Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

There are $2$ maternity hospitals in the town. In the first hospital $50$ children are born every day, in the second $100$ children.

Which hospital will most often experience that the number of new-born boys and new-born girls are the same? Why?

share|cite|improve this question
Huh? \begin{align} \end{align} – Inquest Dec 11 '12 at 19:47
The smaller hospital, by a factor of about $\sqrt{2}$. – André Nicolas Dec 11 '12 at 19:48
Can you calculate the probability the number of boy, and girl new-born be the same for the first and the second hospital? do you have a problem comparing these number? – clark Dec 11 '12 at 19:53

Make the usual (false) assumptions of independence, and equality of probability of girl and boy.

The probability that the smaller hospital hospital has an equal number of girls and boys is $$\binom{50}{25}\left(\frac{1}{2}\right)^{50}.$$

In the bigger hospital, the probability is

We want to compare these rather complicated numbers. The Stirling approximation is overkill, but it yields that the ratio of the first number to the second is about $\sqrt{2}$. (This comes from $\sqrt{100/50}$.) So not only do we know that the probability of equality is greater in the smaller hospital, we also know by roughly how much.

We can do it with less machinery. The idea is to show that if we increase the number of births by $2$, the probability of equality decreases.

With $2n$ births, the probability of equality is $$\binom{2n}{n}\left(\frac{1}{2}\right)^{2n}.$$ With $2n+2$ births, the probability is $$\binom{2n+2}{n+1}\left(\frac{1}{2}\right)^{2n+2}.$$ Calculate the ratio. By expressing the binomial coefficients in terms of factorials, we get a lot of cancellation. The ratio turns out to be $$\frac{4(n+1)(n+1)}{(2n+2)(2n+1)},$$ which simplifies to $1+\dfrac{1}{2n+1}$, a number $\gt 1$.

Remark: If you want to proceed much more informally, calculate the probability of equality of sexes in $2$ births, $4$, $6$, $8$, and $10$. There will be a steady decrease.

A surprising number of people believe that the probability of equality of heads and tails increases as the number of tosses gets large. Completely false!

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.