What are the odds of three people having consecutive birthdays?

Question

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?

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.