#1
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getting a specific hand X times in Y hands
I would like to calculate the probability of that.
Let's say that probability of having a specific hand by the river in a hold'em game is P. I'd say that the probability of having 2 in a row would be P*P. 3 times in a row would be P*P*P Then what is the probability of having this hand X times among Y hands ? Spontaneously, I would say Y*(P^X), but I'm not sure. After a few thinking, I'd rather say : YCX*(P^X) Is that right ? |
#2
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Re: getting a specific hand X times in Y hands
Sounds like a job for the Binomial distribution.
You want the hand to occur X times out of Y. You can pick X from Y in YCX ways (= Y!/(X!(Y-X)!) Probability of getting the hand X times out of Y is therefore Y!/(X!(Y-X)!) x (P^X)((1-P)^Y) I hope that's right |
#3
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Re: getting a specific hand X times in Y hands
[ QUOTE ]
Sounds like a job for the Binomial distribution. You want the hand to occur X times out of Y. You can pick X from Y in YCX ways (= Y!/(X!(Y-X)!) Probability of getting the hand X times out of Y is therefore Y!/(X!(Y-X)!) x (P^X)((1-P)^Y) I hope that's right [/ QUOTE ] Not quite. Y!/(X!(Y-X)!) x (P^X)((1-P)^(Y-X)) |
#4
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Re: getting a specific hand X times in Y hands
Doh!
Yes, you're right, of course. Thank You, Bruce. |
#5
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Re: getting a specific hand X times in Y hands
If you want a less arduous method, you can use a Poisson approximation.
Then you only need p (the probability of making the hand) and y (the number of hands) and you can instantly calculate the (approximate) probability of getting the hand x times. You can read about it here: http://en.wikipedia.org/wiki/Poisson_distribution By the way, the main formula here is (e^(-yp))((yp)^x)/x! which can be calculated relatively quickly for most x. |
#6
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Re: getting a specific hand X times in Y hands
Waouh, nice guys =)
Thx for the answer. May I add something, maybe a little bit more complicated ? I'd like to know the odds of getting : - one specific hand X times - another specific hand Y times out of Z hands. My final point is to calculate the odds of several hands I got recently. Spontaneously, I would say the odds are quite low, but I'd like to calculate it =) Thx for helping guys ! |
#7
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Re: getting a specific hand X times in Y hands
[ QUOTE ]
I'd like to know the odds of getting : - one specific hand X times - another specific hand Y times out of Z hands. [/ QUOTE ] If the probability of one hand being dealt is P1, and the probability of the other hand being dealt is P2, then the probability of exactly X of the first hand and Y of the second hand out of Z total hands is, in your notation (ZCX) * (Z-X)CY * (P1^X)*(P2^Y)*(1-P1-P2)^(Z-X-Y). Note that if you want the probability of at least X of one hand and at least Y of the other, then you must do a double summation of this over all possible numbers of the first hand x from X to Z-Y, and all possible numbers of the second hand y from Y to Z - x. <font class="small">Code:</font><hr /><pre> P(x>=X AND y>=Y) = sum{x = X to Z-Y} sum{y = Y to Z - x} (ZCx) * (Z-x)Cy * (P1^x)*(P2^y)*(1-P1-P2)^(Z-x-y) </pre><hr /> |
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