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{{about||yearly variation in mortality rates|birthday effect|the mathematical brain teaser that was asked in the Math Olympiad|Cheryl's Birthday}}

In [[probability theory]], the '''birthday problem''' or '''birthday paradox''' concerns the [[probability]] that, in a set of {{mvar|n}} [[random]]ly chosen people, some pair of them will have the same [[birthday]]. By the [[pigeonhole principle]], the probability reaches 100% when the number of people reaches 367 (since there are only 366 possible birthdays, including [[February 29]]). However, 99.9% probability is reached with just 70 people, and 50% probability with 23 people. These conclusions are based on the assumption that each day of the year (excluding February 29) is equally probable for a birthday.

Actual birth records show that different numbers of people are born on different days. In this case, it can be shown that the number of people required to reach the 50% threshold is 23 ''or fewer''.<ref name="Borja">{{cite journal |author1 = Mario Cortina Borja |author2 = John Haigh |title = The Birthday Problem |journal = Significance |date = September 2007 |volume = 4 |issue = 3 |pages = 124–127 |publisher = Royal Statistical Society |doi = 10.1111/j.1740-9713.2007.00246.x}}</ref> For example, if half the people were born on one day and the other half on another day, then any ''two'' people would have a 50% chance of sharing a birthday.

It may well seem surprising that a group of just 23 individuals is required to reach a probability of 50% that at least two individuals in the group have the same birthday: this result is perhaps made more plausible by considering that the comparisons of birthday will actually be made between every possible pair of individuals = 23&nbsp;×&nbsp;22/2&nbsp;=&nbsp;253 comparisons, which is well over half the number of days in a year (183 at most), as opposed to fixing on one individual and comparing his or her birthday to everyone else's.

Real-world applications for the birthday paradox include a cryptographic attack called the [[birthday attack]], which uses this probabilistic model to reduce the complexity of finding a [[Collision attack|collision]] for a [[hash function]], as well as calculating the approximate risk of a hash collision existing within the hashes of a given size of population.

The history of the problem is obscure. [[W. W. Rouse Ball]] indicated (without citation) that it was first discussed by [[Harold Davenport]].<ref name="rouse_ball_1960">[[W. W. Rouse Ball]], 1960, ''Other Questions on Probability'', in "Mathematical Recreations and Essays", Macmillan, New York, p 45.</ref> However, [[Richard von Mises]] proposed an earlier version of what is considered today to be the birthday problem.<ref>{{cite book |editor-first1 = P. |editor-last1 = Frank |editor-first2 = S. |editor-last2 = Goldstein |editor-first3 = M. |editor-last3 = Kac |editor-first4 = W. |editor-last4 = Prager |editor-first5 = G. |editor-last5 = Szegö |editor-first6 = G. |editor-last6 = Birkhoff |title = Selected Papers of Richard von Mises |volume = 2 | pages = 313–334 |date=1964 |publisher = Amer. Math. Soc. |location=Providence, Rhode Island}}</ref>
[[Image:Birthday Paradox.svg|thumb|300px|The computed probability of at least two people sharing a birthday versus the number of people]]

==Calculating the probability==
The problem is to compute an approximate probability that in a group of {{mvar|n}} people at least two have the same birthday. For simplicity, variations in the distribution, such as [[leap year]]s, [[twin]]s, seasonal, or weekday variations are disregarded, and it is assumed that all 365 possible birthdays are equally likely. (Real-life birthday distributions are not uniform, since not all dates are equally likely, but these irregularities have little effect on the analysis.{{refn|group=nb|name="nonuniform birthdays"|In reality, birthdays are not evenly distributed throughout the year; there are more births per day in some seasons than in others, but for the purposes of this problem the distribution is treated as uniform. In particular, many children are born in the summer, especially the months of August and September (for the northern hemisphere) [http://scienceworld.wolfram.com/astronomy/LeapDay.html], and in the U.S. it has been noted that many children are conceived around the holidays of [[Christmas]] and [[New Year's Day]].<ref name="Borja" /> Also, because hospitals rarely schedule [[caesarian section]]s and [[induced labor]] on the weekend, more people are born between Tuesday and Friday than on weekends;<ref name="Borja" /> where many of the people share a birth year (e.g. a class in a school), this creates a tendency toward particular dates. In Sweden 9.3% of the population is born in March and 7.3% in November when a uniform distribution would give 8.3% [http://www.scb.se/statistik/BE/BE0101/2006A01a/BE0101_2006A01a_SM_BE12SM0701.pdf Swedish statistics board].  See also:
* {{Cite web
| last = Murphy
| first = Ron
| title = An Analysis of the Distribution of Birthdays in a Calendar Year
| accessdate = 2011-12-27
| url = http://www.panix.com/~murphy/bday.html
}}
* {{Cite journal
| doi = 10.1093/ije/12.3.326
| volume = 12
| issue = 3
| pages = 326–331
| last = Mathers
| first = C D
|author2=R S Harris
| title = Seasonal Distribution of Births in Australia
| journal = International Journal of Epidemiology
| accessdate = 2011-12-27
| year = 1983
| url = http://ije.oxfordjournals.org/content/12/3/326.abstract
| pmid = 6629621
}}
These factors tend to increase the chance of identical birth dates, since a denser subset has more possible pairs (in the extreme case when everyone was born on three days, there would obviously be many identical birthdays). The problem of a non-uniform number of births occurring during each day of the year was first understood by [[Murray Klamkin]] in 1967. A formal proof that the probability of two matching birthdays is least for a uniform distribution of birthdays was given by Bloom {{Harv|Bloom|1973}}.}} Actually, a uniform distribution of birth dates is the worst case.<ref>{{cite book |author-first = J. Michael |author-last = Steele |title = The Cauchy‑Schwarz Master Class | pages = 206, 277 |date=2004 |publisher = Cambridge University Press |location=Cambridge | isbn = 9780521546775 }}</ref>)

The goal is to compute {{math|''P''(''A'')}}, the probability that at least two people in the room have the same birthday. However, it is simpler to calculate {{math|''P''(''A''′)}}, the probability that no two people in the room have the same birthday. Then, because {{math|''A''}} and {{math|''A''′}} are the only two possibilities and are also [[mutually exclusive events|mutually exclusive]], {{math|''P''(''A'') {{=}} 1 − ''P''(''A''′).}}

In deference to widely published solutions{{Which|date=July 2019}} concluding that 23 is the minimum number of people necessary to have a {{math|''P''(''A'')}} that is greater than 50%, the following calculation of {{math|''P''(''A'')}} will use 23 people as an example.  If one numbers the 23 people from 1 to 23, the [[event (probability theory)|event]] that all 23 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 23 does not have the same birthday as any of persons 1 through 22.  Let these events respectively be called "Event 2", "Event 3", and so on.  One may also add an "Event 1", corresponding to the event of person 1 having a birthday, which occurs with probability 1. This conjunction of events may be computed using [[conditional probability]]: the probability of Event 2 is 364/365, as person 2 may have any birthday other than the birthday of person 1.  Similarly, the probability of Event 3 given that Event 2 occurred is 363/365, as person 3 may have any of the birthdays not already taken by persons 1 and 2.  This continues until finally the probability of Event 23 given that all preceding events occurred is 343/365.  Finally, the principle of conditional probability implies that {{math|''P''(''A''′)}} is equal to the product of these individual probabilities:
{{NumBlk|:|<math>P(A')=\frac{365}{365}\times\frac{364}{365}\times\frac{363}{365}\times\frac{362}{365}\times\cdots\times\frac{343}{365}</math>|{{EquationRef|1}}}}

The terms of equation ({{EquationNote|1}}) can be collected to arrive at:
{{NumBlk|:|<math>P(A')=\left(\frac{1}{365}\right)^{23}\times(365\times364\times363\times\cdots\times343)</math>|{{EquationRef|2}}}}

Evaluating equation ({{EquationNote|2}}) gives {{math|''P''(''A''′) ≈ 0.492703}}

Therefore, {{math|''P''(''A'') ≈ 1 − 0.492703 {{=}} 0.507297}}&nbsp;(50.7297%).

This process can be generalized to a group of {{mvar|n}} people, where {{math|''p''(''n'')}} is the probability of at least two of the {{mvar|n}} people sharing a birthday.  It is easier to first calculate the probability {{math|''{{overline|p}}''(''n'')}} that all {{mvar|n}} birthdays are ''different''.  According to the [[pigeonhole principle]], {{math|''{{overline|p}}''(''n'')}} is zero when {{math|''n'' > 365}}.  When {{math|''n''&nbsp;≤&nbsp;365}}:

:<math> \begin{align} \bar p(n) &= 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)!} = \frac{n!\cdot\binom{365}{n}}{365^n} = \frac{_{365}P_n}{365^n}\end{align} </math>

where {{math|!}} is the [[factorial]] operator, {{math|<big><big>(</big></big>{{su|p=365|b=''n''|a=c}}<big><big>)</big></big>}} is the [[binomial coefficient]] and {{math|''<sub>k</sub>P<sub>r</sub>''}} denotes [[permutation]].

The equation expresses the fact that the first person has no one to share a birthday, the second person cannot have the same birthday as the first ({{sfrac|364|365}}), the third cannot have the same birthday as either of the first two ({{sfrac|363|365}}), and in general the {{mvar|n}}th birthday cannot be the same as any of the {{math|''n'' − 1}} preceding birthdays.

The [[event (probability theory)|event]] of at least two of the {{mvar|n}} persons having the same birthday is [[complementary event|complementary]] to all {{mvar|n}} birthdays being different. Therefore, its probability {{math|''p''(''n'')}} is

:<math> p(n) = 1 - \bar p(n). </math>

The following table shows the probability for some other values of {{mvar|n}} (this table ignores the existence of leap years, as described above, as well as assuming that each birthday is equally likely):

[[Image:Birthdaymatch.svg|thumb|right|300px|The probability that no two people share a birthday in a group of {{mvar|n}} people. Note that the vertical scale is logarithmic (each step down is 10<sup>20</sup> times less likely).]]

:{| class="wikitable"
!{{mvar|n}}!!{{math|''p''(''n'')}}
|-
|align=right|1  || {{0}}0.0%
|-
|align=right|5  || {{0}}2.7%
|-
|align=right|10 || 11.7%
|-
|align=right|20 || 41.1%
|-
|align=right|23 || 50.7%
|-
|align=right|30 || 70.6%
|-
|align=right|40 || 89.1%
|-
|align=right|50 || 97.0%
|-
|align=right|60 ||99.4%
|-
|align=right|70 || 99.9%
|-
|align=right|75 || 99.97%
|-
|align=right|100 || {{val|99.99997}}%
|-
|align=right|200 || {{val|99.9999999999999999999999999998}}%
|-
|align=right|300 || (100 − {{val|6|e=-80}})%
|-
|align=right|350 || (100 − {{val|3|e=-129}})%
|-
|align=right|365 || (100 − {{val|1.45|e=-155}})%
|-
|align=right|<!-- per assumptions in text -->≥ 366<!-- leap years are ignored --> || 100%
|}

'''Leap years'''. If we substitute 366 for 365 in the formula for <math> \bar p(n) </math>, a similar calculation shows that for leap years, the number of people required for the probability of a match to be more than 50% is also 23; the probability of a match in this case is 50.6%.

==Approximations==
[[Image:Birthday paradox probability.svg|thumb|right|300px|Graphs showing the approximate probabilities of at least two people sharing a birthday ({{color|red|red}}) and its complementary event ({{color|blue|blue}})]]
[[Image:Birthday paradox approximation.svg|thumb|right|300px|A graph showing the accuracy of the approximation {{math|1 − ''e''<sup>−{{frac|''n''<sup>2</sup>|730}}</sup>}} ({{color|red|red}})]]
The [[Taylor series]] expansion of the [[exponential function]] (the constant {{math|''e'' ≈ {{val|2.718281828}}}})

:<math> e^x = 1 + x + \frac{x^2}{2!}+\cdots </math>

provides a first-order approximation for {{math|''e''<sup>''x''</sup>}} for <math>|x| \ll 1</math>:

:<math> e^x \approx 1 + x.</math>

To apply this approximation to the first expression derived for {{math|''{{overline|p}}''(''n'')}}, set {{math|''x'' {{=}} −{{sfrac|''a''|365}}}}. Thus,
:<math> e^{-a/365} \approx 1 - \frac{a}{365}. </math>

Then, replace {{mvar|a}} with non-negative integers for each term in the formula of {{math|''{{overline|p}}''(''n'')}} until {{math|''a'' {{=}} ''n'' − 1}}, for example, when {{math|''a'' {{=}} 1}},
: <math> e^{-1/365} \approx 1 - \frac{1}{365}. </math>

The first expression derived for {{math|''{{overline|p}}''(''n'')}} can be approximated as

:<math>
\begin{align}
\bar p(n) & \approx 1 \cdot e^{-1/365} \cdot e^{-2/365} \cdots e^{-(n-1)/365} \\[6pt]
& = e^{-\left.\big( 1+2+ \,\cdots\, +(n-1)\big)\right/365} \\[6pt]
& = e^{-(n(n-1)/2)365} = e^{-n(n-1)/730}.
\end{align}
</math>

Therefore,

:<math> p(n) = 1-\bar p(n) \approx 1 - e^{-n(n-1)/730}.</math>

An even coarser approximation is given by

:<math>p(n)\approx 1-e^{-n^2/730},</math>

which, as the graph illustrates, is still fairly accurate.

According to the approximation, the same approach can be applied to any number of "people" and "days". If rather than 365 days there are {{mvar|d}}, if there are {{mvar|n}} persons, and if {{math|''n'' ≪ ''d''}}, then using the same approach as above we achieve the result that if {{math|''p''(''n'', ''d'')}} is the probability that at least two out of {{mvar|n}} people share the same birthday from a set of {{mvar|d}} available days, then:

:<math>\begin{align}
 p(n, d) & \approx 1-e^{-n(n-1)/(2d)} \\[6pt]
& \approx 1-e^{-n^2/(2d)}.
\end{align}</math>

===A simple exponentiation===
The probability of any two people not having the same birthday is {{sfrac|364|365}}. In a room containing ''n'' people, there are {{math|<big><big>(</big></big>{{su|p=''n''|b=2|a=c}}<big><big>)</big></big> {{=}} {{sfrac|''n''(''n''&nbsp;−&nbsp;1)|2}}}} pairs of people, i.e. {{math|<big><big>(</big></big>{{su|p=''n''|b=2|a=c}}<big><big>)</big></big>}} events. The probability of no two people sharing the same birthday can be approximated by assuming that these events are independent and hence by multiplying their probability together. In short {{sfrac|364|365}} can be multiplied by itself {{math|<big><big>(</big></big>{{su|p=''n''|b=2|a=c}}<big><big>)</big></big>}} times, which gives us

:<math>\bar p(n) \approx \left(\frac{364}{365}\right)^\binom{n}{2}.</math>

Since this is the probability of no one having the same birthday, then the probability of someone sharing a birthday is

:<math>p(n) \approx 1 - \left(\frac{364}{365}\right)^\binom{n}{2}.</math>

===Poisson approximation===
Applying the [[Poisson distribution|Poisson]] approximation for the binomial on the group of 23 people,

:<math>\operatorname{Poi}\left(\frac{\binom{23}{2}}{365}\right) =\operatorname{Poi}\left(\frac{253}{365}\right) \approx \operatorname{Poi}(0.6932)</math>

so

:<math>\Pr(X>0)=1-\Pr(X=0) \approx 1-e^{-0.6932} \approx 1-0.499998=0.500002.</math>

The result is over 50% as previous descriptions.  This approximation is the same as the one above based on the Taylor expansion that uses <math> e^x \approx 1 + x</math>.

===Square approximation===
A good [[rule of thumb]] which can be used for [[mental calculation]] is the relation

:<math>p(n) \approx \frac{n^2}{2m}</math>

which can also be written as

:<math>n \approx \sqrt { 2m \times p(n)}</math>

which works well for probabilities less than or equal to {{sfrac|1|2}}. In these equations, {{mvar|m}} is the number of days in a year.

For instance, to estimate the number of people required for a {{sfrac|1|2}} chance of a shared birthday, we get

:<math>n \approx \sqrt{ 2 \times 365 \times \tfrac12} = \sqrt{365} \approx 19</math>

Which is not too far from the correct answer of 23.

===Approximation of number of people===
This can also be approximated using the following formula for the ''number'' of people necessary to have at least a {{sfrac|1|2}} chance of matching:
:<math>n \approx \tfrac{1}{2} + \sqrt{\tfrac{1}{4} + 2 \times \ln(2) \times 365} = 22.999943.</math>
This is a result of the good approximation that an event with {{math|{{sfrac|1|''k''}}}} probability will have a {{sfrac|1|2}} chance of occurring at least once if it is repeated {{math|''k'' [[natural logarithm of 2|ln 2]]}} times.<ref>{{cite journal
 | last = Mathis
 | first = Frank H.
 |date= June 1991
 | title = A Generalized Birthday Problem
 | journal = SIAM Review
 | volume = 33
 | issue = 2
 | pages = 265–270
 | issn = 0036-1445
 | pmc =
 | doi = 10.1137/1033051
 | bibcode =
 | oclc = 37699182
 | jstor = 2031144
 | url = http://http.cs.berkeley.edu/~daw/papers/genbday-crypto02.ps
 }}</ref>

===Probability table===
{{Main|Birthday attack}}

:{| class="wikitable" style="white-space:nowrap;"
|-
! rowspan="2" | length of <br />hex string
! rowspan="2" | no. of<br />bits<br />({{mvar|b}})
! rowspan="2" | hash space<br />size<br />({{math|2<sup>''b''</sup>}})
! colspan="10" | Number of hashed elements such that probability of at least one hash collision&nbsp;≥&nbsp;{{mvar|p}}
|-
! {{mvar|p}} = {{val||e=-18}}
! {{mvar|p}} = {{val||e=-15}}
! {{mvar|p}} = {{val||e=-12}}
! {{mvar|p}} = {{val||e=-9}}
! {{mvar|p}} = {{val||e=-6}}
! {{mvar|p}} = 0.001
! {{mvar|p}} = 0.01
! {{mvar|p}} = 0.25
! {{mvar|p}} = 0.50
! {{mvar|p}} = 0.75
|- align="center"
| bgcolor="#F2F2F2" | 8
| bgcolor="#F2F2F2" | 32
| bgcolor="#F2F2F2" | {{val|4.3|e=9}}
| 2
| 2
| 2
| 2.9
| 93
| {{val|2.9|e=3}}
| {{val|9.3|e=3}}
| {{val|5.0|e=4}}
| {{val|7.7|e=4}}
| {{val|1.1|e=5}}
|- align="center"
| bgcolor="#F2F2F2" | (10)
| bgcolor="#F2F2F2" | (40)
| bgcolor="#F2F2F2" | ({{val|1.1|e=12}})
| 2
| 2
| 2
| 47
| {{val|1.5|e=3}}
| {{val|4.7|e=4}}
| {{val|1.5|e=5}}
| {{val|8.0|e=5}}
| {{val|1.2|e=6}}
| {{val|1.7|e=6}}
|- align="center"
| bgcolor="#F2F2F2" | (12)
| bgcolor="#F2F2F2" | (48)
| bgcolor="#F2F2F2" | ({{val|2.8|e=14}})
| 2
| 2
| 24
| {{val|7.5|e=2}}
| {{val|2.4|e=4}}
| {{val|7.5|e=5}}
| {{val|2.4|e=6}}
| {{val|1.3|e=7}}
| {{val|2.0|e=7}}
| {{val|2.8|e=7}}
|- align="center"
| bgcolor="#F2F2F2" | 16
| bgcolor="#F2F2F2" | 64
| bgcolor="#F2F2F2" | {{val|1.8|e=19}}
| 6.1
| {{val|1.9|e=2}}
| {{val|6.1|e=3}}
| {{val|1.9|e=5}}
| {{val|6.1|e=6}}
| {{val|1.9|e=8}}
| {{val|6.1|e=8}}
| {{val|3.3|e=9}}
| {{val|5.1|e=9}}
| {{val|7.2|e=9}}
|- align="center"
| bgcolor="#F2F2F2" | (24)
| bgcolor="#F2F2F2" | (96)
| bgcolor="#F2F2F2" | ({{val|7.9|e=28}})
| {{val|4.0|e=5}}
| {{val|1.3|e=7}}
| {{val|4.0|e=8}}
| {{val|1.3|e=10}}
| {{val|4.0|e=11}}
| {{val|1.3|e=13}}
| {{val|4.0|e=13}}
| {{val|2.1|e=14}}
| {{val|3.3|e=14}}
| {{val|4.7|e=14}}
|- align="center"
| bgcolor="#F2F2F2" | 32
| bgcolor="#F2F2F2" | 128
| bgcolor="#F2F2F2" | {{val|3.4|e=38}}
| {{val|2.6|e=10}}
| {{val|8.2|e=11}}
| {{val|2.6|e=13}}
| {{val|8.2|e=14}}
| {{val|2.6|e=16}}
| {{val|8.3|e=17}}
| {{val|2.6|e=18}}
| {{val|1.4|e=19}}
| {{val|2.2|e=19}}
| {{val|3.1|e=19}}
|- align="center"
| bgcolor="#F2F2F2" | (48)
| bgcolor="#F2F2F2" | (192)
| bgcolor="#F2F2F2" | ({{val|6.3|e=57}})
| {{val|1.1|e=20}}
| {{val|3.5|e=21}}
| {{val|1.1|e=23}}
| {{val|3.5|e=24}}
| {{val|1.1|e=26}}
| {{val|3.5|e=27}}
| {{val|1.1|e=28}}
| {{val|6.0|e=28}}
| {{val|9.3|e=28}}
| {{val|1.3|e=29}}
|- align="center"
| bgcolor="#F2F2F2" | 64
| bgcolor="#F2F2F2" | 256
| bgcolor="#F2F2F2" | {{val|1.2|e=77}}
| {{val|4.8|e=29}}
| {{val|1.5|e=31}}
| {{val|4.8|e=32}}
| {{val|1.5|e=34}}
| {{val|4.8|e=35}}
| {{val|1.5|e=37}}
| {{val|4.8|e=37}}
| {{val|2.6|e=38}}
| {{val|4.0|e=38}}
| {{val|5.7|e=38}}
|- align="center"
| bgcolor="#F2F2F2" | (96)
| bgcolor="#F2F2F2" | (384)
| bgcolor="#F2F2F2" | ({{val|3.9|e=115}})
| {{val|8.9|e=48}}
| {{val|2.8|e=50}}
| {{val|8.9|e=51}}
| {{val|2.8|e=53}}
| {{val|8.9|e=54}}
| {{val|2.8|e=56}}
| {{val|8.9|e=56}}
| {{val|4.8|e=57}}
| {{val|7.4|e=57}}
| {{val|1.0|e=58}}
|- align="center"
| bgcolor="#F2F2F2" | 128
| bgcolor="#F2F2F2" | 512
| bgcolor="#F2F2F2" | {{val|1.3|e=154}}
| {{val|1.6|e=68}}
| {{val|5.2|e=69}}
| {{val|1.6|e=71}}
| {{val|5.2|e=72}}
| {{val|1.6|e=74}}
| {{val|5.2|e=75}}
| {{val|1.6|e=76}}
| {{val|8.8|e=76}}
| {{val|1.4|e=77}}
| {{val|1.9|e=77}}
|}

The lighter fields in this table show the number of hashes needed to achieve the given probability of collision (column) given a hash space of a certain size in bits (row). Using the birthday analogy: the "hash space size" resembles the "available days", the "probability of collision" resembles the "probability of shared birthday", and the "required number of hashed elements" resembles the "required number of people in a group". One could also use this chart to determine the minimum hash size required (given upper bounds on the hashes and probability of error), or the probability of collision (for fixed number of hashes and probability of error).

For comparison, {{val|e=-18}} to {{val|e=-15}} is the uncorrectable bit error rate of a typical hard disk.<ref>Jim Gray, Catharine van Ingen. [https://arxiv.org/abs/cs/0701166 Empirical Measurements of Disk Failure Rates and Error Rates]</ref> In theory, 128-bit hash functions, such as [[MD5]], should stay within that range until about {{val|8.2|e=11}} documents, even if its possible outputs are many more.

==An upper bound on the probability and a lower bound on the number of people==
The argument below is adapted from an argument of [[Paul Halmos]].{{refn|group=nb|In his autobiography, Halmos criticized the form in which the birthday paradox is often presented, in terms of numerical computation.  He believed that it should be used as an example in the use of more abstract mathematical concepts. He wrote:
<blockquote>The reasoning is based on important tools that all students of mathematics should have ready access to.  The birthday problem used to be a splendid illustration of the advantages of pure thought over mechanical manipulation; the inequalities can be obtained in a minute or two, whereas the multiplications would take much longer, and be much more subject to error, whether the instrument is a pencil or an old-fashioned desk computer. What [[calculator]]s do not yield is understanding, or mathematical facility, or a solid basis for more advanced, generalized theories.</blockquote>}}

As stated above, the probability that no two birthdays coincide is

:<math>1-p(n) = \bar p(n) = \prod_{k=1}^{n-1}\left(1-\frac{k}{365}\right) .</math>

As in earlier paragraphs, interest lies in the smallest {{mvar|n}} such that {{math|''p''(''n'') > {{sfrac|1|2}}}}; or equivalently, the smallest {{mvar|n}} such that {{math|''{{overline|p}}''(''n'') < {{sfrac|1|2}}}}.

Using the inequality {{math|1 − ''x'' < ''e''<sup>−''x''</sup>}} in the above expression we replace {{math|1 − {{sfrac|''k''|365}}}} with {{math|''e''<sup>{{frac|−''k''|365}}</sup>}}. This yields

:<math>\bar p(n) = \prod_{k=1}^{n-1}\left(1-\frac{k}{365}\right) < \prod_{k=1}^{n-1}\left(e^{-k/365}\right) = e^{-n(n-1)/730} .</math>

Therefore, the expression above is not only an approximation, but also an [[upper bound]] of {{math|''{{overline|p}}''(''n'')}}. The inequality

:<math> e^{-n(n-1)/730} < \frac{1}{2}</math>

implies {{math|''{{overline|p}}''(''n'') < {{sfrac|1|2}}}}. Solving for {{mvar|n}} gives

:<math>n^2-n > 730 \ln 2 .</math>

Now, {{math|730 ln 2}} is approximately 505.997, which is barely below 506, the value of {{math|''n''<sup>2</sup> − ''n''}} attained when {{math|''n'' {{=}} 23}}. Therefore, 23 people suffice. Incidentally, solving {{math|''n''<sup>2</sup> − ''n'' {{=}} 730 ln 2}} for ''n'' gives the approximate formula of Frank H. Mathis cited above.

This derivation only shows that ''at most'' 23 people are needed to ensure a birthday match with even chance; it leaves open the possibility that {{mvar|n}} is 22 or less could also work.

==Generalizations==

===The generalized birthday problem===
Given a year with {{mvar|d}} days, the '''generalized birthday problem''' asks for the minimal number {{math|''n''(''d'')}} such that, in a set of {{mvar|n}} randomly chosen people, the probability of a birthday coincidence is at least 50%. In other words, {{math|''n''(''d'')}} is the minimal integer {{mvar|n}} such that

:<math>1-\left(1-\frac{1}{d}\right)\left(1-\frac{2}{d}\right)\cdots\left(1-\frac{n-1}{d}\right)\geq \frac{1}{2}.</math>

The classical birthday problem thus corresponds to determining {{math|''n''(365)}}. The first 99 values of {{math|''n''(''d'')}} are given here:

:{| class="wikitable" style="text-align:center;"
|-
! scope="row" | {{mvar|d}}
| 1–2 || 3–5 || 6–9 || 10–16 || 17–23 || 24–32 || 33–42 || 43–54 || 55–68 || 69–82 || 83–99
|-
! scope="row" | {{math|''n''(''d'')}}
| 2 || 3 || 4 || 5 || 6 || 7 || 8 || 9 || 10 || 11 || 12
|}

A number of bounds and formulas for {{math|''n''(''d'')}} have been published.<ref>{{wikicite|ref={{Harvid|Brink|2012}}|reference=D. Brink, A (probably) exact solution to the Birthday Problem, Ramanujan Journal, 2012, [http://www.springerlink.com/content/1194r3627822841q/ doi: 10.1007/s11139-011-9343-9].}}</ref>
For any {{math|''d'' ≥ 1}}, the number {{math|''n''(''d'')}} satisfies<ref>{{Harvard citations|author=Brink|year=2012|nb=yes|loc=Theorem 2}}</ref>

:<math>\frac{3-2\ln2}{6}<n(d)-\sqrt{2d\ln2}\leq 9-\sqrt{86\ln2}.</math>

These bounds are optimal in the sense that the sequence {{math|''n''(''d'') − {{sqrt|2''d'' ln 2}}}}
gets arbitrarily close to
:<math>\frac{3-2\ln2}{6} \approx 0.27,</math>
while it has
:<math>9-\sqrt{86\ln2}\approx 1.28</math>
as its maximum, taken for {{math|''d'' {{=}} 43}}.

The bounds are sufficiently tight to give the exact value of {{math|''n''(''d'')}} in 99% of all cases, for example {{math|''n''(365) {{=}} 23}}. In general, it follows from these bounds that {{math|''n''(''d'')}} always equals either
:<math>\left\lceil\sqrt{2d\ln2}\,\right\rceil \quad\text{or}\quad \left\lceil\sqrt{2d\ln2}\,\right\rceil+1</math>
where {{math|⌈ · ⌉}} denotes the [[Floor and ceiling functions|ceiling function]].
The formula

:<math>n(d) = \left\lceil\sqrt{2d\ln2}\,\right\rceil</math>

holds for 73% of all integers {{mvar|d}}.<ref name=Brink>{{Harvard citations|author=Brink|year=2012|nb=yes|loc=Theorem 3}}</ref> The formula

:<math>n(d) = \left\lceil\sqrt{2d\ln2}+\frac{3-2\ln2}{6}\right\rceil</math>

holds for [[almost all]] {{mvar|d}}, i.e., for a set of integers {{mvar|d}} with [[asymptotic density]] 1.<ref name=Brink/>

The formula

:<math>n(d)=\left\lceil \sqrt{2d\ln2}+\frac{3-2\ln2}{6}+\frac{9-4(\ln2)^2}{72\sqrt{2d\ln2}}\right\rceil</math>

holds for all {{math|''d'' ≤ {{val|e=18}}}}, but it is conjectured that there are infinitely many counterexamples to this formula.<ref name="ReferenceA">{{Harvard citations|author=Brink|year=2012|nb=yes|loc=Table 3, Conjecture 1}}</ref>

The formula

:<math>n(d)=\left\lceil \sqrt{2d\ln2}+\frac{3-2\ln2}{6}+\frac{9-4(\ln2)^2}{72\sqrt{2d\ln2}}-\frac{2(\ln2)^2}{135d}\right\rceil</math>

holds for all {{math|''d'' ≤ {{val|e=18}}}}, and it is conjectured that this formula holds for all {{mvar|d}}.<ref name="ReferenceA"/>

===Cast as a collision problem===
The birthday problem can be generalized as follows:
:Given {{mvar|n}} random integers drawn from a [[Uniform distribution (discrete)|discrete uniform distribution]] with range {{math|[1,''d'']}}, what is the probability {{math|''p''(''n''; ''d'')}} that at least two numbers are the same? ({{math|''d'' {{=}} 365}} gives the usual birthday problem.)<ref>{{cite conference | title =  Birthday Paradox for Multi-collisions| last1 = Suzuki| first1 = K. 
| last2 = Tonien| first2 = D.|display-authors=et al| date = 2006| publisher = Springer| book-title =  Lecture Notes in Computer Science, vol 4296 | location = Berlin
| id = Information Security and Cryptology – ICISC 2006| editor = Rhee M.S., Lee B. | doi =  10.1007/11927587_5}}</ref>

The generic results can be derived using the same arguments given above.
:<math>\begin{align}
p(n;d) &= \begin{cases} 1-\displaystyle\prod_{k=1}^{n-1}\left(1-\frac{k}{d}\right) & n\le d \\ 1 & n > d \end{cases} \\[8px]
& \approx 1 - e^{-\frac{n(n-1)}{2d}} \\
& \approx 1 - \left( \frac{d-1}{d} \right)^\frac{n(n-1)}{2}
\end{align}</math>

Conversely, if {{math|''n''(''p''; ''d'')}} denotes the number of random integers drawn from {{math|[1,''d'']}} to obtain a probability {{mvar|p}} that at least two numbers are the same, then
:<math>n(p;d)\approx \sqrt{2d \cdot \ln\left(\frac{1}{1-p}\right)}.</math>

The birthday problem in this more generic sense applies to [[hash function]]s: the expected number of {{math|''N''}}-[[bit]] hashes that can be generated before getting a collision is not {{math|2<sup>''N''</sup>}}, but rather only {{math|2<sup>{{frac|''N''|2}}</sup>}}. This is exploited by [[birthday attack]]s on [[cryptographic hash function]]s and is the reason why a small number of collisions in a [[hash table]] are, for all practical purposes, inevitable.

The theory behind the birthday problem was used by Zoe Schnabel<ref>Z. E. Schnabel (1938) ''The Estimation of the Total Fish Population of a Lake'', [[American Mathematical Monthly]] '''45''', 348–352.</ref> under the name of [[mark and recapture|capture-recapture]] statistics to estimate the size of fish population in lakes.

====Generalization to multiple types====
[[File:2d birthday.png|thumb|Plot of the probability of at least one shared birthday between at least one man and one woman]]
The basic problem considers all trials to be of one "type". The  birthday problem has been generalized to consider an arbitrary number of types.<ref>[[Michael Christopher Wendl|M. C. Wendl]] (2003) ''[https://dx.doi.org/10.1016/S0167-7152(03)00168-8 Collision Probability Between Sets of Random Variables]'', Statistics and Probability Letters '''64'''(3), 249–254.</ref> In the simplest extension there are two types of people, say {{mvar|m}} men and {{mvar|n}} women, and the problem becomes characterizing the probability of a shared birthday between at least one man and one woman. (Shared birthdays between two men or two women do not count.) The probability of no shared birthdays here is

:<math>p_0 =\frac{1}{d^{m+n}} \sum_{i=1}^m \sum_{j=1}^n S_2(m,i) S_2(n,j) \prod_{k=0}^{i+j-1} d - k</math>

where {{math|''d'' {{=}} 365}} and {{math|''S''<sub>2</sub>}} are [[Stirling numbers of the second kind]]. Consequently, the desired probability is {{math|1 − ''p''<sub>0</sub>}}.

This variation of the birthday problem is interesting because there is not a unique solution for the total number of people {{math|''m'' + ''n''}}. For example, the usual 50% probability value is realized for both a 32-member group of 16 men and 16 women and a 49-member group of 43 women and 6 men.

==Other birthday problems==

===First match===
A related question is, as people enter a room one at a time, which one is most likely to be the first to have the same birthday as someone already in the room?  That is, for what {{mvar|n}} is {{math|''p''(''n'') − ''p''(''n''&nbsp;−&nbsp;1)}} maximum?  The answer is 20—if there is a prize for first match, the best position in line is 20th.{{citation needed|date=September 2019}}

===Same birthday as you===
[[Image:Birthday paradox.svg|thumb|right|300px|Comparing {{math|''p''(''n'')}} = probability of a birthday match with {{math|''q''(''n'')}} = probability of matching ''your'' birthday]]
In the birthday problem, neither of the two people is chosen in advance. By contrast, the probability {{math|''q''(''n'')}} that someone in a room of {{mvar|n}} other people has the same birthday as a ''particular'' person (for example, you) is given by

: <math> q(n) = 1 - \left( \frac{365-1}{365} \right)^n </math>
and for general {{mvar|d}} by
: <math> q(n;d) = 1 - \left( \frac{d-1}{d} \right)^n. </math>

In the standard case of {{math|''d'' {{=}} 365}}, substituting {{math|''n'' {{=}} 23}} gives about 6.1%, which is less than 1 chance in 16.  For a greater than 50% chance that one person in a roomful of {{mvar|n}} people has the same birthday as ''you'', {{mvar|n}} would need to be at least 253. This number is significantly higher than {{math|{{sfrac|365|2}} {{=}} 182.5}}: the reason is that it is likely that there are some birthday matches among the other people in the room.

It is not a coincidence that {{math|253 {{=}} {{sfrac|23 × (23 − 1)|2}}}}; a similar approximate pattern can be found using a number of possibilities different from 365, or a target probability different from 50%.{{citation needed|date=September 2019}}

===Near matches===
Another generalization is to ask for the probability of finding at least one pair in a group of {{mvar|n}} people with birthdays within {{mvar|k}} calendar days of each other, if there are {{mvar|d}} equally likely birthdays.<ref name="abramson">M. Abramson and W. O. J. Moser (1970) ''More Birthday Surprises'', [[American Mathematical Monthly]] '''77''', 856–858</ref>

:<math> \begin{align} p(n,k,d) &= 1 - \frac{ (d - nk -1)! }{ d^{n-1} \bigl(d - n(k+1)\bigr)!}\end{align} </math>

The number of people required so that the probability that some pair will have a birthday separated by {{mvar|k}} days or fewer will be higher than 50% is given in the following table:

:{| class="wikitable" style="text-align: center"
! {{mvar|''k''}} !! {{mvar|n}}<br />for {{math|''d'' {{=}} 365}}
|-
|0 || 23
|-
|1 || 14
|-
|2 || 11
|-
|3 || 9
|-
|4 || 8
|-
|5 || 8
|-
|6 || 7
|-
|7 || 7
|}

Thus in a group of just seven random people, it is more likely than not that two of them will have a birthday within a week of each other.<ref name="abramson"/>

===Collision counting===
The probability that the {{mvar|k}}th integer randomly chosen from {{math|[1,''d'']}} will repeat at least one previous choice equals {{math|''q''(''k'' − 1; ''d'')}} above.  The expected total number of times a selection will repeat a previous selection as {{mvar|n}} such integers are chosen equals<ref>{{cite web|last1=Might|first1=Matt|title=Collision hash collisions with the birthday paradox|url=http://matt.might.net/articles/counting-hash-collisions/|website=Matt Might's blog|accessdate=17 July 2015}}</ref>

:<math>\sum_{k=1}^n q(k-1;d) = n - d + d \left (\frac {d-1} {d} \right )^n.</math>

===Average number of people===
In an alternative formulation of the birthday problem, one asks the ''average'' number of people required to find a pair with the same birthday. If we consider the probability function Pr[{{mvar|n}} people have at least one shared birthday], this ''average'' is determining the [[mean]] of the distribution, as opposed to the customary formulation, which asks for the [[median]].  The problem is relevant to several [[hash function|hashing algorithms]] analyzed by [[Donald Knuth]] in his book ''[[The Art of Computer Programming]]''. It may be shown<ref name="knuth73">{{cite book |first=D. E. |last=Knuth |title=The Art of Computer Programming |volume=Vol. 3, Sorting and Searching |publisher=Addison-Wesley |location=Reading, Massachusetts |year=1973 |isbn=978-0-201-03803-3 }}</ref><ref name="flajolet95">{{cite journal |first=P. |last=Flajolet |first2=P. J. |last2=Grabner |first3=P. |last3=Kirschenhofer |first4=H. |last4=Prodinger |year=1995 |title=On Ramanujan's Q-Function |journal=Journal of Computational and Applied Mathematics |volume=58 |issue= |pages=103–116 |doi=10.1016/0377-0427(93)E0258-N }}</ref> that if one samples uniformly, with replacement, from a population of size {{math|''M''}}, the number of trials required for the first repeated sampling of ''some'' individual has [[expected value]] {{math|{{overline|''n''}} {{=}} 1 + ''Q''(''M'')}}, where

: <math>Q(M)=\sum_{k=1}^{M} \frac{M!}{(M-k)! M^k}.</math>

The function

: <math>Q(M)= 1 + \frac{M-1}{M} + \frac{(M-1)(M-2)}{M^2} + \cdots + \frac{(M-1)(M-2) \cdots 1}{M^{M-1}}</math>
<!-- Shouldn't this sum be up to M ? Without it the term M!/M^M is missing-->
has been studied by [[Srinivasa Ramanujan]] and has [[asymptotic expansion]]:

: <math>Q(M)\sim\sqrt{\frac{\pi M}{2}}-\frac{1}{3}+\frac{1}{12}\sqrt{\frac{\pi}{2M}}-\frac{4}{135M}+\cdots.</math>

With {{math|''M'' {{=}} 365}} days in a year, the average number of people required to find a pair with the same birthday is {{math|{{overline|''n''}} {{=}} 1 + ''Q''(''M'') ≈ 24.61659}}, somewhat more than 23, the number required for a 50% chance. In the best case, two people will suffice; at worst, the maximum possible number of {{math|''M'' + 1 {{=}} 366}} people is needed; but on average, only 25 people are required

An analysis using indicator random variables can provide a simpler but approximate analysis of this problem.<ref>{{Cite book|title=Introduction to Algorithms|last=Cormen |display-authors=etal |publisher=|year=|isbn=|location=|pages=}}</ref> For each pair (i, j) for k people in a room, we define the indicator random variable X<sub>ij</sub>, for <math>1\leq i \leq j\leq k</math>, by

<math>\begin{alignat}{2} X_{ij} & 
= I \{ \text{person }i\text{ and person }j\text{ have the same birthday} \} \\ & 
= \begin{cases} 
1, & \text{if person }i\text{ and person }j\text{ have the same birthday;} \\ 
0, & \text{otherwise.}
\end{cases}
\end{alignat}</math>

<math>\begin{alignat}{2} E[X_{ij}] & 
= \Pr \{ \text{person }i\text{ and person }j\text{ have the same birthday} \} \\ & 
= 1/n 
\end{alignat}</math>

Let X be a random variable counting the pairs of individuals with the same birthday.

<math>X =\sum_{i=1}^k \sum_{j=i+1}^k X_{ij}</math>

<math>\begin{alignat}{3} E[X] 
& = \sum_{i=1}^k \sum_{j=i+1}^k E[X_{ij}]\\ 
& = \binom{k}{2} \frac{1}{n}\\ 
& = \frac{k(k-1)}{2n}\\
\end{alignat}</math>

For {{math|''n'' {{=}} 365}}, if {{math|''k'' {{=}} 28}}, the expected number of with the same birthday is <math>(28\cdot27)/(2\cdot365)\approx1.0356.</math> Therefore we can expect at least one matching pair with at least 28 people.

An informal demonstration of the problem can be made from the [[list of Prime Ministers of Australia]], of which there have been 29 {{as of|2017|lc=yes}}, in which [[Paul Keating]], the 24th prime minister, and [[Edmund Barton]], the first prime minister, share the same birthday, 18 January.

In the [[2014 FIFA World Cup]], each of the 32 squads had 23 players.  An analysis of the official squad lists suggested that 16 squads had pairs of players sharing birthdays, and of these 5 squads had two pairs: Argentina, France, Iran, South Korea and Switzerland each had two pairs, and Australia, Bosnia and Herzegovina, Brazil, Cameroon, Colombia, Honduras, Netherlands, Nigeria, Russia, Spain and USA each with one pair.<ref>{{cite web |url=https://www.bbc.co.uk/news/magazine-27835311 |title=The birthday paradox at the World Cup |last1=Fletcher |first1=James |date=16 June 2014 |website=bbc.com |publisher=BBC |access-date=27 August 2015 |quote=}}</ref>

Voracek, Tran and [[Anton Formann|Formann]] showed that the majority of people markedly overestimate the number of people that is necessary to achieve a given probability of people having the same birthday, and markedly underestimate the probability of people having the same birthday when a specific sample size is given.<ref>{{cite journal |last=Voracek |first=M. |last2=Tran |first2=U. S. |last3=Formann |first3=A. K. |year=2008 |title=Birthday and birthmate problems: Misconceptions of probability among psychology undergraduates and casino visitors and personnel |journal=Perceptual and Motor Skills |volume=106 |issue=1 |pages=91–103 |doi=10.2466/pms.106.1.91-103 |pmid=18459359 }}</ref> Further results were that psychology students and women did better on the task than casino visitors/personnel or men, but were less confident about their estimates.

===Reverse problem===
The reverse problem is to find, for a fixed probability {{mvar|p}},
the greatest {{mvar|n}} for which the probability {{math|''p''(''n'')}} is smaller than the given {{mvar|p}}, or the smallest {{mvar|n}} for which the probability {{math|''p''(''n'')}} is greater than the given {{mvar|p}}.{{citation needed|date=September 2019}}

Taking the above formula for {{math|''d''&nbsp;{{=}}&nbsp;365}}, one has

:<math>n(p;365)\approx \sqrt{730\ln\left(\frac{1}{1-p}\right)}.</math>

The following table gives some sample calculations.
:{| class="wikitable"
|-----
! {{mvar|p}} || {{mvar|n}}
! {{math|''n''↓}} || {{math|''p''(''n''↓)}} || {{math|''n''↑}} || {{math|''p''(''n''↑)}}
|-----
| <span style="color:maroon">0.01</span>
| 0.14178{{sqrt|365}} = <span style="color:maroon">2.70864</span>
| align="right" | 2 || 0.00274 || align="right" | 3
| <span style="color:maroon">0.00820</span>
|-----
| 0.05 || 0.32029{{sqrt|365}} = 6.11916
| align="right" | 6 || 0.04046 || align="right" | 7 || 0.05624
|-----
| <span style="color:maroon">0.1</span>
| 0.45904{{sqrt|365}} = <span style="color:maroon"> 8.77002</span>
| align="right" | 8 || 0.07434 || align="right" | 9
| <span style="color:maroon">0.09462</span>
|-----
| <span style="color:maroon">0.2</span>
| 0.66805{{sqrt|365}} = <span style="color:maroon">12.76302</span>
| align="right" | 12 || 0.16702 || align="right" | 13
| <span style="color:maroon">0.19441</span>
|-----
| 0.3 || 0.84460{{sqrt|365}} = 16.13607
| align="right" | 16 || 0.28360 || align="right" | 17 || 0.31501
|-----
| 0.5 || 1.17741{{sqrt|365}} = 22.49439
| align="right" | 22 || 0.47570 || align="right" | 23 || 0.50730
|-----
| 0.7 || 1.55176{{sqrt|365}} = 29.64625
| align="right" | 29 || 0.68097 || align="right" | 30 || 0.70632
|-----
| 0.8 || 1.79412{{sqrt|365}} = 34.27666
| align="right" | 34 || 0.79532 || align="right" | 35 || 0.81438
|-----
| 0.9 || 2.14597{{sqrt|365}} = 40.99862
| align="right" | 40 || 0.89123 || align="right" | 41 || 0.90315
|-----
| 0.95 || 2.44775{{sqrt|365}} = 46.76414
| align="right" | 46 || 0.94825 || align="right" | 47 || 0.95477
|-----
| <span style="color:maroon">0.99</span>
| 3.03485{{sqrt|365}} = <span style="color:maroon">57.98081</span>
| align="right" | 57
| <span style="color:maroon">0.99012</span>
| align="right" | 58 || 0.99166
|}
Some values falling outside the bounds have been <span style="color:maroon">colored</span> to show that the approximation is not always exact.

==Partition problem==
A related problem is the [[partition problem]], a variant of the [[knapsack problem]] from operations research. Some weights are put on a [[Weighing scale|balance scale]]; each weight is an integer number of grams randomly chosen between one gram and one million grams (one [[tonne]]). The question is whether one can usually (that is, with probability close to 1) transfer the weights between the left and right arms to balance the scale. (In case the sum of all the weights is an odd number of grams, a discrepancy of one gram is allowed.) If there are only two or three weights, the answer is very clearly no; although there are some combinations which work, the majority of randomly selected combinations of three weights do not. If there are very many weights, the answer is clearly yes. The question is, how many are just sufficient? That is, what is the number of weights such that it is equally likely for it to be possible to balance them as it is to be impossible?

Often, people's intuition is that the answer is above {{val|100000}}. Most people's intuition is that it is in the thousands or tens of thousands, while others feel it should at least be in the hundreds. The correct answer is 23.{{Citation needed|date=December 2016}}

The reason is that the correct comparison is to the number of partitions of the weights into left and right. There are {{math|2<sup>''N'' − 1</sup>}} different partitions for {{math|''N''}} weights, and the left sum minus the right sum can be thought of as a new random quantity for each partition. The distribution of the sum of weights is approximately [[normal distribution|Gaussian]], with a peak at {{math|{{val|1000000}}''N''}} and width {{math|{{val|1000000}}{{sqrt|''N''}}}}, so that when {{math|2<sup>''N'' − 1</sup>}} is approximately equal to {{math|{{val|1000000}}{{sqrt|''N''}}}} the transition occurs.  2<sup>23 − 1</sup> is about 4 million, while the width of the distribution is only 5 million.<ref>{{cite journal |first=C. |last=Borgs |first2=J. |last2=Chayes |first3=B. |last3=Pittel |year=2001 |title=Phase Transition and Finite Size Scaling in the Integer Partition Problem |journal=Random Structures and Algorithms |volume=19 |issue=3–4 |pages=247–288 |doi=10.1002/rsa.10004 }}</ref>

==In fiction==
[[Arthur C. Clarke]]'s novel ''[[A Fall of Moondust]]'', published in 1961, contains a section where the main characters, trapped underground for an indefinite amount of time, are celebrating a birthday and find themselves discussing the validity of the birthday problem. As stated by a physicist passenger: "If you have a group of more than twenty-four people, the odds are better than even that two of them have the same birthday." Eventually, out of 22 present, it is revealed that two characters share the same birthday, May 23.

==Notes==
{{reflist|group=nb}}

==References==
{{Reflist|45em}}

==Bibliography==

* {{cite journal |first=M. |last=Abramson |first2=W. O. J. |last2=Moser |year=1970 |title=More Birthday Surprises |journal=[[American Mathematical Monthly]] |volume=77 |issue= 8|pages=856–858 |doi= 10.2307/2317022|jstor=2317022 }}
* {{cite journal |first=D. |last=Bloom |year=1973 |title=A Birthday Problem |journal=[[American Mathematical Monthly]] |volume=80 |issue= 10|pages=1141–1142 |doi=10.2307/2318556 | jstor=2318556 |ref=harv}}
* {{cite book |first=John G. |last=Kemeny |first2=J. Laurie |last2=Snell |first3=Gerald |last3=Thompson |title=Introduction to Finite Mathematics |location= |edition=First |year=1957 }}
* {{cite journal |first=M. |last=Klamkin |first2=D. |last2=Newman |year=1967 |title=Extensions of the Birthday Surprise |journal=Journal of Combinatorial Theory |volume=3 |issue= 3|pages=279–282 |doi= 10.1016/s0021-9800(67)80075-9}}
* {{cite journal |first=E. H. |last=McKinney |year=1966 |title=Generalized Birthday Problem |journal=[[American Mathematical Monthly]] |volume=73 |issue= 4|pages=385–387 |doi= 10.2307/2315408|jstor=2315408 }}
* {{cite book |author1link=Leila Schneps |first=Leila |last=Schneps |author2link=Coralie Colmez |first2=Coralie |last2=Colmez |title=Math on Trial. How Numbers Get Used and Abused in the Courtroom |location= |publisher=Basic Books |year=2013 |isbn=978-0-465-03292-1 |chapter=Math error number 5. The case of Diana Sylvester: cold hit analysis }}
* {{cite book|author=Sy M. Blinder|title=Guide to Essential Math: A Review for Physics, Chemistry and Engineering Students|url=https://books.google.com/books?id=M7TCNAEACAAJ|year=2013|publisher=Elsevier|isbn=978-0-12-407163-6|pages=5–6}}

==External links==
* [http://www.efgh.com/math/birthday.htm The Birthday Paradox accounting for leap year birthdays]
* {{MathWorld | urlname=BirthdayProblem | title=Birthday Problem}}
* [http://www.damninteresting.com/?p=402 A humorous article explaining the paradox]
* [http://wiki.stat.ucla.edu/socr/index.php/SOCR_EduMaterials_Activities_BirthdayExperiment SOCR EduMaterials activities birthday experiment]
* [http://betterexplained.com/articles/understanding-the-birthday-paradox/  Understanding the Birthday Problem (Better Explained)]
* [http://www.matifutbol.com/en/eurobirthdays.html Eurobirthdays 2012. A birthday problem.] A practical football example of the birthday paradox.
* {{cite web|last=Grime|first=James|title=23: Birthday Probability|url=http://www.numberphile.com/videos/23birthday.html|work=Numberphile|publisher=[[Brady Haran]]|access-date=2013-04-02|archive-url=https://web.archive.org/web/20170225140726/http://www.numberphile.com/videos/23birthday.html|archive-date=2017-02-25|url-status=dead}}
* [https://www.wolframalpha.com/input/?i=birthday+paradox%2C+4+people%2C+100+possible+birthdays Computing the probabilities of the Birthday Problem at WolframAlpha]

{{Portal bar|Mathematics}}

{{DEFAULTSORT:Birthday Problem}}
[[Category:Probability theory paradoxes]]
[[Category:Probability problems]]
[[Category:Applied probability]]
[[Category:Birthdays]]
[[Category:Mathematical problems]]
[[Category:Coincidence]]