Birthday Problem

Consider the probability that no two people out of a group of will have matching birthdays out of equally possible birthdays. Start with an arbitrary person's birthday, then note that the probability that the second person's birthday is different is , that the third person's birthday is different from the first two is , and so on, up through the th person. Explicitly,

(1) (2)

But this can be written in terms of factorials as

(3)

so the probability that two or more people out of a group of do have the same birthday is therefore

(4) (5)

In general, let denote the probability that a birthday is shared by exactly (and no more) people out of a group of people. Then the probability that a birthday is shared by or more people is given by

(6)

In general, can be computed using the recurrence relation

(7)

(Finch 1997). However, the time to compute this recursive function grows exponentially with and so rapidly becomes unwieldy.

If 365-day years have been assumed, i.e., the existence of leap days is ignored, and the distribution of birthdays is assumed to be uniform throughout the year (in actuality, there is a more than 6% increase from the average in September in the United States; Peterson 1998), then the number of people needed for there to be at least a 50% chance that at least two share birthdays is the smallest such that . This is given by , since

(8) (9)

The number of people needed to obtain for , 2, ..., are 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, ... (OEIS A033810). The minimal number of people to give a 50% probability of having at least coincident birthdays is 1, 23, 88, 187, 313, 460, 623, 798, 985, 1181, 1385, 1596, 1813, ... (OEIS A014088; Diaconis and Mosteller 1989).

The probability can be estimated as

(10) (11)

where the latter has error

(12)

(Sayrafiezadeh 1994).

can be computed explicitly as

(13) (14)

where is a binomial coefficient and is a hypergeometric function. This gives the explicit formula for as

(15) (16)

where is a regularized hypergeometric function.

A good approximation to the number of people such that is some given value can be given by solving the equation

(17)

for and taking , where is the ceiling function (Diaconis and Mosteller 1989). For and , 2, 3, ..., this formula gives , 23, 88, 187, 313, 459, 622, 797, 983, 1179, 1382, 1592, 1809, ... (OEIS A050255), which differ from the true values by from 0 to 4. A much simpler but also poorer approximation for such that for is given by

(18)

(Diaconis and Mosteller 1989), which gives 86, 185, 307, 448, 606, 778, 965, 1164, 1376, 1599, 1832, ... for , 4, ... (OEIS A050256).

The "almost" birthday problem, which asks the number of people needed such that two have a birthday within a day of each other, was considered by Abramson and Moser (1970), who showed that 14 people suffice. An approximation for the minimum number of people needed to get a 50-50 chance that two have a match within days out of possible is given by

(19)

(Sevast'yanov 1972, Diaconis and Mosteller 1989).