At least one ace, are both aces?

Question

You are dealt 2 cards from a standard 52-card deck. A friend peeks and truthfully tells you the hand contains at least one ace. Given that, what is the probability both cards are aces?

Accepted Answer

Condition on "at least one ace." Work with two-card hands, of which there are \binom{52}{2} = 1326. Both aces: \binom{4}{2} = 6 hands. At least one ace: easier via the complement, hands with no ace number \binom{48}{2} = 1128, so hands with at least one ace number 1326 - 1128 = 198. The conditional probability is favorable over conditioned total: P(	ext{both aces}\mid 	ext{at least one ace}) = \frac{6}{198} = \frac{1}{33} \approx 0.030. Takeaway. Conditioning restricts the sample space to the 198 hands consistent with "at least one ace," then counts the 6 all-ace hands within it. This is fa…

At least one ace, are both aces?

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