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Is there any implication that the probability that a random permutation is a derangment is $\frac{1}{e}$ when $n->\infty$ ?

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    A basic request: begin to accept some basic answers to the basic questions you asked when these basic answers basically satisfy you. – Did Jun 01 '13 at 10:13

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I assume that by derangement, you mean a permutation not fixing any element of the underlying set.

My probability professor in college offered the following interpretation: $(1-e^{-1})$ is the likelihood of a 'coincidence'.

Here's a practical example (I just realized this very example occurs on the wikipedia article). Your probability class has $n$ students where $n$ is very, very large, and so the professor decides to have students grade each other's homework assignments. A permutation $\sigma \in S_n$ is chosen uniformly; this permutation dictates that the assignment belonging to student $i$ is graded by student $\sigma(i)$. Thus a derangement $\sigma$ is a way of assigning students to papers so that nobody grades their own paper, and a 'coincidence' here is the event 'at least one person grades their own paper'.

For $n$ very large, the likelihood of this coincidence is $(1 - e^{-1})$.

A Blumenthal
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