Common ancestors

(by Christian Robert)

In conjunction with President Obama’s visit to Ireland last week and in particular to his ancestral Irish town, I happened to glance at his family tree and saw that he shared a common ancestor with George W. Bush. (They are 11th cousins, meaning that a 12th-order ancestor is common to both their  family trees.) This sounds at first amazing, but it is another occurrence of the (von Mises) birthday problem. (The fact that it is not that amazing is demonstrated by the simultaneous presence of [French!] ancestors of Dick Cheney in the same tree.) If we consider President Obama’s mother side, the probability that all of her 11th-order ancestors differ from all of George W. Bush’s 12th-order ancestors is

p=\dfrac{(M-2^{12})(M-2^{12}-1)\cdots(M-2^{12}-2^{11}+1)}{M(M-1)\cdots(M-2^{11}+1)}

where M denotes the whole population of potential ancestors at this period. If we consider all those ancestors as coming from the British Isles, then about 1650, the population was about 8 million. This would lead to a probability of p=0.35, i.e. there is a 65% chance that they share a 12th-order ancestor. If instead we consider the whole European population at that time (if only to include German and French ancestors to President Obama), M is about 100 million and the probability increases to p=0.92, so there is then an 8% probability for them to share an ancestor. Obviously, this rough calculation relies on simplifying assumptions, avoiding the issue of inbreeding which means that the potential 2¹¹ ancestors are in fact much less than 2¹¹, and the fact that their ancestors are necessarily emigrants, which reduces the value of M

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