When $\bar s=\bar t$, the prefactor is $1/2$. Therefore \b[$P(r>r_0)\to1/2$ as $r_0\downarrow0$]. The clocks then have identical independent continuous laws, so either is equally likely to finish first. A tie has <probability> zero. The limit is the interruption <probability>, rather than one, because non-interrupted outcomes retain the negative sentinel value.

Solution

Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2011 / Paper 1 / 12 / g / ii / Solution

 Solution (source code)