Parity of determinant of integer matrix 
The elements of a determinant are arbitrary integers. Determine the probability that the value of the determinant is odd.


I tried to compute the probability of the determinant for $2\times2$ matrix as follows. Let
$$A := \begin{bmatrix}a&b\\c&d\\\end{bmatrix}$$
whose determinant is $ad-bc$. From this I found the probability is $37.5\%$
Any hint or solution will be appreciated.
 A: Notice that you can actually reduce the problem mod $2$. Now the probability distribution is well-defined as every element of your matrix has probability $\frac12$ to be congruent to $0$ or $1$ mod $2$. Determinant of a matrix is odd if and only if it is $\,\equiv 1 \pmod{2}$ so we are actually interested in counting $n \times n$ invertible matrices with coefficients in $\mathbb{F}_2$, i.e. the size of $GL_n(\Bbb{F}_2)$.
Invertible matrices are precisely those with linearly independent rows so we construct such a matrix row by row. The first row can be anything but all zeroes, so there are $2^n-1$ possibilities. The second row cannot be all zeroes or equal to the first so there are $2^n-2$ possibilities for it. Third row cannot be a linear combination of the first two rows and there are four such combinations:
$$0 \times \text{first} + 0 \times \text{second},\quad 1 \times \text{first} + 0 \times \text{second}, \quad0 \times \text{first} + 1 \times \text{second}, \quad1 \times \text{first} + 1 \times \text{second}$$
so there are $2^n-4$ possibilities for the third row. In general, for the $i$-th row there are $2^n-2^{i-1}$ possibilities so we conclude
$$|GL_n(\Bbb{F}_2)| = \prod_{i=0}^{n-1} (2^n-2^i).$$
The probability that a random integer matrix has odd determinant is therefore
$$\frac{|GL_n(\Bbb{F}_2)|}{|M_n(\Bbb{F}_2)|} = \frac1{2^{n^2}}\prod_{i=0}^{n-1} (2^n-2^i) = \prod_{i=1}^n \left(1-\frac1{2^i}\right).$$
