Birthday problem

In probability theory, the birthday problem, or birthday paradox,^[1] pertains to the probability that in a set of randomly chosen people some pair of them will have the same birthday. In a group of 23 (or more) randomly chosen people, there is more than 50% probability that some pair of them will have the same birthday. For 57 or more people, the probability is more than 99%, tending toward 100% as the pool of people grows.^[2] The mathematics behind this problem leads to a well-known cryptographic attack called the birthday attack.

A graph showing the 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 23 people have a matching birthday with 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 a list of 23 persons, if you compare the birthday of the first person on the list to the others, you have 22 chances of success, but if you compare each to the others, you have 253 chances. This is because in a group of 23 people there are 23*22/2=253 pairs, which is more than half of the number of days in the year. So the chance that one of these pairs has a matching birthday is not small. For this reason, along with many others we have the Birthday Paradox.

It is easier to figure the probability that the birthdays will be different, such as: with one person they have 365 opportunities to have a different birthday. The second person only has 364 possibilities to have a different birthday than the first person. The third person has 363 days, and so on.

Calculating the probability

To compute the approximate probability that in a room of n people, at least two have the same birthday, we 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.^[3]

It is easier to first calculate the probability p(n) that all n birthdays are different. If n > 365, by the pigeonhole principle this probability is 0. On the other hand, if n ≤ 365, it is given by

Failed to parse (Missing texvc executable; please see math/README to configure.): \bar p(n) = 1 \cdot \left(1-\frac{1}{365}\right) \cdot \left(1-\frac{2}{365}\right) \cdots \left(1-\frac{n-1}{365}\right) = { 365 \cdot 364 \cdots (365-n+1) \over 365^n } = { 365! \over 365^n (365-n)!}

because the 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), etc.

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

Failed to parse (Missing texvc executable; please see math/README to configure.): p(n) = 1 - \bar p(n) .

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):

* excluding leap years.

Approximations

The Taylor series expansion of the exponential function

Failed to parse (Missing texvc executable; please see math/README to configure.): e^x = 1 + x + \frac{x^2}{2!}+\cdots

A graph showing the accuracy of the approximation Failed to parse (Missing texvc executable; please see math/README to configure.): 1-e^{-n^2/{2 \cdot 365}}

provides a first-order approximation for Failed to parse (Missing texvc executable; please see math/README to configure.): e^x

Failed to parse (Missing texvc executable; please see math/README to configure.): e^x \approx 1 + x

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

Failed to parse (Missing texvc executable; please see math/README to configure.): \bar p(n) \approx 1 \cdot e^{-1/365} \cdot e^{-2/365} \cdots e^{-(n-1)/365}

Failed to parse (Missing texvc executable; please see math/README to configure.): = 1 \cdot e^{-(1+2+ \cdots +(n-1))/365}

Failed to parse (Missing texvc executable; please see math/README to configure.): = e^{-(n(n-1))/2 \cdot 365}

Therefore,

Failed to parse (Missing texvc executable; please see math/README to configure.): p(n) = 1-\bar p(n) \approx 1 - e^{-(n(n-1))/2 \cdot 365}

An even coarser approximation is given by

Failed to parse (Missing texvc executable; please see math/README to configure.): p(n)\approx 1-e^{-n^2/{2 \cdot 365}},\,

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

A simple exponentiation

Very basically, the probability of any two people not having the same birthday is 364/365. In a room of people of size N, there are C(N, 2) pairs of people, i.e. C(N, 2) events. We can approximate the probability of no two people sharing the same birthday by assuming that these events are independent and hence by multiplying their probability together. In short we multiply 364/365 by itself C(N, 2) times, which gives us

Failed to parse (Missing texvc executable; please see math/README to configure.): \left(\frac{364}{365}\right)^{C(N,2)}

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

Failed to parse (Missing texvc executable; please see math/README to configure.): p(n)\approx 1 - \left(\frac{364}{365}\right)^{C(N,2)}.

Poisson approximation

Using the Poisson approximation for the binomial,

Failed to parse (Missing texvc executable; please see math/README to configure.): \mathrm{Poi}\left(\frac{C(23, 2)}{365}\right) \approx \mathrm{Poi}\left(\frac{253}{365}\right) \approx \mathrm{Poi}(0.6932)

Failed to parse (Missing texvc executable; please see math/README to configure.): \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

We can also approximate this using the following formula for the number of people necessary to have at least a 50% chance of matching:

Failed to parse (Missing texvc executable; please see math/README to configure.): N = \frac{1}{2} + \sqrt{\frac{1}{4} + 2 \times 365 \times \ln(2)} \approx 22.9999

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.

An upper bound and a different perspective

The argument below is adapted from an argument of Paul Halmos.^[4]

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

Failed to parse (Missing texvc executable; please see math/README to configure.): 1-p(n) = \bar p(n) = \prod_{k=1}^{n-1}\left(1-{k \over 365}\right) .

This can be seen by first counting the number of ways 365 birthdays can be distributed among n people in such a way that no two birthdays are the same, then dividing by the total number of ways 365 birthdays can be distributed among n people:

Failed to parse (Missing texvc executable; please see math/README to configure.): \bar p(n) = \dfrac{365\cdot364\cdots(365-n+1)}{365^n}.

We are interested in the smallest n such that p(n) > 1/2; or equivalently, the smallest n such that p(n) < 1/2.

Replacing 1 − k/365, as above, with e^−k/365, and using the inequality 1 − x < e^−x, we have

Failed to parse (Missing texvc executable; please see math/README to configure.): \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\cdot 365)} .

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

Failed to parse (Missing texvc executable; please see math/README to configure.): e^{-(n(n-1))/(2\cdot 365)} < \frac{1}{2}

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

Failed to parse (Missing texvc executable; please see math/README to configure.): n^2-n > 2\cdot365\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.

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 generalised 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?

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

Failed to parse (Missing texvc executable; please see math/README to configure.): p(n;d) = \begin{cases} 1-\prod_{k=1}^{n-1}\left(1-{k \over d}\right) & n\le d \\ 1 & n > d \end{cases}

Failed to parse (Missing texvc executable; please see math/README to configure.): p(n;d) \approx 1 - e^{-(n(n-1))/2d}

Failed to parse (Missing texvc executable; please see math/README to configure.): q(n;d) = 1 - \left( \frac{d-1}{d} \right)^n

Failed to parse (Missing texvc executable; please see math/README to configure.): n(p;d)\approx \sqrt{2d\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^[5] 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". Michael Wendl^[6] generalized the birthday problem 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

Failed to parse (Missing texvc executable; please see math/README to configure.): 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 we set Failed to parse (Missing texvc executable; please see math/README to configure.): d = 365

and where Failed to parse (Missing texvc executable; please see math/README to configure.): S_2
are Stirling numbers of the second kind. Consequently, the desired probability is Failed to parse (Missing texvc executable; please see math/README to configure.): 1 - p_0

.

This variation of the birthday problem is interesting because there is not a unique solution for the total number of people Failed to parse (Missing texvc executable; please see math/README to configure.): 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.

An approximation to this can be derived by inverting the 'coarser' approximation above:

Failed to parse (Missing texvc executable; please see math/README to configure.): n(p)\approx \sqrt{2\cdot 365\ln\left({1 \over 1-p}\right)}.

Sample calculations

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

Same birthday as you

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

Failed to parse (Missing texvc executable; please see math/README to configure.): q(n) = 1 - \left( \frac{365-1}{365} \right)^n

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.

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 Failed to parse (Missing texvc executable; please see math/README to configure.): k

days will be higher than 50% is:

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.^[7]

Collision counting

The probability that the kth integer randomly chosen from [1, d] will repeat at least one previous choice equals Failed to parse (Missing texvc executable; please see math/README to configure.): 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

Failed to parse (Missing texvc executable; please see math/README to configure.): \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 monumental book The Art of Computer Programming. It may be shown ^[8], ^[9] that if one samples uniformly, with replacement, from a population of size Failed to parse (Missing texvc executable; please see math/README to configure.): M , the number of trials required for the first repeated sampling of some individual has expectation value Failed to parse (Missing texvc executable; please see math/README to configure.): \overline{n}=1+Q(M) , where Failed to parse (Missing texvc executable; please see math/README to configure.): Q(M)=\sum_{k=1}^{M} \frac{M!}{(M-k)! M^k} . The function

Failed to parse (Missing texvc executable; please see math/README to configure.): 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

Failed to parse (Missing texvc executable; please see math/README to configure.): Q(M)\sim\sqrt{\frac{\pi M}{2}}-\frac{1}{3}+\frac{1}{12}\sqrt{\frac{\pi}{2n}}-\frac{4}{135n}+\cdots .

With Failed to parse (Missing texvc executable; please see math/README to configure.): M=365

days in a year, the average number of people required to find a pair with the same birthday is Failed to parse (Missing texvc executable; please see math/README to configure.): \overline{n}=1+Q(M)\approx24.61658

, 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 Failed to parse (Missing texvc executable; please see math/README to configure.): M+1=366

people is needed; but on average, only 25 people are required.

Partition problem

A related problem is the partition problem, a variant of the knapsack problem from computer science. Some weights are put on a balance; each weight is an integer number of grams randomly chosen between one gram and one million grams (one metric ton). The question is whether you can transfer the weights between the left and right arms to balance the scale. If there are only two or three weights, the answer is very clearly no. If there are very many weights, the answer is clearly yes. The question is, how many are just sufficient?

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 Failed to parse (Missing texvc executable; please see math/README to configure.): \scriptstyle 1,000,000\sqrt{N} , so that when 2^N−1 is approximately equal to Failed to parse (Missing texvc executable; please see math/README to configure.): \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[10].

Notes

1. ^ This is not a paradox in the sense of leading to a logical contradiction, but is called a paradox because the mathematical truth contradicts naïve intuition: most people estimate that the chance is much lower than 50%.
2. ^ Note that birthdays are not evenly distributed throughout the year; not only does February 29 occur less than a quarter as often as any other day, but birth rates vary for the other 365 days.
3. ^ In particular, many children are born in the summer, especially the months of August and September (for the northern hemisphere) [1], and in the U.S. it has been noted that many children are conceived around the holidays of Christmas and New Year's Day; and, in environments like classrooms where many people share a birth year, it becomes relevant that due to the way hospitals work, where C-sections and induced labor are not generally scheduled on the weekend, more children are born on Mondays and Tuesdays than on weekends. Both of 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 birthday problem for such non-constant birthday probabilities was tackled 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 D. Bloom (1973)
4. ^ 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:

   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 calculators do not yield is understanding, or mathematical facility, or a solid basis for more advanced, generalized theories.

5. ^ Z. E. Schnabel (1938) The Estimation of the Total Fish Population of a Lake, American Mathematical Monthly 45, 348-352.
6. ^ M. C. Wendl (2003) Collision Probability Between Sets of Random Variables, Statistics and Probability Letters 64(3), 249-254.
7. ^ M. Abramson and W. O. J. Moser (1970) More Birthday Surprises, American Mathematical Monthly 77, 856-858
8. ^ D. E. Knuth; The Art of Computer Programming. Vol. 3, Sorting and Searching (Addison-Wesley, Reading, Massachusetts, 1973)
9. ^ P. Flajolet, P. J. Grabner, P. Kirschenhofer, H. Prodinger (1995), On Ramanujan's Q-Function, Journal of Computational and Applied Mathematics 58, 103-116
10. ^ C. Borgs, J. Chayes, and B. Pittel (2001) Phase Transition and Finite Size Scaling in the Integer Partition Problem, Random Structures and Algorithms 19(3-4), 247-288.

References

• 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.

External links

• http://www.efgh.com/math/birthday.htm
• http://planetmath.org/encyclopedia/BirthdayProblem.html
• Eric W. Weisstein, Birthday Problem at MathWorld.
• Maple vs. birthday paradox
• Probability by Surprise Birthday Applet An animation for simulating the birthday paradox.
• A humorous article explaining the paradox
• The Birthday Problem Spreadsheet
• SOCR EduMaterials Activities BirthdayExperimentcs:Narozeninový problém

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