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Old 10-05-2002, 07:45 AM
BruceZ BruceZ is offline
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Join Date: Sep 2002
Posts: 1,636
Default Correction to 1st part

In part 1 I computed the probability of getting 3 or more aces given that you got 2 aces. We want the probability that we have 2 aces given 1 ace which is just P(2 ace)/P(1 ace) = 26%/48% = 54%, so you are a favorite here too.
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