How deep in the deck the first ace appears

Question

A standard $52$ -card deck is shuffled and cards are flipped over one at a time from the top. What is the expected number of cards flipped up to and including the first ace?

Short answer

$53/5=10.6$ . The four aces split the $48$ other cards into five gaps of equal expected size $\tfrac{48}5$ . The first ace comes after the first gap, so the expected position is $\tfrac{48}{5}+1=\tfrac{53}{5}$ .

Adapted from Zhou, A Practical Guide to Quantitative Finance Interviews, sec. 4.5 Expected Value, Variance & Covariance, 'Card game (first ace)', p.94; question and answer rewritten by SelfMadeQuant.com.

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Lessons behind this question

Expectation, Mean, and Variance
Conditional Expectation and Variance Revisited
Sample Spaces, Events, and the Probability Axioms

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