# Sampling pairs without replacement from two urns with overlapping populations

I have been trying to derive the coefficient of inbreeding lately, but I am at a loss. If I understand correctly it measures the probability of an allele "collision" i.e. getting the same allele from both father and mother.

As an urn problem offspring from two sibling can be stated as having two urns with marbles $\{a, b, c, d\}$ in one and $\{a, e, c, g\}$ in the other, so the intersection between the two are $\{a, c\}$. In each draw you take one marble from each urn without replacement. What is the probability of drawing at least one "collision" over 4 draws, i.e. a situation where either $(a,a)$ or $(c,c)$ comes up? What about the probability of having two collisions?

I have an idea that it is related to the hypergeometric distribution, but I can't figure out how to include that drawing $(a,c)$ removes the possibility of of drawing $(c,c)$.

• You can start by noticing that this is an extension of birthday paradox problem.
– Tim
Commented Nov 21, 2016 at 13:14

Counting will answer this. There are in effect $$4!=24$$ equally likely ways of making the four draws. Of these:
• Drawing $$(a,a)$$ and $$(c,c)$$ can be done $$2$$ ways, with the other draws being $$(b,e)$$ and $$(d,g)$$ or being $$(b,g)$$ and $$(d,e)$$
• Drawing $$(a,a)$$ but not $$(c,c)$$ can be done $$2\times 2=4$$ ways, since each of the $$c$$s can be paired with one of two other marbles fixing the remaining pair
• Similarly drawing $$(c,c)$$ but not $$(a,a)$$ can be done $$4$$ ways
• Leaving $$24-2-4-4=14$$ ways of drawing four non-matching pairs
• $$2$$ matching pairs is $$\frac1{12}$$
• $$1$$ matching pair is $$\frac1{3}$$
• $$0$$ matching pairs is $$\frac7{12}$$