<s>[INST] Probability of sum to be divisible by 7 6 fair dice are thrown simultaneously. What is the probability that the sum of the numbers appeared on dice is divisible by 7 ?
 [/INST] After $n$ rolls, let $p_n$ be the probability that the sum is a multiple of $7$ after $n$ rolls.
Then we get the recursion: $$p_{n+1} = \frac{1}{6}(1-p_n)$$
That's because to get a multiple at roll $n+1$, you have to get a non-multiple of $7$ at roll $n$, and then exactly the right value.
Then start with $p_0=1$, and work from there.
$$\begin{align}p_0&=1\\
p_1&=0\\
p_2&=\frac{1}{6}\\
p_3&=\frac{5}{36}\\
p_4&=\frac{31}{216}\\
p_5&=\frac{185}{6^4}\\
p_6&=\frac{1111}{6^5}
\end{align}$$
(This is essentially a Markov process with two states, much akin to the other answer. You essentially need to compute:
$$\begin{pmatrix}0&\frac{1}{6}\\
1&\frac{5}{6}\end{pmatrix}^6\begin{pmatrix}1\\0\end{pmatrix}$$
The general answer is:
$$p_n=\frac{6^{n-1}-(-1)^{n-1}}{7\cdot 6^{n-1}}$$
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