I'm not sure this qualifies, but here goes.
It doesn't take any fancy enumerations to show that the average number of fixed points is 1: Among the $n!$ permutations of $n$ objects, each object is fixed by $(n-1)!$ permutations, for a total of $n!$ fixed points. So if the probability of having no fixed points tends to 0 (for some subsequence of $n$'s), then the probability of having more than one fixed point must also tend to 0. This seems like it should be easily disprovable nonsense.