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A symmetrical spinning top has five edges number -2, -1, 0, 1, 2. The PGF for the scores when the top is spun once is written $\frac{1-t^5}{5t^2(1-t)}$. Hence find the probability of getting a total score of zero when the top is spun three times.

I cubed the PGF and tried to look for the coefficient of the term $t^0$. Is that right? Also I couldn't really find it because the expression was ugly.

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up vote 1 down vote accepted

Cubing is right. Whether the result is ugly is a matter of taste. The cube of the pgf is $$\frac{1}{125t^6}(1-t^5)^3(1-t)^{-3}.$$ All we need now is the coefficient of $t^6$ in $(1-t^5)^3(1-t)^{-3})$.

Expanding $(1-t^5)^3$ is easy, and we only need the first two terms.

Now we need the coefficients of $t^6$ and $t$ in $(1-t)^{-3}$. These can be written down using the Generalized Binomial Theorem.

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Let PGF of $X_i$ is $P_{X_i} (t)=\frac{1-t^5}{5t^2(1-t)} \forall i=1,2,3$. Also let $Y=X_1+X_2+X_3$.

Now note that $$P_Y(t)=E(t^Y)=E(t^{X_1+X_2+X_3})=\prod_{i=1}^3 E(t^{X_i})=[P_X(t)]^3$$ So, your procedure is right and to get a total score of zero we need the coefficient of the term $t^0$ i.e the constant term in $P_Y(t)$. $$\begin{align} P_Y(t) &=\frac{1}{5^3}(1-t^5)^3\cdot t^{-6}\cdot (1-t)^{-3} \\ &=\frac{1}{5^3} t^{-6}(1-3t^5+3t^{10}-t^{15})(\frac{2 \cdot 1}{2}+\frac{3 \cdot 2}{2}t+\dots+\frac{8 \cdot 7}{2}t^6+\cdots) \end{align}$$ I think now you can get the coefficient of the term $t^0$.

You can note that $$(1-x)^{-3}=\sum_{r=2}^ \infty \frac{r \cdot (r-1)}{2} x^{r-2}$$

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Sorry but $Y\ne3X$ and $E(t^{3X})\ne E(t^X)^3$. – Did Jan 31 '13 at 17:51
@Did : Sorry I don't understand your comment. Can you please explain it. – Argha Jan 31 '13 at 17:55
Sure: why do you think that $Y=3X$? – Did Jan 31 '13 at 17:57
Oh $Y$ will be $X_1+X_2+X_3$ and then I can write $E(t^{\sum_{i=1}^3 X_i})=\prod_{i=1}^3 E(t^{X_i})$, due to independence.Is this correct? Now I edit my answer. – Argha Jan 31 '13 at 18:03
Right. $ $ $ $ $ $ – Did Jan 31 '13 at 19:02

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