Limiting probability that the sum of the values of a die is a multiple of 13

Question

A fair die is thrown repeatedly. Let $X_n$ denote the sum of the $n$ first throws. I have to find $\lim_{n\rightarrow \infty}P(X_n \text{ is multiple of 13})$. Now follows what I tried, which I don't think is really efficient:

We can look modulo 13 and then we have an irreducible Markov chain on a finite state-space which is aperiodic (because for $n>2$ and for all $i,j\in\{1,2,...,13\}$ it is clear that $p_{ij}^{(n)}>0$, this is the $n$-step probability). So by Convergence to equilibrium we have that the asked limit actually exists. Now for calculating it, we have to find an invariant distribution of a matrix of dimensions 13 by 13. The matrix looks like $$\left(\begin{array}{ccccc} 0& 1/6 & 1/6 &1/6 & \dots\\ 0&0&1/6&1/6&\dots\\ 0& 0&0&1/6&\dots\\ \dots&\dots&\dots&\dots&\dots \end{array}\right)$$ This is easily done by a computer, but I think this exercise can be done more efficiently.

Answer 1

The Markov chain of the residues mod $13$ is irreducible and aperiodic on the finite state space $\mathbb Z/13\mathbb Z$ and its transitions are invariant by translation hence the unique stationary distribution is uniform. Thus, for every $i$, $P(X_n=i\pmod{13})\to\frac1{13}$ when $n\to\infty$.

In particular, $P(13\ \text{divides}\ X_n)\to\frac1{13}$ when $n\to\infty$.

Answer 2

Let $p_i$ be the greatest of the steady-state $p_j(j=0..12)$.
$p_i=(p_{i-1}+...+p_{i-6})/6$, so all of $p_{i-1}$ to $p_{i-6}$ must equal $p_i$. (None are greater, and if any were less then $p_i$ would be less as well.)
So $p_i,p_{i-1},...,p_{i-6}$ are equal and the greatest of the $p_j$.
For the same reason, $p_{i-6}=(p_{i-7}+...+p_{i-12})/6$, and the other $p_j$ are the same as well.
So all $p_j$ are equal in the steady state.