[EM] simple question (I think)

[rob brown]

Hmmm, ok.

Obviously though, if A has 8 pairwise wins, B has 7, and everyone else has
less.....A will be the winner no matter what, right? Regardless of how close
A's wins are and how large B's majorities are?

If that isn't true than I must totally misunderstand the concept of a
Condorcet winner.

What system would you prefer? I'm a tad confused, because I thought all
well-known Condorcet methods begin with a pairwise matrix.

-rob


On 11/16/05, Paul Kislanko <kislanko at airmail.net> wrote:
>
> "Candidates A, B and C all have 8 pairwise wins. D has 7. Could D still be
> chosen as the winner by any "reasonable" method? "
>  Sure. D's 7 pairwise wins could be by a large enough majority that the
> "extra" pairwise win that A, B, and C have over the fringe candidate that
> beats D by 1 vote makes the "8" vs "7" irrelevant.
>  This is kind of why I don't like any counting method that BEGINS with the
> pairwise matrix. Some systems would eliminated D when all of A, B, and C are
> real dogs that no majority likes.
>
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