Consider the following tallies from 13 voters with 4 candidates a,b,c, and d, and let the election be to select 3 of the 4 candidates to be winners:

5: d>a>b>c

4: d>c>a>b

4: b>c>a>d

In this case, d is the Condorcet winner, since we have

9: d>a

9: d>b

9: d>c

But a,b, and c form a Condorcet cycle a>b>c>a, since we have

9: a>b

9: b>c

8: c>a

Condorcet would have us break this cycle by rejecting (or reversing) c>a, since it is the weakest link in the cycle, and we would find d>a>b>c. My suggestion is a bit different, although in this case it arrives at the same conclusion. I will try to find another example where it does not, but I just wanted to illustrate the concept.

Essentially, we look directly at the frequencies of each of the triplet orders a>b>c, c>a>b, and b>c>a found in the Condorcet cycle, and evaluate their strengths relative to what would be expected if the pairwise preferences were independently distributed over the electorate.

For example, we have

9: a>b

9: b>c

So, if these preferences were independently distributed over the electorate, we would expect that a (9/13)^2 proportion of the electorate would indicate a>b>c. The true proportion indicating a>b>c in this case is simply 5/13, so that the transitive order a>b>c would be assigned a strength of (5/13)/(9/13)^2. One can repeat this and eliminate the link with the weakest strength.

This method can be extended to higher order cycles as well. For example, if one had a Condorcet cycle a>b>c>d>a, then one could consider each of the quadruplet orders a>b>c>d, d>a>b>c, c>d>a>b, and b>c>d>a in the cycle, and compare their relative frequencies with the expectation based on the relative frequencies of the triplet orders that overlap to provide them under the assumption of independence.

Just as an example, say in a hypothetical election there is a Condorcet cycle a>b>c>d>a, that 3 out of 20 voters indicated a>b>c, and that 4 out of 20 indicated b>c>d. Then the expected frequency of a>b>c>d assuming independence would be (3/20)(4/20)=3/100. If exactly one voter actually indicated a>b>c>d, then that ordering would be assigned a weight of (1/20)/(3/100)=5/3.

]]>The Kemeny-Young rule has some good theoretical properties but it is not resolvable in polynomial time. For a very high-quality cycle resolution method I recommend ranked pairs, or for something simpler take Black's method.

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