The problem is to compute the approximate probability that in a room of n people, at least two have the same birthday. For simplicity, disregard variations in the distribution, such as leap years, twins, seasonal variations, etc., and assume that the 365 possible birthdays are equally likely. Real-life birthday distributions are not uniform since not all dates are equally likely.
If P(A) is the probability of at least two people in the room having the same birthday, it may be simpler to calculate P(A’), the probability of there not being any two people having the same birthday. Then, because P(A) and P(A’) are the only two possibilities and are also mutually exclusive, P(A’) = 1 − P(A).
In deference to widely published solutions concluding that 23 is the number of people necessary to have a P(A) that is greater than 50%, the following calculation of P(A) will use 23 people as an example.
When events are independent of each other, the probability of all of the events occurring is equal to a product of the probabilities of each of the events occurring. Therefore, if P(A’) can be described as 23 independent events, P(A’) could be calculated as P(1) × P(2) × P(3) × … × P(23).
The 23 independent events correspond to the 23 people, and can be defined in order. Each event can be defined as the corresponding person not sharing their birthday with any of the previously analyzed people. For Event 1, there are no previously analyzed people. Therefore, the probability, P(1), that person number 1 does not share his/her birthday with previously analyzed people is 1, or 100%. Ignoring leap years for this analysis, the probability of 1 can also be written as 365/365, for reasons that will become clear below.
For Event 2, the only previously analyzed people are Person 1. Assuming that birthdays are equally likely to happen on each of the 365 days of the year, the probability, P(2), that Person 2 has a different birthday than Person 1 is 364/365. This is because, if Person 2 was born on any of the other 364 days of the year, Persons 1 and 2 will not share the same birthday.
Similarly, if Person 3 is born on any of the 363 days of the year other than the birthdays of Persons 1 and 2, Person 3 will not share their birthday. This makes the probability P(3) = 363/365.
This analysis continues until Person 23 is reached, whose probability of not sharing their birthday with people analyzed before, P(23), is 343/365.
P(A’) is equal to the product of these individual probabilities:
(1) P(A’) = 365/365 × 364/365 × 363/365 × 362/365 × … × 343/365
The terms of equation (1) can be collected to arrive at:
(2) P(A’) = (1/365)²³ × (365 × 364 × 363 × … × 343)
Evaluating equation (2) gives P(A’) = 0.492703
Therefore, P(A) = 1 − 0.492703 = 0.507297 (50.7297%) or the chances are better than 1 in 2 that two people in a group of 23 will share a birthday.
