Ace up your Sleeve

Solution

Answer: 0.456

Therefore, among the remaining cards there are \[ 4 - 2 = 2 \text{ aces} \] left in the deck. Since you removed \(13\) cards from the deck, the number of remaining cards is \[ 52 - 13 = 39. \] Among these \(39\) cards: \[ 2 \text{ are aces}, \qquad 37 \text{ are non-aces}. \] Your friend draws \(13\) cards from these \(39\). Let \(X\) be the number of aces your friend receives. Then \[ X \sim \text{Hypergeometric}(N=39, K=2, n=13), \] where \[ N = 39 \quad (\text{total cards}), \quad K = 2 \quad (\text{aces}), \quad n = 13 \quad (\text{cards drawn}). \] The hypergeometric probability formula is \[ P(X=k) = \frac{\binom{K}{k}\binom{N-K}{n-k}} {\binom{N}{n}}. \] We want the probability that your friend has exactly one ace: \[ P(X=1). \] Substitute the values: \[ P(X=1) = \frac{\binom{2}{1}\binom{37}{12}} {\binom{39}{13}}. \] \[ P(X=1) = \frac{2\binom{37}{12}}{\binom{39}{13}}. \] \[ P(X=1) \approx 0.456. \]