matrices - Expected determinant of a random NxN matrix.

As everyone above has pointed out, the expected value is $0$.

I expect that the original poster might have wanted to know about how big the determinant is. A good way to approach this is to compute $sqrt{E((det A)^2)}$, so there will be no cancellation.

Now, $(det A)^2$ is the sum over all pairs $v$ and $w$ of permutations in $S_n$ of

$$(-1)^{ell(v) + ell(w)} (1/2)^{2n-# { i : v(i) = w(i) }}$$

Group together pairs $(v,w)$ according to $u := w^{-1} v$. We want to compute

$$(n!) sum_{u in S_n} (-1)^{ell(u)} (1/2)^{2n-# (mbox{Fixed points of }i)}$$

This is $(n!)^2/2^{2n}$ times the coefficient of $x^n$ in

$$e^{2x-x^2/2+x^3/3 - x^4/4 + cdots} = e^x (1+x).$$

So $sqrt{E((det A)^2)}$ is

$$sqrt{(n!)^2/2^{2n} left(1/n! + 1/(n-1)! right)} = sqrt{(n+1)!}/ 2^n$$

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