SittingBull
05-12-2003, 05:24 PM
On the river in a Hold'em game,there are 4 of one suit showing and I have one of the suit. Then the Probability that my heads-up opponent has at least one of the same suit is***
1-(C(37,2)/C(45,2)) /forums/images/icons/shocked.gif
Any fallacy in this thinking?
Just wondering,
SittingBull
1-(C(37,2)/C(45,2)) /forums/images/icons/shocked.gif
Any fallacy in this thinking?
Just wondering,
SittingBull