Re: flush
Stephen J. Herschkorn wrote:
bob@xxxxxxxxxxxxxx wrote:
Does anyone here know how to calculate the probability of a flush in
Texas Holdem using combinatorics?
Yes.
I guess you want to know how to calculate that probability. This is
the probability that in a random seven-card hand, at least five of the
cards are of the same suit. Let C(n,m) denote the number of ways of
selecting m objects from n distinct objects; C(n,m) = n! / [m!
(n-m)!]. The probability is then
4 [C(13,5) C(39,2) + C(13,6) 39 + C(13,7)] / C(52,7)
= 78639 / 2572780 = .0306, approximately.
Or, if you want the conditional probability of making a flush given that
your two cards are of the same suit, this is
[C(11,3) C(39,2) + C(11,4) 39 + C(11,5)] / C(50,5)
= 19371 / 302680 = .0640, approximately.
--
Stephen J. Herschkorn sjherschko@xxxxxxxxxxxx
Math Tutor in Central New Jersey and Manhattan
.
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(rec.games.bridge) |
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