How to go about counting the sample set of rolling dice based off previous rolls. I'm in need of some help. I can't seem to wrap my mind around this question. 
Here's the task:
I have this problem. It goes like this:
I am to do an experiment with rolling a dice. There are 3 steps:
1) You roll the dice once. You assign the value rolled to 'x'.
2) You then roll the dice 'x' times (independently). You assign smallest value of the 'x' rolls to 'y'. 
3) You roll the dice 'y' times (independently). You assign the largest value of 'y' rolls to 'z'.
Based on this, I'm suppose to find the probability of certain events (such as: Pr(x=1 and y= 5 and z=1). However,  I'm having trouble of first pinning down the size of the sample space (number of different possible outcomes). Any hints to guide me along the way?
 A: You might not need to count the number of possible outcomes, but instead use conditional probabilities.
The distribution of X is $P(X = x) = \frac16$ for $x=1$, $2 \dots$
You can find the distribution of Y given a particular value of $X$, by thinking carefully about what it means for the minimum of all of the rolls to be a particular number. The reasoning goes like:
$P(Y \geq 1 | x) = P( \text{all rolls are at least 1} | x) = 1$.
$P(Y \geq 2 | x) = P(\text{all rolls are at least 2} |x) = (\frac56)^x$.
$P(Y \geq 3 | x) = P(\text{all rolls are at least 3} | x) = (\frac46)^x$.
and so on...
Thus we get the following:
$P(Y = 1 | x) = P(Y \geq 1|x) - P(Y\geq 2|x) = 1 - (\frac56)^x$
$P(Y = 2 | x) = P(Y \geq 2 |x) - P(Y \geq 3|x) = (\frac56)^x - (\frac46)^x$.
and so on...
You can find the distribution of Z, given particular values of x and y by thinking similarly to Y but in the opposite direction:
$P(Z \leq 6 | y | x) =  P( \text{all y rolls are at most 6} | y | x) = 1$
$P(Z \leq 5 | y | x) = P(\text{all y rolls are at most 5} | y|x) = (5/6)^y$.
and so on...
Thus we get the following:
$P(Z = 6 | y |x) = P(Y \leq 6|y|x) - P(Y\leq 5|y|x) = 1 - (\frac56)^y$
$P(Y = 5 | y | x) = P(Y \leq 5 |y|x) - P(Y \leq 4|y|x) = (\frac56)^y - (\frac46)^y$.
and so on...
Now if we use conditional probabilities we can figure out any combination of values for $X$, $Y$ and $Z$:
$P(X = x, Y = y, Z = z) = P(X=x)P(Y=y |X=x)P(Z=z|Y=y| X=X)$.
For example:
$$
\begin{align}
P(X=3, Y=5, Z=4) &= P(X=3)P(Y=5|X=3)P(Z=4|Y=5|X=3)\\
&= \frac16 \left[ (\frac26)^3 - (\frac16)^3\right]\left[(\frac46)^5 - (\frac36)^5\right]
\end{align}
$$
Your particular example of $P(X=1, Y=5, z=1)$ is particularly simple because if $X=1$ you only roll the die once, so $Y=5$ means the maximum of one roll is $5$ whose probability is $1/6$. Then you need the minimum of 5 rolls to be 1, which is only possible if all 5 are 1, whose probability is $(\frac16)^5$. So the probability is $\frac16\cdot \frac16 \cdot (\frac16)^5 = (\frac16)^7$.
A: To roll a value of 1 on $1$ toss can be done in $1$ way of the $6$ possible rolls of the die.
To roll a minumum value of 5 on all of $1$ toss, count the ways to roll $1$ results of 5, out of $6^1$ possible outcomes. 
Similarly to roll a maximum value of $1$ on all of $5$ tosses, count the ways to roll $5$ results of 1, out of $6^5$ possible outcomes.
$$\begin{align}
 \mathsf P(X=1) & = \frac \Box 6
\\[1ex]
 \mathsf P(Y=5\mid X=1) & = \frac{\Box}{6}
\\[2ex]
\mathsf P(Z=1\mid Y=5) & = \frac{\Box}{6^5}
\\[3ex]
 \mathsf P(X=1, Y=5, Z=1)
 & = \mathsf P(X=1)\;\mathsf P(Y=5\mid X=1)\;\mathsf P(Z=1\mid Y=5)
\\[1ex] & = \frac{\Box}{6^7}
\end{align}$$
Edit: See also http:.../questions/1025244/independent-probability-example
