It's a fairly well-known fact that if we pick a permutation of the set [n] at random, then the expected number of cycles of length k in that permutation is 1/k, for 1 ≤ k ≤ n. (Somewhat more memorably, the average number of elements in a permutation of [n] which are members of a k-cycle is 1.)
So what would you expect to happen if you just considered permutations have cycles of length 1 and 2 -- that is, involutions -- and sampled uniformly at random from them? If you're like me, your first thought is that the average involution will have twice as many 1-cycles as 2-cycles, and the same number of elements in 1-cycles as 2-cycles -- that is, the average involution on [n] will have n/2 1-cycles (i. e. fixed points) and n/4 2-cycles, for a total of n/2 fixed points and n/2 elements in 2-cycles.
But then you look at "n/2 fixed points" and think that that seems awfully large...
it turns out the average number of fixed points of an involution chosen uniformly at random from all involutions is about n1/2. This follows from standard generating function arguments. The exponential generating function of the number of involutions marked for their number of fixed points is exp(uz+z2/2); that is, the coefficient of znuk in that function is n! times the number of involutions on [n] with k fixed points. Standard methods from, say, Flajolet and Sedgewick (which I will probably buy when it comes out in print later this year, because I seem to cite it constantly) gives that the expected number of fixed points is
and this can actually be rewritten as nan-1/an, where an is the number of involutions on [n], that is, . (There's a nice interpretation for this -- an-1/an is the probability that any given element of an involution is actually a fixed point -- although it's hard to say exactly why this should be true.)
Then, if you're still like me, you think "alas, I have forgotten how to figure out the asymptotics of coefficients of entire functions". But the asymptotic number of involutions is the last example of Herbert Wilf's generatingfunctionology. After some work, the asymptotic formula he gives for an gives that the expected number of fixed points in an involution is n1/2 - 1/2 + o(1)
Once you know that fixed points are rare, then it's not hard to guess that their distribution should be approximately Poisson, and thus variance should be of the same order of magnitude as the mean -- and the variance result turns out to be true. (I don't know about the Poisson result.) The variance is, I believe, n1/2 - 1 + o(1), although this is only from numerical evidence. (The generating-function way to calculate the variance relies on the definition of the variance as the mean of the square minus the square of the mean; this means I need better asymptotics in order to verify this. The better asymptotics are certainly achievable, but they're not at my fingertips.)
The result is a bit surprising, though -- why does cutting out cycles of length 3 and greater so drastically change the relative numbers of 1-cycles and 2-cycles? But involutions make up a vanishingly small proportion of all permutations, and weird things can happen in these asymptotically negligible sets without the bulk of the population caring at all.