Let n dice be rolled. Let $S_{i}$ be the sum of the first $i$ rolls for $i=1...n$
Find $Prob($All $S_{i}$ are composite) as $n$ tends to ∞
My guess is 0 but how can I prove this? Or if I'm wrong how do I proceed?
Probability of sum of dice is composite
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probability
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0Welcome to MSE. Can you please explain what you mean by *composite*? You are also saying that "$n$ dice" are rolled. Do you mean that a die is rolled $n$ times ($i=1,\ldots,n$)? Although your problem statement is not very clear to me, you might be looking for [Kolmogorov's 0-1 law](https://en.wikipedia.org/wiki/Kolmogorov%27s_zero%E2%80%93one_law) – 2017-02-24
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0Thanks. Some clarifications: Composite as in not prime. There is no difference between rolling 1 dice n times and taking the sum and rolling n dice and taking their sum – 2017-02-24
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0This is just intuition talking but why not try and find the probability $P_0=P(x\in\mathbb P>0)$ such that $x \in S_i$, and $\mathbb P$ is the set of prime numbers. So then you just need to compute $1-P_0$ . – 2017-02-24
1 Answers
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Consider the first $n$ rolls for sufficiently large $n$. Let $$\mathbb{P} = \{0\}\cup\{ 1\leq i \leq n-1\mid \text{there is at least one prime in the range }[S_{i} + 1, S_{i} + 6]\}$$ Since there are more than $\frac{n}{\ln n}$ primes in the range $[1, n]$ for $n \geq 17$ (see here), the size of $\mathbb{P}$ is more than $\frac{n}{6\ln n}$. Hence, \begin{align} \Pr(S_1, S_2, \cdots, S_n \text{ are all composite})~\leq~\left(\frac{5}{6}\right)^{n / (6\ln n)} \end{align} When $n$ goes to $\infty$, the probability goes to $0$.