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I'm working on a variant of the coat check problem (each person checks their coat, and at the end of the night each person is handed a random coat) in which each person has a friend who has an identical coat, so in total they are $2n$ people and $2$ coats that they consider 'their coat'.

It seems fairly straight forward, except for one thing - when I'm trying to calculate the variance, (given by $\sum\sum \mathrm{Var}(X_i,X_j)-\mathrm{Var}(X_i)\mathrm{Var}(X_j)$, where $i\neq j$ and $X$ is an indicator variable), I'm calculating $\mathrm{Var}(X_i,X_j)$ as $\frac{2}{2n} \cdot \frac{2}{2n-1}$... What i'm not sure of is if the $E(X_j)=\frac{2}{2n-1}$ is correct; since it does not account for the chance that $X_i$ is the other person with $X_j$'s coat.

Any thoughts?

Thanks!

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I'm assuming $X_i = 1$ if person $i$ gets a correct coat. If $i$ and $j$ are not friends, then $E[X_i X_j] = P(X_i=1) P(X_j = 1 | X_i = 1) = \frac{2}{2n} \frac{2}{2n-1}$ and then $\text{Cov}(X_i, X_j) = E[X_i X_j] - E[X_i] E[X_j]$. Don't forget to do separately the case where $i$ and $j$ are friends.

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    Indeed! Thanks for the help :)2012-11-03