Fishburn//Copeland//Condorcet

[Mike Ossipoff]

Though I don't know if Fishburn would get rid of the Rich Parties
problem, I do know that there's absolutely no reason to use
Copeland instead of Smith. Smith//Condorcet avoids the rich parties
problem, and meets the Condorcet Loser Criterion, the Smith Criterion,
the Mutual Majority Critrerion, the Majority Loser Criterion, and
all the criteria by which Bruce has attempted to attack plain Condorcet.
It also has all the Condorcet properties I've described--it meets
GMC, LO2E-1, LO2E-2, and has the obvious benefit of a votes-against
count. Doing Copeland before Condorcet, you lose the Condorcet
properties. 

Has Bruce given a reason why Fishburn//Copeland//Condorcet is
better than Smith//Condorcet? No? That's because it isn't better,
and isn't as good, and Bruce knows that.

As for whether Fishburn//Copeland//Condorcet avoids Copeland's
Rich Parties problem, I rather doubt that it would. I don't
suppose Bruce offered a demonstration to back up that statement?
I emphasikze that even if it's true, about the Rich Parties 
problem, that doesn't even begin to justify 
Fishburn//Copeland//Condorcet. Smith//Condorcet, as I've repeatedlyl
explained to Bruce, has Bruce's methods completely dominated, when
it comes to criteria. That's one of the things that Bruce, 
reminiscent of Don & Demorep, has never replied to. Bruce
keeps on pushing Copeland, while never replying to the
statement that his Copeland versions are all completely dominated
by Smith//Condorcet. 

I didn't check the header to find out if this is going to the
whole list, but, if it is, then, Bruce, what do you say
about that? Nothing? That's what I expected. In the case of
Don & Demorep, I agree with Steve in putting it down to
cluelessness. In the case of Bruce, it isn't that easy to
explain. Bruce is a mathematician, and must be fully aware
that he's pushing something that he won't defend. Why?

Mike





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