I cant see a clever method to do this, so i'd do it the old fashioned way.

there are only 18 possible outcomes, each with probability 1/18.

As an example, from theCode:Chosen | X | Prob 1,1,1 | 2 | 1/18 1,1,2 | 4 | 1/18 1,1,3 | 6 | 1/18 2,1,1 | 2 | 1/18 ..... etcincompletetable above, we have:

P(X=2) = 2/18

(it appears twice, each with probability 1/18)

When you do the full table it will appear more times.

Use the same method to work out the chance of X taking all possible values.