Steve Eppley seppley at alumni.caltech.edu
Thu Nov 7 11:09:15 PST 1996

```Mike O wrote:
>Bruce said that the posted wordings of Condorcet's method assumed
>that if a candidate does not pairwise defeat all the other
>candidates, then that candidate must be pairwise defeated by at
>least 1 of the others. Bruce said that is a false assumption.
-snip-

I think what Bruce is getting at is that if there isn't a candidate
who pairwise defeated all the others, there might (theoretically)
be one (or more) candidate(s) with no pairlosses and at least one
pair-tie.  As literally written, such a candidate is NOT the one
with the smallest largest pairloss.

If a candidate has no pairlosses, we can intuit that he would have a
Condorcet score which is lower than anyone else's.  (This can make
the sentence about the beats-all winner unnecessary, replaceable by
defining the Condorcet score of candidates with no pairlosses as
zero, and in both cases electing the candidates with the smallest
Condorcet score.)  But this intuition isn't rigorous.

Arguably it's not necessary that "Intro To Condorcet" be rigorous,
but we should only be inaccurate if the tradeoff is worthwhile.
We could modify the wording to say "if there is a candidate with
no pair-defeats, this candidate wins, else..."  Sounds simple,
and sounds like it won't disturb Bruce.

I'm curious about the pair-ties.  As defined, a pair-tie doesn't
count toward a candidate's Condorcet score.  Should it, and have
we been over this before?  My memory is tingling...

By the way, I've received one private feedback about the wording so
far, from Kevin H.  He also prefers wording #1.

---Steve     (Steve Eppley    seppley at alumni.caltech.edu)

```