Birthday problem

File:Birthday Paradox.svg

The computed probability of at least two people sharing the same birthday versus the number of people

In probability theory, the birthday problem asks for the probability that, in a set of Template:Mvar randomly chosen people, at least two will share the same birthday. The birthday paradox is the counterintuitive fact that only 23 people are needed for that probability to exceed 50%.

The birthday paradox is a veridical paradox: it seems wrong at first glance but is, in fact, true. While it may seem surprising that only 23 individuals are required to reach a 50% probability of a shared birthday, this result is made more intuitive by considering that the birthday comparisons will be made between every possible pair of individuals. With 23 individuals, there are Template:Sfrac = 253 pairs to consider.

Real-world applications for the birthday problem 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 problem is generally attributed to Harold Davenport in about 1927, though he did not publish it at the time. Davenport did not claim to be its discoverer "because he could not believe that it had not been stated earlier". The first publication of a version of the birthday problem was by Richard von Mises in 1939.

Calculating the probability

Consider the event Template:Mvar that a group of Template:Mvar people does not have any repeated birthdays, and let the complementary event Template:Mvar be that of a group of Template:Mvar people contains at least two people who share a birthday. Then the probabilities P(A) and P(B) of the two events are related by the equation P(B) = 1 − P(A). The probability P(A) can be computed using the perspective of permutations, as follows. Let $V_{nr}$ be the total number of ways that Template:Mvar can have distinct birthdays, and let $V_t$ be the total number of ways Template:Mvar people can have birthdays arranged, including possibly repeated birthdays. The probability P(A) is the ratio of these two quantities, $V_{nr}$ divided by $V_t$. When $k=23$, the two counts are given by $$V_{nr}=365\cdot 364\cdot 363\cdots (365-22)=\frac{365!}{(365-23)!}$$ and $$V_t=365\cdot 365\cdot 365\cdots 365=365^{23},$$ their ratio is $P(A)=\frac{V_{nr}}{V_t}\approx 0.492703$, and so $$P(B)=1-P(A)\approx 1-0.492703=0.507297(50.7297\mathrm{\%}).$$

Another way the birthday problem can be solved is by asking for an approximate probability that in a group of Template:Mvar people at least two have the same birthday. For simplicity, leap years, twins, selection bias, and seasonal and weekly variations in birth ratesTemplate:Refn are generally disregarded, and instead it is assumed that there are 365 possible birthdays, and that each person's birthday is equally likely to be any of these days, independent of the other people in the group.

For independent birthdays, a uniform distribution of birthdays minimizes the probability of two people in a group having the same birthday. Any unevenness increases the likelihood of two people sharing a birthday. However real-world birthdays are not sufficiently uneven to make much change: the real-world group size necessary to have a greater than 50% chance of a shared birthday is 23, as in the theoretical uniform distribution.

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

Here is the calculation of P(B) for 23 people. Let the 23 people be numbered 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 22. Let these events be called Event 2, Event 3, and so on. Event 1 is 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 Template:Sfrac, 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 Template:Sfrac, 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 Template:Sfrac. Finally, the principle of conditional probability implies that P(A′) is equal to the product of these individual probabilities: Template:NumBlk

The terms of equation (1) can be collected to arrive at: Template:NumBlk

Evaluating equation (2) gives P(A′) ≈ 0.492703

Therefore, P(B) ≈ 1 − 0.492703 = 0.507297 (50.7297%).

This process can be generalized to a group of Template:Mvar people, where p(n) is the probability of at least two of the Template:Mvar people sharing a birthday. It is easier to first calculate the probability p(n) that all Template:Mvar birthdays are different. According to the pigeonhole principle, p(n) is zero when n > 365. When n ≤ 365:

$\begin{array}{rl}\overline{p}(n)& =1\times (1-\frac{1}{365})\times (1-\frac{2}{365})\times \cdots \times (1-\frac{n-1}{365})\\ & =\frac{365\times 364\times \cdots \times (365-n+1)}{365^n}\\ & =\frac{365!}{365^n(365-n)!}=\frac{n!\cdot (\genfrac{}{}{0ex}{}{365}{n})}{365^n}=\frac{{}_{365}{P}_n}{365^n}\end{array}$

where ! is the factorial operator, Template:Pars is the binomial coefficient and _kP_r 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 Template:Pars, the third cannot have the same birthday as either of the first two Template:Pars, and in general the Template:Mvarth birthday cannot be the same as any of the n − 1 preceding birthdays.

The event of at least two of the Template:Mvar persons having the same birthday is complementary to all Template:Mvar birthdays being different. Therefore, its probability p(n) is

$p(n)=1-\overline{p}(n).$

The following table shows the probability for some other values of Template:Mvar (for this table, the existence of leap years is ignored, and each birthday is assumed to be equally likely):

File:Birthdaymatch.svg

The probability that no two people share a birthday in a group of Template:Mvar people. Note that the vertical scale is logarithmic (each step down is 10²⁰ times less likely).

Template:Mvar	p(n)
1	00.0%
5	02.7%
10	11.7%
20	41.1%
23	50.7%
30	70.6%
40	89.1%
50	97.0%
60	99.4%
70	99.9%
75	99.97%
100	%
200	(100 − )%
300	(100 − )%
350	(100 − )%
365	(100 − )%
≥ 366	100%

Approximations

File:Birthday paradox probability.svg

Graphs showing the approximate probabilities of at least two people sharing a birthday (red) and its complementary event (blue)

File:Birthday paradox approximation.svg

A graph showing the accuracy of the approximation 1 − e^−n2/730 (red)

The Taylor series expansion of the exponential function (the constant e ≈ )

$e^x=1+x+\frac{x^2}{2!}+\cdots$

provides a first-order approximation for e^x for $|x|\ll 1$:

$e^x\approx 1+x.$

To apply this approximation to the first expression derived for p(n), set x = −Template:Sfrac. Thus,

$e^{-a/365}\approx 1-\frac{a}{365}.$

Then, replace Template:Mvar with non-negative integers for each term in the formula of p(n) until a = n − 1, for example, when a = 1,

$e^{-1/365}\approx 1-\frac{1}{365}.$

The first expression derived for p(n) can be approximated as

$\begin{array}{rl}\overline{p}(n)& \approx 1\cdot e^{-1/365}\cdot e^{-2/365}\cdots e^{-(n-1)/365}\\ & =e^{-(1+2+\cdots +(n-1))/365}\\ & =e^{-\frac{n(n-1)/2}{365}}=e^{-\frac{n(n-1)}{730}}.\end{array}$

Therefore,

$p(n)=1-\overline{p}(n)\approx 1-e^{-\frac{n(n-1)}{730}}.$

An even coarser approximation is given by

$p(n)\approx 1-e^{-\frac{n^2}{730}},$

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 Template:Mvar, if there are Template:Mvar persons, and if n ≪ d, then using the same approach as above we achieve the result that if p(n, d) is the probability that at least two out of Template:Mvar people share the same birthday from a set of Template:Mvar available days, then:

$\begin{array}{rl}p(n,d)& \approx 1-e^{-\frac{n(n-1)}{2d}}\\ & \approx 1-e^{-\frac{n^2}{2d}}.\end{array}$

Simple exponentiation

The probability of any two people not having the same birthday is Template:Sfrac. In a room containing n people, there are Template:Pars = Template:Sfrac pairs of people, i.e. Template:Pars 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. Being independent would be equivalent to picking with replacement, any pair of people in the world, not just in a room. In short Template:Sfrac can be multiplied by itself Template:Pars times, which gives us

$\overline{p}(n)\approx {\left(\frac{364}{365}\right)}^{(\genfrac{}{}{0ex}{}{n}{2})}.$

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

$p(n)\approx 1-{\left(\frac{364}{365}\right)}^{(\genfrac{}{}{0ex}{}{n}{2})}.$

And for the group of 23 people, the probability of sharing is

$p(23)\approx 1-{\left(\frac{364}{365}\right)}^{(\genfrac{}{}{0ex}{}{23}{2})}=1-{\left(\frac{364}{365}\right)}^{253}\approx 0.500477.$

Poisson approximation

Applying the Poisson approximation for the binomial on the group of 23 people,

$\mathrm{Poi}\left(\frac{(\genfrac{}{}{0ex}{}{23}{2})}{365}\right)=\mathrm{Poi}\left(\frac{253}{365}\right)\approx \mathrm{Poi}(0.6932)$

so

$\mathrm{Pr}(X>0)=1-\mathrm{Pr}(X=0)\approx 1-e^{-0.6932}\approx 1-0.499998=0.500002.$

The result is over 50% as previous descriptions. This approximation is the same as the one above based on the Taylor expansion that uses e^x ≈ 1 + x.

Square approximation

A good rule of thumb which can be used for mental calculation is the relation

$p(n,d)\approx \frac{n^2}{2d}$

which can also be written as

$n\approx \sqrt{2d\times p(n)}$

which works well for probabilities less than or equal to Template:Sfrac. In these equations, Template:Mvar is the number of days in a year.

For instance, to estimate the number of people required for a Template:Sfrac chance of a shared birthday, we get

$n\approx \sqrt{2\times 365\times \frac{1}{2}}=\sqrt{365}\approx 19$

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 Template:Sfrac chance of matching:

$n\ge \frac{1}{2}+\sqrt{\frac{1}{4}+2\times \ln (2)\times 365}=22.999943.$

This is a result of the good approximation that an event with Template:Sfrac probability will have a Template:Sfrac chance of occurring at least once if it is repeated k ln 2 times.

Probability table

length of hex string	no. of bits (Template:Mvar)	hash space size (2b)	Number of hashed elements such that probability of at least one hash collision ≥ Template:Mvar
			Template:Mvar =	Template:Mvar =	Template:Mvar =	Template:Mvar =	Template:Mvar =	Template:Mvar = 0.001	Template:Mvar = 0.01	Template:Mvar = 0.25	Template:Mvar = 0.50	Template:Mvar = 0.75
8	32		2	2	2	2.9	93
(10)	(40)	()	2	2	2	47
(12)	(48)	()	2	2	24
16	64		6.1
(24)	(96)	()
32	128
(48)	(192)	()
64	256
(96)	(384)	()
128	512

File:Birthday attack vs paradox.svg

Comparison of the birthday problem (1) and birthday attack (2):Template:Parabreak In (1), collisions are found within one set, in this case, 3 out of 276 pairings of the 24 lunar astronauts.Template:Parabreak In (2), collisions are found between two sets, in this case, 1 out of 256 pairings of only the first bytes of SHA-256 hashes of 16 variants each of benign and harmful contracts.

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, to is the uncorrectable bit error rate of a typical hard disk. In theory, 128-bit hash functions, such as MD5, should stay within that range until about 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.Template:Refn

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

$1-p(n)=\overline{p}(n)=\prod _{k=1}^{n-1}(1-\frac{k}{365}).$

As in earlier paragraphs, interest lies in the smallest Template:Mvar such that p(n) > Template:Sfrac; or equivalently, the smallest Template:Mvar such that p(n) < Template:Sfrac.

Using the inequality 1 − x < e^−x in the above expression we replace 1 − Template:Sfrac with e^−k⁄365. This yields

$\overline{p}(n)=\prod _{k=1}^{n-1}(1-\frac{k}{365})<\prod _{k=1}^{n-1}\left(e^{-\frac{k}{365}}\right)=e^{-\frac{n(n-1)}{730}}.$

Therefore, the expression above is not only an approximation, but also an upper bound of p(n). The inequality

$e^{-\frac{n(n-1)}{730}}<\frac{1}{2}$

implies p(n) < Template:Sfrac. Solving for Template:Mvar gives

$n^2-n>730\ln 2.$

Now, 730 ln 2 is approximately 505.997, which is barely below 506, the value of n² − n attained when n = 23. Therefore, 23 people suffice. Incidentally, solving n² − 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 the chances of a birthday match are at least even; it leaves open the possibility that Template:Mvar is 22 or less could also work.

Generalizations

Arbitrary number of days

Given a year with Template:Mvar days, the generalized birthday problem asks for the minimal number n(d) such that, in a set of Template:Mvar randomly chosen people, the probability of a birthday coincidence is at least 50%. In other words, n(d) is the minimal integer Template:Mvar such that

$1-(1-\frac{1}{d})(1-\frac{2}{d})\cdots (1-\frac{n-1}{d})\ge \frac{1}{2}.$

The classical birthday problem thus corresponds to determining n(365). The first 99 values of n(d) are given here (sequence A033810 in the OEIS):

Template:Mvar	1–2	3–5	6–9	10–16	17–23	24–32	33–42	43–54	55–68	69–82	83–99
n(d)	2	3	4	5	6	7	8	9	10	11	12

A similar calculation shows that n(d) = 23 when Template:Mvar is in the range 341–372.

A number of bounds and formulas for n(d) have been published. For any d ≥ 1, the number n(d) satisfies

$\frac{3-2\ln 2}{6}<n(d)-\sqrt{2d\ln 2}\le 9-\sqrt{86\ln 2}.$

These bounds are optimal in the sense that the sequence n(d) −

1. REDIRECT Template:Radic

Template:Rcat shell gets arbitrarily close to

$\frac{3-2\ln 2}{6}\approx 0.27,$

while it has

$9-\sqrt{86\ln 2}\approx 1.28$

as its maximum, taken for d = 43.

The bounds are sufficiently tight to give the exact value of n(d) in most of the cases. For example, for d = 365 these bounds imply that 22.7633 < n(365) < 23.7736 and 23 is the only integer in that range. In general, it follows from these bounds that n(d) always equals either

$\lceil \sqrt{2d\ln 2}\rceil \text{or}\lceil \sqrt{2d\ln 2}\rceil +1$

where ⌈ · ⌉ denotes the ceiling function. The formula

$n(d)=\lceil \sqrt{2d\ln 2}\rceil$

holds for 73% of all integers Template:Mvar. The formula

$n(d)=\lceil \sqrt{2d\ln 2}+\frac{3-2\ln 2}{6}\rceil$

holds for almost all Template:Mvar, i.e., for a set of integers Template:Mvar with asymptotic density 1.

The formula

$n(d)=\lceil \sqrt{2d\ln 2}+\frac{3-2\ln 2}{6}+\frac{9-4(\ln 2{)}^2}{72\sqrt{2d\ln 2}}\rceil$

holds for all d ≤ , but it is conjectured that there are infinitely many counterexamples to this formula.

The formula

$n(d)=\lceil \sqrt{2d\ln 2}+\frac{3-2\ln 2}{6}+\frac{9-4(\ln 2{)}^2}{72\sqrt{2d\ln 2}}-\frac{2(\ln 2{)}^2}{135d}\rceil$

holds for all d ≤ , and it is conjectured that this formula holds for all Template:Mvar.

More than two people sharing a birthday

It is possible to extend the problem to ask how many people in a group are necessary for there to be a greater than 50% probability that at least 3, 4, 5, etc. of the group share the same birthday.

The first few values are as follows: >50% probability of 3 people sharing a birthday - 88 people; >50% probability of 4 people sharing a birthday - 187 people (sequence A014088 in the OEIS).

Everyone shares a birthday

The strong birthday problem asks for the number of people that need to be gathered together before there is a 50% chance that everyone in the gathering shares their birthday with at least one other person. For d=365 days the answer is 3,064 people.

The number of people needed for arbitrary number of days is given by (sequence A380129 in the OEIS)

Probability of a shared birthday (collision)

The birthday problem can be generalized as follows:

Given Template:Mvar random integers drawn from a discrete uniform distribution with range [1,d], what is the probability p(n; d) that at least two numbers are the same? (d = 365 gives the usual birthday problem.)

The generic results can be derived using the same arguments given above.

$\begin{array}{rl}p(n;d)& =\{\begin{array}{ll}1-\prod _{k=1}^{n-1}(1-\frac{k}{d})& n\le d\\ 1& n>d\end{array}\\ [8px]& \approx 1-e^{-\frac{n(n-1)}{2d}}\\ & \approx 1-{\left(\frac{d-1}{d}\right)}^{\frac{n(n-1)}{2}}\end{array}$

Conversely, if n(p; d) denotes the number of random integers drawn from [1,d] to obtain a probability Template:Mvar that at least two numbers are the same, then

$n(p;d)\approx \sqrt{2d\cdot \ln \left(\frac{1}{1-p}\right)}.$

The birthday problem in this more generic sense applies to hash functions: the expected number of N-bit hashes that can be generated before getting a collision is not 2^N, but rather only 2^N⁄2. This is exploited by birthday attacks on cryptographic hash functions 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 under the name of 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.

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 Template:Mvar people there is exactly one pair of individuals with a matching birthday given Template:Mvar possible days is

$p_2(n;d)=\frac{(\genfrac{}{}{0ex}{}{n}{2})}{d-n+1}(1-p(n;d))$

Unlike the standard birthday problem, as Template:Mvar increases the probability reaches a maximum value before decreasing. For example, for d = 365, the probability of a unique match has a maximum value of 0.3864 occurring when n = 28.

Generalization to multiple types of people

File:2d birthday.png

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. In the simplest extension there are two types of people, say Template:Mvar men and Template:Mvar 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

$p_0=\frac{1}{d^{m+n}}\sum _{i=1}^m\sum _{j=1}^n{S}_2(m,i)S_2(n,j)\prod _{k=0}^{i+j-1}d-k$

where d = 365 and S₂ are Stirling numbers of the second kind. Consequently, the desired probability is 1 − p₀.

This variation of the birthday problem is interesting because there is not a unique solution for the total number of people 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 Template:Mvar is p(n) − p(n − 1) maximum? The answer is 20—if there is a prize for first match, the best position in line is 20th.

Same birthday as you

File:Birthday paradox.svg

Comparing p(n) = probability of a birthday match with q(n) = probability of matching your birthday

In the birthday problem, neither of the two people is chosen in advance. By contrast, the probability q(n) that at least one other person in a room of Template:Mvar other people has the same birthday as a particular person (for example, you) is given by

$q(n)=1-{\left(\frac{365-1}{365}\right)}^n$

and for general Template:Mvar by

$q(n;d)=1-{\left(\frac{d-1}{d}\right)}^n.$

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

Number of people with a shared birthday

For any one person in a group of n people the probability that he or she shares his birthday with someone else is $q(n-1;d)$, as explained above. The expected number of people with a shared (non-unique) birthday can now be calculated easily by multiplying that probability by the number of people (n), so it is:

$n(1-{\left(\frac{d-1}{d}\right)}^{n-1})$

(This multiplication can be done this way because of the linearity of the expected value of indicator variables). This implies that the expected number of people with a non-shared (unique) birthday is:

$n{\left(\frac{d-1}{d}\right)}^{n-1}$

Similar formulas can be derived for the expected number of people who share with three, four, etc. other people.

Number of people until every birthday is achieved

The expected number of people needed until every birthday is achieved is called the Coupon collector's problem. It can be calculated by nH_n, where H_n is the Template:Mvarth harmonic number. For 365 possible dates (the birthday problem), the answer is 2365.

Near matches

Another generalization is to ask for the probability of finding at least one pair in a group of Template:Mvar people with birthdays within Template:Mvar calendar days of each other, if there are Template:Mvar equally likely birthdays.

$\begin{array}{rl}p(n,k,d)& =1-\frac{(d-nk-1)!}{d^{n-1}(d-n(k+1))!}\end{array}$

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

Template:Mvar	Template:Mvar for 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.

Number of days with a certain number of birthdays

Number of days with at least one birthday

The expected number of different birthdays, i.e. the number of days that are at least one person's birthday, is:

$d-d{\left(\frac{d-1}{d}\right)}^n$

This follows from the expected number of days that are no one's birthday:

$d{\left(\frac{d-1}{d}\right)}^n$

which follows from the probability that a particular day is no one's birthday, Template:Pars, easily summed because of the linearity of the expected value.

For instance, with Template:Var = 365, you should expect about 21 different birthdays when there are 22 people, or 46 different birthdays when there are 50 people. When there are 1000 people, there will be around 341 different birthdays (24 unclaimed birthdays).

Number of days with at least two birthdays

The above can be generalized from the distribution of the number of people with their birthday on any particular day, which is a Binomial distribution with probability Template:Sfrac. Multiplying the relevant probability by Template:Mvar will then give the expected number of days. For example, the expected number of days which are shared; i.e. which are at least two (i.e. not zero and not one) people's birthday is:

$$d-d{\left(\frac{d-1}{d}\right)}^n-d\cdot (\genfrac{}{}{0ex}{}{n}{1}){\left(\frac{1}{d}\right)}^1{\left(\frac{d-1}{d}\right)}^{n-1}=d-d{\left(\frac{d-1}{d}\right)}^n-n{\left(\frac{d-1}{d}\right)}^{n-1}$$

Number of people who repeat a birthday

The probability that the Template:Mvarth integer randomly chosen from [1,d] will repeat at least one previous choice equals q(k − 1; d) above. The expected total number of times a selection will repeat a previous selection as Template:Mvar such integers are chosen equals

$\sum _{k=1}^{n}q(k-1;d)=n-d+d{\left(\frac{d-1}{d}\right)}^n$

This can be seen to equal the number of people minus the expected number of different birthdays.

The distribution of the random variable reporting the number Template:Mvar of integers to be chosen in order to get exactly Template:Mvar repeats (for a constant Template:Mvar) converges to a chi-distributed random variable as $d\to \mathrm{\infty }$.

Average number of people to get at least one shared birthday

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[[[:Template:Mvar]] 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 hashing algorithms analyzed by Donald Knuth in his book The Art of Computer Programming. It may be shown that if one samples uniformly, with replacement, from a population of size M, the number of trials required for the first repeated sampling of some individual has expected value n = 1 + Q(M), where

$Q(M)=\sum _{k=1}^M\frac{M!}{(M-k)!M^k}.$

The function

$Q(M)=1+\frac{M-1}{M}+\frac{(M-1)(M-2)}{M^2}+\cdots +\frac{(M-1)(M-2)\cdots 1}{M^{M-1}}$

has been studied by Srinivasa Ramanujan and has asymptotic expansion:

$Q(M)\sim \sqrt{\frac{\pi M}{2}}-\frac{1}{3}+\frac{1}{12}\sqrt{\frac{\pi }{2M}}-\frac{4}{135M}+\cdots .$

With M = 365 days in a year, the average number of people required to find a pair with the same birthday is 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 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. For each pair (i, j) for k people in a room, we define the indicator random variable X_ij, for $1\le i\le j\le k$, by

$$\begin{array}{rl}{X}_{ij}& =I\{\text{person }i\text{ and person }j\text{ have the same birthday}\}\\ & =\{\begin{array}{ll}1,& \text{if person }i\text{ and person }j\text{ have the same birthday;}\\ 0,& \text{otherwise.}\end{array}\end{array}$$

$$\begin{array}{rl}E[X_{ij}]& =\mathrm{Pr}\{\text{person }i\text{ and person }j\text{ have the same birthday}\}=\frac{1}{n}.\end{array}$$

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

$$X=\sum _{i=1}^k\sum _{j=i+1}^k{X}_{ij}$$

$$\begin{array}{rl}E[X]& =\sum _{i=1}^k\sum _{j=i+1}^{k}E[X_{ij}]\\ & =(\genfrac{}{}{0ex}{}{k}{2})\frac{1}{n}\\ & =\frac{k(k-1)}{2n}\end{array}$$

For n = 365, if k = 28, the expected number of pairs of individuals with the same birthday is Template:Sfrac ≈ 1.0356. Therefore, we can expect at least one matching pair with at least 28 people.

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.

Voracek, Tran and 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. Further results showed 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 Template:Mvar, the greatest Template:Mvar for which the probability p(n) is smaller than the given Template:Mvar, or the smallest Template:Mvar for which the probability p(n) is greater than the given Template:Mvar.

Taking the above formula for d = 365, one has

$n(p;365)\approx \sqrt{730\ln \left(\frac{1}{1-p}\right)}.$

The following table gives some sample calculations.

Template:Mvar	Template:Mvar	n↓	p(n↓)	n↑	p(n↑)
0.01	0.14178 REDIRECT Template:Radic Template:Rcat shell = 2.70864	2	0.00274	3	0.00820
0.05	0.32029 REDIRECT Template:Radic Template:Rcat shell = 6.11916	6	0.04046	7	0.05624
0.1	0.45904 REDIRECT Template:Radic Template:Rcat shell = 8.77002	8	0.07434	9	0.09462
0.2	0.66805 REDIRECT Template:Radic Template:Rcat shell = 12.76302	12	0.16702	13	0.19441
0.3	0.84460 REDIRECT Template:Radic Template:Rcat shell = 16.13607	16	0.28360	17	0.31501
0.5	1.17741 REDIRECT Template:Radic Template:Rcat shell = 22.49439	22	0.47570	23	0.50730
0.7	1.55176 REDIRECT Template:Radic Template:Rcat shell = 29.64625	29	0.68097	30	0.70632
0.8	1.79412 REDIRECT Template:Radic Template:Rcat shell = 34.27666	34	0.79532	35	0.81438
0.9	2.14597 REDIRECT Template:Radic Template:Rcat shell = 40.99862	40	0.89123	41	0.90315
0.95	2.44775 REDIRECT Template:Radic Template:Rcat shell = 46.76414	46	0.94825	47	0.95477
0.99	3.03485 REDIRECT Template:Radic Template:Rcat shell = 57.98081	57	0.99012	58	0.99166

Some values falling outside the bounds have been colored 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 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 . 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.

The reason is that the correct comparison is to the number of partitions of the weights into left and right. There are 2^{N − 1} different partitions for 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 Gaussian, with a peak at N and width

1. REDIRECT Template:Radic

Template:Rcat shell, so that when 2^{N − 1} is approximately equal to

1. REDIRECT Template:Radic

Template:Rcat shell the transition occurs. 2^{23 − 1} is about 4 million, while the width of the distribution is only 5 million.

In fiction

Arthur C. Clarke's 1961 novel A Fall of Moondust 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.

Bibliography

• Reprinted in

External links

• The Birthday Paradox accounting for leap year birthdays
• A humorous article explaining the paradox
• SOCR EduMaterials activities birthday experiment
• Understanding the Birthday Problem (Better Explained)
• Eurobirthdays 2012. A birthday problem. A practical football example of the birthday paradox.
• Computing the probabilities of the Birthday Problem at WolframAlpha