Each member of a group and $n$ players roll a fair die. For every pair of players who throw the same number, the group scores 1 point. Find the mean and variance of the total score of the group.
Here is what I tried:
Let $S_i$ denote the score obtained by the players who throw $i, i=1,2,\ldots,6$, and let $X_i$ be the number of people who throw $i$, and $1_{ij}$ be the indicator function which is 1 only if the $j^{th}$ person throws $i$. So, we have $X_i=\sum_{j=1}^n 1_{ij}$. I computed $E[S_i|X_i]=\frac{X_i(X_i-1)}{2}$ and so $E[S_i]=\frac{n^2-n}{72}$ and therefore $E[S]=\frac{n^2-n}{12}$, where $S=\sum_{i=1}^6 S_i$ is the total score.
In the same manner, by using variance-covariance expansion and conditional variance, I tried to compute $\mathrm{var}(S)$, but at some point I got $\mathrm{var}(\sum_{j<k}1_{ij}1_{ik})$ and I can't compute this variance because of dependence of summands. Also, I'm not sure if my solution so far is correct.
Appreciate any help!
Best Answer
Community wiki answer so the question can be marked as answered:
As pointed out in the comments, the indicator variables $Y_{ij}$ that are $1$ if $i$ and $j$ throw the same number are pairwise independent and can thus be used to find both the expectation and the variance. The expectation of $Y_{ij}$ is $\frac16$ and the variance is $\frac16\cdot\left(1-\frac16\right)=\frac5{36}$, so the expectation of the score is $\binom n2\cdot\frac16=\frac{n(n-1)}{12}$ and the variance of the score is $\binom n2\cdot\frac5{36}=\frac{5n(n-1)}{72}$.