Editing Birthday problem (section)

===Probability of a shared birthday (collision)===
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. The birthday problem and its generalizations are also useful tools for modelling coincidences.<ref name="Pollanen">M. Pollanen (2024) ''A Double Birthday Paradox in the Study of Coincidences'', [[Mathematics]] '''23'''(24), 3882. https://doi.org/10.3390/math12243882</ref>

====Probability of a unique collision====

The classic birthday problem allows for more than two people to share a particular birthday or for there to be matches on multiple days. The probability that among {{mvar|n}} people there is exactly one pair of individuals with a matching birthday given {{mvar|d}} possible days is<ref name="Pollanen"/>

: <math> p_2(n; d) =   \frac{{n \choose 2}}{d-n+1} (1-p(n; d)) </math>

Unlike the standard birthday problem, as {{mvar|n}} increases the probability reaches a maximum value before decreasing. For example, for {{math|''d'' {{=}} 365}}, the probability of a unique match has a maximum value of 0.3864 occurring when {{math|''n'' {{=}} 28}}.

====Generalization to multiple types of people====
[[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.