$n$ players roll a die. For every pair rolling the same number, the group scores that number. Find the variance of the total score. This is problem 3.3.3.(b) in Probability and Random Processes by Grimmett and Stirzaker.

Here's my attempted solution: 
We introduce the random variables $\{X_{ij}\}$, denoting the scores of each pair (player $i$ and player $j$), and the total score $Y = \sum_{i<j}X_{ij}$. We calculate the expected value of $Y$: 
$$ 
\mathbb{E}(Y) = \sum_{i<j}\mathbb{E}(X_{ij}) = {n\choose{2}}\mathbb{E}(X_{12}) =  
\frac{7}{12}{n\choose{2}}.
$$
Now, let's determine the variance of $Y$:
$$
\mathrm{var}(Y) = \mathbb{E}(Y^2) - \mathbb{E}(Y)^2 
= \mathbb{E} \left\{ \left( \sum_{i<j}X_{ij} \right)^2 \right\} - \mathbb{E}(Y)^2
= \mathbb{E} \left( \sum_{i<j,\space k<l}X_{ij}X_{kl} \right) - \mathbb{E}(Y)^2 .
$$
Further, we look closer at the sum 
$\sum_{i<j,\space k<l}X_{ij}X_{kl}$.
Here $X_{ij}$ and $X_{kl}$ are independent whenever $i\neq k$, $i\neq l$, $j\neq k$ and $j\neq l$. 
When both $i = k$ and $j = l$ we get the random variables $\{X_{ij}^2\}_{i<j}$, each having expected value 
$$
\mathbb{E}(X_{ij}^2) = \mathbb{E}(X_{12}^2) = \sum_{m=1}^6\frac{m^2}{36}=\frac{91}{36}.
$$
 When only three of the four inequalities above hold we get a random variable on one of the following forms: $X_{ij}X_{jl}$, $X_{ij}X_{il}$, $X_{ij}X_{ki}$ or $X_{ij}X_{kj}$. These have expected value 
$$
\mathbb{E}(X_{ij}X_{il}) = \mathbb{E}(X_{12}X_{13}) 
= \sum_{m=1}^6\frac{m^2}{216}=\frac{91}{216}.
$$
We note that each triple $\{a,b,c\}$, such that $1\leq a < b < c \leq n$, is associated to the following six terms $X_{ab}X_{ac}$, $X_{ac}X_{ab}$, $X_{ab}X_{bc}$, $X_{bc}X_{ab}$, $X_{ac}X_{bc}$, $X_{bc}X_{ac}$, all with the above expected value. Clearly there are ${n\choose{3}}$ such triples.
This gives us the following:
$$
\mathrm{var}(Y) = 
\mathbb{E} \left( \sum_{i<j,\space k<l}X_{ij}X_{kl} \right) - \mathbb{E}(Y)^2
$$
$$
= {n\choose{2}}\mathbb{E}(X_{12}^2) + 
6{n\choose{3}}\mathbb{E}(X_{12}X_{13})+
\left\{{n\choose{2}}^2 - {n\choose{2}} - 6{n\choose{3}} \right\}\mathbb{E}(X_{12})^2 - \left\{{n\choose{2}}\mathbb{E}(X_{12})\right\}^2 
$$
$$
= \frac{35}{16}{n\choose{2}} + \frac{35}{72}{n\choose{3}}
$$

The book of solutions gives the answer $\frac{35}{16}{n\choose{2}} + \frac{35}{432}{n\choose{3}}$ though, which is what I would get if each triple $\{a,b,c\}$ only was associated with one term $X_{ab}X_{bc}$. So I guess I'm asking why(/if) that is the case!

Your solution appears to be correct. If $n=3$, there are $216$ equally likely outcomes of the dice. They're easy to enumerate, and the variance of the scores comes out to be $\frac{1015}{144}$. This agrees with your answer, not the book's. 
