= Solution
Label the two paternal ancestral marker copies $P_1,P_2$ and the two maternal copies $M_1,M_2$. Treat these as four distinct ancestral copies in an <outbred pedigree>; equal observed <alleles> need not imply <identity by descent>. By <Mendelian segregation>, each child independently receives either paternal copy with <probability> $1/2$. Thus the indicator $I_P$ that the two <full siblings> inherit the same paternal copy satisfies
$$
P(I_P=1)=P(P_1,P_1)+P(P_2,P_2)=\frac14+\frac14=\frac12.
$$
The same argument gives $P(I_M=1)=1/2$ for the maternal copy. Paternal and maternal <meioses> are independent, so $I_P$ and $I_M$ are independent. Consequently the number $J=I_P+I_M$ of copies shared <identical by descent> has a <binomial distribution>, giving
$$
\boxed{P(J=0)=\frac14,\qquad P(J=1)=\frac12,\qquad P(J=2)=\frac14.}
$$
This is <Mendelian IBD sharing of full siblings>. It describes an ordinary, unselected sib pair: conditioning on a shared disease <phenotype> can change the distribution if there is <genetic linkage> to its causal locus.
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