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In probability theory , the birthday problem or birthday paradox concerns the probability that, in a set of n randomly chosen people, some pair of them will have the same birthday. However, 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.

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 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. Rouse Ball indicated without citation that it was first discussed by Harold Davenport.

The problem is to compute an approximate probability that in a group of n people at least two have the same birthday. For simplicity, variations in the distribution, such as leap years , twins , seasonal, or weekday variations are disregarded, and it is assumed that all 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. The goal is to compute P A , the probability that at least two people in the room have the same birthday.

In deference to widely published solutions [ which? If one numbers the 23 people from 1 to 23, the 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 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 process can be generalized to a group of n people, where p n is the probability of at least two of the n people sharing a birthday.

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It is easier to first calculate the probability p n that all n birthdays are different. The event of at least two of the n persons having the same birthday is complementary to all n birthdays being different.

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Therefore, its probability p n is. The following table shows the probability for some other values of n this table ignores the existence of leap years, as described above, as well as assuming that each birthday is equally likely :. Leap years. The first expression derived for p n can be approximated as. According to the approximation, the same approach can be applied to any number of "people" and "days". 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.

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

Applying the Poisson approximation for the binomial on the group of 23 people,. A good rule of thumb which can be used for mental calculation is the relation. In these equations, m is the number of days in a year. 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.

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## Birthday problem

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. The argument below is adapted from an argument of Paul Halmos. This yields. Therefore, the expression above is not only an approximation, but also an upper bound of p n.

The inequality.

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Solving for n gives. Now, ln 2 is approximately Therefore, 23 people suffice. 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 n is 22 or less could also work.

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In other words, n d is the minimal integer n such that. The classical birthday problem thus corresponds to determining n You can search by as much or as little input as you want Unknown fields can be left blank. If the Date of Birth is unknown, try searching by Name.

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