Mean value of the determinant of a $2n \times 2n$ skew-symmetric matrix with random entries 
Let $A$ be a $2n \times 2n$ matrix with entries chosen independently at random. Each entry is chosen to be $0$ or $1$, each with probability $1/2$. Find the expected value of $\det(A - A^T)$ as a function of $n$.

Appeared in a class that prepares students for the Putnam exam, but we did not get to it.

Let $B = A - A^T$, we have
$$\verb/E/[\det B] = \sum_{\sigma \in S_{2n}}(-1)^{\sigma}\verb/E/\left[\prod_{k=1}^{2n} B_{k\sigma(k)}\right]$$
where the sum on RHS is over $S_{2n}$, the collection of permutation $\sigma$ over $2n$ symbols. 
By definition, the diagonal entries of $B$ are zero. Since the entries of $A$ are i.i.d., any two off-diagonal entries of $B$,
$B_{ij}$ and $B_{k\ell}$, will be independent unless $(i,j) = (k,\ell)$ or $(\ell,k)$.
For any $\sigma \in S_{2n}$ such that $\sigma(\sigma(1)) \ne 1$, $B_{1\sigma(1)}$ will be independent of all entries appear in $\prod\limits_{k=2}^{2n} B_{k\sigma(k)}$. This implies
$$
\verb/E/\left[\prod_{k=1}^{2n} B_{k\sigma(k)}\right]
= \verb/E/[B_{1\sigma(1)}]\cdot\verb/E/\left[\prod_{k=2}^{2n} B_{k\sigma(k)}\right]
= 0\cdot \verb/E/\left[\prod_{k=2}^{2n} B_{k\sigma(k)}\right] = 0
$$
Apply similar argument to other $B_{k\sigma(k)}$ in the product $\prod\limits_{k=1}^{2n} B_{k\sigma(k)}$, we find the expectation value of the product is zero unless $\sigma$ is a product of $n$ transpositions.
There are $(2n-1)!!$ such permutations. It is easy to see all of them contribute same value to the determinant. One of such permutation  $\sigma_0 = (12)(34)(56)\cdots(2n-1,n)$. We have
$$\verb/E/[B] = (-1)^{\sigma_0} (2n-1)!!\;\verb/E/\left[\prod_{k=1}^{2n}B_{k\sigma_0(k)}\right]
= (-1)^n (2n-1)!!\; \verb/E/[B_{12}B_{21}]^n$$
Since 
$$\begin{align}
\verb/E/[B_{12}B_{21}] &= \verb/E/[(A_{12}-A_{21})(A_{21}-A_{12})]
= 2\left(\verb/E/[A_{12}]^2 - \verb/E/[A_{12}^2]\right)\\
&= 2\left(\frac1{2^2} - \frac12\right) = -\frac12\end{align}$$
We obtain 
$$\verb/E/[B] = \frac{(2n-1)!!}{2^n} = \frac{(2n)!}{4^nn!}$$
In particular, for $n = 2$ and $3$, this reduces to $\verb/E/[B] = \frac34$ and $\frac{15}{8}$, reproducing  the result in another answer.
