Game Theory and Poker

[JaredL]

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In poker other players are far from rational. The clearest example of this is a player calling on 7th street in Stud without being able to beat the other players board. A point I make in the article, and this will be expanded upon in the future, is that you can still use some of the techniques from game theory to help you determine the best play even if your opponents' are not playing optimally.

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Actually, game theory is about studying the strategies when all players are rational. If you're going to study scenarios where only one of the players is "rational", then you have to build a model of the supposed "irrational" behavior of the other players. Once you've done that, you don't need game theory anymore - you just need the (earlier) discipline that game theory was built on: decision theory.

Very much of the poker literature that discusses game theory is really discussing decision theory. When you see "assume he would raise with any high pair and call with any suited ace" or similar hypothetical scenarios, you've built a model of the other player, who's no longer a "player" in the game-theoretic sense, but now just a stochastic process that you've statistically modelled.

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There are equilibrium refinements that rely on players assuming that other players are rational. One of them involves forward induction, which conveniently is discussed in my next article currently up. For the standard Nash Equilibrium that's not the case. When you show that whatever strategy n-tuple you've found is an equilibrium you start by assuming that players 1 through n-1 play their equilibrium strategy and show that a best response for n is her equilibrium strategy. When you assume that 1-(n-1) are playing their equilibrium strategy, you don't assume anything about rationality, just that they play their equilibrium strategy. Rationality comes into play only when assessing whether or not n's strategy is a best response.

It's like that in poker as well. You assume that your opponents are playing some strategy and then if you are Ray Zee you will always play a best response to what they are playing. In solving games you put yourself in player n's shoes looking at what the other players are doing and respond accordingly. In poker you actually are that player n. The only difference is that in poker the opponents aren't rational and won't be playing a best response to what you're doing. That doesn't affect the analysis.

You are correct that it's not technically a game theory problem because our opponents aren't rational, but we don't care about them because we are (attempting to be) rational. The thought process is identical.

edited part in bold

edit #2: just want to clarify from the first paragraph that eventually you will need to assume that players 1 through n-1 are rational but only one at a time when you are showing that their NE strategy is a best response to the other players' strategies. So you basically assume one at a time that they are rational. In poker we're always trying to be the rational one which is why the thinking is so similar.

[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.