Combinatorics question with variable number of possibilities

Question

The problem goes as follows: you roll a die with 6 sides, and whatever number comes up, the next die has as many sides. You then roll a second die and the third one has the number of sides that came up on the second die. You finally roll the third die. Calculate the possibility of the situation when the sum of the numbers equals 12. I tried to calculate all the possible outcomes by trying to come with an equation but had no luck. I am stuck at the beginning of the problem.

Answer 1

You can start by reducing a little, if you realize that the first roll needs to be 4 or more. In fact if you roll a 4, then you must roll three fours. The odds of that is $$\frac{1}{6} \cdot \frac{1}{4} \cdot \frac{1}{4}.$$ We only have two more cases to consider, that of a first roll of 5, and that of a first roll of 6.

We want to count triplets which are non-increasing and sum to 12. For the five case we have $(5,5,2),(5,4,3)$ and for the six case $(6,5,1),(6,4,2),(6,3,3)$.

The odds of each triple are $\frac{1}{6}$ times the reciprocal of the first number of the triple, times the reciprocal of the second number of the triple. We then get: $$\frac{1}{6} \cdot \frac{1}{4} \cdot \frac{1}{4} +\frac{1}{6} \cdot \frac{1}{5} \cdot \frac{1}{5} +\frac{1}{6} \cdot \frac{1}{5} \cdot \frac{1}{4} +$$ $$\frac{1}{6} \cdot \frac{1}{6} \cdot \frac{1}{5} +\frac{1}{6} \cdot \frac{1}{6} \cdot \frac{1}{4} +\frac{1}{6} \cdot \frac{1}{6} \cdot \frac{1}{3}= $$ $$\frac{1}{6 \cdot 4 \cdot 4} +\frac{1}{6 \cdot 5 \cdot 5} +\frac{1}{6 \cdot 5 \cdot 4} +\frac{1}{6 \cdot 6 \cdot 5} +\frac{1}{6 \cdot 6 \cdot 4} +\frac{1}{6 \cdot 6 \cdot 3} = \frac{1019}{21600}.$$