Birthday paradox

The birthday coincidence, also known as the birthday paradox or the birthday problem, results when in a small group of people, there are two that celebrate their birthday on the same day of the year. It is surprising to most people, how small a group of people are needed to have a 50-50 probability of having a matching birthday. (In this article leap year is ignored.)

To show how to calculate the probability of a group finding such a match, it is simpler to first find the probability of all the birthdays being different. Consider a group of two people. The first person can have been born on any of the 365 days of the year, while the second must have been born on one of the other 364 days in order to not match. The first person has a probability of ${\frac {365}{365}}$ , which equals 1.0, and the second has a probability of ${\frac {364}{365}}$ which is 0.9973. Multiplying these probabilities together gives a net probability of 0.9973 for having different birthdays. Subtracting this number from 1.0 gives 0.0027 probability of having the same birthday.

The third person in the group, having to miss the birthdays of the first two people, has only 363 days, with a probability of ${\frac {363}{365}}$ which is 0.9945. Multiplying this on to the previous result, gives 0.9918.

For four in the group the fractions are

${\frac {365}{365}}\cdot {\frac {364}{365}}\cdot {\frac {363}{365}}\cdot {\frac {362}{365}}=0.9836\dots .$

And the probability of a pair matching birthdays is 1 − 0.9836 = 0.0164, over 1 percent.

As more and more people come into the group, the number of days available, not matching any of the previous ones, decreases; and the probability of everyone having a different birthday decreases. Thus the probability of there being two with the same birthday increases. Continuing to multiply more and more, smaller and smaller fractions, eventually a point near 0.5 is exceeded as soon as the group consists of 23 people.

The formula for the probability of having a matching birthday in a group of n people is

$p(n)=1-{\frac {365!}{(365-n)!\cdot 365^{n}}}$