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

What are the odds of a mother giving birth to 3 children on 3 consecutive days, different years? Example: June 7, 1960, June 8, 1964, June 9, 1968 These were all natural births, no induced deliveries.

share|cite|improve this question
That needs some pretty specific assumptions about the probability distribution of the time between two consecutive births. Lots of explicit choice on the part of the parents go into the calculation too. – Henning Makholm Jun 2 '12 at 20:29
If those are all C-sections then the probability is pretty good. – Asaf Karagila Jun 2 '12 at 20:36
The question cannot be answered without knowledge of a vast array of real world data. Is an average baby equally likely to be born on any given day of the year? How many mothers have three children at all? – Zev Chonoles Jun 2 '12 at 20:40
Perhaps her husband worked was a sailor and came visit home the first days of every september...? things like these can affect probability in a serious way. – DonAntonio Jun 3 '12 at 2:08

One draws uniformly and independently three elements $x_1$, $x_2$ and $x_3$ of $\mathbb Z/N\mathbb Z$. Consider the event $A$ that there exists some $k$ in $\mathbb Z/N\mathbb Z$ such that $\{x_1,x_2,x_3\}=\{k,k+1,k+2\}$. Assume that $x_1=i$. Then, $A$ can happen in several ways:

  • If $x_2=i+1$, then $x_3=i+2$ or $x_3=i-1$: this happens with probability $\frac1N\frac2N$.
  • If $x_2=i-1$, then $x_3=i-2$ or $x_3=i+1$: this happens with probability $\frac1N\frac2N$.
  • If $x_2=i+2$, then $x_3=i+1$: this happens with probability $\frac1N\frac1N$.
  • If $x_2=i-2$, then $x_3=i-1$: this happens with probability $\frac1N\frac1N$.

Thus, the probability that $A$ happens is $\frac6{N^2}$.

In the present case, $N=365$ but note that the result only assumes that all the points $i-2$, $i-1$, $i$, $i+1$ and $i+2$ are different in $\mathbb Z/N\mathbb Z$, that is, that $N\geqslant5$. This also considers that December 31st and January 1st are consecutive days. More importantly, as explained in the comments, this assumes at the onset that some nontrivial and pretty unrealistic modeling hypothesis hold: the uniform distribution of each birth date and the independence of the different birth dates.

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.