TITLE: Number of invertible {0,1} real matrices?
QUESTION [17 upvotes]: This question is inspired from here, where it was asked what possible determinants an $n \times n$ matrix with entries in {0,1} can have over $\mathbb{R}$.  
My question is: how many such matrices have non-zero determinant?  
If we instead view the matrix as over $\mathbb{F}_2$ instead of $\mathbb{R}$, then the answer is 
$(2^n-1)(2^n-2)(2^n-2^2) \dots (2^n-2^{n-1}).$
This formula generalizes to all finite fields $\mathbb{F}_q$, which leads us to the more general question of how many $n \times n$ matrices with entries in { $0, \dots, q-1$ } have non-zero determinant over $\mathbb{R}$?

REPLY [2 votes]: Lurking around MO, I found a question which is related to the second part of my question.  Namely, Greg Martin and Erick B. Wong prove that assuming that the entries of an $n \times n$ matrix are chosen randomly with respect to a uniform distribution from the set $\{-k,  \dots, k\}$, then the probability that the resulting matrix will be singular is $\ll k^{-2 + \epsilon}$.  
See this MO question (where the above paragraph is plagarized from) and also here for the link to the Martin & Wong paper.