THE BIRTHDAY PROBLEM

There are k people in a room. What is the probability that at least two of them have the SAME birthday, that is, month and day?

The common assumption is that every day is equally likely, that is, no twins, and no power blackouts so there was no unusual frequency of sex on a particular day, etc. We also ignore February 29 and assume a 365 day year.

For k=2, the probability they have DIFFERENT birthdays is (365 X 364) / (365 X 365) = .997. In the numerator, there are 365 possible birthdays for the first person, and once that birthday is chosen there are 364 birthdays for the second person. In the denominator, there are are 365 possible birthdays for the first person and 365 for the second person.

For k=2, the probability they have the SAME birthday is 1 - (365 X 364) / (365 X 365) = .003.

For arbitrary k, the probability they have DIFFERENT birthdays is (365 X 364 X … X (365 – k + 1)) / (365 ^ k). The carat "^" is the symbol for exponentiation., that is "365 raised to the k power." The numerator is denoted "365Pk," or "the number of permutations of 365 things, taken k at a time."

For arbitrary k, the probability that at least two of them have the SAME birthday is 1 - (365 X 364 X … X (365 – k + 1)) / (365 ^ k), or 1 – (365Pk) / (365 ^ k).

For k=23, the probability that at least two of them have the SAME birthday equals .507, which is better than a 50:50 chance. For k=30, the probability is .706.

Here is a table of some of these values:

k | Different | Same |

2 | .997 | .003 |

10 | .883 | .117 |

15 | .747 | .253 |

20 | .589 | .411 |

23 | .493 | .507 |

25 | .431 | .569 |

27 | .373 | .627 |

30 | .294 | .706 |

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