[kdotsky]

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Its seems as though that a perfect game theory is unbeatable the psychology but not unbeatable (through perhaps sheer luck, or another optimal game theory), does that mean there’s a more specific game theory is maybe more optimal?

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The game theory strategy, by proof, is unbeatable, regardless of the opponent (think of the rock-paper-scissors example). There may be more than one optimal strategy, but by definition if they played against each other neither would loose. And by loose we mean have a negative long run expectation -- meaning that yes one may win by sheer luck as you mentioned.
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I guess my main question is that one of the arguments is stating that perfect game theory is unbeatable, that mean against a psychologist, correct?

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No. It's not an argument for one -- it's mathematically proven. The strategy would be the exact same against any opponent. Keep in mind, though, that though we have an algorithm to calculate this strategy, it would take a computer an impractical amount of time to compute.
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My next question is, is a game theory always constant whether full table, short table, or heads up, then does it take into affect stack size or is this all part of the optimal game strategy?

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The optimal strategy would change if you added more players to the table. It does take into account stack sizes. It calculates the optimal strategy (e.g. raise 30%, fold 30%, call 60% of the time) for every possible scenario you can be in in the game. The opponent have different stack sizes, or there being different numbers of opponents, are different scenarios.

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And lastly is optimal game strategy for efficiency or consistency?

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I have no idea what this means. I would call it for "security". You are guaranteed not to lose (which means you will most likely win), but against worse players you could win a lot more playing an exploitive strategy (though you would loose a lot more doing this with a better opponent).