[EM] MinMax variant

[Forest Simmons]

I had a simpler example in mind:

Three teams in a round robin tournament compile the following scores:

A beats B four to three, B beats C two to one, and A beats C two to one.

A is the Condorcet winner while C is not only the Condorcet loser, but
also the method winner under this variant of MinMax, since this team never
gave up more than one point per game, while each of the other teams had at
least two points scored against it in one game or another.

Forest

On Fri, 7 Mar 2003, Markus Schulze wrote:

> Dear Forest,
>
> you wrote (7 March 2003):
> > Has anybody ever proposed minimizing the maximum opposition rather than
> > minimizing the maximum defeat? I know that theoretically this could elect
> > the Condorcet Loser, but it seems very unlikely that it would do so.
>
> Of course, the fact that the MinMax method can elect the Condorcet Loser
> has nothing to do with the way the strength of a pairwise comparison is
> measured.
>
> Example:
>
>    20 ABCD
>    20 BCAD
>    20 CABD
>    13 DABC
>    13 DBCA
>    13 DCAB
>
> Markus Schulze
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