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You hold 2 suited cards and the board has 3 more of your suit (along with 2 more offsuit cards). What are the chances another person has a flush (assuming your opp plays all cards, even 23s)
Cobra
05-02-2005, 12:51 AM
This is an inclusion/exclusion problem. There are seven known cards, 5 of your suit and two other. There are 45 unseen cards with 8 of your suit and 37 other. Assuming your opponent plays any suited hand the probability of one or more opponents having a flush are:
=9*(8c2)/(45c2)-(9c2)*(8c4)/(45c4)+(9c3)*(8c6)/(45c6)-(9c4)*
(8c8)/(45c8) = 23.79% for nine opponents.
with six opponents = 16.27%
with four opponents = 11.03%
Cobra
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