Probability of Flush Given 3 clubs on flop
 

This was inspired by a post in HSMTT. The problem I'm posting here is slightly different. Seven handed game, all see the flop, assume random distribution of cards for all seven (that is, don't use conditional probabilities based on the fact that they chose to stay in to see the flop). Flop comes out 3 clubs. What is the probability that one or more players has a flush?

I came up with 24.75652%. It took fairly serious number crunching to get it. I want to see if I got it right, and see if anyone comes up with a solution that arrives at an answer more efficiently than mine did.

Re: Probability of Flush Given 3 clubs on flop
 

Seems right:

P(first player has flush) = C(10,2)/C(49,2) = 45/1176
P(1st and 2nd player has flush) = [C(10,2)/C(49,2)][C(8,2)/C(47,2)] ~ 0.0009911457645 (call this p)
P(1st, 2nd and 3rd player has flush) = p x C(6,2)/C(45,2) = p/66
P(1st, 2nd, 3rd and 4th player has flush) = p/66 x C(4,2)/C(43,2) = p/9933

Using inclusion-exclusion, the probability is approximately

7(45/1176) - C(7,2)p + C(7,3)p/66 - C(7,4)p/9933

~0.2475652.

Re: Probability of Flush Given 3 clubs on flop
 

[ QUOTE ]
Seems right:

P(first player has flush) = C(10,2)/C(49,2) = 45/1176
P(1st and 2nd player has flush) = [C(10,2)/C(49,2)][C(8,2)/C(47,2)] ~ 0.0009911457645 (call this p)
P(1st, 2nd and 3rd player has flush) = p x C(6,2)/C(45,2) = p/66
P(1st, 2nd, 3rd and 4th player has flush) = p/66 x C(4,2)/C(43,2) = p/9933

Using inclusion-exclusion, the probability is approximately

7(45/1176) - C(7,2)p + C(7,3)p/66 - C(7,4)p/9933

~0.2475652.

[/ QUOTE ]

Very cool, I didn't even know what inclusion-exclusion was prior to your post (or if I did, I had forgotten about it in the many years since I last did stuff like this). I did this the hard way, calculating the probability that no one had a flush (it would be more effort than it's worth to try to post the formula I came up with) and subtracting from 1.