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BoxTree
08-30-2005, 05:20 PM
I have 87s.

There are five to the flop.

I flop a four-flush.

I turn the flush.

Given all of these events, what are the odds that at least one other player also has a flush?

DyessMan89
08-30-2005, 05:56 PM
Not likley.

Kaeser
08-30-2005, 06:41 PM
Well there are 8 of your suit unaccounted for. so a players odds of being dealt both of them is 8/46*7/45 which is about 1 in 33.

BigBiceps
08-31-2005, 12:53 PM
So the odds of at least one other player having a flush -- assuming that the starting cards of the players are no more likely to be suited than a random hand is ~ 1 - (1-.02705314)^^4 = 0.10.

So against 4 other players at least one other person will also have a flush 10% of the time under the assumptions above.

LetYouDown
08-31-2005, 02:16 PM
Pretty quickly...so correct any errors at your leisure.

Flopping a 4 flush:
C(11,2) * 39/C(50,3) = 2145/19600 = 10.94%

Turning the flush:
9/47 = 19.15% (assuming 4 flush flopped)

Someone else has it as well:
C(8,2)/C(46,2) = 28/1035 = 2.7%

Note: The two prior odds ignore that someone else has 2 of your suit.

Cobra
08-31-2005, 04:14 PM
I donīt have excel or a calculator in front of me but I can give you the formula you are looking for. You have to assume that your four opponents are playing four random hands. This obviously is not a good assumption. Given that you have hit a flush by the turn the probability of one or more of your four opponents having a flush also is:

4*(8c2)/(46c2)-(4c2)*(8c4)/(46c4)+(4c3)*(8c6)/(46c6)

Cobra

mostsmooth
09-02-2005, 09:11 PM
[ QUOTE ]
I have 87s.

There are five to the flop.

I flop a four-flush.

I turn the flush.

Given all of these events, what are the odds that at least one other player also has a flush?

[/ QUOTE ]
not enough information to answer the question