[BruceZ]

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Hey aaron. Will you explain where I went wrong? Thanks.

The probability of just exactly 2 players geting distinct PPs is:
13(4,2)/(52,2)*12(4,2)/(48,2) *100 =0.346%

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The (48,2) should be (50,2), but your number is right.

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At a 9 player table you can estimate the probability of 2 players out of 9 getting a distinct PP by [C(9,2)*0.00346]*100= 12.456%

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Note that Aaron's 4.4% = 9*6/C(50,2) is the approximate probability that one of 9 opponents at a 10-player table has a particular higher pocket pair when YOU have a pocket pair. You are computing the probabilty that ANY 2 players at a 9-player table both have distinct pocket pairs, which is a very different thing.


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The probability of a player flopping a set is 11.76%. The probability of 2 players who start with a PP flopping a set is (0.1176^2)*100= 1.38%

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You can't just square 11.76% because the 2 sets are not independent. When one player flops a set, the probability that the other player flops a set will be smaller than 11.76% because there are only 2 remaining flop cards. The probability that both pairs flop sets is:

2*2*44/C(48,3) =~ 1.02%

OR

4/48 * 2/47 * 44/46 * 3 =~ 1.02%

where we multiply by 3 since the non-set card can come in any of 3 positions. This does not include full houses/quads.

Aaron did this for the whole board, not just the flop. This gives:

2*2*C(44,3)/C(48,5) =~ 3.1%.

This includes boards that pair to make full houses for both players, but not boards that pair to make a full house for one player and quads for the other. These would add another 2*2*1*C(44,2)/C(48,5) =~ 0.2%, half of which would allow the smaller pair to regain the pot.


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The probability on any give hand that 2 players will flop a set is ~ (0.12456*0.0138)*100= 0.17%

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~(0.12456*0.0102)*100= 0.13%