getting a specific hand X times in Y hands

[pirlouit]

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 ?

[BritNewbie]

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

[BruceZ]

[ 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))

[BritNewbie]

Doh!

Yes, you're right, of course.
Thank You, Bruce.

[T50_Omaha8]

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.

[pirlouit]

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 !

[BruceZ]

[ 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&gt;=X AND y&gt;=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 />