Where's the Ace

Let $p_k$ be the probability that the first ace appears at the $k$ th card.

For the first ace to appear at the $k$ th card, the first $k-1$ cards must be non-aces, the next card an ace, and the final $3$ aces must be distributed amongst the remaining $52-k$ cards. Thus there are $\binom{52-k}{3}$ ways of choosing the positions of the $4$ aces.; If we regard the four aces as indistinguishable, and the $48$ other cards as indistinguishable, there are then $\binom{52-k}{3}$ ways of shuffling the pack and have the first ace at the $k$ th card, out of a total of $\binom{52}{4}$ possible shuffles. Hence $p_k \; = \; \frac{\binom{52-k}{3}}{\binom{52}{4}} \; = \; \frac{(52-k)(51-k)(50-k)}{13 \times 49 \times 50 \times 51} \hspace{2cm} 1 \le k \le 49$ It is clear that $p_k$ is a decreasing function of $k$ , and so $p_1$ is the largest probability. The answer is $\boxed{1}$ .

Let $P(n)$ be the probability that the first ace is the $n$ -th card drawn, then we want to find $\arg \max_n P(n)$ ; that is, $n$ such that $P(n)$ is maximized.

Note that $P(n)$ is simply the product of the probabilities that the first $n-1$ cards drawn are not aces, and the next card drawn is an ace. But we can evaluate this in a simpler manner: if, instead, we begin by opening the $n$ -th card, followed by the $n-1$ cards above, we obtain that $P(n)$ is the probability that the first card we open (which would be $n$ -th down the deck) is an ace, followed by the next $n-1$ cards we open (which would be the original first $n-1$ cards in the deck) are not aces. The first portion is constant (1/13) and the latter portion decreases with larger $n$ (it's more likely that we'll encounter some ace in a larger portion of the deck than a smaller one), thus to maximize the probability clearly we need $n = \boxed{1}$ , the minimum possible.