Shuffled deck, expected position of the last ace

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You shuffle a standard 52-card deck and number the positions 1 (top) through 52 (bottom). On average, at what position does the last (fourth) ace appear?

Accepted Answer

The cleanest route is a mirror-image symmetry argument. Set up the picture. There are 4 aces and 48 non-aces. As before, the 4 aces split the 48 non-aces into 5 equal gaps on average, each holding \frac{48}{5} = 9.6 non-aces. Non-aces after the last ace. By symmetry, the gap after the last ace also holds 9.6 non-aces on average, exactly like the gap before the first ace. Position of the last ace. Counting from the top, the last ace is followed by those 9.6 trailing non-aces out of 52 cards, so its position is: EV = 52 - 9.6 = 42.4 = \frac{212}{5}. Another way to see it: the deck is symmetric …

Shuffled deck, expected position of the last ace

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