Here the question is:
Each member of a group on $n$ players rolls a die:
Find the mean and variance of the total score ($S_n$) if any players who threw the same number scores that number.
The solution is as follows:
Let $X_{ij}$ be the common score of players $i$ and $j,$ so that $X_{ij}=0$ if there scores are different. The first part I follow, the expected total score is:
$$E[S_n] =\binom{n}{2}E(X_{12})=\binom{n}{2}\frac{1}{6}\frac{7}{2}$$
I don't see how they get the expression for the variance:
\begin{align} \operatorname{var}(S) &= E \left \{\left(\sum_{i<j}X_{ij}\right)^2 \right \}-E(S)^2 \\
& = \binom{n}{2}E(X_{12}^2) + \binom{n}{3}E(X_{12}X_{23})+\left\{ \binom{n}{2}^2-\binom{n}{2} – \binom{n}{3}\right \}E(X_{12})^2-(\frac{7}{12})^2 (\binom{n}{2})^2 \end{align}
I would only expect to see the first and last terms in the last line, where do there other terms come from?
Best Answer
When you expand out $(\sum_{i<j} X_{i,j})^2=(X_{12}+X_{13}+X_{23}+X_{14}+\dots)^2$, there are three types of terms which appear:
However, I think the solution has a mistake in the second bullet. The three numbers $i,j,k$ are not unordered; the $j$ is special because it is repeated, and $i,k$ are distinct since $i$ comes from the first factor. Indeed, when $n=3$, the expansion $(X_{12}+X_{13}+X_{23})^2$ has six factors of the form $X_{ij}X_{jk}$, not one. Therefore, you should replace $\color{red}{\binom{n}3}$ with $n(n-1)(n-2)$.