TITLE: Non-enumerative proof that there are many derangements?
QUESTION [47 upvotes]: Recall that a derangement is a permutation $\pi: \{1,\ldots,n\} \to \{1,\ldots,n\}$ with no fixed points: $\pi(j) \neq j$ for all $j$.  A classical application of the inclusion-exclusion principle tells us that out of all the $n!$ permutations, a proportion $1/e + o(1)$ of them will be derangements.  Indeed, by computing moments (or factorial moments) or using generating function methods, one can establish the stronger result that the number of fixed points in a random permutation is asymptotically distributed according to a Poisson process of intensity 1.
In particular, we have:
Corollary: the proportion of permutations that are derangements is bounded away from zero in the limit $n \to \infty$.
My (somewhat vague) question is whether there is a "non-enumerative" proof of this corollary that does not rely so much on exact combinatorial formulae.  For instance, a proof using the Lovasz Local Lemma would qualify, although after playing with that lemma for a while I concluded that there was not quite enough independence in the problem to make that lemma useful for this problem.
Ideally, the non-enumerative proof should have a robust, "analytic" nature to it, so that it would be applicable to other situations in which one wants to lower bound the probability that a large number of weakly correlated, individually unlikely events do not happen (much in the spirit of the local lemma).  My original motivation, actually, was to find a non-enumerative proof of a strengthening of the above corollary, namely that given $l$ permutations $\pi_1,\ldots,\pi_l: \{1,\ldots,n\} \to \{1,\ldots,n\}$ chosen uniformly and independently at random, where $l$ is fixed and $n$ is large, the probability 
that these $l$ permutations form a $2l$-regular graph is bounded away from zero in the limit $n \to \infty$.  There is a standard argument (which I found in Bollobas's book) that establishes this fact by the moment method (basically, showing that the number of repeated edges or loops is distributed according to a Poisson process), but I consider this an enumerative proof as it requires a precise computation of the main term in the moment expansion.

REPLY [17 votes]: L. Lu and L. A. Szekely have successfully applied the Lopsided (i.e. Negative Dependency Graph) Lovasz Local Lemma to this problem.
A negative dependency graph is as a dependency graph except that independence is replaced with the inequality
$$
\Pr\left(A_k \middle\vert \bigwedge_{i \in S} A_i\right) \leq \Pr(A_k)
$$
for any fixed event $A_k$ and collection $S$ of non-neighbors of $A_k$. Erdos and Spencer showed the Lovasz Local Lemma holds with negative dependency graphs in place of dependency graphs.
Let $\Omega$ be the probability space of all perfect matchings of $K_{n,n}$ equipped with the uniform distribution. For a partial matching $M$ of $K_{n,n}$, define the event
$$
A_M = \{M^\prime \in \Omega \mid M \subseteq M^\prime\}
$$
(all perfect matching that contain the partial matching $M$).
Given a collection $\mathcal{M}$ of partial matchings of $K_{n,n}$, construct a graph with vertex set $\{A_M \mid M \in \mathcal{M}\}$ and set two matchings adjacent if their union is not again a matching. L. Lu and L. A. Szekely showed this graph is a negative dependency graph.
Finally we can address the problem at hand. Let the partite sets of $K_{n,n}$ be $\{1, \dots, n\}$ and $\{1^\prime, \dots, n^\prime\}$. For $1 \leq i \leq n$, let $M_i$ be the one-edge matching $ii^\prime$. Viewing perfect matchings of $K_{n,n}$ as permutations of an $n$-element set, the event $\bigwedge_{i=1}^n \overline{A_{M_i}}$ contains precisely those permutations not having a fixed point. Choosing $x_i = \frac{1}{n}$ for the purposes of the Lopsided Lovasz Local Lemma, we get
$$
\Pr\left( \bigwedge_{i=1}^n \overline{A_{M_i}} \right) \geq \left(1 - \frac{1}{n}\right)^n,
$$
which converges to $\frac{1}{e}$ as $n \rightarrow \infty$.