Shuffled suit: how many times does the next card climb?

Question

Take the 13 cards of a single suit (ranks 1 through 13) and shuffle them into a random order laid left to right.

On average, how many neighboring pairs have the later card ranked strictly higher than the earlier one?

Accepted Answer

Thirteen cards in a row make 12 neighboring pairs. Put one indicator on each pair and add. Fix a neighboring pair. The two ranks there are distinct, and by symmetry the later card is equally likely to outrank the earlier one or not, so P(	ext{later card is higher}) = \frac12. Let Xi = 1 if neighboring pair i climbs; E[Xi] = 	frac12. With 12 pairs, E[	ext{climbing pairs}] = 12 	imes \frac12 = 6. Takeaway. This is the ascent count for a shuffle of 13 items: each of the 13 - 1 = 12 neighboring pairs climbs with probability 	frac12, giving 6 on average.

Shuffled suit: how many times does the next card climb?

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