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Probability and Cumulative Dice Sums

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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.

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