Probability and Cumulative Dice Sums

Let a die be labeled with increasing positive integers $a_1,\ldots , a_n$, and let the probability of getting $a_i$ be $p_i>0$. We start at 0 and roll the die, adding whatever number we get to the current total. If ${\rm Pr}(N)$ is the probability that at some point we achieve the sum $N$, then $\lim_{N \to \infty} {\rm Pr}(N)$ exists and equals $1/\rm{E}(X)$ iff $(a_1, \ldots, a_n) = 1$.

The direction $\implies$ is obvious. Now, if the limit exists it must equal $1/{\rm E}(X)$ by Chebyshev's inequality, so we only need to show that the limit exists assuming that $(a_1, \ldots, a_n) = 1$.

We have the recursive relationship $${\rm Pr}(N) = p_1 {\rm Pr}(N-a_1) + \ldots + p_n {\rm Pr}(N-a_n);$$ the characteristic polynomial is therefore $$f(x) = x^{a_n} - \left(p_1 x^{(a_n-a_1)} + \ldots + p_n\right).$$

This clearly has the root $x=1$. Next note $$f'(1) = a_n - \sum_{i=1}^{n} p_i a_n + \sum_{i=1}^{n} p_i a_i = \rm{E}(X) > 0 ,$$ hence this root is also unique.

 I'll now show that all other roots have absolute value $< 1$.

We have $$x^{a_n} - \left( p_1 x^{(a_n-a_1)} + \ldots + p_n \right) = 0 \iff x^{a_1} = p_1 + \ldots  + \frac{p_n}{x^{(a_n-a_1)}} .$$

If $|x|>1$ then $$|x^{a_1}| = \left|p_1 + \ldots + \frac{p_n}{x^{(a_n-a_1)}}\right| \leq |p_1| + \ldots + \left|\frac{p_n}{x^{(a_n-a_1)}}\right| \leq p_1 + \ldots + p_n = 1;$$ contradiction.

If $|x|=1$ $$|x^{a_1}| = \left|p_1 + \ldots + \frac{p_n}{x^{(a_n-a_1)}}\right| < 1$$ unless $x^{(a_1 - a_i)} = 1$ for all $i \implies x^d = 1$ where $d = (a_1 - a_2, \ldots, a_1 - a_n)$.

Assume $x^d = 1$; then $$x^{a_1} = p_1 + \ldots + \frac{p_n}{x^{(a_n-a_1)}} = p_1 + \ldots + p_n = 1 \implies x^{a_1} = 1 \implies x^e = 1$$ where $e = (d, a_1) = (a_1, \ldots, a_n) = 1$ by assumption, hence $x^1 = 1 \implies x = 1$ is the only root of $f(x)$ such that $|x| \geq 1 \implies \lim_{N \to \infty} {\rm Pr}(N)$ exists.