidates are A, B, C, D and E. Now suppose that candidate C suddenly became drastically ill and has to drop out of the race immediately before the election ends. Number of votes 4 12 8 2 1st Choice A B C D E 2nd Choice 3rd Choice 4th Choice 5th Choice

Example: Suppose we are holding a election for a committee post and the candidates are A, B, C, D and E. Since candidate C cannot serve, we remove him/her from the ballots and the preference schedule now looks like the one below. Number of votes 4 12 8 2 1st Choice A B D E 2nd Choice 3rd Choice 4th Choice

Example: Suppose we are holding a election for a committee post and the candidates are A, B, C, D and E. Since candidate C cannot serve, we remove him/her from the ballots and the preference schedule now looks like the one below. If we hold another Pairwise Comparison election after removing candidate C the comparisons come out as: A versus B: 14 votes to 30 votes. B gets 1 point. A versus D: 26 votes to 18 votes. A gets 1 point. A versus E: 36 votes to 8 votes. A gets 1 point. B versus D: 22 votes to 22 votes. B and D get 1/2 point. B versus E: 28 votes to 16 votes. B gets 1 point. D versus E: 36 votes to 8 votes. D gets 1 point. Number of votes 4 12 8 2 1st Choice A B D E 2nd Choice 3rd Choice 4th Choice

Example: Suppose we are holding a election for a committee post and the candidates are A, B, C, D and E. Since candidate C cannot serve, we remove him/her from the ballots and the preference schedule now looks like the one below. The points then tally as. . . A - 2 points B - 2 1/2 points D - 1 1/2 points E - 0 points . . .and B would be our new winner. Number of votes 4 12 8 2 1st Choice A B D E 2nd Choice 3rd Choice 4th Choice

The Independence-of-Irrelevant-Alternatives Criterion If choice X is a winner of an election and one or more of the other choices is disqualified and the ballots are recounted, then X should still be a winner of the election.

The Pairwise Comparison Method What’s wrong with this method? It violates the Independence-of-Irrelevant-Alternative Criterion. . .

The Pairwise Comparison Method What’s wrong with this method? It violates the Independence-of-Irrelevant-Alternative Criterion. . . . . .for elections with too many candidates it is not practical. (There are a lot of comparisons to be made!)

How Many Pairwise Comparisons Are There? When we had 4 candidates there were 6 comparisons we had to examine. When we had 5 candidates there were 10. How can we arrive at these numbers without actually making all of the comparisons?

How Many Pairwise Comparisons Are There? Let’s look at the case with 10 candidates. The first candidate needs to be compared with each of the nine others--9 pairwise comparisons. The second candidate needs to compared to eight other candidates (we have already compared the second to the first)--8 pairwise comparisons. The third candidate needs to be compared against seven other candidates--7 pairwise comparisons.

How Many Pairwise Comparisons Are There? Continuing as we have done so far. . . The ninth candidate needs to be compared to every candidate except for the first eight (those comparisons have already been made). In fact, the ninth candidate only needs to be compared with the tenth--1 pairwise comparison.

How Many Pairwise Comparisons Are There? We can see that the total number of pairwise comparisons in this case should be: 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45

How Many Pairwise Comparisons Are There? In general, if we have N candidates in a given election the total number of pairwise comparisons will be: 1 + 2 + 3 + 4 + . . . + (N - 1) = (N - 1) N 2 = N C 2