Probability of common ancestors after n generations from pool of x

Question

I'm no mathematician, so a simple answer with omitted technicalities could still be appreciated. While discussing with a friend, I've came to that question :

Say that there is initially a pool of 10 000 people, 50% being male and 50% female. Assume for simplicity that there is no immigration, and that, on average, a couple of two individuals have 4 children. Assume for simplicity that every generation have children at the same age.

How many generations would it take for the probability of two individual from the resulting population of having ancestors in common to reach 0.5? What about 0.8?

Thanks a lot.

Answer 1

Say we're on generation N, so that there are $P_N=10000\cdot 2^N$ people. The probability that two people in this generation don't share parents is $\frac{P_N-3}{P_N}$. Given that they don't share parents, the probability that they don't share grandparents is $\frac{3}{P_{N-1}}\left(\frac{(P_{N-1}-4)(P_{N-1}-5)}{(P_{N-1}-2)(P_{N-1}-3)}\right)+\frac{P_{N-1}-3}{P_{N-1}}\left(\frac{P_{N-1}-8)(P_{N-1}-9)}{(P_{N-1}-2)(P_{N-1}-3)}\right)$, where here we are conditioning on the event that one of the parents are themselves from the same family. After that the calculations get a bit hairy, unless you assume no direct inbreeding, in which case the probability of sharing no grandparents given distinct parents is now $\frac{(P_{N-1}-8)(P_{N-1}-12)}{6(P_{N-1}-8)+(P_{N-1}-8)(P_{N-1}-12)}$. If you can see where im going with these probabilities then the probably of sharing no ancestors is the product of all these probabilities from N back to 1. If you can set this product greater than, say 0.8 to find N, although you may need a computer program. Interestingly I think as your parents are having more than 2 children I think you could find that some probabilities will never be reached for any $N$.

Answer 2

Say Ted and Jane are in the later generation. Seven generations ago, they have 128 ancestors each. Maybe fewer, due to inbreeding.

There are $128\times128=16384$ chances of a match. Each chance is 0.0001, and the chance of a miss is 0.9999.

The odds of no match, or all misses, is $0.9999^{16384}=0.194$. Ted and Jane probably have a common ancestor.

Eight generations ago, there are $256$ ancestors each, $256\times256=65536$ and $0.9999^{65536}=0.0014$, so they are almost certainly related.

Six generations ago, $0.9999^{4096}=0.664$, and only one chance in three.

In general, compare the number of ancestors with the square-root of the population size. In a well-mixed city of a million, things become likely around the tenth generation.

I tried to take account of inbreeding, and got the following numbers; as the comments below say, they are probably wrong.

$$\begin{array}{c|c}N&P\\1&0.9996\\2&0.9984\\ 3&0.9936\\4&0.9748\\5&0.9030\\6&0.6656\\7&0.1984\\8&0.0017 \end{array}$$

EDIT:

I did a simulation. Marriage has the effect of replacing people with couples. The population starts with 5000 couples rather than 10000 people, but people have half as many ancestors (two couples of grandparents rather than four individual grandparents.) Seven generations ago, the chance of no common ancestor is more like $0.9998^{4096}=0.441$ instead of 0.194, so it is less likely they have a common ancestor. Inbreeding also reduces the chance of common ancestors.