Birthday problem

In probability theory, the birthday problem or birthday paradox pertains to the probability that, in a set of n randomly 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 366. However, 99% probability is reached with just 57 people, and 50% probability with 23 people. These conclusions are based on the assumption that each day of the year (except February 29) is equally probable for a birthday.

The mathematics behind this problem led to a well-known cryptographic attack called the birthday attack, which uses this probabilistic model to reduce the complexity of cracking a hash function.

A graph showing the approximate probability of at least two people sharing a birthday amongst a certain number of people.

Understanding the problem

The birthday problem asks whether any of the people in a given group has a birthday matching any of the others — not one in particular. (See "Same birthday as you" below for an analysis of this much less surprising alternative problem.)

In the example given earlier, a list of 23 people, comparing the birthday of the first person on the list to the others allows 22 chances for a matching birthday (The second person on the list to the others allows 21 chances for a matching birthday, third person has 20 chances, and so on. Hence 22+21+20+....+1 = 253), but comparing every person to all of the others allows 253 distinct chances (combinations): in a group of 23 people there are ${25 \choose 2}={\frac {25\cdot 22}{2}}=253$ pairs.

Presuming all birthdays are equally probable, the probability of a given birthday for a person chosen from the entire population at random is 1/365 (ignoring Leap Day, February 29). Although the pairings in a group of 23 people are not statistically equivalent to 253 pairs chosen independently, the birthday paradox becomes less surprising if a group is thought of in terms of the number of possible pairs, rather than as the number of individuals.

Calculating the probability

The problem is to compute the approximate probability that in a room of n people, at least two have the same birthday. For simplicity, disregard variations in the distribution, such as leap years, twins, seasonal or weekday variations, and assume that the 365 possible birthdays are equally likely. Real-life birthday distributions are not uniform since not all dates are equally likely.

If P(A) is the probability of at least two people in the room having the same birthday, it may be simpler to calculate P(A'), the probability of there not being any two people having the same birthday. Then, because P(A) and P(A') are the only two possibilities and are also mutually exclusive, P(A') = 1 − P(A).

In deference to widely published solutions concluding that 25 is the number of people necessary to have a P(A) that is greater than 50%, the following calculation of P(A) will use 25 people as an example.

When events are independent of each other, the probability of all of the events occurring is equal to a product of the probabilities of each of the events occurring. Therefore, if P(A') can be described as 23 independent events, P(A') could be calculated as P(1) × P(2) × P(3) × ... × P(25).

The 23 independent events correspond to the 23 people, and can be defined in order. Each event can be defined as the corresponding person not sharing their birthday with any of the previously analyzed people. For Event 1, there are no previously analyzed people. Therefore, the probability, P(1), that person number 1 does not share his/her birthday with previously analyzed people is 1, or 100%. Ignoring leap years for this analysis, the probability of 1 can also be written as 365/365, for reasons that will become clear below.

For Event 2, the only previously analyzed people are Person 1. Assuming that birthdays are equally likely to happen on each of the 365 days of the year, the probability, P(2), that Person 2 has a different birthday than Person 1 is 364/365. This is because, if Person 2 was born on any of the other 364 days of the year, Persons 1 and 2 will not share the same birthday.

Similarly, if Person 3 is born on any of the 363 days of the year other than the birthdays of Persons 1 and 2, Person 3 will not share their birthday. This makes the probability P(3) = 363/365.

This analysis continues until Person 25 is reached, whose probability of not sharing their birthday with people analyzed before, P(25), is 343/365.

P(A') is equal to the product of these individual probabilities:

(1) P(A') = 365/365 × 364/365 × 363/365 × 362/365 × ... × 343/365

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

(2) P(A') = (1/365)²³ × (365 × 364 × 363 × ... × 343)

Evaluating equation (2) gives P(A') = 0.492703

Therefore, P(A) = 1 − 0.492703 = 0.507297 (50.7297%)

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. It is easier to first calculate the probability p(n) that all n birthdays are different. According to the pigeonhole principle, p(n) is zero when n > 365. When n ≤ 365:

${\begin{aligned}{\bar {p}}(n)&=1\times \left(1-{\frac {1}{365}}\right)\times \left(1-{\frac {2}{365}}\right)\times \cdots \times \left(1-{\frac {n-1}{365}}\right)\\&={365\times 364\times \cdots \times (365-n+1) \over 365^{n}}\\&={365! \over 365^{n}(365-n)!}={\frac {n!\cdot {365 \choose n}}{365^{n}}}\end{aligned}}$

where ' ! ' is the factorial operator.

The equation expresses the fact that for no persons to share a birthday, a second person cannot have the same birthday as the first (364/365), the third cannot have the same birthday as the first two (363/365), and in general the n^th birthday cannot be the same as any of the n − 1 preceding birthdays.

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

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

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

This probability surpasses 1/2 for n = 23 (with value about 50.7%). The following table shows the probability for some other values of n (this table ignores the existence of leap years, as described above):

n	p(n)
10	11.7%
20	41.1%
23	50.7%
30	70.6%
50	97.0%
57	99.0%
100	99.99997%
200	99.9999999999999999999999999998%
300	(100 − (6×10−80))%
350	(100 − (3×10−129))%
365	(100 − (1.45×10−155))%
366	100%

Approximations

The Taylor series expansion of the exponential function (the constant e = 2.718281828, approximately)

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

A graph showing the accuracy of the approximation $1-e^{-n^{2}/(2\times 365)}$

provides a first-order approximation for e^x for x << 1:

$e^{x}\approx 1+x.\$

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

${\begin{aligned}{\bar {p}}(n)&\approx 1\times e^{-1/365}\times e^{-2/365}\cdots e^{-(n-1)/365}\\&=1\times e^{-(1+2+\cdots +(n-1))/365}\\&=e^{-(n(n-1)/2)/365}.\end{aligned}}$

Therefore,

$p(n)=1-{\bar {p}}(n)\approx 1-e^{-n(n-1)/(2\times 365)}.$

An even coarser approximation is given by

$p(n)\approx 1-e^{-n^{2}/(2\times 365)},\,$

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

A simple exponentiation

The probability of any two people not having the same birthday is 364/365. In a room containing n people, there are C(n, 2)=n(n-1)/2 pairs of people, i.e. C(n, 2) 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 364/365 can be multiplied by itself C(n, 2) times, which gives us

$\left({\frac {364}{365}}\right)^{C(n,2)}.$

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

$p(n)\approx 1-\left({\frac {364}{365}}\right)^{C(n,2)}.$

Poisson approximation

Using the Poisson approximation for the binomial,

$\mathrm {Poi} \left({\frac {C(23,2)}{365}}\right)=\mathrm {Poi} \left({\frac {253}{365}}\right)\approx \mathrm {Poi} (0.6932)$

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

Again, this is over 50%.

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 50% chance of matching:

$n\approx {\frac {1}{2}}+{\sqrt {{\frac {1}{4}}-2\times 365\times \ln(0.5)}}=22.999943.$

This is a result of the good approximation that an event with 1 in k probability will have a 50% chance of occurring at least once if it is repeated k ln 2 times.

Probability table

Main article: Birthday attack

length of hex string	#bits	hash space size (2#bits)	Number of hashed elements such that (probability of at least one hash collision) = p
			p = 10−18	p = 10−15	p = 10−12	p = 10−9	p = 10−6	p = 0.1%	p = 1%	p = 25%	p = 50%	p = 75%
8	32	4.3 × 109	2	2	2	2.9	93	2.9 × 103	9.3 × 103	5.0 × 104	7.7 × 104	1.1 × 105
16	64	1.8 × 1019	6.1	1.9 × 102	6.1 × 103	1.9 × 105	6.1 × 106	1.9 × 108	6.1 × 108	3.3 × 109	5.1 × 109	7.2 × 109
32	128	3.4 × 1038	2.6 × 1010	8.2 × 1011	2.6 × 1013	8.2 × 1014	2.6 × 1016	8.3 × 1017	2.6 × 1018	1.4 × 1019	2.2 × 1019	3.1 × 1019
64	256	1.2 × 1077	4.8 × 1029	1.5 × 1031	4.8 × 1032	1.5 × 1034	4.8 × 1035	1.5 × 1037	4.8 × 1037	2.6 × 1038	4.0 × 1038	5.7 × 1038
(96)	(384)	(3.9 × 10115)	8.9 × 1048	2.8 × 1050	8.9 × 1051	2.8 × 1053	8.9 × 1054	2.8 × 1056	8.9 × 1056	4.8 × 1057	7.4 × 1057	1.0 × 1058
128	512	1.3 × 10154	1.6 × 1068	5.2 × 1069	1.6 × 1071	5.2 × 1072	1.6 × 1074	5.2 × 1075	1.6 × 1076	8.8 × 1076	1.4 × 1077	1.9 × 1077

The white squares in this table show the number of hashes needed to achieve the given probability of collision (column) given a hashspace of a certain size in bits (row). (Using the birthday analogy: the "hash space size"(row) would be "365 days", the "probability of collision"(column) would be "50%", and the "required number of people" would be "26"(row-col intersection).) One could of course 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, 10⁻¹⁸ to 10⁻¹⁵ is the uncorrectable bit error rate of a typical hard disk [2]. In theory, MD5, 128 bits, should stay within that range until about 820 billion documents, even if its possible outputs are many more.

An upper bound

The argument below is adapted from an argument of Paul Halmos.

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

$1-p(n)={\bar {p}}(n)=\prod _{k=1}^{n-1}\left(1-{k \over 365}\right).$

As in earlier paragraphs, interest lies in the smallest n such that p(n) > 1/2; or equivalently, the smallest n such that p(n) < 1/2.

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

${\bar {p}}(n)=\prod _{k=1}^{n-1}\left(1-{k \over 365}\right)<\prod _{k=1}^{n-1}\left(e^{-k/365}\right)=e^{-(n(n-1))/(2\times 365)}.$

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

$e^{-(n(n-1))/(2\cdot 365)}<{\frac {1}{2}}$

implies p(n) < 1/2. Solving for n gives

$n^{2}-n>2\times 365\ln 2\,\!.$

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. Solving n² − n = 2 · 365 · ln 2 for n gives, by the way, 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, say, n = 22 could also work.

Generalizations

Cast as a collision problem

The birthday problem can be generalized as follows: given n 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.

$p(n;d)={\begin{cases}1-\prod _{k=1}^{n-1}\left(1-{k \over d}\right)&n\leq d\\1&n>d\end{cases}}$

$p(n;d)\approx 1-e^{-n(n-1)/(2d)}$

$p(n;d)\approx 1-\left({\frac {d-1}{d}}\right)^{n(n-1)/2}$

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

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

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.

Generalization to multiple types

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 just two types, say m "men" and n "women", and the problem becomes characterizing the probability of a shared birthday between at least one man and one woman. (Shared birthdays between, say, two women do not count.) The probability of no (i.e. zero) shared birthdays here is

$p_{0}=1-{\frac {1}{d^{m+n}}}\sum _{i=1}^{m}\sum _{j=1}^{n}S_{2}(m,i)S_{2}(n,j)\prod _{k=0}^{i+j-1}d-k$

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 0.5 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

Reverse problem

For a fixed probability p:

• Find the greatest n for which the probability p(n) is smaller than the given p, or
• Find the smallest n for which the probability p(n) is greater than the given p.

Taking the above formula for d = 365 we have:

$n(p;365)\approx {\sqrt {2\times 365\ln \left({1 \over 1-p}\right)}}.$

Sample calculations

p	n	n↓	p(n↓)	n↑	p(n↑)
0.01	0.14178√365 = 2.70864	2	0.00274	3	0.00820
0.05	0.32029√365 = 6.11916	6	0.04046	7	0.05624
0.1	0.45904√365 = 8.77002	8	0.07434	9	0.09462
0.2	0.66805√365 = 12.76302	12	0.16702	13	0.19441
0.3	0.84460√365 = 16.13607	16	0.28360	17	0.31501
0.5	1.17741√365 = 22.49439	22	0.47570	23	0.50730
0.7	1.55176√365 = 29.64625	29	0.68097	30	0.70632
0.8	1.79412√365 = 34.27666	34	0.79532	35	0.81438
0.9	2.14597√365 = 40.99862	40	0.89123	41	0.90315
0.95	2.44775√365 = 46.76414	46	0.94825	47	0.95477
0.99	3.03485√365 = 57.98081	57	0.99012	58	0.99166

Note: some values falling outside the bounds have been colored to show that the approximation is not always exact.

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 n is p(n) − p(n − 1) maximum? The answer is 20—if there's a prize for first match, the best position in line is 20th.

Same birthday as you

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

Note that in the birthday problem, neither of the two people is chosen in advance. By way of contrast, the probability q(n) that someone in a room of n 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 d 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 one person in a roomful of n people has the same birthday as you, n would need to be at least 253. Note that this number is significantly higher than 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 $253={\frac {23\times (23-1)}{2}}$; a similar approximate pattern can be found using a number of possibilities different from 365, or a target probability different from 50%.

Near matches

Another generalization is to ask how many people are needed in order to have a better than 50% chance that two people have a birthday within one day of each other, or within two, three, etc., days of each other. This is a more difficult problem and requires use of the inclusion-exclusion principle. The number of people required so that the probability that some pair will have a birthday separated by fewer than k days will be higher than 50% is:

k	# people required
1	23
2	14
3	11
4	9
5	8
6	8
7	7
8	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.

Collision counting

The probability that the kth 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 n such integers are chosen equals

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

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. 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 $\scriptstyle {\overline {n}}\,=\,1+Q(M)$, 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 $\scriptstyle {\overline {n}}\,=\,1+Q(M)\approx 24.61658$, slightly more than 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 informal demonstration of the problem can be made from the list of Prime Ministers of Australia, in which Paul Keating, the 24th Prime Minister, and Edmund Barton, the first Prime Minister, share same birthday i.e. 18 January.

James K. Polk and Warren G. Harding, the 11th and 29th Presidents of the United States, were both born on November 2.

Sir John A. Macdonald and Jean Chrétien, the 1st and 20th Prime Ministers of Canada, were both born on January 11.

Of the 73 male actors to win the Academy Award for Best Actor, there are six pairs of actors who share the same birthday.

Of the 67 actresses to win the Academy Award for Best Actress, there are three pairs of actresses who share the same birthday.

Of the 61 directors to win the Academy Award for Best Director, there are five pairs of directors who share the same birthday.

Of the 52 people to serve as Prime Minister of the United Kingdom, there are two pairs of men who share the same birthday.

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 metric ton). 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?

Some people's intuition is that the answer is above 100,000. 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 approximately 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 1,000,000 N and width $\scriptstyle 1,000,000{\sqrt {N}}$, so that when 2^N−1 is approximately equal to $\scriptstyle 1,000,000{\sqrt {N}}$ the transition occurs. 2²³⁻¹ is about 4 million, while the width of the distribution is only 5 million.

References

• John G. Kemeny, J. Laurie Snell, and Gerald Thompson Introduction to Finite Mathematics . The first edition, 1957
• E. H. McKinney (1966) Generalized Birthday Problem, American Mathematical Monthly 73, 385–387.
• M. Klamkin and D. Newman (1967) Extensions of the Birthday Surprise, Journal of Combinatorial Theory 3, 279–282.
• M. Abramson and W. O. J. Moser (1970) More Birthday Surprises, American Mathematical Monthly 77, 856–858
• D. Bloom (1973) A Birthday Problem, American Mathematical Monthly 80, 1141–1142.
• Shirky, Clay Here Comes Everybody: The Power of Organizing Without Organizations, (2008.) New York. 25–27.

External links

• Coincidences: the truth is out there Experimental test of the Birthday Paradox and other coincidences
• http://www.efgh.com/math/birthday.htm
• http://planetmath.org/encyclopedia/BirthdayProblem.html
• Weisstein, Eric W. "Birthday Problem". MathWorld.
• Maple vs. birthday paradox
• A humorous article explaining the paradox
• SOCR EduMaterials activities birthday experiment
• Understanding the Birthday Problem (Better Explained)