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?
Random NxN Matrix – Expected Determinant
determinantsmatricesrandom matrices
Related Solutions
I have given some detail in a comment to another answer. I have a proof that the number of determinants is greater than 4 times the nth Fibonacci number for (n+1)x(n+1) (0,1) matrices, and I conjecture that for large n the number of distinct determinants approaches a constant times n^(n/2). Math Overflow has some hints of the proof in answers I made on other questions.
I am interested in your idea for an application, and am willing to share more information on this subject.
Gerhard "Ask Me About System Design" Paseman, 2010.03.19
Here is a solution to Hipstpaka's question about $\det(A)^2$. I don't have enough space to answer in a comment. I don't know where a statement appears in the literature, but the proof uses a standard technique discussed for instance in Enumerative Combinatorics, vol. 2, Exercise 5.64.
Write $\varepsilon_w$ for the sign of the permutation $w\in S_n$. Then \begin{eqnarray*} \sum_A \det(A)^2 & = & \sum_{i,j=1}^n\sum_{a_{ij}=0,1} \left(\sum_{w\in S_n} \varepsilon_w a_{1,w(1)}\cdots a_{n,w(n)}\right)^2\\ & = & \sum_{i,j=1}^n\sum_{a_{ij}=0,1} \sum_{u,v\in S_n} \varepsilon_u\varepsilon_v a_{1,u(1)}\cdots a_{n,u(n)} a_{1,v(1)}\cdots a_{n,v(n)}. \end{eqnarray*} Let $f(u,v)$ be the number of distinct variables occurring among $a_{1,u(1)},\dots, a_{n,u(n)},a_{1,v(1)},\dots, a_{n,v(n)}$. The product $a_{1,u(1)}\cdots a_{n,u(n)}a_{1,v(1)}\cdots a_{n,v(n)}$ is 1 with probability $p^{f(u,v)}$ and is otherwise 0. Moreover, $f(u,v)= 2n-\mathrm{fix}(uv^{-1})$, where $\mathrm{fix}(uv^{-1})$ denotes the number of fixed points of $uv^{-1}$. If $E$ denotes expectation, then we get $$ E(\det(A)^2) = \sum_{u,v\in S_n}\varepsilon_u\varepsilon_v p^{2n-\mathrm{fix}(uv^{-1})}. $$ Setting $w=uv^{-1}$ and noting that $\varepsilon_u\varepsilon_{wu^{-1}} = \varepsilon_w$, we get \begin{eqnarray*} E(\det(A)^2) & = & \sum_{w\in S_n} p^{2n-\mathrm{fix(w)}} \sum_u\varepsilon_u\varepsilon_{wu^{-1}}\\ & = & n!p^{2n}\sum_{w\in S_n}p^{-\mathrm{fix(w)}}\varepsilon_w. \end{eqnarray*} Let $g(n)=\sum_{w\in S_n}p^{-\mathrm{fix(w)}}\varepsilon_w$. By standard generating function techniques (Enumerative Combinatorics, vol. 2, Section 5.2) we have \begin{eqnarray*} \sum_{n\geq 0} g(n)\frac{x^n}{n!} & = & \exp\left( \frac 1p x-\frac{x^2}{2}+\frac{x^3}{3} -\frac{x^4}{4}+\cdots\right)\\ & = & (1+x)\exp \left( \frac 1p-1\right)x. \end{eqnarray*} It is now routine to extract the coefficient of $x^n$ and then compute $$ E(\det(A)^2)= n!\,p^n(p-1)^{n-1}(1+(n-1)p). $$
In general we don't have $E(\det(A)^2)=E(|\det(A)|)^2$, so this does not directly answer the original question.
For a related question (and answer) see Expected determinant of a random NxN matrix.
Best Answer
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 $$\sqrt{(n!)^2/2^{2n} \left(1/n! + 1/(n-1)! \right)} = \sqrt{(n+1)!}/ 2^n$$