Suppose you shuffle a deck of cards. How likely is it that there are two cards that were next to each other before the shuffle are still next to each other after the shuffle?
It depends on how well you shuffle the cards. If you do what’s called a “faro shuffle” then the probability of a pair of cards remaining neighbors is 0. In a faro shuffle, if the cards were originally numbered 1 through 52, the new arrangement will be 1, 27, 2, 28, 3, 29, …, 26, 52. All pairs are split up.
But if you shuffle the cards so well that the new arrangement is a random permutation of the original, there’s about an 86% chance that at least one pair of neighbors remain neighbors. To be more precise, consider permutations of n items. As n increases, the probability of no two cards remaining consecutive converges to exp(-2), and so the probability of at least two cards remaining next to each other converges to 1 – exp(-2).
By the way, this considers pairs staying next to each other in either order. For example, if the cards were numbered consecutively, then either a permutation with 7 followed by 8 or 8 followed by 7 would count.
More generally, for large n, the probability of k pairs remaining consecutive after a shuffle is 2ke-2 / k!.
One application of this result would be testing. If you randomly permute a list of things to break up consecutive pairs, it probably won’t work. Instead, you might want to split your list in half, randomize each half separately, then interleave the two half lists as in the faro shuffle.
Another application would be fraud detection. If you’re suspicious that a supposedly random process isn’t splitting up neighbors, you could use the result presented here to calibrate your expectations. Maybe what you’re seeing is consistent with good randomization. Or maybe not. Note that the result here looks at any pair remaining together. If a particular pair remains together consistently, that’s more suspicious.
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Reference: David P. Robbins, The Probability that Neighbors Remain Neighbors After Random Rearrangements. The American Mathematical Monthly, Vol. 87, No. 2, pp. 122-124
Proobability that any given pair remains (in either order) can be easily shown to be 1/26. (2*(51*50!))/52!