Thread: Probability of 2 PP's at 9 handed table?
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Old 10-10-2007, 07:17 AM
tshort tshort is offline
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Join Date: May 2005
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Default Re: Probability of 2 PP\'s at 9 handed table?
If you assume any given player has a 3/51 chance of being dealt a PP:

C(9,2) * (3/51)^2 * (48/51)^7 = 0.0815

Binomial Distribution

This should be close enough to the correct answer, but isn't exactly correct because the events are not independent.

I see no way of working around the numerous cases of conditional probabilities to get the exact solution.
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