One quarter of couples in a society have no children. The other three quarters have exactly three children, with each child being equally likely to be a boy or girl. What is the probability that the male line of descent (let's identify this by "last name") of a particular husband will die out?

My assumption is that all males of initial population have different last names. I have: Probability that a last name survives a generation is: $$0.25\cdot0+0.75(1- (1/2)^3)$$ that is, the probability that at least one boy is born.

since the expected value of boys is 7/8, any male line of descent will die out...

  • $\begingroup$ Not sure the population is shrinking: each 8 people (four couples) produce 9 children. $\endgroup$ – Ethan Bolker Apr 8 '16 at 17:00
  • $\begingroup$ thanks, you're right, ill edit question accordingly. $\endgroup$ – ak87 Apr 8 '16 at 17:05
  • $\begingroup$ Only one surname will survive, wichever the probability, as you can't take back an extinct surname. $\endgroup$ – N74 Apr 8 '16 at 17:26

A rather patriarchal question.

The husband has no male descendants with probability $\frac14$ and $k$ male descendants with probability $\frac34\cdot\frac18\cdot\binom3k$. Thus the extinction probability $p$ satisfies

$$ p=\frac14+\frac3{32}\left(1+3p+3p^2+p^3\right)\;. $$

Since one solution of such an equation is always $p=1$ but this solution doesn't apply here because the average number of male descendants is $\gt1$, we can divide through by $p-1$ to find

$$ 3p^2+12p-11=0 $$

with solutions $p=-2\pm\sqrt{23/3}$. The solution in $[0,1]$ is $p=-2+\sqrt{23/3}\approx0.76887$.


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