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#11
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[ QUOTE ]
Pretend we have two players that play only two hands: AK or AA. First player: AK - 16 combos, AA 6 combos, total 22 First Players chance of playing AK is 16/22 First Players chance of playing AA is 6/22 Now assign the second player: If player 1 was assigned AK (p = 16/22), then Player 2: (Baysian updating) Plays AK with 9/12 probability Plays AA with 3/12 probability If player 1 was assigned AA (p = 6/22), then Player 2: (Baysian updating) Plays AK with 8/9 probability Plays AA with 1/9 probability Now what is Player 2's chance of playing AA (16/22)*(3/12) + (6/22)*(1/9) = 7/33 Now the problem: Player 1 and Player 2 had the same distributions. But 7/33 != 6/22. The problem is that assigning PLAYER 1 his distribution frist, gives player 2 an unequal chance of getting AA. The player most likely to be dealt pocket pairs, is the one where all the aces are in deck when he gets a card on his distribution. A correct algorithm, would at the very least, have to give both players (with equal distributions) an equal chance of getting AA. Thus, dealing everyone random hands, and discarding until everyone gets a hand on their distribution is an unbiased, but albeit extremely slow method of getting the right answer. [/ QUOTE ] You are correct. If only you had written this earlier it would have saved me a few days of thought/work two years ago. [img]/images/graemlins/smile.gif[/img] I came up with a (relatively) fast way to generate random razz hands, but could not prove it was correct. So I in fact implemented the extremely boneheaded (but unambiguosly correct) algorithm you outlined above for comparison purposes. Took billions of hands to converge within 0.01% in some cases, but converge they did. - bachfan |
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