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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?

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The probability generating function has the property that $G_{X+Y}(\eta)=G_X(\eta)G_Y(\eta)$ for random variables $X$ and $Y$ (this is kind of the whole point). So $G_{X+X+X+X+X}(\eta)=G_X(\eta)^5$. – mjqxxxx Jan 29 '13 at 21:11
up vote 0 down vote accepted

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

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hence $$ G(-1)=-1+2\sum_{n\,\text{even}}\mathbb P(\text{score}=n) $$ I don't understand this. – bbr4in Jan 31 '13 at 15:59
Please answer this comment. – bbr4in Jan 31 '13 at 16:06
Note that sum of probabilities of even scores + sum of probabilities of odd scores = 1. – Did Jan 31 '13 at 17:34

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:


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.

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