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PaperclipPerfector  3mo ago (text post)   2806 thread views

Small-Scale Question Sunday for November 23, 2025

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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 is 1/(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:

Draw three squares of equal area 1/3, two squares representing you being a carrier and one not. Then, carve out a 1/2N-fraction of the two carrier squares and color those blue (this represents the probability that your children all have brown eyes AND you are a carrier), and take the entire non-carrier square and color it brown (this represents the probability that your children all have brown eyes AND you are not a carrier). The probability that you are a carrier is the blue area divided by the blue + brown areas, which is [2/3 * 1/2N]/[2/3 * 1/2N + 1/3] = 1/(2N-1 + 1)

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