Can you be 1/12th Cherokee?

I was watching an old Daily Show clip and someone self-identified as “one twelfth Cherokee”. It sounded peculiar, as people usually state they’re “1/16th”, or generally $1/2^n, n \in \mathbb{N}$.

Obviously you could also be some summation of same to achieve 3/32nds, etc., but will an irreducible fraction with the numerator 1 always need a power-of-two denominator? More generally does it always require a power-of-two denominator?

Assumptions (as per comments):

  1. Nobody can trace their lineage infinitely
  2. Incest is OK, including transgenerational, but someone can’t be their own parent.

Answer

This depends on the model. Instead of arguing that we have only $46$ chromosomes and cross-overs or whatever the mechanism is called are not that common, let us assume a continuous model.
That is, a priori, everybody can be $\alpha$ Cherokee for any $\alpha\in[0,1]$ and the rules are as follows

  • Everybody has exactly two parents.
  • If the parents have Cherokee coefficients $\alpha_m, \alpha_f$, then the child has $\alpha=\frac{\alpha_m+\alpha_f}2$
  • In a sufficiently large but finite number of generations ago, people had $\alpha\in\{0,1\}$

It follows by induction, that $\alpha$ can always be expressed as $\alpha=\frac{k}{2^n}$ with $k,n\in\mathbb N_0$ and $0\le k\le 2^n$. Consequently $\alpha=\frac1{12}$ is not possible exactly (though for example $\frac{85}{1024}\approx\frac1{12}$ would be possible).
It doesn’t matter if there is any type of inbreeding taking place anywhere in the tree (or then not-tree) of ancestors.
The only way to obtain $\alpha$ not of this form would involve time-travel and genealogical paradoxes: If you travel back in time and paradoxically become your own grandparent and one of the other three grandparents is $\frac14$-Cherokee and the others are $0$-Cherokee, you end up as a solution to $\alpha=\frac{\frac14+\alpha}4$, i.e. $\alpha=\frac1{12}$.

Attribution
Source : Link , Question Author : Nick T , Answer Author : Hagen von Eitzen

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