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Starting from 0, each time one rolls 2 fair dice and moves forward by $k$ steps, $k$ is the sum of the two dice, what is the probability of ending at $N$? as $N$ towards infinity.

I guess this can be shown as recursion, depending on the last roll (sum $k_n = 2, 3, ... 12$), $$P(N) = \sum_{k_n = 2}^{12}P(N-k_n)P(k_n)$$ and take limits on both sides to solve for $P(N)$ but seems not right...

Morty19
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  • Empirically it looks like it might be $1/7$, which kind of corresponds with the average roll being 7. (If you roll exactly 7 every time, and your starting point is uniform on $1,2,...,7$, then $P(N)=1/7$. I think this is roughly what the $N \to \infty$ limit of your setup looks like.) – angryavian May 14 '22 at 03:50
  • I guess it's $1/7$ as well, trying to solve it by linear recursion and characteristic function, it has a root of 1. The probability as n towards infinity shall be the coefficient of root 1 – Morty19 May 14 '22 at 03:58
  • Found a similar question here, very helpful https://math.stackexchange.com/questions/1092426/probability-of-rolling-a-sum-of-n-with-up-to-infinite-rolls-of-a-die – Morty19 May 14 '22 at 04:09

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