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

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• ? -jinjing- ♀ (0 bytes) () 03/31/2010 postreply 17:30:00