What is the expected value of the determinant over the uniform distribution of all possible 1-0 NxN matrices? What does this expected value tend to as the matrix size N approaches infinity?



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
$$(n!) \sum\_{u \in S\_n} (-1)^{\ell(u)} (1/2)^{2n-\# (\mbox{Fixed points of }u)}$$


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


