Can you prove a random walk might never hit 0, given a probability system that only uses the finite additivity axiom (rather than the standard countable additivity axiom)?
Specifically: Imagine the standard random walk problem on the nonnegative integers: We start at integer location $i>0$. Every step we independently move left with probability $\theta$, and right with probability $1-\theta$. Assume $0<\theta < 1/2$. Let $q$ be the probability that we ever hit 0, given we start at location 1. By the repeated structure of the problem, we can infer: $$ q=\theta + q^2(1-\theta) $$ The difficulty is that this quadratic has two roots: $q=1$ and $q=\frac{\theta}{1-\theta}$. If we can prove that $q<1$, then we infer $q=\frac{\theta}{1-\theta}$.
The answer is $q=\theta/(1-\theta)$ under the standard probability axioms, which includes the countable additivity axiom. Is this provable with only finite additivity, i.e., $P[\cup_{i=1}^k A_i] = \sum_{i=1}^k P[A_i]$ for finite integers $k$ and for disjoint events $A_i$? Or is it undecidable?
Note: The union bound $P[\cup_{i=1}^{\infty} A_i] \leq \sum_{i=1}^{\infty} P[A_i]$ is unprovable without countable additivity.