Re: Conditional probability question - Hold \'em
The probability that those 18 cards contain no ace is 28*29/(47*46)= 37.6% (I think!).
Correct. That's C(45,18)/C(47,18) = 29*28/(47*46). He asked about 4 opponents, and that would be
C(45,8)/C(47,8) = 39*38/(47*46) = 68.5% that they do NOT have an A. You don't need to use Bayes' theorem.
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