Consider the outcome space
\begin{equation*}
S = {(1,1), (1,2), (1,3), (1,4), (1,4), (1,5), (1, 5), \\
(1,6), (1,6), (1,7), (1,8), (1,9), (2,1) ... (3,9)}
\end{equation*}
Then P(5) = |(1,4), (1,4), (2,3), (3,2) |/36 = 4/36. Compare this to the exercise with regular dice performed above. Similarly, compute the remaining probabilities.