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The Hardy-Weinberg principle describes the genotype proportions produced by independently combining two parental gametes from the same allele pool. For alleles $A,a$ with allele frequencies $p,q$, where $p+q=1$, the Hardy-Weinberg proportions areThe two ordered heterozygote outcomes contribute $pq+qp$. These proportions sum to $(p+q{)}^2=1$, and their implied $A$ allele frequency is $p^2+(2pq)/2=p$. Reshuffling copies therefore preserves their frequencies.

For stable Hardy-Weinberg equilibrium across generations, use a large population with random mating, equal allele frequencies in the parental pools, ordinary Mendelian segregation, and no natural selection, mutation, gene flow or genetic drift. Negligible genetic drift is an idealization of large population size. The one-generation random-mating equilibrium needs independent union of gametes; persistence additionally requires processes that change allele frequencies to be absent.

At $p=0.70,q=0.30$, the predicted genotype proportions are $0.49,0.42,0.09$. For 100 sampled individuals the expected values of the counts are $49,42,9$, rather than the $25,50,25$ of a specified $Aa\times Aa$ cross. The latter fixes both parental genotypes; Hardy-Weinberg proportions average over population mating.

An observed population need not have these proportions. For example, $70AA$, no $Aa$, and $30aa$ among 100 individuals still gives $p=0.70$, but has no heterozygotes. Random mating predicts $0.42$ heterozygosity in the next generation under the stated model. A finite sample can also depart by chance, so a discrepancy is a starting point for investigating inbreeding, population stratification or genotyping error, rather than proof of one cause.

Hardy's 1908 note made the conceptual payoff explicit: a dominant phenotype does not by itself make its allele spread. Complete dominance describes expression; changes in frequency require a separate mechanism such as natural selection. See Hardy, Mendelian Proportions in a Mixed Population, Science 28 (1908), 49–50.

Let the initial genotype frequencies be arbitrary $u,v,w$ for $a/a,a/b,b/b$, with $u+v+w=1$. Mendelian segregation makes the allele frequency among gametesIn random mating, the two parents are selected independently. If $g(G)$ is a parent's transmission probability for allele $a$, then independence gives $\mathbb{E}[g(G_1)g(G_2)]=\mathbb{E}g(G_1)\mathbb{E}g(G_2)=p^2$. Thus the two independently united gametes have allele probabilities $p,1-p$, even though the two alleles within an original parent need not have been independent. The offspring frequencies are thereforeNo initial condition $v^2=4uw$ was used. This proves one-generation random-mating equilibrium. The usual assumptions are a common allele pool in the two sexes, random union of gametes, ordinary Mendelian segregation, and no intervening mutation or genotype-dependent survival selection. Unequal male and female allele frequencies, for example, would instead give $p_m{p}_f$ for $a/a$, not generally the square of a single common frequency.