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Let $(\Omega, P)$ be a laplacian probability space with $\Omega = \{1, 2, 3, 4, 5, 6\}^3$, and let $X: \Omega \rightarrow \Bbb Z$, $(i, j, k) \rightarrow i - j + k$ be a random variable. Specify $P_x$ with $P_x(F) := P(\{X \in F\})$ for $F \in \mathscr P(\Omega)$.

I already calculated the following things:

There are $6^3 = 216$ possible combinations of the $(i, j, k) \in \Omega$, and the possible results range from $-4$ to $11$. In order to specify $P_x$, I have to find the number of all possible $(i, j, k) \in \Omega$ such that $i - j + k = l$ for $l \in \{-4, ..., 11\}$. I have written it down for the numbers $11, 10, 9, 8$ and $7$. I calculated that there is one combination of $(i, j, k)$ such that $i - j + k = 11$, three combinations for $10$, six combinations for $9$, ten combinations for $8$ and fifteen combinations for $7$.

I guess that there is a much more easy way to calculate all of these possible combinations, but I wasn't able to see a pattern yet. Plus, what would be a good way to define $P_x$? I thought about something like splitting it up in the sets $\{-4, ..., -1\}$, $\{0, ..., 6\}$ and $\{7, ..., 11\}$.

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Generating functions work nicely here.

Let $p=\frac{1}{6}\left(x+x^2+x^3+x^4+x^5+x^6\right)$.

Let $q=\frac{1}{6}\left(x^{-1}+x^{-2}+x^{-3}+x^{-4}+x^{-5}+x^{-6}\right)=\frac{p}{x^7}$.

Then the distribution you seek is given by $p^2q = \frac{p^3}{x^7}$: \begin{align*} p^2q =&\frac{1}{216}\left(x^{11} + 3 x^{10} + 6 x^{9} + 10 x^{8} + 15 x^{7} + 21 x^{6} + 25 x^5 \\ + 27 x^4 + 27 x^3 + 25 x^2 + 21 x^1 + 15 x^0 + 10 x^{-1} + 6 x^{-2} + 3 x^{-3} + x^{-4}\right) \end{align*} Thus, for example, $P_9=\frac{6}{216}$ and $P_5=\frac{25}{216}$.

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