Let $\sigma$ be the time a nearest neighbor random walk started at 1 that has probability $p>1/2$ of moving left reaches $0$. Let $\sigma'$ be an independent copy of $\sigma$. Let $(X_k)_1^\infty$ be iid unit Exponential random variables, and let $(Y_k)_1^\infty$ be iid Exponentials with mean $v$ (i.e. $P(Y_1 \geq t) = e^{-t/v}$).

We are interested in if there is a closed form in terms of $p$ and $v$ for the probability $$P \left( \sum_1^\sigma X_k < \sum_1^{\sigma '} Y_k \right).$$

Conditioning on the value of either sum gives a messy expression that isn't obvious how to simplify.

A reformulation of the problem is to think of this as a race to reach 0 by two continuous time random walks with rates $1$ and $1/v$. Using the memoryless property, the probability the rate-1 walk advances at a jump time is $q=v/(1+v)$. Otherwise the rate-$1/v$ walk advances. Let $Z(r,q)$ be the number of successes before $r$ failures occur in iid trials with success probability $q$ (i.e. negative binomial). If we think of the rate-$1$ walk advancing as a success, we can rewrite the above probability as $$P( Z(\sigma,q) > \sigma').$$ Condition on the values of $\sigma$ and $\sigma'$ and use the distribution for a negative binomial to write this as $$\sum_{i,j \geq 0} C_i C_j p^2(p(1-p))^{i+j} \sum_{k\geq 2i+1} \binom{2j +k}{k} q^k (1-q)^{2j+1} .$$ Here $C_i$ is the $i$th Catalan number. It does not look easy to evaluate exactly. We are happier with this though because it is easier to numerically approximate (though we would prefer a closed form).