What is the probability of two permutations on set X of size m (i.e. X=m) having at least n points of intersection? By this I mean that if two permutations, which I'll call g(x) and h(x), map a member x to g(x) for the first permutation and x to h(x) for the second permutation, what is the chance that g(x)=h(x) for at least n values of x?

You can use inclusionexclusion to show that the number of permutations in $S_m$ having at least $n$ fixed points is $$\sum_{k=n}^m (1)^{kn}{k1\choose n1}{m\choose k}(mk)! = m! \sum_{k=n}^m (1)^{kn}\frac{1}{k!}{k1\choose n1}$$ We describe now how to obtain the lefthand expression. The ${m\choose k}$ comes from choosing $k$ fixed points, the $(mk)!$ counts permutations in $S_m$ having these $k$ fixed points, and then $(1)^{kn}{k1\choose n1}$ is an inclusionexclusion counting coefficient, namely the Möbius function $\mu (\hat{0},\hat{1})$ on the subposet of the Boolean algebra of subsets of $\{ 1,\dots ,k \} $ where we exclude the subsets having size $1\le i \le n1$. One way to calculate this Möbius function is to use that each rankselection of the Boolean algebra is lexicographically shellable. The desired Möbius function will be $(1)^{kn}$ multiplied by the number of socalled ``descending chains'' in the lexicographic shelling, which in this case is the number of permutations in $S_k$ that are ascending in the first $n$ letters and then descending after that, which in particular forces the letter $k$ to be the $n$th letter in the permutation (in oneline notation). This includes the wellknown special case (usually phrased in terms of derangements) that the number of permutations in $S_m$ with at least one fixed point is $\sum_{k = 1}^m (1)^{k1} {m\choose k}(mk)! $ which equals $  (m! + \sum_{k= 0}^m(1)^k {m\choose k}(mk)!) $. As $m$ goes to infinity, this approaches $ m!(1 + 1/e) = m!(11/e)$. A good reference for the $n=1$ case is chapter 2 of Enumerative Combinatorics, Volume 1, by Richard Stanley. The original source for lexicographic shellability is Anders Björner's paper ``Shellable and CohenMacaulay partially ordered sets''. Added later: the comments above mention the recontres numbers. These give an approach for obtaining the $n>1$ case as a consequence of the $n=1$ case  by choosing your fixed point set and then counting derangements on the remaining letters, summing over the possible fixed point sets. This results in a double sum, with an alternating sum as the inner sum. 

