Re: Which poker variation is most exploitable by pure math strategy?
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I want to add something to my statement above (and I am no longer free to edit my post apparently).
With regards to Chinese Poker, it is explotiable in the sense I can develop a strategy that is optimal, meaning that any deviation from my strategy results in a loss of expected value. That is it. My edge will always be small and variance will always be high, but mathematically an optimal solution does exist (that could also be derived and/or approximated closely) and mixing up play is totally irrelevant (because as I have stated there are no betting rounds and nothing you do will allow your opponent to gauge your hand particularly well). Such is NOT the case with other forms of poker even if one can have a high-expected-value/low-variance on average. However, mathematical exploitable != highest EV. In those cases a very good strategy could be guessed at, but it isn't mathematically soluble.
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No, this is wrong. Headsup Chinese could (I guess) be solved for an equilibrium strategy, but without some mixed strategies your nemesis EV will be higher than it would be if you were playing equilibrium (with mixed strategies).
Anyway, solving the game for equilibrium purposes is stupid because there are like 10^21 hand matchups which is totally computationally intractable and you can just approximate down to really small nem-ev anyway. I'm comfortable with the idea that my opponents could exploit me for .0001/h if they knew my strategy.
Also, you seem to be confused about what is possible in other forms of poker. ALL zero-sum two player games have equilibria in the domain of mixed strategies.
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