Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
1  
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

2 Answers 2

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

share|improve this answer
    
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:

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

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.