birthday problem

The birthday problemMathworldPlanetmath is: What is the probability of two or more people out of a of n do have the same birthday?

Solution. Let A be the event two or more people out of a group of n to have the same birthday. It is easy to find first the complement of A, Ac, which is that no two people out of a group of n will have matching birthdays out of 365 (number of days of one year) equally possible birthdays. We start with an arbitrary person’s birthday. So the second person’s birthday in order to be different from the first one gives us the probability, 1-1365. Consecutively, the third person’s birthday is different from the first two and we have (1-1365)(1-2365). Finally for the the nth person’s birthday we get the probability

P[Ac] =(1-1365)(1-2365)(1-n-1365)

By the last expression of the estimated probability we can see that there exists another easier approach to find the probability which is the following:

  1. 1.

    We set the possible ways for 1n birthdays, 365n as the denominator.

  2. 2.

    Using the binomial coefficientDlmfDlmfMathworldPlanetmath (365n), which represents the number of ways of picking n unordered birthdays from 365 days multiplied by the factorialMathworldPlanetmath n! used for the number of orders, in that way we determine the numerator.

Therefore we have




The table below shows the probability P[A] for some values of n

n P[A]
10 0,11695
20 0,41144
23 0,50730
30 0,70632
40 0,89123
50 0,97037
Title birthday problem
Canonical name BirthdayProblem
Date of creation 2013-03-22 16:09:38
Last modified on 2013-03-22 16:09:38
Owner PrimeFan (13766)
Last modified by PrimeFan (13766)
Numerical id 16
Author PrimeFan (13766)
Entry type Definition
Classification msc 05A10
Classification msc 60C05
Classification msc 60-00
Synonym birthday paradox