(Unspecified) (cheeser1) wrote in innumeracy,


So this happened some time ago, but I'm new to the community and I thought I should share.

After discussing my lineage with someone, an acquaintance, he commented that he too was part Irish. 1/8th Irish. He then commented that he was 1/3 Jewish.

I spent a great deal of time trying, as much as I could, to explain that this was impossible. He insisted (somehow) that he was in fact, exactly* 1/3 Jewish. I decided that whatever the case was, he was 0/3 good at math.

*added for clarity.

Further clarity:

I just re-read the community info, and it says maybe to provide explanation (I don't know if this counts as "college level"). So if anybody is wondering, no matter what part of your ancestry something is, it must be representable in least terms as k/2^n (for nonnegative integers k,n). The reason is this: at some level of your ancestry, presumably, there is a set of ancestors that are all "100%" something. There must be 2^n of them.* So you count up k of them to be something (eg Jewish), and that means you are k/2^n parts Jewish, by simply summing.

For example, if your father is 100% Italian and your mother is half Italian, half Chinese (with a 100% Italian parent, say her mother), then you skip up to that generation:

father's father: 100% Italian
father's mother: 100% Italian
mother's father: 100% Chinese
mother's mother: 100% Italian

There are 3 full-Italians, and 4 total. You are 3/4 Italian.

*regardless of if some are repeated - eg your grandfather is your mother's father and your father's father, he will essentially function as two distinct grandfathers in the summation, and will contribute two terms of 1/4.

You could also argue this inductively. Assume someone's parents are all k/2^n of anything (for various values of k,n). The child is simply the average of the two. So if the father is k/2^n of X, and the mother is p/2^q, then the child must be:

k/2^n + p/2^q = (k 2^q + p 2^n) / 2^(q+n)

While that is not in least terms, you can at best cancel out more twos. Note that this also accounts for cloning, in the sense that if the child is cloned, the inductive step involves no averaging and becomes trivial.
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