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Is it possible to have a mixed parentage such that you are, for example, 10% Dutch and 90% English? One generation up your mother would be 100% English and your father 20% Dutch and 80% English etc etc. How do you figure this out back up to the ancestors that were 100% Dutch / 100% English?

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5  
It depends how you count. If the mother's lineage and the father's lineage both include Genghis Khan (which in some parts of the world is very likely, he and his sons were enthusiastic breeders), how many times shall he be counted? And to what country shall he be assigned? Ethnicity is a concept that tends to evaporate under closer examination. –  André Nicolas Nov 28 '11 at 14:54
    
@AndréNicolas You're completely right of course! I was just wondering how to solve it mathematically. :) –  c4urself Nov 28 '11 at 15:10
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Let's work with dogs instead of people, since for dog breeds there is an official concept of pure-bred (which is ridiculous, but at least well-defined). If $A$ and $B$ are poodle by reduced fractions $x$ and $y$, then their litter is poodle fraction $\frac{x+y}{2}$. Suppose that we are looking at the (mixed) distant descendants of pure-breds. Then starting poodle fractions of the ancestors in every line are $1$ or $0$, so all subsequent poodle fractions are of the shape $\frac{k}{2^n}$, where $k$ is an integer. And $\frac{1}{10}$ is not such a fraction. –  André Nicolas Nov 28 '11 at 15:26

2 Answers 2

up vote 6 down vote accepted

First I claim that all the proportions we can reach starting from ancestors who were purely English or purely Dutch are of the form $(\sum 2^{\eta_i}, \sum 2^{\delta_i})$ where both sums contain finitely many (possibly zero) terms. This can be proven by induction: Suppose that parents of some person have proportions of form $(\sum 2^{\eta^1_i}, \sum 2^{\delta^1_i})$ and $(\sum 2^{\eta^2_i}, \sum 2^{\delta^2_i})$. Then the person has proportions of $(\sum 2^{\eta^1_i - 1} + \sum 2^{\eta_i^2 - 1}, \sum 2^{\delta_i^1 - 1} + \sum 2^{\delta_i^2 - 1})$, which are of the wanted form.

Next I'll show that $1/10$ can't be represented in this form. Suppose in contrary that there exist integers $a_i > 0$, $i=1,\dots,m$, such that $$\frac{1}{10} = \sum_{i=1}^m 2^{-a_i}.$$ Let $a_m$ be the largest of $a_i$. Then we get $$2^{a_m - 1} = 5 \sum_{i=1}^m 2^{a_m - a_i}.$$ But this is a contradiction because now both sides are integers and the right hand side is divisible by $5$ whereas the left hand side is not.

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J. J.'s answer and André Nicolas's second comment have shown why you cannot have exactly "10% Dutch, 90% English". This is because anyone born to parents who are a fraction $y$ Dutch and fraction $z$ Dutch respectively is $\displaystyle x = \frac{y+z}{2}$ Dutch, so we can inductively prove (assuming we start in the sufficiently distant past with ancestors who are either 100% Dutch or 0% Dutch) that $x$ must be a rational number whose denominator (in lowest terms) is a power of $2$. And in the fraction $\frac{1}{10}$, the denominator $10$ is not a power of $2$.

However, note that you can get arbitrarily close to "10% Dutch, 90% English", with enough generations. Specifically, with $k$ generations, since you want $\displaystyle \frac{p}{2^k} \approx \frac{1}{10}$, you can pick $p$ to be the closest integer to $\displaystyle \frac{2^k}{10}$. So you can have, after

  • 3 generations: $(1/8, 7/8) = (12.5\%, 87.5\%)$
  • 5 generations: $(3/32, 29/32) = (9.375\%, 90.625\%)$
  • 7 generations: $(13/128, 115/128) = (10.15625\%, 89.84375\%)$
  • 9 generations: $(51/512 , 461/512) = (9.9609375\%, 90.0390625\%)$
  • 11 generations: $(205/2048 , 1843/2048) = (10.009765625\%, 89.990234375\%)$
  • 13 generations: $(819/8192 , 7373/8192) = (9.99755859375\%, 90.0024414062\%)$

etc.

(Aside: Perhaps there's a way to do these approximations using continued fractions, but I can't think of anything.)

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