# I don't contest precision issue, but...

Mike Ossipoff dfb at bbs.cruzio.com
Sat Nov 9 15:12:44 PST 1996

```I don't oppose making the introductory definition of Condorcet's
method more precise if that can be done without complication. You
surely agree about the importance of non-complication, considering
how all the IRO advocates insist that Condorcet is too complicated.
Anyway, I don't want to seem a stubborn holdout on that question. I
have no objection to improved precision.

But I still claim that the meaning of #1 is precise rather than
intuitive. If you have no defeats, then the number of votes against
you in a defeat is zero.

If it had been worded as in #2, to refer to your greatest defeat,
as measured by how many voters ranked the defeating candidate over
you, then yes, that, literally speaking, doesn't say anything about
your score if you have no defeats. But I claim that #1 is precise,
for the reason in the paragraph before this one.

And #1 is the favorite of these wordings, according to 3 people,
including me.

Anyway, saying that an alternative wins if it is unbeaten, instead
of if it beats everything else, would, as you said, avoid those
imprecisions in #s 2 & 4. And would also remove any controversy

Saying that an alternative wins if it is unbeaten is ok
when there are several unbeaten alternatives, in terms of the
way the academics word things, because they speak of there
being several winners when there's a tie.

So after it says that an alternative wins if it's unbeaten, it
could say "If there's no unbeaten alterntive, then:...", followed
by one of the wordings for the circular tie solution.

Maybe it's a matter of opinion about the precision of #1, as it
stands, but it seems certain to me that if you have no defeats,
then, without a defeat, you can't have any votes against you
in a defeat. And so, for you, the number is zero, rather than
being un-defined.

If it said "...in one of the defeats that it has", then it would
only apply to those having a defeat, but that isn't how it's worded.

Anyway, as I said, I'm not being a stubborn holdout about that,
and I'm just saying my opinion about #1's precision. I have no
objection if others don't agree on that, meaning that I have
no objection to applying the fix that Steve suggested, even if
only #1 is used to define the circular tie solution.

***

I get 2 copies of most postings for some reason. I believe that
I'm replying to a new posting, with this letter, but there could
be a chance that I'm replying to the 2nd copy of a posting that

As worded, Condorcet doesn't count votes-against in pairwise ties.
There's no need for it to, though need can be a sugjective judgement.
Unbeaten alternatives are in a special class above all the other
alternatives, and so there's important reason _not_ to count

But I was saying that the first way I'd distinguissh between unbeaten
alternatives would be by limiting the choice to those unbeaten
alternatives that beat something, if there be such. That's what
I've called the Beat-Something tie-breaker.

If there are several unbeaten alternatives that beat something,
what then? Go on to the next tie-breaker, such as (in my list
of tie-breakers) Fishburn? Or determine the smallest votes-against
in a pairwise tie to choose from among the unbeaten alternatives
that beat something?

I'd go on to the next tie-breaker, which would be Fishburn in
its job, even just by returning a tie, and needn't be used more.

So we can instead look at an alternative's status as judged by
what beats it. If kids get their status from whom they can beat
up, what could be a more embarrassing loss of status than being
beaten up by a kid that the other tough kids can beat up?

If the people have said, by the pairwise comparison result, that
alternative Z is better than alternative B, but not better than
alternative A, then, indirectly the people can be interpreted as
having said that A is better than B. That's why Fishburn appeals
to me as the next tie-breaker after Beat-Something.

But I emphasize that all this about tie-breakers is unimportant,
not only because ties won't happen in a public political election,
but because, even in an EM vote, where ties, including pairwise
ties, aren't unlikely, they don't really matter, because the
main method has done its job, even when it returns a tie, and
everything in that tie is a winner, and nothing in that tie

Mike

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