P0 = (P12 + P11 + P10 + P9 + P8 + P7) / 6

If such limit does exist, let Pi be the probability of the remainder being i when Xn (n goes to infinity) is divided by 13.

We have
P0 = (P12 + P11 + P10 + P9 + P8 + P7) / 6
P1 = (P0 + P12 + P11 + P10 + P9 + P8) / 6
...
and so on.

Also,
P0 + P1 + ... + P12 = 1

The solution to the equations is
P0 = P1 = ... = P12 = 1/13