Solution
Holding the Ace of Spades
Show the problem again
A 13-card bridge hand is dealt uniformly at random from a standard 52-card deck. You learn that the hand contains the ace of spades. What is the probability that it contains at least two aces? Give the exact reduced fraction, or a decimal rounded to four places.
Worked solution
The answer is 11686/20825 ≈ 0.5612. Given the ace of spades, the other 12 cards are a uniform random 12-card subset of the remaining 51 cards, which include 3 aces. The hand has no second ace with probability C(48,12)/C(51,12) = 9139/20825, so the probability of at least two aces is 1 − 9139/20825 = 11686/20825. The paradox is in the comparison. If you only know that the hand contains "at least one ace", the probability of at least two aces is about 0.370, much lower. Naming a particular ace changes the answer because hands with several aces are proportionally more likely to include any given ace. "At least one ace" lets in many one-ace hands, while "the ace of spades" gives a multi-ace hand several ways to qualify.
Source: Classic 'two aces' paradox of conditional probability, discussed in many probability texts and puzzle collections. Statement written for AxiomIQ.