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Max Weinberg
09-14-2003, 02:11 AM
I'm sorry if this post has been answered in the past, but I searched through old posts and the archives and couldn't find much on it.

Here's the situation, 9 people limp to me with 2 /images/graemlins/club.gif 3 /images/graemlins/club.gif in the BB. The flop has two clubs on it. What are the odds of there being a higher flush draw out there?

I only ask this because I kept raising for value and want to know the EV implications of raising the worst possible flush draw for value.

BruceZ
09-14-2003, 03:28 AM
You can use inclusion-exclusion for the exact answer. Note that at most 4 other players can hold a pair of clubs, so there are 4 terms in the exact solution:

9*C(9,2) / C(47,2) -
C(9,2)*C(9,4) / C(47,4) +
C(9,3)*C(9,6) / C(47,6) -
C(9,4)*C(9,8) /C(47,8)

= 27.5%

If you search for "inclusion-exclusion", you can find a million posts I wrote about this method, so I won't explain it further here. For example, click this link (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&amp;Board=probability&amp;Number=272 265&amp;Forum=probability&amp;Words=simpler&amp;Match=Entire%2 0Phrase&amp;Searchpage=0&amp;Limit=25&amp;Old=6months&amp;Main=270 079&amp;Search=true#Post272265) for some examples with an explanation. Some of the problems I worked in the past were not done as simply as this one, i.e., I've gotten better at it after using it so much.

Nice problem. You can easily modify this for any higher draw that you may hold. Of course this is assuming that the players hold random hands.