[EM] MinMax(ao)

Simmons, Forest simmonfo at up.edu
Fri Sep 9 18:58:39 PDT 2005

I wrote ...
> Yes, this is the measure of defeat strength used in AWP, but here's the question:  Is AWP based
> on River the same as AWP based on MinMax?  
> Somewhat surprisingly, the answer to the analogous question for "winning approval" as a measure
> of defeat strength is affirmative:  River(wa) = MinMax(wa)= Ranked Pairs(wa)= Beatpath(wa)=DMC.

Kevin replied ...

I think the answer for AWP is no. All these methods using WA are the same probably
because any given candidate's wins are all of the same strength. That isn't true
with AWP.

An interesting question: What if all one's *losses* are the same strength? Same

I answer ...
I'm more interested in whether or not MinMax(AO) is clone proof like MinMax(WA) or fails clone winner like the better know MinMax versions MinMax(wv) and MinMax(margins).
If MinMax(AO) turns out to satisfy clone winner, then I think it would be the best AWP proposal, because it would be simpler to explain than Beatpath or Ranked Pairs, and no harder to explain than Copeland, especially when you take into account that MinMax has just been proposed by Rob Lanphier as a Copeland tiebreaker.
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