Hello! How\'s my Thinking On This One?--Hmmm...
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)) [img]/forums/images/icons/shocked.gif[/img]
Any fallacy in this thinking?
Just wondering,
SittingBull
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