pssandwich
Pro-life, MRA, Social Democrat, Environmentalist, Atheist
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User ID: 2755
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Pro-life, MRA, Social Democrat, Environmentalist, Atheist
User ID: 2755
Pro-life, MRA, Social Democrat, Environmentalist, Atheist
No bio...
User ID: 2755
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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