Re: Ranking is hardsniffnoyOctober 26 2015, 18:45:47 UTC
Oh man, I hadn't even thought of this in terms of voting! I suppose you could think of this as a voting system, too -- a candidate's "matchup" against another candidate is how many people vote for the one candidate over the other. (Except that you'd have to normalize, because otherwise it might not be zero-sum, if some people only ranked some of the candidates.) It would be an odd voting system, that's for sure. I'm not sure it would be Condorcet. (Maybe it is? Is it in general true that if there's character with no disadvantageous matchups, and at least one advantageous one, then the game is dominance-solvable with them as the winner? I hadn't considered that. I'll have to think about that, or look it up...)
EDIT: OK, that's definitely false, because we have a counterexample above: In SSBM, Fox and Falco both have no disadvantageous matchups, but obviously the optimal strategy can't both be "only play Fox" and "only play Falco"! Of course, that's a pretty complicated counterexample, I should come up with a simpler one
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Re: Ranking is hardsniffnoyOctober 26 2015, 21:25:49 UTC
OK, here's a simple example: Suppose you have four strategies, A, B, C, and D. A beats the others by 1, while B, C, and D exist in a rock-paper-scissors relation, each beating the other by 2. A has only positive matchups, but nothing dominates anything else.
However, plugging this into a solver, apparently the optimal strategy is in fact pure A. So finding one where it's mixed will be harder (assuming that's possible, which I don't see why it wouldn't be). Still, the SSBM example illustrates that in general having no bad matchups doesn't mean you're part of the optimal strategy.
Re: Ranking is hardsniffnoyNovember 21 2015, 23:25:14 UTC
OK, here's a nice small sort-of-example, with four strategies. Actually, it's trivial to come up with examples if you allow two of the strategies to be identical (and then you can do it with three), but let's assume we don't allow that
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EDIT: OK, that's definitely false, because we have a counterexample above: In SSBM, Fox and Falco both have no disadvantageous matchups, but obviously the optimal strategy can't both be "only play Fox" and "only play Falco"! Of course, that's a pretty complicated counterexample, I should come up with a simpler one ( ... )
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However, plugging this into a solver, apparently the optimal strategy is in fact pure A. So finding one where it's mixed will be harder (assuming that's possible, which I don't see why it wouldn't be). Still, the SSBM example illustrates that in general having no bad matchups doesn't mean you're part of the optimal strategy.
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