3 suited card flops: made flush?
 

if the board came all clubs, what would be the probability that somebody had a made flush if there were:

1 other player (besides me)
2 other players
3 other players
4 other players
5 other players
6 other players
7 other players
8 other players
9 other players?

-Lane

Re: 3 suited card flops: made flush?
 

The trouble with this question is it depends on which hands people are likely to play.

If you deal one hand face down, then the board comes up with three clubs, the chance that the facedown hand makes a flush is (10/49)*(9/48) = 15/392 = 3.8%. For N hands, the chance is approximately 1 - (1 - 15/392)^N.

Re: 3 suited card flops: made flush?
 

[ QUOTE ]
1 - (1 - 15/392)^N.

[/ QUOTE ]

Could you please explain the thinking behind this formula?? I understand that (1 - 15/392) is calculating the probability that one face down hand doesn't have the flush, but I forget why we have to approach the problem from this angle.
Cheers

Re: 3 suited card flops: made flush?
 

[ QUOTE ]
[ QUOTE ]
1 - (1 - 15/392)^N.

[/ QUOTE ]

Could you please explain the thinking behind this formula?? I understand that (1 - 15/392) is calculating the probability that one face down hand doesn't have the flush, but I forget why we have to approach the problem from this angle.
Cheers

[/ QUOTE ]

If you take the probability that 1 hand does not have a flush and multiply it by itself N times (raise it to the Nth power), this gives you approximately the probability that N hands do not have a flush. This would be exact if the hands were independent, but actually they are not since if one hand has a flush, it changes slightly the probability that the other hands have a flush.

For the exact answer, you need the inclusion-exclusion principle. The last example in that post has a similar calculation. If you only compute 2 terms of inclusion-exclusion, the approximation is better than the independence approximation. The data below shows a comparison of these two approximations for this problem. The first column is the number of hands N. The second column gives the exact probability that at least 1 hand has a flush to 6 decimal places computed by the inclusion-exclusion principle. The second column gives the result of the first 2 terms of inclusion-exclusion which are N*15/396 - C(N,2)*C(10,4)/C(49,4). The last column gives the result of the independence approximation 1 - (1 - 15/396)^N. As you can see, the error gets worse as the number of hands increases, but it increases much more slowly for the inclusion-exclusion approximation than for the independence approximation.

<font class="small">Code:</font><hr /><pre>
# hands P(flush) exact P(flush) 2 terms P(flush) independence

1 3.826531% 3.826531% 3.826531%
2 7.553947% 7.553947% 7.506638%
3 11.183750% 11.182248% 11.045925%
4 14.717432% 14.711435% 14.449780%
5 18.156475% 18.141507% 17.723385%
6 21.502350% 21.472465% 20.871725%
7 24.756520% 24.704308% 23.899593%
8 27.920436% 27.837037% 26.811598%
9 30.995541% 30.870651% 29.612175%
10 33.983266% 33.805150% 32.305586%
11 36.885034% 36.640535% 34.895934%
</pre><hr />

Re: 3 suited card flops: made flush?
 

Cheers BruceZ!