Probability generating function $5$

Question

In a game of chance one of the numbers $1,2,3$ appears. The game is played five times and the total score is recorded. If the probs of $1,2,3$ are $\frac{1}{6}, \frac{1}{3}$ and $\frac{1}{2}$ respectively, write down an expression for the PGF, $G(t)$, for the possible total scores. Evaluate $G(-1)$ and hence find the prob that the total score is even.

I don't know what to write if it's done five times, rather than once. Also what has $G(-1)$ got to do with even numbers?

Answer 1

$$G(-1)=\sum_n\mathbb P(\text{score}=n)(-1)^n=\sum_{n\,\text{even}}\mathbb P(\text{score}=n)-\sum_{n\,\text{odd}}\mathbb P(\text{score}=n) $$ hence $$ G(-1)=-1+2\sum_{n\,\text{even}}\mathbb P(\text{score}=n) $$

Answer 2

Let $a,b,c$ denote the number of times $1,2,3$ occur respectively. Then after 5 times, the distribution of $a+b+c$ has PGF:

$$G(t)=\left(\frac{1}{6}t^1+\frac{1}{3}t^2+\frac{1}{2}t^3\right)^5$$.

Indeed, when you expand the above, you are summing combinations of $1,2,3$ a total of 5 times in the exponent of $t$, and each occurance gets a weight of the product of probabilities. As well, different combinations of $1,2,3$ give the same sum, they are added since they correspond to same power of $t$.

As Did mentions in his answer, $G(-1)$ indeed corresponds to just the even number occurances.