Game Theory and Poker

[sjb]

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... game theory doesn't tell you how to win. It tells you how to not lose.

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Actually an equilibrium strategy doesn't even accomplish the minimal goal of not losing when there are more than two players. If everyone follows the equilibrium strategy than the EV of each player across an entire orbit will be zero and you won't lose. But one player not following the equilibrium strategy can easily place another "innocent" player in an unavoidable negative EV situation.

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Yes, and no. If player A departs from the equilibrium strategy in such a way that an "innocent" player (B) is in a "negative EV situation", then by definition, B has a different strategy they could follow that would be better than the equilibrium strategy they were originally following, and so can obtain a non-negative EV - this shifts the negative EV to some other "innocent", who can also switch to a better strategy, and so on.

Ultimately, all of the players can (in theory) switch strategies so that the negative EV falls back on the player who's playing a non-equilibrium strategy. The other players may not be able to guarantee a positive EV in their new equilibrium, but they can at least remain non-negative.

[StellarWind]

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Yes, and no. If player A departs from the equilibrium strategy in such a way that an "innocent" player (B) is in a "negative EV situation", then by definition, B has a different strategy they could follow that would be better than the equilibrium strategy they were originally following, and so can obtain a non-negative EV - this shifts the negative EV to some other "innocent", who can also switch to a better strategy, and so on.

Ultimately, all of the players can (in theory) switch strategies so that the negative EV falls back on the player who's playing a non-equilibrium strategy. The other players may not be able to guarantee a positive EV in their new equilibrium, but they can at least remain non-negative.

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This is not correct. If player A deviates from the equilibrium strategy in a multiplayer poker game then player B may be screwed. He may have no strategy that avoids negative EV. Players A and B both lose and the other players gain.

I believe I gave some valid poker examples of this in my last post, but poker is complicated and proof can be elusive.

Here is a non-poker example that illustrates my point. Three players play a game in which each player antes $1. The first player chooses a whole number between one and three and annouces it out loud. Then the second player chooses a number between one and three and announces it out loud. Finally the last player chooses his number. Choosing duplicate numbers is allowed.

Now a "3-sided die" is rolled to generate a random number between 1 and 3. Each player who chose that number gets an equal share of the pot. If no one picked the number the game is a chop and the antes are returned.

Clearly it is bad to have the same number as someone else. Optimal strategy for each player is to pick a number that no one else has picked yet. This is also the Nash Equilibrium strategy for all players and it yields EV=0 for each player if everyone follows it.

I'm player "A". Whenever I am second to act my strategy is to choose the same number announced by player B sitting on my right. This is a really bad strategy and it is going to cost me a lot of money. But notice that Player B has the same EV as I do. He's played perfectly but he is no better off than I am and there is nothing he can do to improve his situation. We're both losing money to player C on my left.

[jogsxyz]

http://www.cardplayer.com/poker_maga...amp;m_id=65572

sjb, read Matt Matros' article on game theory.