A standard fair 6-sided dice is rolled $n$ times. Let $X_k$ be the spot which faced up inthe k-th round. What is the probability that $X_1+X_2+...+X_n$ is divisible by 7?
I tried to solve it by translating the problem into: Counting the number of integral solutions of $X_1+...+X_n=7k$, subjected to the constrients $1 \leq X_i \leq 6, k \geq 1$. But then I dont know how to continue, please helps.
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Define $R_{k}$ by the conditions that $R_{k}\in\left\{ 0,1,2,3,4,5,6\right\} $ and $X_{k}=7n+R_{k}$ for some integer $n$. Then you are looking for $P\left\{ R_{k}=0\right\} $.
It is obvious that $P\left\{ R_{1}=0\right\} =0$ and if $k>1$ then: $P\left\{ R_{k}=0\right\} =\sum_{i=0}^{6}P\left\{ R_{k}=0\mid R_{k-1}=i\right\} P\left\{ R_{k-1}=i\right\} =\frac{1}{6}\left(1-P\left\{ R_{k-1}=0\right\} \right)$. This recursion enables you to find $P\left\{ R_{k}=0\right\} $.
Let $P(n)$ be the chance that $X_1+\dots+X_n$ is a multiple of 7.
If $X_1+\dots+X_{n-1}$ is a multiple of 7, then $X_1+\dots+X_{n}$ is not. Otherwise, there is a 1 in 6 chance.
So can you find a recursion for $P(n)$ in terms of $P(n-1)$?