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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!

user120005
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    Looks right. I chose to let $Y_{i,j}=1$ if players $i$ and $j$ throw the same number, and let $Y_{i,j}=0$ otherwise. Then the score is the sum of the $Y_{i,j}$ over all choices of two distinct $i,j$. We have $\Pr(Y_{i,j}=1)=\frac{1}{6}$ so the mean is $\binom{n}{2}\cdot\frac{1}{6}$. – André Nicolas Jan 19 '15 at 04:02
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    If I’m not mistaken, André’s random variables are independent, and you can calculate the variance their sum. – Brian M. Scott Jan 19 '15 at 19:10
  • Yes it works! I could show they're independent. Thanks! – user120005 Jan 20 '15 at 03:09

1 Answers1

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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 $\binom n2\cdot\frac5{36}=\frac{5n(n-1)}{72}$.

joriki
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