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

As $r_0\to +\infty$, the exponential factor tends to zero, so $P(r>r_0)\to 0$. Requiring an arbitrarily large amount of remaining time must have vanishing probability for this finite-mean distribution.

When $\bar{s}=\bar{t}$, the prefactor is $1/2$. Therefore $P(r>r_0)\to 1/2$ as $r_0\downarrow 0$. 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.