Re: 7-Stud odd
# of aces chance of at least one pair of aces
1 11.53%
2 15.51%
3 12.24%
The way I derived these numbers is by using combinatoric probability of there being no aces amongst the down card and substracting that from one to determine the chances of there being at least one ace among the unknown cards. For example in the case with 3 up aces the chance for no aces amongst the 6 down cards is 1 - (C(48,6)/C(49,6)). Where C(N,K) = N!/(K!*(N-K)!))
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