[EM] Condorcet Borda?

[Forest Simmons]

Generalized Borda requires a probability distribution (i.e. a lottery on
the candidates). Ordinary Borda is the ccase where the distribution is
uniform ... which is why it fails Clone Independece. Borda based on a clone
free, proportional lottery will be clone free, and Borda based on a
Condorcet lottery will be Condorcet efficient:

In fact, in Generalized Borda the score for alternative X is the sum over
the ballots B of the total Prob of the alternatives ranked behind X minus
the total Prob of the alternatives ranked abead of X.  If X is a Condorcet
Candidate and Prob is a Condorcet lottery, then Prob(X) is one. In this
immediate context all other candidates have zero probability. So the Borda
Score of X is zero and the Borda score for any Y not equal to X is the
number of ballots on which Y is ranked ahead of X minus the number of
ballot on which X is ranked ahead of Y.

This difference has to be negative because X beats Y pairwise. Since X has
the only non-negative Borda Score, it has to be the Borda winner ...QED!
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