Expected value of determinant of simple infinite random matrix 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.
 A: 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.
A: Hadamard inequality , implies
$$
\mathbb{E}(|\mathrm{det}(A)|) \leq (pn)^{n/2} =e ^{\frac {n}2\log pn}.
$$
A: Very nice problem!
Let me recall you that the determinant of $n \times n$ matrices with entries in $\{0,1\}$  is related to the one of $n+1 \times n+1$ matrices with entries in $\{-1,+1\}$: replace the zeros  by $-1$'s and add a row of $-1$'s on the top and a column of ones on the right (you may want to read this arXiv). 
I can now tell you that, besides the trivial cases $p=0$ and $p=1,$ your problem is solved for $p=\frac{1}{2}$.
Indeed it was studied by T. Tao and V. Vu in arXiv. They proved that for the matrix $M_n$ of size $n \times n$ where the entries are i.i.d. Bernoulli random variables in $\{-1,+1\},$ the probability that $|\det(M_n)|$ is close to $\sqrt{n!}$ tends to one:
$$P \left(|\det (M_n)|\geq \sqrt{n!}\exp(-29n^{1/2}\ln^{1/2}n)\right)=1-o(1).$$
A: What follows is very heuristic [and updated according to first comment].
If we divide by $pn$, the matrix $B=\frac{1}{pn}A$ is, on average, a stochastic matrix. In the generic case, the Perron-Frobenius theorem says $B$ has one eigenvalue equal to $1$ and all the others have modulus smaller than $1$.
Therefore, it would be reasonable to expect there exists some $0<b(p)<1$ such that $\langle |{\rm det}(B)|\rangle\sim b(p)^n$. The function $b(p)$ is probably decreasing with $p$ since as $p\to 1$ the lines of $B$ become equal and the determinant must vanish.
This would give $\langle |{\rm det}(A)|\rangle\sim (pnb(p))^n=e^{n{\rm log}(pnb(p))}$.
A: I agree with user39115!
I will give a heuristic from random matrix theory because we know the global behaviour of the eigenvalue. First
$$A=p 1 +\sqrt{N(p-p^2)}\frac{B}{\sqrt{N}} $$ where $1$ is the matrix with only 1 entries and $$B_{i,j}=\begin{cases} \frac{1-p}{\sqrt{(p-p^2)}} &\text{with probability } p\\ \frac{-p}{\sqrt{(p-p^2)}} &\text{ with probability }1-p \end{cases} $$
$B$ is a random matrices independent entries with zero mean and variance 1 so we are exactly in set up of generalise Wigner Matrices https://terrytao.files.wordpress.com/2011/02/matrix-book.pdf. The eigenvalue of $B/\sqrt{N}$ converge to the Uniform law on the unit circle. $$\mu_n = \frac{1}{N}\sum_{\lambda\in \sigma(B/\sqrt{N})}\delta_\lambda \rightarrow \frac{1}{\pi}\mathcal{U}_{x^2+y^2\leq 1}$$
$A$ is just a rank one perturbation of $B$ so it does not change the global distribution of the eigenvalue of $B$ only add the larger eigenvalue $\lambda_1\sim pN$. 
$$\log(|det(A)|)\sim \log(pN)+N\log(\sqrt{N(p-p^2)})+\log(|det(B/\sqrt{N})|)$$
and we believe that $$\frac{1}{N}\log(|det(B/\sqrt{N})|=\frac{1}{N}\sum_{\lambda\in\sigma(B/\sqrt{N})} \log(|\lambda|)\sim \frac{1}{\pi}\int_{x^2+y^2\leq 1}\log(\sqrt{x^2+y^2})dxdy$$
and the last integral is equal to $-1/2$.Therefore we should get $$\log(|det(A)|)\sim N\log(\sqrt{\frac{N(p-p^2)}{e}}) $$
note that we recover $\sqrt{N!}$ of user39115.
So this is my heuristic (which rigourous proof is probably extremly hard.)
