If you throw a dice 5 times, what is the expected value of the square of the median?

Let $X_1,\ldots, X_n$ be i.i.d. r.v. If we order the previous list of r.v. by it values we get r.v. $X_{(1)},\ldots ,X_{(n)}$ named the ranks of the list $X_1,\ldots,X_n$. Now, observe that the distribution of the $k$-th rank of $n$ i.i.d. r.v. is given by

$$ \Pr [X_{(k)}\leqslant c]=\Pr \left[\bigcup_{r\geqslant k}(\{X_{(j)}\leqslant c, j\leqslant r\}\cap \{X_{(j)}>c,j\geqslant r+1\})\right]\\ =\Pr \left[\bigcup_{r\geqslant k}\bigcup_{A\in \mathcal{A}_r}(\{X_j\leqslant c, j\in A\}\cap \{X_j>c,j\in A^\complement \})\right]\\ =\sum_{r\geqslant k}\binom{n}{r}F_{X_1}(c)^r(1-F_{X_1}(c))^{n-r} $$

with the convention that $0^0:=1$ and where $\mathcal{A}_r:=\{B\subset \{1,\ldots,n\}: |B|=r\}$. Now observe that the median, $M$, of five i.i.d. r.v. is just $X_{(3)}$ and

$$ \mathrm{E}[M^2]=\sum_{k= 1}^6 k^2p_M(k)=\sum_{k=1}^{6}k^2(F_M(k)-F_M(k-1)) $$

Therefore, putting all together and using a CAS we get the result $\mathrm{E}[M^2]=\frac{2207}{162}\approx 13.62$.