A standard 52-card deck is shuffled uniformly. What is the probability that no two of the four aces end up adjacent to each other?

A standard 52-card deck is shuffled uniformly. What is the probability that no two of the four aces end up adjacent to each other?

Approach: Count positions rather than full arrangements. Lay out the 48 non-aces first and choose which of the gaps between them receive aces, at most one ace per gap.

4324/5525. Only the set of four ace positions matters, and there are C(52,4) = 270725 equally likely sets. Lay out the 48 non-aces, which creates 49 gaps between cards including the two ends, and non-adjacent placement means putting the four aces in four distinct gaps, which can be done in C(49,4) = 211876 ways. The count gives 211876/270725 = 4324/5525 = 0.7826. In general k non-adjacent items among n positions can be placed in C(n - k + 1, k) ways. As a check, the expected number of adjacent ace pairs is 51 * (4 * 3)/(52 * 51) = 3/13, and the Poisson heuristic exp(-3/13) then gives 0.794, close to the exact figure for a deck this size.

Follow-up: What is the probability that no two aces are within three cards of each other?

Key concepts: gaps between cards, combinatorial counting, non-adjacent placement, poisson heuristic.