Suppose we have a matrix $A \in \mathbb{R}^{n\times n}$ where
$$A_{ij} = \begin{cases} 1 & \text{with probability} \quad p\\ 0 &\text{with probability} \quad1-p\end{cases}$$
I would like to know the following expected value
$$\lim_{n \rightarrow \infty} \mathbb E(| \det (A) |)$$
i.e., the asymptotic behavior as $n$ becomes large.
What I tried so far
It feels like this has been done already, but after searching for quite a while I performed simulations and it looks like
$$\mathbb E(|\det(A)|) \propto e^{n f(p) }$$
where $f$ is some function of the probability $p$.
$p$." />I would be very happy if someone knows the result or a good reference where I could look it up.
Edit: I changed the figure and added plots of the solutions
$$ \text{log} \mathbb E(|\det(A)|) \propto \frac{n}{2} \text{log} \frac{n p (1-p)}{e}$$
given by Richard Stanley and RaphaelB4 which coincide up to multiplicative terms if the factorial in Richard Stanleys solution is replaced by stirlings formula.
Best Answer
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.