One-generation random-mating equilibrium ($p=u+v/2,(p^2,2p(1-p),(1-p{)}^2)$)

If parental genotype frequencies are $(u,v,w)$ and the two parental gametes are independently sampled from the same allele pool, Mendelian segregation gives allele frequency $p=u+v/2$. Their independent union has the displayed Hardy-Weinberg proportions, irrespective of parental genotype proportions. Equal allele frequencies in the two sexes and no transmission or viability distortion are part of this conclusion.