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#1
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How do I compute this?
I've seen 30 flops with 30 pocket pairs and only hit sets twice. Would that be (1/8.5)^2 x (7.5/8.5)^28? |
#2
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[ QUOTE ]
How do I compute this? I've seen 30 flops with 30 pocket pairs and only hit sets twice. Would that be (1/8.5)^2 x (7.5/8.5)^28? [/ QUOTE ] For exactly twice: C(30,2) x (1/8.5)^2 x (7.5/8.5)^28 where C(30,2) = 30*29/2 or use Excel function =BINOMDIST(2,30,1/8.5,false) = 18%. For 0, 1 or 2 times: (7.5/8.5)^30 + 30 x (1/8.5) x (7.5/8.5)^29 + C(30,2) x (1/8.5)^2 x (7.5/8.5)^28 or use Excel function =BINOMDIST(2,30,1/8.5,true) = 30%. |
#3
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beat me by a minute you bastard
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#4
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First of all, you are probably looking for the probability of hitting two or fewer sets in 30 flops. That number is more important intuitively because 1 minus that number gives you the probability of hitting more than 2 sets in 30 flops. In that case the answer is:
[(1/8.5)^2]*[(7.5/8.5)^28]*[(30!)/(28!2!)] + [(1/8.5)^1]*[(7.5/8.5)^29]*[(30!)/(29!1!)] + [(1/8.5)^0]*[(7.5/8.5)^30]*[(30!)/(30!0!)] = Answer If you really just want the probability of hitting 2 sets, then just compute the first term. |
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