zero. We find in this limit for the expected value of $n_{\rm
  fix}$: $$\langle n_{\rm fix}(-\infty)\rangle = \pi_1(-\infty) + \pi_2(-\infty) = \pi\,.$$ In the limit $\Delta t\rightarrow \infty$, on the other hand, we find $$\begin{aligned}
  \langle n_{\rm fix}(\infty)\rangle &= \pi_1(-\infty) + \pi_2(-\infty)
+P(s_1,u)P\Big(\frac{1+s_2}{(1+s_1)(1-u)}-1,u\Big)\notag \\
& = \pi + P(s_1,u)P\Big(\frac{1+s_2}{(1+s_1)(1-u)}-1,u\Big)\,.\end{aligned}$$ Clearly, $\langle n_{\rm fix}(\infty)\rangle\geq\langle n_{\rm
fix}(-\infty)\rangle$. This means that if the mutant with the smaller selective advantage appears before the mutant with the larger selective advantage, then the expected number of mutants that go to fixation is larger than if the mutant with the larger selective advantage appears first. However, at the same time the expected increase in average fitness is smaller, as we saw in the previous paragraph.

Numerical Simulation
====================

In order to measure fixation probabilities, we carried out $100,000$ replicates of the simulation for each set of parameters, and recorded the final outcome (all individuals unmarked, or marked as descendants of either advantageous mutant). We studied a population of size $N=1000$ and mutation rates $u=0.0$, 0.01, 0.02, 0.03, 0.05, 0.07, 0.1, 0.15, 0.20, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8. The selective advantages were $s_{1}=0.1$ and $s_{2}=0.2$, $s_{1}=0.1$ and $s_{2}=0.5$, $s_{1}=0.05$ and $s_{2}=0.2$. We also studied a population of 