Count "votes against" in pair-ties? (was Re: I don't contest pre
seppley at alumni.caltech.edu
Tue Nov 12 13:43:40 PST 1996
Mike O wrote:
>As worded, Condorcet doesn't count votes-against in pairwise ties.
>There's no need for it to, though need can be a subjective
>judgement. Unbeaten alternatives are in a special class above all
>the other alternatives, and so there's important reason _not_ to
>count votes-against in a pairwise tie.
Isn't it possible for the votes against an unbeaten candidate in a
pair-tie to be worse than the Condorcet score of a beaten candidate?
A B C
A 50= 2L
B 50= 35
C 1 45L
---- ---- ----
? ? 2 <--- Maximum votes against. Count the ties?
A (and only A) is unbeaten and wins by our definition. But A's 50
in the tie is larger than C's 2 in C's only loss. Is A really less
beaten than C?
I hope I'm not muddying the waters. The odds against a pair-tie in a
large election are negligible.
---Steve (Steve Eppley seppley at alumni.caltech.edu)
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