Table 3.

Probability of observing genotypes given IBD sharing.

Genotypes lj,1, lj,2	IBD=0	IBD=1	IBD=2
AA,AA	$P\left(AAAA\right)=p_A^4$	$P\left(AAA\right)=p_A^3$	$P\left(AA\right)=p_A^2$
AA,AB	$2P\left(AAAB\right)=2p_A^3{p}_B$	$P\left(AAB\right)=p_A^2{p}_B$	0
AA,BB	$P\left(AABB\right)=p_A^2{p}_B^2$	0	0
AA,BC	$2P\left(AABC\right)=2p_A^2{p}_B{p}_C$	0	0
AB,AB	$4P\left(AABB\right)=4p_A^2{p}_B^2$	P(AAB) + (ABB) = pApB(pA + pB)	2P(AB) = 2pApB
AB,AC	$4P\left(AABC\right)=4p_A^2{p}_B{p}_C$	P(ABC) = pApBpC	0
AB,CD	4P(ABCD) = 4pApBpCpD	0	0