[EM] Top Three Condorcet

[Forest Simmons]

If I understand correctly, Beat Path, Ranked Pairs, MinMax, and all of the
other serious Condorcet methods are in agreement (except for the
margins/wv debate) when there are only three candidates: if one of them
beats each of the others pairwise, then that candidate is the winner.
Otherwise, the cycle is broken at the weakest link.

So why not take advantage of this agreement by using some simple but
reasonable method to eliminate all but three candidates and then among
those three

    If there is a cycle
       Then break it at the weakest link
       Else go with the one who beats the other two.


Elimination methods that eliminate all the way down to two candidates
offer too much order reversal incentive, but if there is room for three
finalists, then that incentive may be negligible.

Here's a more specific proposal along these lines:

Use grade ballots.  The three finalists are A the candidate with the
greatest number of top grades, B the candidate with the highest grade
point average, and C the candidate with greatest number of passing grades.

If all three of these turn out to be the same candidate, then this
candidate wins.

If the set {A,B,C} has only two distinct members, then whichever wins
pairwise between them is the method winner.

If all three are distinct, and one of them beats the other two pairwise,
then that one is the winner.

If there is a three way cycle, then the cycle is broken at the weakest
link.

This method is summable, easy to understand, and hard to criticize, though
I'm sure the purists will have plenty to say :')

The main disadvantages I see are (1) the controversy over margins versus
winning votes, and (2) some folks think that it is too hard to grade the
candidates.

Any other proposals for Top Three Condorcet?

Forest