Independent probability example I'd like some input for this question.
The question says 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).
Here's how I tackled the question:
Let x be a random number between 1 and 6. The sample space for y would be such that there are X elements and each element can be the numbers 1 through 6. Since each event in sample space has a (1/6) chance of being 2, the Pr(y = 2) would be $$(1/6)^x$$
Likewise for z, the Pr(z = 3) is $$(1/6)^y$$
So the Pr(x = 1 and y = 2 and z = 3) is $$(1/6)*(1/6)^x*(1/6)^y$$
 A: The individual die results are independent, but the random variables are clearly not.
To roll a value of $x$ on $1$ toss can be done in $1$ way of the $6$ possible rolls of the die.
To roll a minumum value of $y$ on $x$ tosses, count the ways to roll $x$ results of no less than $y$ minus the ways to roll $x$ results of no less than $y+1$, out of $6^x$ possible outcomes.
Similarly to roll a maximum value of $z$ on $y$ tosses, count the ways to roll $y$ results of no more than $z$ minus the ways to roll $y$ results of no more than $z-1$, out of $6^y$ possible outcomes.
$$\begin{align}
 \mathsf P(X=x) & = \frac 1 6
\\[2ex]
 \mathsf P(Y=y\mid X=x) & = \frac{(7-y)^x - (6-y)^x}{6^x}
\\[1ex]
 \mathsf P(Y=2\mid X=1) & = \frac{1}{6}
\\[3ex]
 \mathsf P(Z=z\mid Y=y) & = \frac{z^y-(z-1)^y}{6^y}
 \\[2ex]
\mathsf P(Z=3\mid Y=2) & = \frac{3^2-2^2}{6^2} 
\\[1ex]
 & = \frac{5}{6^2}
\\[3ex]
 \mathsf P(X=x, Y=y, Z=z)
 & = \mathsf P(X=1)\;\mathsf P(Y=2\mid X=1)\;\mathsf P(Z=3\mid Y=2)
\\[1ex]
 & = \frac{5}{1296}
\end{align}$$
