= Haploid selection recurrence
{title2=$p'=\frac{wp}{wp+(1-p)}$}
Consider an explicitly idealized <haploid> population with two inherited types $A,B$, one <allele> per individual at the tracked <genetic locus>, and discrete, nonoverlapping generations. Assume faithful type transmission, no <mutation> or <gene flow>, a constant environment, and reproductive contributions independent of type frequency. Let the <expected values> of offspring surviving to the next census be $m_A,m_B>0$, with relative <genotype fitness> $w=m_A/m_B$.
For $A$ frequency $p$, expected offspring contributions are proportional to $pm_A$ and $(1-p)m_B$. Normalizing gives the deterministic large-population <mathematical model>
$$
p'=\frac{pm_A}{pm_A+(1-p)m_B}=\frac{wp}{wp+(1-p)}.
$$
The denominator is mean relative <genotype fitness>. For $w=1.2$ and $p_0=0.2$, $p_1=0.24/1.04=3/13\approx0.23077$. Equivalently, 200 $A$ and 800 $B$ parents with $m_A=1.2,m_B=1$ contribute expected counts 240 and 800. Fractional per-parent <expected values> are averages, not literal fractional offspring. The ratio of these expectations is a deterministic approximation, not generally the expectation of a random ratio.
The odds multiply particularly simply:
$$
\frac{p_{t+1}}{1-p_{t+1}}=w\frac{p_t}{1-p_t},\qquad p_t=\frac{w^tp_0}{1-p_0+w^tp_0}.
$$
Thus $p_{10}\approx0.60753$: repeated modest advantage can substantially change composition. For $w>1$ and $0<p<1$, $p'-p=(w-1)p(1-p)/(1+(w-1)p)>0$. The deterministic boundary $p=0$ remains unchanged: <natural selection> needs existing variation.
\Image[/natural-selection-frequency.png]
{title=Deterministic selection and stochastic sampling in a finite population}
{description=Both panels start at $p_0=0.2$. The left compares the deterministic <haploid selection recurrence> with eight simulated finite-population paths at $w=1.2$; the right shows eight neutral <genetic drift> paths at $w=1$. Each simulation samples $N=50$ offspring independently per generation. These are illustrative <mathematical models>, not experimental measurements.}
A distinct stochastic extension fixes $N$ offspring and samples their types independently using a <binomial distribution>: $K\sim\operatorname{Binomial}(N,q)$, where $q=wp/(wp+1-p)$, then $p'=K/N$. Its conditional <expected value> is $q$ and conditional <variance> is $q(1-q)/N$. With $w=1$, the conditional <expected value> remains $p$, yet realized frequencies wander through <genetic drift>. With $w>1$, <genetic drift> can still lower $p$ or lose $A$ entirely. Over several generations, the deterministic curve need not equal the mean stochastic trajectory, because the recurrence is nonlinear.
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