This online Birthday Paradox Calculator performs calculation of the probability that two or more persons in given group of people will have the same birthday.
Birthday Paradox
In probability theory and statistics, the birthday problem or birthday paradox concerns the probability that, in a group of \(n\) randomly chosen people, at least two of them will have the same birthday.
The source of confusion within the birthday paradox is that the probability grows with the number of possible pairings of people in the group, rather than the group’s size. The number of pairings grows with respect to the square of the number of participants. For example, in a group of 23 people, the probability of a shared birthday is 50%, while a group of 70 has a 99.9% chance of a shared birthday.
It’s not difficult to compute the probability \(P(A)\) that in a group of \(n\) people at least two have the same birthday. For simplicity assume that the year consists of 365 days and all 365 possible birthdays are equally likely. Then, obviously, \(P(A)\) = 1 in case of \(n\) > 365. So, in what follows we assume that \(n\) ≤ 365.
It is simpler to calculate first \(P(A′)\), the probability that no two people in the group have the same birthday. The event that all \(n\) people have different birthdays is the same as the event that person 2 does not have the same birthday as person 1, and that person 3 does not have the same birthday as either person 1 or person 2, and so on, and finally that person \(n\) does not have the same birthday as any of persons 1 through \(n\) – 1.
This conjunction of events may be computed using conditional probability. The probability of the first event is \(364/365\), as person 2 may have any birthday other than the birthday of person 1. Similarly, the probability of the second event given that the first event occurred is \(363/365\), as person 3 may have any of the birthdays not already taken by persons 1 and 2. And finally the probability of the last event given that all preceding events occurred is \((366-n)/365\). Finally, the principle of conditional probability implies that \(P(A′)\) is equal to the product of these individual probabilities:
$$\begin{align} P(A’) &= 1 \times \left(1-\frac{1}{365}\right) \times \left(1-\frac{2}{365}\right) \times \cdots \times \left(1-\frac{n-1}{365}\right) \\[6pt] &= \frac{ 365 \times 364 \times \cdots \times (365-n+1) }{ 365^n } \\[6pt] &= \frac{ 365! }{ 365^n (365-n)!}. \end{align}$$
The event of at least two of the \(n\) persons having the same birthday is complementary to all \(n\) birthdays being different. Therefore, its probability \(P(A)\) is:
$$P(A) = 1 – P(A’).$$
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