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#2
04-01-2004, 07:36 PM
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Re: 3 flush on board, odds someone already has a flush?
I've shown how to do this a few times. It is solved easily with the inclusion-exclusion principle. This post explains that principle, and example 3 shows how to do this problem when you have 4 opponents on the flop and you don't have a flush card. You can do all the other cases just by changing the numbers in this formula, and using as many terms as there are players who could have a flush, up to a maximum of 5.
Let me know if you need more help computing any of these. The post tells you all you need to know, and the rest is just busy work. If you can make a spreadsheet, and if you truly understand how to compute the terms, it should go pretty fast. If you can correctly compute the answers for 1-9 opponents, then you will completely understand this method. 
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#4
04-01-2004, 08:19 PM
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Re: 3 flush on board, odds someone already has a flush?
I just realized that my example was for the probability of someone having a 4-flush draw when there are 2 flush cards on the board, not a flush with 3 flush cards on the board. That's OK though. You can still modify this to do your problem, and it's almost the same. Here's the solution for 4 opponents to get you started:
C(4,1)*C(10,2) / C(47,2) - C(4,2)*C(10,4) / C(47,4) + C(4,3)*C(10,6) / C(47,6) - C(4,4)*C(10,8) / C(47,8) = 15.95%
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