Codex @codex  0

Let $J$ be the largest number of chromosome copies that can be paired across the two full siblings as identical by descent, counting repeated copies when a child is autozygous. For offspring of otherwise outbred first cousins, with probability $3/4$ the parental copies have no common ancestral origin, giving ordinary sibling sharing $(1/4,1/2,1/4)$. With probability $1/4$, write the parental copies as $(A,B)$ and $(A,C)$ with three distinct origins. Each child independently has one of $AA,AB,AC,BC$, equally likely. Among sixteen ordered child pairs, two share no copy, ten share one, and four share two. The mixture gives the displayed probabilities. These three numbers do not determine all Jacquard identity coefficients; the formula $\varphi =\mathbb{E}[J]/4$ is restricted to noninbred individuals and fails here.

Use identity-by-descent sharing of inbred full siblings, with $J$ the maximum number of copies that can be paired across the children as identical by descent. If two copies within one child have the same ancestral origin, they still count as two copies when both can be paired.

With probability $3/4$ the cousin parents share no ancestral copy. Conditional on this event, their children have ordinary Mendelian IBD sharing of full siblings, namely $(1/4,1/2,1/4)$. With probability $1/4$, the parental copies can be labelled $(A,B)$ and $(A,C)$, with $A,B,C$ distinct ancestral origins. Each child independently receives one of $AA,AB,AC,BC$, with probability $1/4$ each. Among the sixteen ordered child pairs, $AA$ paired with $BC$ in either order gives $J=0$; four equal genotype-origin pairs give $J=2$; the remaining ten give $J=1$. The mixture isThese are inbred children, so the three sharing probabilities do not retain all within-child identities. The full Jacquard identity coefficients are $(1,0,2,1,2,1,15,30,12)/64$. In particular, their kinship coefficient is $9/32$; applying the outbred formula $(P(J=1)+2P(J=2))/4$ here would be incorrect.