Re: How would you calc probability of opponents hands?
I don't know about a formula, but here is the thought process:
Opponents holds:
Jacks: 3 ways / c(50,2)
2pair: 54 ways / c(50,2) ie JT,J8,J4,T8,T4,84
Str: 32 ways / c(50,2) ie Q9 and 97
Flush: 120 ways / c(50,2)
So you would sum all of these if you wanted the probablity that he has any one of the hands, but make sure you aren't double counting the straights and the flushes...
-Gryph
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