Problem source

\begin{questionparts}
    \item In a game, each member of a team of $n$ players rolls a fair six-sided die.
 
    The total score of the team is the number of pairs of players rolling the same number. For example, if $7$ players roll $3, 3, 3, 3, 6, 6, 2$ the total score is $7$, as six different pairs of players both score $3$ and one pair of players both score $6$.
 
    Let $X_{ij}$, for $1 \leqslant i < j \leqslant n$, be the random variable that takes the value $1$ if players $i$ and $j$ roll the same number and the value $0$ otherwise.
 
    Show that $X_{12}$ is independent of $X_{23}$.
 
    Hence find the mean and variance of the team's total score.
 
    \item Show that, if $Y_i$, for $1 \leqslant i \leqslant m$, are random variables with mean zero, then
    \[
        \mathrm{Var}(Y_1 + Y_2 + \cdots + Y_m) = \sum_{i=1}^{m} \mathrm{E}(Y_i^2) + 2\sum_{i=1}^{m-1}\sum_{j=i+1}^{m} \mathrm{E}(Y_i Y_j).
    \]
 
    \item In a different game, each member of a team of $n$ players rolls a fair six-sided die.
 
    The total score of the team is the number of pairs of players rolling the same even number minus the number of pairs of players rolling the same odd number. For example, if $7$ players roll $3, 3, 3, 3, 6, 6, 2$ the total score is $-5$.
 
    Let $Z_{ij}$, for $1 \leqslant i < j \leqslant n$, be the random variable that takes the value $1$ if players $i$ and $j$ roll the same even number, the value $-1$ if players $i$ and $j$ roll the same odd number and the value $0$ otherwise.
 
    Show that $Z_{12}$ is not independent of $Z_{23}$.
 
    Find the mean of the team's total score and show that the variance of the team's total score is $\dfrac{1}{36}n(n^2 - 1)$.
\end{questionparts}

Solution source

\begin{questionparts}
\item First note that $\mathbb{P}(X_{ij} = 1) = \frac16$ since it doesn't matter what $i$ rolls, it only matters that $j$ rolls the same thing, which happens $1/6$ of the time.
 \begin{align*}
&& \mathbb{P}(X_{12} = 1, X_{23} = 1) &= \mathbb{P}(1, 2\text{ and }3\text{ all roll the same})\\
&&&= \frac{6}{6^3}= \frac1{6^2} \\
&&&= \mathbb{P}(X_{12} = 1)\mathbb{P}(X_{23} = 1) \\
&& \mathbb{P}(X_{12} = 1, X_{23} = 0) &=  \mathbb{P}(1, 2\text{ roll the same and }3\text{ rolls different}) \\
&&&= \frac{6 \cdot 1 \cdot 5}{6^3} = \frac{5}{6^2} \\
&&&= \mathbb{P}(X_{12} = 1)\mathbb{P}(X_{23} = 0) \\
&& \mathbb{P}(X_{12} = 0, X_{23} = 0) &= \mathbb{P}(2, 3 \text{ roll different to} 2)\\
&&&= \frac{6 \cdot 5 \cdot 5}{6^3}= \frac{5^2}{6^2} \\
&&&= \mathbb{P}(X_{12} = 0)\mathbb{P}(X_{23} = 0) 
\end{align*}
Therefore they are independent (the final case is clear by symmetry from case 2).

Note that the score is $S = \sum_{i \neq j} X_{ij}$ so

\begin{align*}
&& \E[S] &= \E \left [ \sum_{i \neq j} X_{ij} \right] \\
&&&= \sum_{i \neq j} \E \left [ X_{ij} \right] \\
&&&=  \sum_{i \neq j} \frac16 \\
&&&= \binom{n}{2} \frac16 = \frac{n(n-1)}{12} \\
\\
&& \var[S] &= \var \left [ \sum_{i \neq j} X_{ij} \right] \\
&&&  \sum_{i \neq j} \var \left [X_{ij} \right]  \tag{pairwise ind.} \\
&&&= \binom{n}{2} \frac{5}{36} = \frac{5n(n-1)}{72}
\end{align*}

\item Note that $\mathbb{P}(Z_{ij} = 1)=\mathbb{P}(Z_{ij} = -1) = \frac{3}{6^2} = \frac{1}{12}$ but that $\mathbb{P}(Z_{12} = 1, Z_{23} = -1) = 0$.

Notice that $Z_{12}Z_{23}$ is either $1$ or $0$ (since $2$ can't be both odd and even). $\mathbb{P}(Z_{12}Z_{23} = 1) = \frac{1}{36}$. Notice that $Z_{ij}, Z_{kl}$ are independent if $i \neq j \neq k \neq l$ and so

\begin{align*}
&& \E[T] &= \E \left [ \sum_{i \neq j} Z_{ij} \right] \\
&&&= \sum_{i \neq j}\E \left [  Z_{ij} \right] \\
&&&= 0 \\
\\
&& \E[T^2] &= \E \left [ \left ( \sum_{i \neq j} Z_{ij} \right)^2 \right] \\
&&&= \E \left [ \sum_{i \neq j} Z_{ij}^2 + \sum_{i \neq j \neq k} Z_{ij}Z_{jk} + \sum_{i \neq j \neq k \neq l} Z_{ij}Z_{kl}\right] \\
&&&= \binom{n}{2} \frac{1}{6} + 2\frac{n(n-1)(n-2)}{2} \frac{1}{36} + 0 \\
&&&= \frac{n(n-1)}{12} + \frac{n(n-1)(n-2)}{6} \\
&&&= \frac{n(n-1)[3 + (n-2)]}{36} \\
&&&= \frac{n(n^2-1)}{36}
\end{align*}

\end{questionparts}