Game Theory and Poker

[sjb]

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Thank you for the feedback, I'm glad you liked the article.

You said:
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in all forms of poker I know of (certainly HE, Omaha, and Stud, and regardless of limit structure or high/low splits) and in the absence of a rake, the EV of any Nash equilibrium strategy is exactly zero. With a rake, the EV is negative and equal to the total amount of the rake, divided by the number of players.

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This isn't quite true. The EV (Expected Value, or average earn) of the Nash Equilibrium will be zero minus rake, but the EV of the NE strategies won't necesarily be. If the other players aren't playing their NE strategy then it could be different. In the bluffing example the bluffers EV doesn't depend on the other guy's strategy but in other cases it might. For example, the caller in the bluffing situation should call or fold with the right frequency to make the bettor indifferent between bluffing and checking. If the bettor changes her strategy to, for example, always betting then the caller is no longer breaking even with the Nash strategy and will have won a positive amount in expectation.

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Maybe I didn't state it clearly, but when I was talking about the EV of the equilibrium strategy being zero (less the rake), I was talking about the game-theoretic "value" of the game - when players deviate from the strategy dictated by the equilibrium, then of course it's possible (even likely) for the EV to go up. But then you're no longer talking about the EV of the equilibrium strategy - you're talking about the EV of some non-equilibrium. You're not in equilibrium unless both players are playing that way.