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An unbiased six-faced die is rolled r times. The probability generating function for the total score is

$[\frac{t(1-t^6)}{6(1-t)}]^r$

Hence show that the probability of the total score being (r+3) is

$\frac{1}{6}^{r+1}r(r+1)(r+2)$

I appreciate any help anyone can provide and this is a statistics question in a Further Maths A-Level style paper for context on how they want the question to be solved.

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Let $[t^n] f(t)$ denote the coefficient of $t^n$ in $f(t)$. (I'm not sure how standard this notation is but it's very useful.) There is a factor of $t^r$ in the generating function.

$$[t^{r+3}] \left( t (1-t^6) \over 6(1-t) \right)^r = [t^3] \left( 1-t^6 \over 6(1-t) \right)^r $$

and you can pull out a factor of $(1/6)^r$ as well to get

$$ [t^3] \left( {1-t^6 \over 1-t} \right)^r $$

At this point there are two possible methods.

(1) Note that

$$\left( {1-t^6 \over 1-t} \right)^r = (1 + t + t^2 + t^3 + t^4 + t^5)^r$$

and think about how multiplication of polynomials works to get the coefficient of $t^3$ - you can get a factor of $t^3$ from one factor of $t^3$ and the rest 1, or a $t^2$ and a $t$, or three copies of $t$. (This isn't all that different from solving the problem without generating functions.)

(2) Take the third derivative and evaluate at zero. Since you have an $r$ in the exponent this might be a bit tricky - I'd try logarithmic differentiation. I have not actually done this.

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  • $\begingroup$ thanks for you help, I ended up using your first method and using permutations and combinations to find out how many ways you can get $t^3$ and got the right answer. I think method 2 is beyond my knowledge but thank you anyway! $\endgroup$ – Dev Vadhwana Feb 20 '19 at 16:30
  • $\begingroup$ You're welcome. I'm only slightly familiar with the UK system so I wasn't sure what was reasonable to assume in the Further Maths context. $\endgroup$ – Michael Lugo Feb 20 '19 at 16:55
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Picking up from where Michael Lugo left off, we want to compute $$[t^3]{(1-t^6)^r \over (1-t)^r}.$$ Expand the numerator via the binomial theorem and drop any terms that involve a power of $t$ greater than $3$: $$[t^3]{(1-t^6)^r \over (1-t)^r} = [t^3]{1\over (1-t)^r}.$$ Now use the identity $${1\over(1-s)^{k+1}} = \sum_{n=0}^\infty \binom{n+k}k s^n$$ to get $$[t^3]{1\over (1-t)^r} = \binom{r+2}{r-1} = \binom{r+2}3 = {r(r+1)(r+2)\over6}.$$ Multiply this by the factor of $1/6^r$ that was pulled out earlier to get the final result.

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  • $\begingroup$ thanks for continuing the answer. this is a much more concise method than the one i ended up using so thanks for sharing this method! $\endgroup$ – Dev Vadhwana Feb 25 '19 at 20:11

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