How can I find the probability of getting a Yahtzee using probability generating functions for each roll? I was recently taught the concept of Probability Generating Functions (PGFs) and while revising said concept, I came across a question about the game Yahtzee. The question was as follows:

Find the probability of achieving a Yahtzee made entirely of the number $k$ on five 6-sided dice after 3 rolls. You may assume the player achieving the Yahtzee removes all of the dice which rolled $k$ after each of the 3 rolls.

Most of the answers I googled after attempting to solve this consisted of combinatorics and/or simply doing it manually or by using other ways I don't understand; however, while attempting to solve this using PGFs, I came across a curious problem in my method which I was unsure could be worked with, leading me to ask this question, this problem is explained below:

Let $X$ be the number of 6-sided dice in a game of Yahtzee that land on the number $k$ in one roll.
$X$ can be modelled binomially as $X \sim B(n,\frac{1}{6})$ for each throw where $n$ is the number of dice in each throw.
The probability generating function for $X$ is $G_X(t) = (\frac{5}{6}+\frac{1}{6}t)^n$
Thus, for the first roll, the PGF is as follows:
$G_{X_1}(t)=(\frac{5}{6}+\frac{1}{6}t)^5$
However, given the second roll is dependent on the amount of successes of the first roll, I would have to implement the first roll's outcome into the second roll's PGF:
$G_{X_2}(t)=(\frac{5}{6}+\frac{1}{6}t)^{5-X_1}$
And the third roll would implement the same concept:
$G_{X_3}(t)=(\frac{5}{6}+\frac{1}{6}t)^{5-X_1-X_2}$
This is not the answer I'm looking for since I would like $G_{X_3}(t)$ and $G_{X_2}(t)$ to be solely in terms of $t$ to then solve the question by finding the coefficient of $t^5$.
I would like to have the skill to manipulate embedded random variables in PGFs, thus my question is:
If possible, how can I implement the PGF of a random variable into the PGF of another random variable? (A method which I can use to solve the above question for example)
 A: I will say up front, that this is a really bad idea.  If ever a problem was designed to be solved Combinatorically, this is it.  How I would handle it would be to assume that the dice are labeled $D-1, D-2, \cdots D-5$.  Then, in each round, I would roll all $5$ dice.  Then, I would examine the $6^{15}$ combinations possible, to determine which of them resulted in each of $D-k$ being hit at least once.
I suspect that I would use Inclusion-Exclusion, and take advantage of the symmetrical aspects of the situation.
However, there is no reason that the problem can not be conquered by computing the appropriate probability function.

Edit
I should also mention that it is unclear to me whether the OP (i.e. original poster) will even regard this response as on point.  I have no formal knowledge of Probability Generating Functions, so I don't know if the analysis below is what the OP is looking for.
Edit
It just hit me.  I just took a very easy problem and made it very difficult.  All of the analysis from Old Work is valid, but unnecessary.  I left it in, because

*

*it might still have value

*it is an example of my not seeing the forest for the trees.

Mathematically, the problem is unchanged, if you never remove any dice.  Dice don't talk to each other.  You roll all $5$ dice all $3$ times.  The question becomes, what is the probability that all $5$ dice are each hit at least once.
This is clearly
$$\left[\frac{6^3 - 5^3}{6^3}\right]^5.$$
That is, you have $5$ independent events of each die being hit at least once.

Old Work
Taking the probability function approach, in my opinion, the best strategy is to work backwards.
At the point just before the 3rd roll, you will have $r$ dice remaining, where $r$ is some element in $\{0,1,2,3,4,5\}$.
Let $f(r)$ denote the probability of achieving the Yahtzee, as a function of $r$.
Clearly, $f(0) = 1$, which represents that you have achieved the Yahtzee before the 3rd roll.
For $r \in \{1,2,3,4,5\}$, you have that $f(r) = 6^{-r}.$ 
This formula also works for $r = 0$.

Now, consider the situation at the point just before the 2nd roll.
At the point just before the 2nd roll, you will have $r$ dice remaining, where $r$ is some element in $\{0,1,2,3,4,5\}$.
Let $g(r)$ denote the probability of achieving the Yahtzee, as a function of $r$.
Clearly, $g(0) = 1$, which represents that you have achieved the Yahtzee before the 2nd roll.
For $r \in \{1,2,3,4,5\}$, you have that there are $(r+1)$ different outcomes possible.  That is, for a specific value of $r$, the 2nd roll will result in their being $s$ dice left, where $s$ is some element in $\{0,1,2,\cdots,r\}$
What is necessary here is to compute, for a specific value of $r$, the probability of each event $s$ occurring.  Then, with each event $s$ occurring, your probability of success on the 3rd roll will be $f(s) = 6^{-s}.$

I find it helpful here to stretch intuition, by looking at a specific example.
Suppose that just before the 2nd roll, you have $3$ dice remaining.
First, see binomial distribution,
specifically $\displaystyle \binom{n}{k} p^k q^{n-k}.$
If there are $3$ dice remaining just before the 2nd roll, then the probability of having $s$ dice remaining after the 2nd roll ($s \in \{0,1,2,3\}$) is
$$\binom{3}{s}\left(\frac{5}{6}\right)^s \left(\frac{1}{6}\right)^{3-s}
= \binom{3}{s}\frac{5^s}{6^3}.$$
For each such value of $s$, you then multiply the probability of the above event occurring by $f(s)$.
Therefore,
$$g(3) = \sum_{s=0}^3 \binom{3}{s}\frac{5^s}{6^3} \times \frac{1}{6^s}
= \sum_{s=0}^3 \binom{3}{s}\frac{5^s}{6^{3+s}}.$$
Consequently, it becomes easy to generalize:
$$g(r) = \sum_{s=0}^r \binom{r}{s}\frac{5^s}{6^{r+s}}. \tag1 $$
Note that this formula (also) works, for $r = 0.$

Now, imagine that you are at the point just before the 1st roll.
At the end of the 1st roll, there will be $r$ dice left, where $r$ is some element in $\{0,1,2,3,4,5\}.$
Let $h(r)$ denote the probability of having $r$ dice remaining, after the first roll.
Then, the probability of a Yahtzee is
$$\sum_{r=0}^5 \left[h(r) \times g(r)\right].$$
$g(r)$ is given by (1) above.  A similar use of the binomial theorem will allow $h(r)$ to be computed.
$$h(r) = \binom{5}{r}\left(\frac{5^r}{6^5}\right).$$
Putting this all together, starting from scratch, the probability of the Yahtzee is
$$\sum_{r=0}^5 \left\{ ~\binom{5}{r}\left(\frac{5^r}{6^5}\right) \times 
\left[\sum_{s=0}^r \binom{r}{s}\frac{5^s}{6^{r+s}}\right]
 ~\right\}
.$$
