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 ?
 A: 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:
$$p_n=\frac{6^{n-1}-(-1)^{n-1}}{7\cdot 6^{n-1}}$$
A: One way is to add the coefficients of $x^7+x^{14}+x^{21}+x^{28}+x^{35}$ in the expression $(x+x^2+x^3+x^4+x^5+x^6)^6$ which will be symmetrical around the middle.
Another way is to use stars and bars, and apply inclusion-exclusion by preplacing $6$ in one or more of the $6$ cells, e.g. for a sum of $21$,
$\binom{20}{5} - \binom61\binom{14}{5} + \binom85 = 4332$
Once you have the number of favorable ways, you can compute $Pr$ 
A: We could approach this via markov chains and matrices.
$A=\begin{bmatrix}0&\frac{1}{6}&\frac{1}{6}&\frac{1}{6}&\frac{1}{6}&\frac{1}{6}&\frac{1}{6}\\
\frac{1}{6}&0&\frac{1}{6}&\frac{1}{6}&\frac{1}{6}&\frac{1}{6}&\frac{1}{6}\\
\frac{1}{6}&\frac{1}{6}&0&\frac{1}{6}&\frac{1}{6}&\frac{1}{6}&\frac{1}{6}\\
\frac{1}{6}&\frac{1}{6}&\frac{1}{6}&0&\frac{1}{6}&\frac{1}{6}&\frac{1}{6}\\
\frac{1}{6}&\frac{1}{6}&\frac{1}{6}&\frac{1}{6}&0&\frac{1}{6}&\frac{1}{6}\\
\frac{1}{6}&\frac{1}{6}&\frac{1}{6}&\frac{1}{6}&\frac{1}{6}&0&\frac{1}{6}\\
\frac{1}{6}&\frac{1}{6}&\frac{1}{6}&\frac{1}{6}&\frac{1}{6}&\frac{1}{6}&0\end{bmatrix}$
Each entry $A_{i,j}$ corresponds to the probability of moving from the $j^{th}$ state to the $i^{th}$ state, where each state corresponds to the remainder of the sum of the rolls modulo $7$.
With initial state vector $v=\begin{bmatrix}1\\0\\0\\0\\0\\0\\0\end{bmatrix}$ we have:
The $i^{th}$ entry of $A^nv$ denotes the probability that after $n$ dice are tossed that the total is $i-1$ more than a multiple of $7$.  In particular, the first entry of $A^6v$ will be the probability that the sum of six dice is a multiple of seven.

As @ThomasAndrews points out, the tedium of this process can be cut down dramatically by noting that we could just as easily have described it as a two-state markov chain.  Either it is or it isn't a multiple of seven.
To see how this looks, we have:
$A=\begin{bmatrix}0&\frac{1}{6}\\1&\frac{5}{6}\end{bmatrix}$ with initial state $v=\begin{bmatrix}1\\0\end{bmatrix}$
We have then the first entry of $A^n v$ will correspond to the probability that after $n$ rolls of the dice the sum is a multiple of $7$, and the second entry corresponds to not being a multiple of $7$.  The original matrix could give us additional information such as probabilities of having specifically remainder $3$ or $4$ modulo $7$ which this simplified matrix could not, but such information is irrelevant to the stated problem.
