Count "votes against" in pair-ties? (was Re: I don't contest pre

[Steve Eppley]

Mike O wrote:
-snip-
>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.
-snip-

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)