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#1
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Flopping a set heads up
I understand a similar situation was posted before, but the situation im looking for is a bit different.
How do you calculate the probability that player A and player B both flop a set, assuming it's heads up and they both have different pocket pairs. is it as simple as just 2*2*48/48C3?? |
#2
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Re: Flopping a set heads up
i know that P(player a floping a set) = (2* 48C2/48C3) but then i get dizzy cuz i dont think that player B also flopping a set is independent of player A flopping a set.
i.e., if player A flops one player B only has 2 of the 3 flop cards to match with his pocket pair. EDIT: Sorry, I edited this by mistake, but I changed it back to what you had originally. -BZ |
#3
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Re: Flopping a set heads up
[ QUOTE ]
i know that P(player a floping a set) = (2* 48C2/48C3). [/ QUOTE ] That should be 2*C(46,2)/C(48,3) including full houses. For exactly a set it should be (2*44*40/2 + 2*2*44)/C(48,3), where the 2*2*44 corresponds to both players flopping a set. For a set, full house, or quads, it would be [2*C(46,2) + 48]/C(48,3), where the 48 corresponds to the quads. If we don't know player B's cards, and if we also include full houses, then it would be 2* C(48,2)/C(50,3). [ QUOTE ] but then i get dizzy cuz i dont think that player B also flopping a set is independent of player A flopping a set. i.e., if player A flops one player B only has 2 of the 3 flop cards to match with his pocket pair. [/ QUOTE ] They are not independent. That's why the answer is not [P(player A flopping a set)]^2. |
#4
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Re: Flopping a set heads up
[ QUOTE ]
I understand a similar situation was posted before, but the situation im looking for is a bit different. How do you calculate the probability that player A and player B both flop a set, assuming it's heads up and they both have different pocket pairs. is it as simple as just 2*2*48/48C3?? [/ QUOTE ] 2*2*44/C(48,3) for both flopping exactly a set. |
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