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A couple has four kids already, all girls. the couple would like to have a son and would like to give it another try. what is the probability that the next kid will be a girl?

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    $\begingroup$ half because it's independent of the gender of the previous 4 kids $\endgroup$ Mar 6 '14 at 17:19
  • $\begingroup$ On the assumption of independence, the events are independent. Parenthetically, even without sex determination interference, more boys are born than girls. (But quite soon there are more girls: males are more fragile.) $\endgroup$ Mar 6 '14 at 17:25
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    $\begingroup$ Be sure to condition the probability on the husband being Henry VIII. $\endgroup$
    – Emily
    Mar 6 '14 at 17:29
  • $\begingroup$ With data, we could verify if events are really independent. It is not obvious to me $\endgroup$ Mar 6 '14 at 17:30
  • $\begingroup$ This is actually a somewhat profound question, 18th century mathematician Pierre-Simon Laplace put it in another context: en.wikipedia.org/wiki/Sunrise_problem $\endgroup$
    – DanielV
    Mar 6 '14 at 21:50
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If we presume that we know nothing about the actual probability of having a girl, $p$, then we can use Hinkley's predictive likelihood for the binomial to predict the probability that the next birth will also be a girl, assuming births are conditionally independent, given $p$. This is sensible, since the probability that a couple will have a girl or boy is affected primarily by the X/Y ratio of the male gametes, and to some extent selective factors in the female. Therefore, while overall the bias is slightly toward male births, there will be overdispersion around this mean due to individual propensities that are at odds with the overall average.

Let $g$ be the number of girls born to date and $n$ be the number of children born to-date. For general prediction, we want to know the probability that there will be $r$ girls in the next $s\geq r$ births.

$P(r|g,n,s) = K\frac{{n \choose g}{s \choose r}}{n+s\choose g+r}$, where $K$ is a normalizing constant to make $\sum\limits_{r=0}^s\frac{{n \choose g}{s \choose r}}{n+s\choose g+r} = 1$

In your case, we are dealing with $s=1$ and $n=g=4$, which greatly simplifies things, since there are only two outcomes ($r=0,r=1$) and ${1\choose 0} = {1 \choose 1} = 1$. The simplified formula is:

$P(r|4,4,1) = K\frac{{4 \choose 4}{1 \choose r}}{5\choose 4+r}= K\frac{(4+r)!(r)!}{120} = K\left( \frac{1}{5}+\frac{4r}{5}\right)$, (since $r$ can only be $1$ or $0$)

In this case, there are only two values of $r$ so getting $K$ is easy: $P(r=0|4,4,1) + P(r=1|4,4,1) = 1 \rightarrow K \left( \frac{1}{5} + 1\right) = 1 \therefore K=\frac{5}{6}$

Given the above, the probability that the next child is a girl, given that all 4 previous children were girls, is: $P(r=1|4,4,1)=\frac{5}{6} \approx 83\%$

A more detailed exposition using more sophisticated models can be found here - it studies possible models for the overdispersion of actual sex ratios in the human population, which lends evidence that the probability of having a girl can vary either between families or even within a single family over time.

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    $\begingroup$ How could meiosis possibly be changed to produce a different X/Y ratio? That alters the fundamental aspects of the process. $\endgroup$
    – qwr
    Mar 6 '14 at 21:40
  • $\begingroup$ $\frac 5 6$ is the answer you get if you assume a uniform distribution of bias towards sex, but here you seem to assume a binomial distribution. Is that a coincidence or just two different ways of saying the same thing? $\endgroup$
    – DanielV
    Mar 6 '14 at 21:47
  • $\begingroup$ @qwr I don't know the mechanism or why this happens, but the link I showed you observed overdispersion in the intra-family sex ratios, hence there is some sort of pair-effect happening. If there werent, then the undispersed binomial model would do just fine. $\endgroup$
    – user76844
    Mar 6 '14 at 21:49
  • $\begingroup$ @DanielV you are referring to the Bayesian formulation. Hinkley used a pure likelihood approach, which does not use a prior, but it give you the same answer as a Bayesian estimate with a uniform prior. $\endgroup$
    – user76844
    Mar 6 '14 at 21:50
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    $\begingroup$ @gina If you are indeed using a textbook problem, then I'd suggest qwr's answer, as it uses the prior knowldge that births are likely to be near 50% boys. Mine makes no prior assumptions and so has the potential for more variability. $\endgroup$
    – user76844
    Mar 6 '14 at 22:00
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i agree. having the 4 girls in the beginning means nothing: it does not affect if the next child will be a girl or a boy. So it is independent from everything and the chances is 50%

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This problem seems to be homework, so I assume the problem is not a complex biology problem and there's a 50/50 chance of boy or girl. We can assume the probabilities are independent to each other, so the probability is $50\%$.

Notes: The average male/female ratio in the world is about $1:1$, and it's safe to assume half of sperm contain an X chromosome and other half contain Y from meiosis. I think it's also safe to assume fertilization is independent.

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