Major Probability Question

I'll write a sketch as I don't have time to do a complete solution. the generating function for the score on one die is for $t\neq1$ $g(t)=\frac{t^{7}-t}{6(t-1)}$ So by the convolution theorem our generating function for the sum of the scores is for (t\neq1) $G(t)=(\frac{t^{7}-t}{6(t-1)})^{n}$ Now to get the probability our score is divisible by 5 we need the coefficient on all powers of 5 in the expansion. Note that if we compute $G(1)+G(w)+...+G(w^{4})$ ,where $w$ is the first complex fifth root of unity, then the coefficient of every $t$ who's power isn't divisible by 5 goes to zero as $1+w+...+w^{4}=0$ and we are left with 5x the sum of the coefficients of those $t$ who's power is a multiple of 5 (as $w^{5k}=1$ for integer $k$ ). Hence our probability is (noting $G(1)=1$ as it's the sum of all the probabilities, and that $w_{i}^{n}=1$ )

$\frac{1}{5} (1+ \sum_{i=1}^4 (\frac{w_{i}^{6}-1}{6(w_{i}-1)})^{n})$

Noting that $w_{i}^{6}=w_{i}$ and that this becomes

$\frac{1}{5} (1+\frac{4}{6^{n}})=\frac{6^{n}+4}{5 \times 6^{n}}$

Hence our answer is $\boxed{21}$

$\textbf{Lemma:}$ If $p$ is a prime number and $a_0, a_1, ... , a_{p-1}$ are rational numbers satisfying $a_0 + a_1\epsilon + ... + a_{p-1}\epsilon^{p-1} = 0,$ where $\epsilon = e^{\frac{2\pi}{p}i} ,$ then $a_0 = a_1 = ... = a_{p-1}$ .

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$\textbf{Proof:}$ Observe that the polynomials $f(x) = a_0 + a_1x + a_2x^2 + ... + a_{p-1}x^{p-1}$ and $g(x) = 1 + x + x^2 + ... + x^n$ are not relatively prime because they share a common root. Since $g(x)$ is irreducible over $\mathbb{Q}$ , $g$ must divide $f$ , which can only happen when $a_0 = a_1 = ... = a_{p-1}$ .

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Let $a_n^{(k)}$ be the number of die rollings such that the sum of the numbers is $k$ modulo $5$ .

Let $\epsilon = e^{\frac{2\pi}{5}i}$ It is clear that $\sum_{k=0}^{4}{a_n^{(k)}\epsilon^k} = (\epsilon + \epsilon^2 + \epsilon^3 + \epsilon^4 + \epsilon^5 + \epsilon^6)^n = \epsilon^n = 1$

Due to the lemma, we have $a_n^{0} - 1 = a_n^{1} = a_n^{2} = a_n^{3} = a_n^{4} = q$

Then $5q = \sum_{k = 0}^{4}{a_n^{k}} - 1 = 6^n - 1 \implies q = \frac{6^n - 1}{5} \implies a_n^{0} = \frac{6^n + 4}{5}$

Thus the probability is $\frac{\frac{6^n + 4}{5}}{6^n} = \frac{6^n + 4}{5 \cdot 6^n} \implies a+b+c+d = \boxed{21}$

Prob that 1-5 nos are on top= 5^n÷6^n And prob that sum of no on top =1/5 Now prob that all are six =1/6^n Now also in remaining cases prob=1/5 So required prob = (5/6)^n+1/6^n + 6^n-5^n-1/5*6^n

Let $a_{n,k}$ be the number of ordered $n$ -tuples out of $\{ 1, 2, 3, 4, 5, 6\}^n$ for which the sum is equal to $k$ (modulo 5).

We will prove by induction that $a_{n,k} = \frac{6^n-1}5 + \delta_{nk},$ where $\delta_{nk} = 1$ if $n \equiv k$ mod 5, and $= 0$ otherwise.

Initial step: When rolling $n = 1$ dice, there are two ways for the sum to be equal to 1 mod 5 (namely, rolling 1 and rolling 6). This shows $a_{1,1} = 2$ . For $k\not\equiv 1$ , $a_{1,k} = 1$ . This satisfies the equation.

Induction step: Suppose that the formula is proven up to a certain value $n$ . Choose $1 \leq m \leq n$ and let $n' = n+m$ . We will prove that the formula also holds for $n'$ .

We have $a_{n',k} = \sum_{i + j \equiv k}^{\text{(5 terms)}} a_{n,i}a_{m,j} = \sum_{i = 0}^4 \left( \frac{6^n-1}5 + \delta_{ni}\right) \left( \frac{6^m-1}5 + \delta_{mj}\right).$ If $k\not\equiv n'$ we must account for the $\delta$ s in two of the terms, and find: $a_{n',k} = 5\cdot \frac{6^{n+m}-6^n-6^m+1}{25} + \frac{6^n-1}{5} + \frac{6^m-1}{5},$ and if $k\equiv n'$ , the non-zero $\delta$ s occur in the same term: $a_{n',k} = 5\cdot \frac{6^{n+m}-6^n-6^m+1}{25} + \frac{6^n-1}{5} + \frac{6^m-1}{5} + 1.$ In either case, we have $a_{n',k} = \frac{6^{n+m}-6^n-6^m+1}{5} + \frac{6^n-1}{5} + \frac{6^m-1}{5} + \delta_{n+m,k}$ $= \frac{6^{n+m}-1}5 + \delta_{n+m,k}.$ This completes the induction step.

Finally, if $5|n$ , $n \equiv 0$ mod 5. The probability of rolling a sum that is a multiple of 5 is $\frac{a_{n,0}}{6^n} = \frac{6^n-1}{5\cdot 6^n}.$