Probability of selecting a $4\times4$ matrix (having elements $1$ & $-1$) whose each $2\times2$ submatrix contains two $1$s and two $-1$s This is a problem from a test I have taken

Suppose set $X$ contains all $4\times4$ matrices that can be formed using elements $1$ and $-1$. Find the probability of selecting an element of set $X$ whose each $2\times2$ submatrix contains two $1$s and two $-1$s.

I have no idea where to start on this problem. There is the obvious case of alternating $1$s and $-1$s but beyond that, I am not sure how to find the other matrices. Any help would be great.

There are a total of $2^{16}$ matrices. So, to compute the requested probability we need to count the number of matrices whose each $2\times 2$ submatrix contains two 1s and two -1s. Let's try to do that.
A $4\times 4$ matrix contains nine $2\times 2$ submatrices. Consider the one in the middle (call it $B$).
If $B=[1, 1; -1, -1]$, then you can build at most 4 matrices with the desired property, and the same holds if $B=[-1, -1; 1, 1]$. To see this, note that the mid-north and midh south matrices are uniquely determined by $B$ and the first and fourth column can onlytake two values each.
If $B=[-1, 1; -1, 1]$, there are still 4 matrices with the desired property (by symmetry with respect to the argument above), and the same holds if $B=[1, -1; 1, -1]$.
If $B=[1, -1; -1, 1]$, then you can build at most 9 matrices with the desired property, and the same holds if $B=[-1, 1; 1, -1]$. To see this, assume that the $B_{2,2}=1$, and see that there at most 3 possibilities for the south-east submatrix and three possibilities for the north-west submatrix, with the remaining part of the matrix uniquely determined.
To sum up, we count 4+4+4+4+9+9 matrices with the desired property and the probability you look for is $\frac{34}{2^{16}}$
