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The disease-ascertained transmission probability follows directly from Bayes' theorem. Assume that, given the child's genotype $g$, its disease probability is a common baseline penetrance times the genotype relative risk $R_g$, with no additional dependence on the parental genotypes. Write $q_g=P(g\mid g_f,g_m)$ for the Mendelian segregation probability. ThenThere are four equally likely labelled parental transmission outcomes. For one specified outcome $t$, this becomes $P(t\mid g_f,g_m,D)=R_{g(t)}/{\sum }_{t^{\prime }}{R}_{g(t^{\prime })}$, which is the four-outcome interpretation of the displayed formula.

For an unordered child genotype, let $m_g$ be the number of those outcomes producing it. Its probability is insteadFor example, two heterozygous parents have multiplicities $(1,2,1)$ for $11,12,22$. A heterozygous child therefore has probability $2R_{12}/(R_{11}+2R_{12}+R_{22})$. The four possible outcomes must be counted with multiplicity; treating their unordered genotypes as four distinct equally likely values would be incorrect.

Let $A=R_{11}+2R_{12}+R_{22}$. Among the ten complete trios, six have two heterozygous parents: three have child $11$, two have child $12$ and one has child $22$. Their conditional likelihood contribution isTwo more trios have a heterozygous parent and a $22$ parent, with child $12$ in each; they contribute $[R_{12}/(R_{12}+R_{22}){]}^2$. The remaining two parent pairs can produce only child $12$, so their conditional probabilities are one. Multiplying givesThe factor $4$ does not depend on the genotype relative risks, so the conventional proportional likelihood function isThis is the expression in the original PDF, up to the parameter-independent constant. Its last denominator is $R_{12}+R_{22}$; the converted TeX's $R_{11}+R_{22}$ is a transcription error. The factor $4$ in the exact probability comes from the two heterozygous children of two heterozygous parents, each of which has two labelled transmission routes.