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Depending on which mutant goes to fixation, the average fitness of the final population is either 1, $1+\gamma_1$, or $1+\gamma_2$, with $\gamma$ as defined following Eq. (\[pieq\]). The expected fitness increase $\langle \gamma(\Delta t)\rangle$ after the introduction of the two mutants is therefore $$\label{expectedinc}
  \langle \gamma(\Delta t)\rangle = \gamma_1 \pi_1(\Delta t) + \gamma_2
  \pi_2(\Delta t)\,.$$ Since we know that $\pi_1(\Delta t)+\pi_2(\Delta t)$ is constant, and $\gamma_2 \geq \gamma_1$ as a direct consequence of our assumption $s_2>s_1$, it follows that $\langle\gamma(-\infty)\rangle \geq \langle
\gamma(\infty)\rangle$. If we write $\pi_1(\infty)=\pi_1(-\infty)+\Delta
\pi$, and $\pi_2(-\infty)=\pi_2(\infty)+\Delta \pi$, then we find $$\langle \gamma(-\infty)\rangle - \langle \gamma(\infty)\rangle =
    (\gamma_2-\gamma_1)\Delta \pi\geq 0\,.$$ Thus, the expected fitness increase is larger if we introduce the mutant with the higher selective advantage first.

Let $n_{\rm fix}$ denote the number of fixed mutants, that is, we have $n_{\rm
fix}=0$ if none of the mutants reach fixation, $n_{\rm fix}=1$ if exactly one of the mutants reaches fixation, and $n_{\rm fix}=2$ if first mutant 1 reaches fixation and is later replaced by mutant 2. For $\Delta t\rightarrow-\infty$ and $N$ large, $n_{\rm fix}$ can never be larger than one, because the probability that mutant 1 goes to fixation in the background of mutant 2 is 