TC3 → Stan Brown → Statistics → Shared Birthdays
revised 25 May 2003

Probability of Shared Birthdays
or, How to Win Money in Bar Bets

Copyright © 2001–2014 by Stan Brown, Oak Road Systems

Summary: In a group of 30 people, would you be surprised if two of them have the same birthday? As it turns out, you should be more surprised if they don’t.

There are 365 possible birthdays. (To keep the numbers simpler, we’ll ignore leap years.) The key to assigning the probability is to think in terms of complements: “Two (or more) people share a birthday” is the complement of “All people in the group have different birthdays.” Each probability is 1 minus the other.

(a) What is the probability that any two people have different birthdays? The first person could have any birthday (p = 365÷365 = 1), and the second person could then have any of the other 364 birthdays (p = 364÷365). Multiply those two and you have about 0.9973 as the probability that any two people have different birthdays, or 1−0.9973 = 0.0027 as the probability that they have the same birthday.

(b) Now add a third person. What is the probability that her birthday is different from the other two? Since there are 363 days still “unused” out of 365, we have p = 363÷365 = about 0.9945. Multiply that by the 0.9973 for two people and you have about 0.9918, the probability that three randomly selected people will have different birthdays.

(c) Now add a fourth person, and a fifth, and so on until you have 22 people with different birthdays (p ≈ 52.4%). When you add the 23rd person, you should have p ≈ 49.3%.

(d) If the probability that 23 randomly selected people have different birthdays is 49.3%, what is the probability that two or more of them have the same birthday? 1−0.493 = 0.507 or 50.7%. In a randomly selected group of 23 people, it is slightly more likely than not that two or more of them share a birthday.

For n people (n ≤ 365), your chain of n fractions would be

Probability of n birthdays all different is 365-P-n divided by 365 to the nth power

and therefore

Probability of at least 2 dupes in n birthdays 1 minus the above

On your TI-83, to get 365Pn you first enter the 365, then press [MATH] [] to get the PRB menu and [2] for nPr, then enter the second number. In Excel, it’s PERMUT(365,n).

What if n > 365? In this case there is no need for any calculations (and in fact the above formula won’t work). If there are 366 or more people, but only 365 possible birthdays disregarding leap year, then two or more of them must share a birthday.

Here are some sample results:

Probability in a group of n people
that 2 or more have the same birthday
100.117
200.411
220.476
230.507
300.706
400.891
500.970

You can see that the dividing line is between 22 and 23 people. In a group of 22 people, the odds are less than 50–50 that two share a birthday; in a group of 23, the odds are better than 50–50. In a bar with even a small crowd, if you can get someone to take your bet that two people share a birthday, you’ll win more often than you lose.

In a randomly selected group of 50 people or more, it is nearly certain that two or more will share a birthday (p ≥ 97%). On a crowded Friday night you can really clean up — if nobody else in the bar knows probability!


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