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I have a genetics problem I don't know if my math/analysis is right because I haven't had to do math or Punnett squares in approximately 20 years and I was always kinda shit at statistics. I nerd baited myself I guess because I've been thinking about this for a week:
My wife has blue eyes. I have brown eyes as did both my parents, but both my grandmothers had blue eyes. So I know my parents were both carriers for blue eyes. The chances of me also being a carrier is 2/3 then, so the chances of my first child having blue eyes is 1/3. Our oldest came out with blue eyes so now the chances of having a kid with blue eyes is 1/2 because that confirms I'm a carrier. But what if my oldest came out with brown eyes? What would be the chances for every subsequent child to have blue eyes if all their older siblings had brown eyes?
My initial guess was 2/3s of whatever the last probability was. So kid #2 would be 2/9, then #3 would be 4/27, but that seems to drop off way too quickly. Doesn't pass the smell test.
I thought maybe I have to evaluate the probability that I'm a carrier before bringing the wife into it. So that would mean if I had a brown eyed first born, my chances of being a carrier are the chances I had a blue eyed kid for #1 [without knowing my status yet], divided by the chances that I had a blue eyed kid for #1 [without knowing my status yet] plus the chances that I had a brown eyed kid [in the event that I'm not a carrier, which is 1/3 atm]. I reasoned this because having a brown eyed kid will lower the chances of subsequent blue eyed kids so it goes in the denominator. That way, it still maintains the possibility while making it subsequently less likely.
This would be (2/3 x 1/2)/[(2/3 x 1/2)+(1/3)] which is (1/3)/(2/3) which is 1/2, making the chances of us having a blue eyed kid for #2 now 1/4. Kids number 3 and then 4 would be 1/6 and 1/10 respectively. This seems way more reasonable, but my equations are just going on vibes here. I have no idea if its right so can any math people clue me in?
This is a standard application of Bayes's theorem. The probability that you are a carrier given that you have N consecutive children with brown eyes and zero with blue eyes is1/(2N-1 + 1) so the probability that your next child will have blue eyes is 1/(2N+2) . When N = 0, this agrees with your correct statement that the probability that the first child has blue eyes will be 1/3.
Bayes's theorem says that
P(Carrier|N children with brown eyes) = P(N chlidren with brown eyes | carrier) P(carrier)/P(N children with brown eyes).
You are correct that the a priori probability that you are a carrier is 2/3. Clearly P(N children with brown eyes | carrier) = 1/2N. To compute the probability that you have N children with brown eyes unconditionally, you need to take P(N children with brown eyes | Carrier) P(Carrier) + P(N children with brown eyes | Not Carrier)P(Not Carrier) = 1/2N * 2/3 + 1 * 1/3 = 1/3 * [(2N-1 + 1)/2N-1].
Hence Bayes's theorem gives
P(Carrier|N children with brown eyes) = [1/2N * 2/3]/[1/3 * [(2N-1+1)/2N-1]] = 1/(2N-1+1).
A nice visualization of this answer:
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