# Probability of Shared Birthdays

or, How to Win Money in Bar Bets

Copyright © 2001–2020 by Stan Brown

or, How to Win Money in Bar Bets

Copyright © 2001–2020 by Stan Brown

**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

and therefore

On your TI-83, to get _{365}P_{n} you first
enter the 365, then press [`MATH`

] [`◄`

] to get the PRB menu
and [`2`

] for nPr, then enter the second number (*n*). 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 peoplethat 2 or more have the same birthday | |
---|---|

10 | 0.117 |

20 | 0.411 |

22 | 0.476 |

23 | 0.507 |

30 | 0.706 |

40 | 0.891 |

50 | 0.970 |

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!

You can download an Excel workbook to compute the probability of shared birthdays for any size group.

The workbook uses two methods to compute these probabilities.
**Method 1** uses the formula shown above.
You’ll notice #NUM errors for groups larger than
*n* = 120, because 365^{121} is bigger than the
largest number Excel can handle. In this particular case that’s
not a problem, because the probability is effectively 1 for
*n* > 118 anyway, but still all those #NUM cells
look strange.

**Method 2** solves this by computing the probability for each
size group from the probability for a group one person smaller, just
as I did in steps (a) through (d) above. That way
Excel never has to deal with numbers bigger than 365 in any one
step, and the #NUM errors are avoided.

**19 Sept 2020**: Added an Excel workbook.**25 May 2003**: New article.

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