Q:

Probability of rolling a die

You have a fair die and do the following experiment:
    - Roll the die once; let x be the outcome.
    - Roll the die x times (independently); let y be the smallest outcome of these x rolls.
    - Roll the die y times (independently); let z be the largest outcome of these y rolls.
Determine:
     Pr(x = 1 and y = 2 and z = 3):

I've been trying to solve it for hours and I can't find a solution. Maybe I'm overthinking it and it's actually fairly simple.
I would like to get some hints on how to approach/solve this problem.
Thanks!

A:

Think about the probability of getting a particular value for each variable.
For $x=1$ we need to roll a 1 with a single roll. So that's $\frac{1}{6}$.
For $y=2$ we need to roll a 2 with $x$ dice rolls. So we need to get a 2 with 1 roll. That's $\frac{1}{6}$.
For $z=3$ we need to roll a 3 with one of the rolls, and then something smaller than or equal to a 3 on the next roll. (We only have 2 rolls since $y=2$.) Case 1: The first roll is a 3, and the second roll is either 1, 2, or 3. The probability of this happening is $\frac{1}{6} \times \frac{1}{2}$. Case 2: The first roll is either 1 or 2; and the second roll is a 3. (If the first roll is 3 this is covered in the first case.) The probability of this happening is $\frac{1}{3} \times \frac{1}{6}$. Since either can happen the probability of $z=3$ is 
$(\frac{1}{6} \times \frac{1}{2}) + (\frac{1}{3} \times \frac{1}{6}) = \frac{5}{36}$.
So: P(x=1, y=2, z=3) = $\frac{1}{6} \times \frac{1}{6} \times \frac{5}{36} = \frac{5}{1296}$.

