Probability of a large random integer Matrix to have zero determinant Suppose we have a matrix $A \in \{0,1\}^{n \times n}$ where
$$A_{ij} = \begin{cases} 1 & \text{with probability} \quad p\\ 0 &\text{with probability} \quad 1-p\end{cases}$$ 
I would like to know the probability
$$\mathbb P( \det (A) =0) \ \text{ for large }n,$$
i.e., the asymptotic probability of the determinant being zero as $n$ becomes large.

I know that the probability 
$$\lim_{n \rightarrow \infty} \mathbb P \left( \det (A) =0 \right) = 0$$ if $p \neq 0,1$ but could not find results for finite $n$/the asymptotic behavior.
Edit: I changed the symbol $\mathbb{E}$ to $\mathbb{P}$ since I meant the probability and not the expectation value as some comments assumed correctly.
 A: A nice survey is given by Voigt and Ziegler (but they only address the $p=\frac12$ case explicitly).
Voigt, Thomas; Ziegler, Günter M., Singular 0/1-matrices, and the hyperplanes spanned by random 0/1-vectors, Comb. Probab. Comput. 15, No. 3, 463-471 (2006). ZBL1165.15018.
A: 
NOTE: This answer was posted in response to an earlier version of the
  question that was about the expected value and not the probability. So
  it has now become irrelevant. I decided not to delete it because in a
  comment there is some useful general result provided. If moderators
  think the answer should be deleted, no problem.


It appears I am missing something here, but as long as I cannot find what I am missing:  
"$\det$" is an operator on all the elements of a matrix that results in an expression that combines them by using addition, subtraction and multiplication. Then, due to the linearity of the expected value we have 
$$\mathbb E[A_{ij} + A_{k\ell}] = \mathbb E[A_{ij}] + \mathbb E[A_{k\ell}] \tag{1}$$
Moreover, in the OP's case as explained in a comment, the elements of the matrix are i.i.d. random variables. Due to the statistical independence assumption we have
$$\mathbb E[A_{ij} A_{k\ell}] = \mathbb E[A_{ij}] \mathbb E[A_{k\ell}] \tag{2}$$
But in such a case we have
$$\mathbb E[\det(A)] = \det(\mathbb E[A]) \tag{3}$$
$\mathbb E(A)$ is a matrix with all elements identical and equal to $p$ (due to the "identically distributed" assumption). 
Then $\det(\mathbb E[A]) = 0 = \mathbb E[\det(A)]$ for all $n$, and so
$$\Pr\{\mathbb E[\det(A)]=0\} = 1, n\geq 2$$
