Dealt AA HU in the SB, whats the chance BB has: a) AK b) KQ?

[jukofyork]

Is my thought process here flawed and does permuting the ordering of card assignment make for an unbiased selection of hands?

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

[jukofyork]

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Is my thought process here flawed and does permuting the ordering of card assignment make for an unbiased selection of hands?

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I didn't phrase my question very well and it's not really what I was trying to ask (as I already know I have AA and obviously P(BB|AK)=8/1225 and P(BB|KQ)=16/1225), so I'll have another try at re-phrasing it:

"If I am observing a game where I know player 1 only plays AA and player 2 only plays AK and KQ, and I observe then both playing, then obviously P1 has AA, but what is the probability player 2 has a) AK and b) KQ?"

Here is what I think is the answer:

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Imagine if player #1 only plays AA and player #2 only plays AK and KQ. If you don't permute the order then the fact that player #1 has taken away two of the aces every time before player #2 gets his selection would mean that he would end up with KQ far more often than he should (as there will always only be 2 aces left to choose from). On the other hand, if you permute the player order then 50% of the time player #2 gets to choose when there are 4 aces still available and 50% of the time he has just 2 aces left to choose from.

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There is no bias introduced with a partial resample on collision.
When you say 'has 2 aces left to choose from' - what you really mean is '50% of the time the algorithm will encounter collisions if P2 chooses an Ace which was already used (50% x 50%)'

P2 by definition within your constraints must be dealt KQ far more often than AK.
Consider this - when P2 is first, 50% of the time P1 only has 3 aces to choose from. He still gets AA every time but the algorithm will encounter collisions with 1 of the aces when it has been used. The only variable is which player's cards collide and how many collisions occur.

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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.

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My second question is this:

"If I always select a hand for P1 based on his distribution, then select a hand for P2 based on his then assuming what I wrote above is correct this will lead to a biased selection. If however I permute the ordering of the selection so that P1 and P2 get an equal chance at being assigned cards first, does this full remove the bias? Also, does removing the bias by permuting the order of hand assignments extend to more than 2 players?"

There is a discussion going on about this here also.

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

[jukofyork]

Nvm, it seems that the idea of permuting the player orderings to select from is also flawed and doesn't produce the same results as you would get for a full re-sample.

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

[Cobra]

If you are observing a game and player one only plays AA and player two only plays AK or KQ and both players have bet. Player two has AK 8/24 and KQ 16/24, or put it another way he will have KQ twice as ofter as AK.

I don't understand what you are asking in number two.

Cobra

[jukofyork]

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If you are observing a game and player one only plays AA and player two only plays AK or KQ and both players have bet. Player two has AK 8/24 and KQ 16/24, or put it another way he will have KQ twice as ofter as AK.

I don't understand what you are asking in number two.

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Number two was todo with how to assign cards for each player for monte-carlo simulation without having to do a full re-sample each time their card's collide, but the idea I had seems to give P(2|AK)=5/12 and P(2|KQ)=7/12 and is biased.

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