Probability & Brainteasers · Interview Question
Drawing one at a time from a shuffled 52-card deck, what is the expected position of the first Ace?
How to answer
10.6. The 4 Aces split the 48 non-Aces into 5 gaps of expected size 48/5 = 9.6; the first Ace sits just after the first gap, so E = 9.6 + 1 = 10.6. General: for k items among n, E[position of first] = (n+1)/(k+1) = 53/5 = 10.6.
Key idea: Computing 52/4 = 13 (mean spacing BETWEEN Aces) instead of the position of the FIRST Ace.
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