Under Hardy-Weinberg equilibrium, the two parental allele copies are sampled independently from the same pool. Multiplying their frequencies, and adding the two orderings for a heterozygote, gives
Their sum is . The genotypes are unordered even though the paternal and maternal transmissions may be distinguished in the calculation.
Treat the two parents as unordered, so three possible genotypes give six mating types. A homozygous parent transmits its allele with probability one; a heterozygous parent transmits either allele with probability one half by Mendelian segregation. Independent parental transmissions give the following complete offspring distributions:
Each row sums to one. In particular, heterozygote-by-heterozygote mating can produce all three offspring genotypes, whereas the two opposite homozygotes produce only heterozygotes. These conditional offspring probabilities do not require Hardy-Weinberg equilibrium of the parents.
Let the initial genotype frequencies be arbitrary for , with . Mendelian segregation makes the allele frequency among gametes
In random mating, the two parents are selected independently. If is a parent's transmission probability for allele , then independence gives . Thus the two independently united gametes have allele probabilities , even though the two alleles within an original parent need not have been independent. The offspring frequencies are therefore
No initial condition 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 for , not generally the square of a single common frequency.

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