Derangement is a specific type of problems: $n$ pairs, all messed up. The count is denoted as $D_n (n\ge 2)$: $$D_n = n!\left(\frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!}-\cdots + (-1)^n\frac{1}{n!}\right)$$

Correspondingly, the probability of occurring is $$\frac{D_n}{n!} = \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \cdots +(-1)^n\frac{1}{n!}$$

When $n$ approaches infinity, this probability approaches $\frac{1}{e}\approx 0.3679\cdots$.