Jonah11
asked on
Game Theory Nash Equilibirum Question
Hi,
I'm using the open-source game-theory program Gambit (http://gambit.sourceforge.net/) to solve a simple toy poker game. But Gambit is giving me a solution that does not make sense.
Let me first describe this 2 player game. Each player is a dealt a card 1, 2, or 3 -- 3 being best. Player 1 and Player 2 both have a 1/3 chance of getting any card, and it's possible for both to be dealt the same card (the deal events are independent). Each player antes 2 chips. Player 1 then has the option of raising 1 chip more (making the pot 5 total), or folding. Player 2 can then fold, call, or re-raise (making the pot 7 total). Player 1 can then fold, call, or re-raise again. At that point the raising is over and Player 2 can only call or fold. If both players have the same card at showdown, they split the pot.
Now, if I tell Gambit to solve a linear complementarity system, I get a "rational" solution:
http://img168.imageshack.us/img168/1055/linearcomplementaritysy.jpg
However, if I solve for a solution using a linear program OR using logit trace, I still get something nonsensical:
http://img192.imageshack.us/img192/5692/linearprogram.jpg
In particular, the solution still fails to re-raise as Player 1 100% of the time when holding a 3, despite the fact that player 2 will call that raise 50% of the time when holding a 2. Thus, player 1 can still unilaterally improve so I don't see how this can be a NE.
Thanks,
Jonah
For those wiho have or download gambit, here is that file:
http://filebin.ca/mpcmeu/test.gbt
I'm using the open-source game-theory program Gambit (http://gambit.sourceforge.net/) to solve a simple toy poker game. But Gambit is giving me a solution that does not make sense.
Let me first describe this 2 player game. Each player is a dealt a card 1, 2, or 3 -- 3 being best. Player 1 and Player 2 both have a 1/3 chance of getting any card, and it's possible for both to be dealt the same card (the deal events are independent). Each player antes 2 chips. Player 1 then has the option of raising 1 chip more (making the pot 5 total), or folding. Player 2 can then fold, call, or re-raise (making the pot 7 total). Player 1 can then fold, call, or re-raise again. At that point the raising is over and Player 2 can only call or fold. If both players have the same card at showdown, they split the pot.
Now, if I tell Gambit to solve a linear complementarity system, I get a "rational" solution:
http://img168.imageshack.us/img168/1055/linearcomplementaritysy.jpg
However, if I solve for a solution using a linear program OR using logit trace, I still get something nonsensical:
http://img192.imageshack.us/img192/5692/linearprogram.jpg
In particular, the solution still fails to re-raise as Player 1 100% of the time when holding a 3, despite the fact that player 2 will call that raise 50% of the time when holding a 2. Thus, player 1 can still unilaterally improve so I don't see how this can be a NE.
Thanks,
Jonah
For those wiho have or download gambit, here is that file:
http://filebin.ca/mpcmeu/test.gbt
Brian Utterback
I think that the issue is that calling on the second round 50% of the time results in an outcome that is the same as raising. Obviously, raising results in a 50% chance of a better outcome and is the best strategy. However, Gambit is probably attacking the problem from the other side in the second case.
ASKER
blu:
can you clarify what you mean by gambit it attacking the problem from the other side? i'm not exactly following you... also, are you suggesting that this is some kind of bug?
can you clarify what you mean by gambit it attacking the problem from the other side? i'm not exactly following you... also, are you suggesting that this is some kind of bug?
ASKER CERTIFIED SOLUTION
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ASKER
Yes, I agree with your sentiments about multiple models. It would also not be surprising to me if this game had multiple equilibrium points.
What IS suprising is that one of the Gambit methods would arrive at an answer which is clearly wrong. To use your example, if the algorithm was an iterative one that waited until stabilization occured, I don't see how this answer could be a stable one.....
What IS suprising is that one of the Gambit methods would arrive at an answer which is clearly wrong. To use your example, if the algorithm was an iterative one that waited until stabilization occured, I don't see how this answer could be a stable one.....
ASKER
blu:
I think I may have figured this out, maybe you can confirm:
The linear program solution, which I thought was a bug, actually makes sense:
http://img192.imageshack.us/img192/5692/linearprogram.jpg
My issue was that we are not reraising a 3 100% of the time, even though our opponent calls that re-raise with a 2 (bottom 2 branches of the tree). However, this doesn't matter, since with a 2 he is ALWAYS only calling the first raise, so he can never arrive at this portion of the graph that my problem depends on. Also, he never calls a re-raise with a 1, so basically when we re-raise our 3, no worse hand is ever calling, so we aren't actually giving anything up. It can be considered an irrelevant side effect of the solution. Does that make sense?
I think I may have figured this out, maybe you can confirm:
The linear program solution, which I thought was a bug, actually makes sense:
http://img192.imageshack.us/img192/5692/linearprogram.jpg
My issue was that we are not reraising a 3 100% of the time, even though our opponent calls that re-raise with a 2 (bottom 2 branches of the tree). However, this doesn't matter, since with a 2 he is ALWAYS only calling the first raise, so he can never arrive at this portion of the graph that my problem depends on. Also, he never calls a re-raise with a 1, so basically when we re-raise our 3, no worse hand is ever calling, so we aren't actually giving anything up. It can be considered an irrelevant side effect of the solution. Does that make sense?
SOLUTION
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Thank goodness your AI doesn't slow play that 3 or bluff with a 1 because that would be just down right mean to the other player >:)
ASKER
blue,
it couldn't do that, otherwise we would know that a 4bet was a bluff with a 1.
it couldn't do that, otherwise we would know that a 4bet was a bluff with a 1.
ASKER
Hey blu, I'm just awarding you the points for helping me out. thanks.