The famous 'birthday problem' finds that it takes 23 people to have a 50% chance of some two people having the same birthday.

However, how does the question change if given that person 1's birthday is January 1st, how many people would it take to have 50% chance of another January 1st birthday?

I think that the second question would require more people because we are looking for a specific pair while the original is looking for any pair. But exactly how are the numbers affected?


If you have $n$ people, number one has a birthday on the first day of the year and the remaining $n-1$ of their birthdays are independent and uniformly distributed across a 365 day year, then the chances that someone else also has a birthday on the first day of the year equals one less the chance that none of the other $n-1$ have a birthday on the first day of the year.

So, you want to figure out the smallest $n$ such that $1-(\frac{364}{365})^{n-1} \ge \frac{1}{2}$. This means we need $n \geq 1 + \frac{\ln \frac{1}{2}}{\ln \frac{364}{365}} \approx 253.65$, so we need $n = 254$.


You are correct that the number will be higher. In fact, it will be much higher.

Ignoring leap years, each person has a $1\over365$ chance of having a January 1 birthday, and a $364\over365$ chance of having some other birthday.

Given a group of $n$ people, the chance that none of them have a January 1 birthday, then, is $\frac{364}{365}^n$, so the chance that at least one has a January 1 birthday is $1-\frac{364}{365}^n$.

You asked how large $n$ must be to make this chance over $\frac12$, or equivalently how large $n$ must be to make $\frac{364}{365}^n < \frac12$.

Since $\frac{364}{365}^{252} = 0.5008\dots$, and $\frac{364}{365}^{253} = 0.4995\dots$, the minimum number of people to get a $50\%$ chance of a January 1 birthday is $253$.

  • $\begingroup$ You need $n-1$ in the exponent, the first person is on the first of Jan., so there are $n-1$ remaining. This will make the answer 254, not 253. $\endgroup$ – copper.hat Sep 25 '13 at 6:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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