Cards Between the Aces

Question

A standard 52-card deck is shuffled. In expectation, how many cards lie strictly between the first ace and the second ace?

Accepted Answer

The four aces create five gaps: before the first ace, three between consecutive aces, and after the last. For any fixed non-ace card, consider the relative order of just five cards, that card and the four aces. Its position among them is uniform over the five slots, so it lands in any specific gap with probability frac15. By linearity over the 48 non-aces, the gap between the first and second ace holds 48 cdot frac15 = frac485 = 9.6 text cards in expectation. Zero-computation check: the five gaps are exchangeable and their sizes always sum to 48, so each must average frac485. Done. What this buys you: the gap sizes are strongly dependent (they're constrained to sum to 48), but expectations don't care, exchangeability plus linearity dispatches the whole family: position of the first ace (9.6 + 1 = 10.6), of the last ace (52 - 9.6 = 42.4), expected cards before the first spade, and so on. The standard time sink is to attack this with draw-by-draw conditional probabilities, it works, painfully. Recognizing "uniform placement into exchangeable slots" is the entire problem, and it's the same symmetry that prices random insertion positions in order queues: a new arrival is uniform over t

Cards Between the Aces

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