Reply: Bruce's incorrect statement

[Mike Ossipoff]

Here's what I meant to say in reply to Bruce's statement:

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.

This amounts to 2 statements by Bruce. The 1st statement is
outright false:

Those wordings don't assume that a candidate who doesn't beat
everyone must be beaten. Say an unbeaten candidate ties a
beaten candidate. Because no one beats everyone, Condorcet's
count rule is used, and the unbeaten candidate wins, if he's
the only unbeaten candidate.

Now, suppose the unbeaten candidate ties another unbeaten candidate.
Then, as before, since no one beats everyone, Condorcet's count
rule is used, and the 2 unbeaten candidates tie according to that
rule. What's the problem? Nothing in the rule implied that there
could never be a tie.

Of course as pairwise tie is as unlikely in a public election
as a Plurality tie would be now. 

Of course the fewest-votes-against-in-a-defeat rule could be
initially used instead of first looking for a beats-all candidate.
A candidate who beats each one of the others would automatically
win that count.

***

Now, about Bruce's 2nd statement: He said that an assumption that
a candidate who doesn't beat everyone is beaten is a false assumption.

Though those proposed wordings don't make that assumption, it's actually
a very safe assumption in public political election, due to the
large number of voters. Its really a sure thing that there won't
be any ties, pairwise ties or any other kind except for Copeland
ties :-)

So though that's a false assumption in the sense that there's
a probability, even just in California, that would likely be
a fraction of a trillionth, and which would be _lots_ smaller
in a national election than in a state election--that there
could be a pairwise tie, it's something that can safely be
ignored. As I said, the only recognition that ties need is
to leave in whatever tie-breaking language might already
exist. Pairwise ties? No special language is needed. In the
very unlikely event that it happened, it wouldn't cause a
Condorcet tie unless there are 2 unbeaten candidates. And
then it's just a tie like any other, and no more likely than
a Pluralitly tie. Well I haven't actually compared the liklihood
of those 2 kinds of ties, but they're both virtually certain
to not happen in a public election.

Mike





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