Probability Generating Functions with Three Dice

Question

Three identical dice are thrown. The dice are fair, that is, for all three dice the probability of turning up face $j$ is $1/6$, $1 \le j \le 6$. Let $X_1,\ X_2,\ X_3$ be the independent random variables representing the numbers on dice $1$, $2$, and $3$ respectively.

1. Find the probability generating function (pgf) of $X_1$.
2. Let $G_Y(z)$ be the pgf of $Y = X_1 + X_2 + X_3$. Show that $$ G_Y(z) = \dfrac{z^3 (1 - z^6)^3}{216} \sum_{k = 0}^{\infty} \binom{k + 2}{2} z^k. $$ You need to specify every rule you have used.
3. Hence find $P(Y = 9)$.

I got the first two parts. I am having issues with the third part.

My thought is to sum the coefficients of $z^9$?

Is this the correct approach? If not, could you please demonstrate the correct approach please?

PS: Please show how to do part $3$ from part $2$ instead of computing the probability directly.

Answer 1

From Part $2$, we have $$G_Y(z) = \dfrac{z^3(1 - z^6)^3}{216} \sum_{k = 0}^{\infty} \binom{k + 2}{2}z^k.$$

$P(Y = 9)$ is the coefficient of $z^9$ in $G_Y(z)$. As the least degee term of $G_Y(z)$ is $z^3$, we can ignore the $z^3$ in the above formula and find the coefficient of $z^6$ from the remaining terms $(1 - z^6)^3$ and $z^k$ (for appropriate $k$).

Now, $(1 - z^6)^3 = 1 - 3z^6 + \ldots$ (where $\ldots$ are terms of degree higher than $6$). To get $z^6$, the first term $1$ can be multiplied by $z^6$ (by taking $k = 6$ in $z^k$) and the second can be multiplied by $1$ (by taking $k = 0$). Thus, the coefficient of $z^9$ in $G_Y(z)$ is

$$ \dfrac{1}{216} \left[ \binom{8}{2} - 3\binom{2}{2} \right] = \dfrac{25}{216} $$ $\therefore \boxed{P(Y = 9) = \dfrac{25}{216}}$

Answer 2

Write $$G_Y(z)=\frac{1}{216}z^3(1-3z^6+\cdots)\left(\binom22+\cdots+\binom82z^6+\cdots\right)\ .$$ The coefficient of $z^9$ is $$\frac{\binom82-3\binom22}{216}=\frac{25}{216}$$ and this is the required probability.