Using Probability Generating Function to find probability

Question

so I've come across a Probability generating function (pgf) question that requires me to find $f(21)$.The PGF is defined as follows. $$G(t)=\left(\frac{t(1-t^6)}{6(1-t)}\right)^6$$

As mentioned before the task is to find $f(21)$.

I've used differenced of squares and difference of cubes to simplify $G(t)$ as follows:

$$G(t)= \frac1{6^6}[t + t^2 + t^3 + t^4 + t^5 + t^6]^6$$

Expanding the terms in the brackets $6$ times is not an efficient method! I'm struggling to find a way to get to the term $f(21)$ so that I can find the coefficient and hence the probability.

Any help is highly appreciated!

Answer 1

EDIT In the original post I omitted a sign (shown in $\color{red}{\text{red}}$ below.) I've corrected that and carried out the arithmetic.

I'll use the notation $[t^n]G(t)$ to mean the coefficient of $t^n$ in in $G(t)$ so that I understand you to be looking for $[t^{21}]G(t)$. We have $$ \begin{align} [t^{21}]G(t)&=[t^{21}]\left(\frac{t(1-t^6)}{6(1-t)}\right)^6\\ &=6^{-6}[t^{15}]\left(\frac{(1-t^6)}{1-t}\right)^6\\ &=6^{-6}[t^{15}]\sum_{n=0}^6\color{red}{(-1)^n}\binom{6}{n}t^{6n}\sum_{n=0}^\infty(-1)^n\binom{-6}{n}t^n\\ &=6^{-6}\sum_{n=0}^2\binom{6}{n}\binom{-6}{15-6n}(-1)^{15-5n}\\ &=6^{-6}\left(-\binom60\binom{-6}{15}+\binom61\binom{-6}{9}-\binom62\binom{-6}{3}\right)\\ &=6^{-6}\left(\binom60\binom{20}{15}-\binom61\binom{14}{9}+\binom62\binom{8}{3}\right)\\ &=\frac{4332}{46656}=\frac{361}{3888}\approx0.09285 \end{align}$$

Answer 2

You don't need to do the full expansion; you just need to see which combination of terms will correspond to $t^{21}$. This is equivalent to the number of ways to choose $x_1, x_2, x_3, x_4, x_5, x_6$ from the set $\{1, 2, 3, 4, 5, 6\}$ with repetitions allowed, such that $x_1 + \cdots + x_6 = 21$. For example, $$(x_1, \ldots, x_6) = (1, 1, 3, 5, 6, 5)$$ works, and is distinct from $$(x_1, \ldots, x_6) = (1, 3, 5, 6, 1, 5)$$ because in your case, order counts. But this seems tedious and prone to error. Notably, there are $32$ partitions of $21$ into six positive integers from $\{1, \ldots, 6\}$, but they do not all have the same number of permutations; e.g., $\{6,6,6,1,1,1\}$ has $20$ but $\{6,5,4,3,2,1\}$ has $6! = 720$. Is there a better way?

Well, one alternative is to instead compute the coefficient of $t^{15}$ in $$\frac{1 - 6t^6 + 15t^{12} - 20t^{18} + 15t^{24} - 6t^{30} + t^{36}}{1 - 6t + 15t^2 - 20t^3 + 15t^4 - 6t^5 + t^6}$$ via synthetic division. But this seems hardly satisfactory, either.

One more possibility is to compute the $21^{\rm st}$ derivative of $G$ evaluated at $0$. This of course is masochistic and not anywhere in the realm of reality for hand computation.

Of the three, I think the second approach is most likely to be successful. It is also worth noting that your question is essentially asking for the probability of rolling a sum of exactly $21$ on six fair dice.