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,
But this can be written in terms of factorials
as
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(3)
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so the probability that two or more people out of
a group of do have the same birthday is therefore
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(4)
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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
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(5)
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In general, can be computed using the recurrence relation
![Q_k(n,d)=sum_(i=1)^(|_n/k_|)[(n!d!)/(d^(ik)i!(k!)^i(n-ik)!(d-i)!)sum_(j=1)^(k-1)Q_j(n-ik,d-i)((d-i)^(n-ik))/(d^(n-ik))]](/images/equations/BirthdayProblem/NumberedEquation4.gif) |
(6)
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(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
The number of people needed to obtain
for , 2, ..., are 2, 2, 3, 3, 3, 4, 4,
4, 4, 5, ... (Sloane's 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, ... (Sloane's A014088; Diaconis and Mosteller 1989).
The probability can be estimated as
where the latter has error
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(11)
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(Sayrafiezadeh 1994).
can be computed explicitly as
where is a binomial coefficient, is a gamma function, and
is a hypergeometric function.
This gives the explicit formula for as
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
![ne^(-n/(dk))=[d^(k-1)k!ln(1/(1-p))(1-n/(d(k+1)))]^(1/k)](/images/equations/BirthdayProblem/NumberedEquation6.gif) |
(17)
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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, ... (Sloane's 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
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(18)
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(Diaconis and Mosteller 1989), which gives 86, 185, 307, 448, 606, 778, 965, 1164, 1376, 1599, 1832, ... for , 4, ... (Sloane's
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
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(19)
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