Five friends grab phones at random

Question

Five friends drop their five look-alike phones into a bowl. Each friend then blindly grabs one phone, so the phones come back in a uniformly random order.

What is the probability that not a single friend gets their own phone back?

Accepted Answer

An arrangement in which nobody gets their own item is called a derangement. We need the fraction of the 5! = 120 possible orderings that are derangements. There is a clean counting formula. Using inclusion-exclusion (start with all orderings, subtract those where at least one person matches, add back double counts, and so on), the number of derangements of n items is Dn = n!\left(1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \cdots + \frac{(-1)^n}{n!}ight). For n = 5: D5 = 120\left(1 - 1 + 	frac12 - 	frac16 + 	frac{1}{24} - 	frac{1}{120}ight) = 120 \cdot 	frac{44}{120} = 44. So th…

Five friends grab phones at random

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