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Old 05-09-2007, 10:19 AM
jukofyork jukofyork is offline
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Join Date: Sep 2004
Location: Leeds, UK.
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Default Re: Selecting a valid hand configuration for Monte-Carlo simulation

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If P2 acts after P1 there will be: Ax, Ay, Kh, Ks, Kc, Kd, Qh, Qs, Qc, and Qd left for him to get dealt without causing a collision and if you keep re-sampling each time one of player 1's aces are chosen you end up with 8xAK and 16xKQ hands for P2 to get dealt, therefore P(P2|AK)=1/3 and P(P2|KQ)=2/3.

If P2 acts first he will have all 4 aces, kings and queens free, so he will end up with P(P2|AK)=1/2 and P(P2|KQ)=1/2 and since P1 only plays AA he will just re-sample until he gets AA whatever P1 chooses before him.

If you were to sample from all possible configurations by doing a full "discard and re-sample" each time then P(P2|AK)=(1/3+1/2)/2=5/12 and P(P2|KQ)=(2/3+1/2)/2=7/12.

Juk [img]/images/graemlins/smile.gif[/img]

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All these approaches seem broken. If you have AA and villain has AK or KQ, doesn't a plain bayesian analysis 'prove' villain has P(P2|AK)=8/24=1/3
The logic from your example results in 5/12.
I'm thinking we cannot simply "ignore" the frequency of collisions.
(walks away muttering something about independent events while looking for a statistician) [img]/images/graemlins/confused.gif[/img]

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Hehe, this has got me thinking too. I agree that if you are dealt AA then the probability the opponent has AK=8/(50*49)=8/1225 and the probability that he has KQ=16/(50*49)=16/1225, so if your opponent only chooses to play AK and KQ P(AK)=1/3 and P(KQ)=2/3, but the problem arises now from his perspective: If he has AK then the probability you have AA=3/1225 and if he has KQ the probability you have AA=6/1225.

Now if you were observing this and couldn't see any of the player's cards, but knew that P1 only plays AA and P2 only plays AK and KQ and you saw them both decide to play, then I'm thinking that either of the two cases above must have happened with equal chances of either happening.

I'm hoping my post in the Probability forums gets and answer one way or another, but I agree in practice the differences caused by this bias (if it exists) are tiny and I have myself used the deal P1 some cards, then deal P2 some cards, etc before without really thinking/caring about this too much but it's only recently when I went hunting for an efficient algorithm to do the selections that I though harder about the possible bias this was introducing and came up with the idea of permuting the player orderings to cancel it out.

Juk [img]/images/graemlins/smile.gif[/img]
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