Someone brought up in conversation that three of her friends had consecutive birthdays (such as November 10, 11, and 12), and I wanted to figure out how likely that is for any randomly selected three people, assuming that birthdays are randomly distributed and the birthdays of two people in a sample are independent. My answer:

= possible arrangement of consecutive birthdays / possible arrangements all birthdays
= 365 / 365^3
= 0.0000075 

Does that sound about right? Or am I missing something?

  • 1
    $\begingroup$ Your problem is ill-posed. The answer depends on the number of people the 3 were randomly selected from. $\endgroup$ Sep 10, 2012 at 3:18
  • $\begingroup$ combinatorics/probability $\endgroup$
    – pyCthon
    Sep 28, 2012 at 17:21
  • $\begingroup$ My daughter, her father and brother are right in a row April 16,17,18..im curious if there is anybody else like this? Brother is 16th, dad is 18th and daughter right in the middle 17th $\endgroup$
    – Rachelle
    Sep 3, 2017 at 20:55
  • $\begingroup$ @Rachelle first, this is not an answer so I turned it into a comment. Second, it is unrelated to statistics. Your question is basically: is it possible that a strange coincidence will ever happen? Yes, they happen all the time. $\endgroup$
    – Tim
    Sep 3, 2017 at 22:07

1 Answer 1


For simplicity, ignore leap days and that the distribution of birthdays is not uniform.

There are $365$ sets of consecutive triples of days. We can index them by their first day.

There are $3! = 6$ ways the $3$ people can have a particular triple of distinct birthdays.

There are $365^3$ ways the people can have birthdays, which we are assuming are equally likely.

So, the chance that three random people have consecutive birthdays is $\frac {6 \times 365}{365^3} = \frac {6}{365^2} \approx 0.0045\% \approx 1/22,000.$

Of course, if you have $60$ friends, there are ${60 \choose 3} = 34,220$ ways to choose $3$ of them, and so the average number of triples with consecutive birthdays among your friends is about $1.5$, even if you disregard the chance that the real pattern was a superset such as "consecutive or equal" or "within 2 days of each other." If this is counterintuitive, look up the Birthday Problem.

  • $\begingroup$ I had doubts about the necessity of the 3! term, so I wrote a program to select three random number between 0-364 and test if they are consecutive (including wraps). Roughly 1:22000 accurately represents my results. $\endgroup$
    – Octopus
    Aug 21, 2017 at 16:47
  • 1
    $\begingroup$ In hindsight, the 3! explains why I had to sort the order every time I tested. $\endgroup$
    – Octopus
    Aug 21, 2017 at 16:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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