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The Method of Pairwise Comparisons, a voting system proposed by Marie Jean Antoine Nicolas de Caritat, marquis de Condorcet. The system involves comparing each pair of candidates and awarding points based on the winner of each comparison. The candidate with the most points wins the election. The document also discusses the advantages and disadvantages of this system, including its satisfaction of the Majority, Condorcet, Public-Enemy, and Monotonicity criteria.
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Suggestion from a Math 105 student (8/31/11): Hold a knockout tournament between candidates.
I (^) This satisfies the Condorcet Criterion! A Condorcet candidate will win all his/her matches, and therefore win the tournament. (Better yet, seeding doesn’t matter!)
I (^) But, if there is no Condorcet candidate, then it’s not clear what will happen.
I (^) Using preference ballots, we can actually hold a round-robin tournament instead of a knockout.
The Method of Pairwise Comparisons
Proposed by Marie Jean Antoine Nicolas de Caritat, marquis de Condorcet (1743–1794)
I (^) Compare each two candidates head-to-head. I (^) Award each candidate one point for each head-to-head victory. I (^) The candidate with the most points wins.
Number of Voters 14 10 8 4 1 1st choice A C D B C 2nd choice B B C D D 3rd choice C D B C B 4th choice D A A A A
Compare A to B. I (^) 14 voters prefer A. I (^) 10+8+4+1 = 23 voters prefer B. I (^) B wins the pairwise comparison and gets 1 point.
Number of Voters 14 10 8 4 1 1st choice A C D B C 2nd choice B B C D D 3rd choice C D B C B 4th choice D A A A A
Compare C to D: I (^) 14+10+1 = 25 voters prefer C. I (^) 8+4 = 12 voters prefer D. I (^) C wins the pairwise comparison and gets 1 point.
Number of Voters 14 10 8 4 1 1st choice A C D B C 2nd choice B B C D D 3rd choice C D B C B 4th choice D A A A A
A B C D Wins Losses Points A B C D
Number of Voters 14 10 8 4 1 1st choice A C D B C 2nd choice B B C D D 3rd choice C D B C B 4th choice D A A A A
A B C D Wins Losses Points A 14 B 23 C D
Number of Voters 14 10 8 4 1 1st choice A C D B C 2nd choice B B C D D 3rd choice C D B C B 4th choice D A A A A
A B C D Wins Losses Points A 14 14 14 B 23 18 C 23 19 D 23
Number of Voters 14 10 8 4 1 1st choice A C D B C 2nd choice B B C D D 3rd choice C D B C B 4th choice D A A A A
A B C D Wins Losses Points A 14 14 14 B 23 18 28 C 23 19 25 D 23 9 12
Number of Voters 14 10 8 4 1 1st choice A C D B C 2nd choice B B C D D 3rd choice C D B C B 4th choice D A A A A
A B C D Wins Losses Points A 14 14 14 — B,C,D 0 B 23 18 28 A,C D 2 C 23 19 25 A,B,D — 3 Winner! D 23 9 12 A B,C 1
I (^) The Method of Pairwise Comparisons satisfies the Majority Criterion. (A majority candidate will win every pairwise comparison.)
I (^) The Method of Pairwise Comparisons satisfies the Public-Enemy Criterion. (If there is a public enemy, s/he will lose every pairwise comparison.)
I (^) The Method of Pairwise Comparisons satisfies the Public-Enemy Criterion. (If there is a public enemy, s/he will lose every pairwise comparison.)
I (^) The Method of Pairwise Comparisons satisfies the Monotonicity Criterion. (Ranking Candidate X higher can only help X in pairwise comparisons.)
Problem #1: It’s somewhat inefficient. How many pairwise comparisons are necessary if there are N candidates? How many spaces are there in the crosstable?
I (^) N^2 squares in crosstable I (^) N squares on the main diagonal don’t count I (^) Other squares all come in pairs
Number of comparisons =
