Expected position of the first ace

Question

A standard 52-card deck is shuffled uniformly. You turn cards over one at a time. What is the expected position of the first ace?

Accepted Answer

The 4 aces divide the 48 non-aces into 5 gaps: before the first ace, between consecutive aces, and after the last ace. By symmetry, each non-ace is equally likely to fall in any of the 5 gaps, so each gap contains \frac{48}{5} cards in expectation. The first ace sits right after the first gap: E[	ext{position}] = \frac{48}{5} + 1 = \frac{53}{5} = 10.6. Indicator check. Card c (a specific non-ace) precedes the first ace iff, among the 5 cards consisting of c and the 4 aces, c comes first, probability \frac{1}{5}. So the expected number of non-aces before the first ace is 48 \cdot \frac15 = \f…

Expected position of the first ace

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