Where the first ace shows up in a shuffled deck

Question

A standard 52-card deck is shuffled uniformly at random. You deal cards face up one at a time until the first Ace appears. On average, what is the position of the first Ace (the number of cards dealt, including that Ace)?

Accepted Answer

This is a waiting time without replacement, so it is not geometric. A clean way to see it: the 4 Aces split the other 48 cards into 5 gaps: before the first Ace, between consecutive Aces, and after the last Ace. By symmetry, each of the 48 non-Aces is equally likely to fall in any of these 5 gaps, so on average \frac{48}{5} = 9.6 non-Aces sit before the first Ace. The first Ace is the very next card, so its expected position is 9.6 + 1 = 10.6 	ext{ cards.} Equivalently, for m special items among N total in random order, the expected position of the first special item is \frac{N+1}{m+1}; here…

Where the first ace shows up in a shuffled deck

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