Birthday Problem - Wikipedia, The Free Encyclopedia

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

Birthday problemFrom Wikipedia, the free encyclopedia

In probability theory, the birthday problem or birthday paradox[1] concerns the probability that, in a set of n randomly chosen people, some pair of themwill have the same birthday. By the pigeonhole principle, the probability reaches 100% when the number of people reaches 367 (since there are 366possible birthdays, including February 29th). However, 99% probability is reached with just 57 people, and 50% probability with 23 people. Theseconclusions 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 thecomplexity of cracking a hash function.

Contents

1 Understanding the problem2 Calculating the probability3 Approximations

3.1 A simple exponentiation3.2 Poisson approximation3.3 Approximation of number of people3.4 Probability table

4 An upper bound5 Generalizations

5.1 The generalized birthday problem5.2 Cast as a collision problem

5.2.1 Generalization to multiple types6 Other birthday problems

6.1 Reverse problem6.1.1 Sample calculations

6.2 First match6.3 Same birthday as you6.4 Near matches6.5 Collision counting6.6 Average number of people

7 Partition problem8 Notes9 References10 External links

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 "Samebirthday 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 matchingbirthday, 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 total chancesare: 22+21+20+....+1 = 253, so comparing every person to all of the others allows 253 distinct chances (combinations): in a group of 23 people there are

pairs.

Presuming all birthdays are equally probable,[2][3][4] 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, thebirthday 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 variationsin the distribution, such as leap years, twins, seasonal or weekday variations, and assume that the 365 possible birthdays are equally likely. Real-lifebirthday distributions are not uniform since not all dates are equally likely.[5]

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 beingany two people having the same birthday. Then, because A and A' are the only two possibilities and are also mutually exclusive, P(A') = 1 − P(A).

In deference to widely published solutions concluding that 23 is the number of people necessary to have a P(A) that is greater than 50%, the followingcalculation of P(A) will use 23 people as an example.

Birthday problem - Wikipedia, the free encyclopedia http://en.wikipedia.org/wiki/Birthday_problem

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

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 eventsoccurring. Therefore, if P(A') can be described as 23 independent events, P(A') could be calculated as P(1) × P(2) × P(3) × ... × P(23).

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 sharinghis/her birthday with any of the previously analyzed people. For Event 1, there are no previously analyzed people. Therefore, the probability, P(1), thatperson 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 1can 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 ofthe 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. Thismakes the probability P(3) = 363/365.

This analysis continues until Person 23 is reached, whose probability of not sharing his/her birthday with people analyzed before, P(23), 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 firstcalculate 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:

where ' ! ' is the factorial operator, is the binomial coefficient and denotes permutation.

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 thirdcannot have the same birthday as the first two (363/365), and in general the nth 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

This probability surpasses 1/2 for n = 23 (with value about 50.7%). Thefollowing table shows the probability for some other values of n (this tableignores 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))%


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less likely).

A graph showing the accuracy of the approximation

365 (100 − (1.45×10−155))%366 100%

Approximations

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

provides a first-order approximation for ex for x << 1:

To apply this approximation to the first expression derived for p(n) set . Then, Then for each term in the formula for p(n) .

For i = 1,

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

Therefore,

An even coarser approximation is given by

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

It is easy to see that the same approach can be applied to any number of "people" and "days". If rather than 365 days there are n, if there are m persons,and if m<<n, then using the same approach as above we achieve the result that if Pr[(n, m)] is the probability that at least two out of m people share thesame birthday from a set of n available days, then:

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 henceby multiplying their probability together. In short 364/365 can be multiplied by itself C(n, 2) times, which gives us

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

Poisson approximation

Using the Poisson approximation for the binomial,

Again, this is over 50%.


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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:

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 2times.[6]

Probability table

Main article: Birthday attack

length ofhex string #bits

hash spacesize

(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 acertain 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 "23"(row-col intersection).) One could of course also use this chart to determinethe minimum hash size required (given upper bounds on the hashes and probability of error), or the probability of collision (for fixed number ofhashes and probability of error).For comparison, 10−18 to 10−15 is the uncorrectable bit error rate of a typical hard disk [2] (http://arxiv.org/abs/cs/0701166) . 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.[7]

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

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

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

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

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 = 22could also work.


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Generalizations

The generalized birthday problem

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

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

d 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 number of bounds and formulas for n(d) have been published.[8] For any d≥1, the number n(d) satisfies[9]

These bounds are optimal in the sense that the sequence gets arbitrarily close to , while it has as its maximum, taken for d=43. The bounds are sufficiently tight to give the exact value of n(d) in 99% of all cases, for example

n(365)=23. In general, it follows from these bounds that n(d) always equals either or where denotes the ceiling

function. The formula

holds for 73% of all integers d.[10] The formula

holds for almost all d, i.e., for a set of integers d with asymptotic density 1.[10] The formula

holds for all d up to 1018, but it is conjectured that there are infinitely many counter-examples to this formula.[11] The formula

holds too for all d up to 1018, and it is conjectured that this formula holds for all d.[12]

Cast as a collision problem

The birthday problem can be generalized as follows[citation needed]: 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.

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


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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 acollision is not 2N, but rather only 2N/2. This is exploited by birthday attacks on cryptographic hash functions and is the reason why a small number ofcollisions in a hash table are, for all practical purposes, inevitable.

The theory behind the birthday problem was used by Zoe Schnabel[13] under the name of capture-recapture statistics to estimate the size of fish populationin 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.[14] In thesimplest extension there are just two types, say m "men" and n "women", and the problem becomes characterizing the probability of a shared birthdaybetween at least one man and one woman. (Shared birthdays between, say, two women do not count.) The probability of no (i.e. zero) shared birthdayshere is

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 0.5probability 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, orFind the smallest n for which the probability p(n) is greater than the given p.

Taking the above formula for d = 365 we have:

Sample calculations

p n n↓ p(n↓) n↑ p(n↑)0.01 0.14178√365 = 2.70864 2 0.00274 3 0.008200.05 0.32029√365 = 6.11916 6 0.04046 7 0.056240.1 0.45904√365 = 8.77002 8 0.07434 9 0.094620.2 0.66805√365 = 12.76302 12 0.16702 13 0.194410.3 0.84460√365 = 16.13607 16 0.28360 17 0.315010.5 1.17741√365 = 22.49439 22 0.47570 23 0.507300.7 1.55176√365 = 29.64625 29 0.68097 30 0.706320.8 1.79412√365 = 34.27666 34 0.79532 35 0.814380.9 2.14597√365 = 40.99862 40 0.89123 41 0.903150.95 2.44775√365 = 46.76414 46 0.94825 47 0.954770.99 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 theroom? 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

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 nother people has the same birthday as a particular person (for example, you), is given by


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

and for general d by

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, nwould need to be at least 253. Note that this number is significantly higher than 365/2 = 182.5: the reasonis that it is likely that there are some birthday matches among the other people in the room.

It is not a coincidence that ; 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 what is the probability of finding at least one pair in a group of n people with birthdays within k calendar days of eachother's, if there are m equally likely birthdays.

[15]

The number of people required so that the probability that some pair will have a birthday separated by k days or fewer will be higher than 50% is:

k # people required(i.e. n) when m=3650 231 142 113 94 85 86 77 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.[15]

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 totalnumber of times a selection will repeat a previous selection as n such integers are chosen equals[citation needed]

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 isrelevant to several hashing algorithms analyzed by Donald Knuth in his book The Art of Computer Programming. It may be shown[16][17] that if onesamples uniformly, with replacement, from a population of size M, the number of trials required for the first repeated sampling of some individual hasexpected value , where

The function

has been studied by Srinivasa Ramanujan and has asymptotic expansion:


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With M = 365 days in a year, the average number of people required to find a pair with the same birthday is , slightly more than thenumber 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; buton 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, andEdmund 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.[18]

Of the 67 actresses to win the Academy Award for Best Actress, there are three pairs of actresses who share the same birthday.[19]

Of the 61 directors to win the Academy Award for Best Director, there are five pairs of directors who share the same birthday.[20]

Of the 52 people to serve as Prime Minister of the United Kingdom, there are two pairs of men who share the same birthday.[21]

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; eachweight 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 oddnumber 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 somecombinations which work, the majority of randomly selected combinations of three weights do not. If there are very many weights, the answer is clearlyyes. 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 themas 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 itshould 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 2N−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 approximatelyGaussian, with a peak at 1,000,000 N and width , so that when 2N−1 is approximately equal to the transition occurs. 223−1 is about4 million, while the width of the distribution is only 5 million.[22]

Notes^ 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: mostpeople estimate that the chance of two individuals sharing the same birthday in a group of 23 is much lower than 50%.

1.

^ In reality, birthdays are not evenly distributed throughout the year; there are more births per day in some seasons than in others, but for the purposes of thisproblem the distribution is treated as uniform.[citation needed]

2.

^ Murphy, Ron. "An Analysis of the Distribution of Birthdays in a Calendar Year" (http://www.panix.com/~murphy/bday.html) . http://www.panix.com/~murphy/bday.html. Retrieved 2011-12-27.

3.

^ MATHERS, C D; R S HARRIS (1983). "Seasonal Distribution of Births in Australia" (http://ije.oxfordjournals.org/content/12/3/326.abstract) . InternationalJournal of Epidemiology 12 (3): 326–331. doi:10.1093/ije/12.3.326 (http://dx.doi.org/10.1093%2Fije%2F12.3.326) . http://ije.oxfordjournals.org/content/12/3/326.abstract. Retrieved 2011-12-27.

4.

^ In particular, many children are born in the summer, especially the months of August and September (for the northern hemisphere) [1](http://scienceworld.wolfram.com/astronomy/LeapDay.html) , and in the U.S. it has been noted that many children are conceived around the holidays of Christmasand New Year's Day[citation needed]. Also, because hospitals rarely schedule C-sections and induced labor on the weekend, more Americans are born on Mondaysand Tuesdays than on weekends[citation needed]; where many of the people share a birth year (e.g. a class in a school), this creates a tendency toward particulardates. In Sweden 9.3% of the population is born in March and 7.3% in November when a uniform distribution would give 8.3% Swedish statistics board(http://www.scb.se/statistik/BE/BE0101/2006A01a/BE0101_2006A01a_SM_BE12SM0701.pdf) Both of these factors tend to increase the chance of identical birthdates, since a denser subset has more possible pairs (in the extreme case when everyone was born on three days, there would obviously be many identicalbirthdays). The birthday problem for such non-constant birthday probabilities was first understood by Murray Klamkin in 1967. A formal proof that the probabilityof two matching birthdays is least for a uniform distribution of birthdays was given by D. Bloom (1973)

5.

^ Mathis, Frank H. (June 1991). "A Generalized Birthday Problem". SIAM Review (Society for Industrial and Applied Mathematics) 33 (2): 265–270.doi:10.1137/1033051 (http://dx.doi.org/10.1137%2F1033051) . ISSN 0036-1445 (//www.worldcat.org/issn/0036-1445) . JSTOR 2031144 (http://www.jstor.org/stable/2031144) . OCLC 37699182 (//www.worldcat.org/oclc/37699182) .

6.

^ In his autobiography, Halmos criticized the form in which the birthday paradox is often presented, in terms of numerical computation. He believed that it shouldbe 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 splendidillustration of the advantages of pure thought over mechanical manipulation; the inequalities can be obtained in a minute or two, whereas themultiplications 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.

7.


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^ D. Brink, A (probably) exact solution to the Birthday Problem, Ramanujan Journal, 2012, doi: 10.1007/s11139-011-9343-9 (http://www.springerlink.com/content/1194r3627822841q/) .

8.

^ Brink 2012, Theorem 29.^ a b Brink 2012, Theorem 310.^ Brink 2012, Table 3, Conjecture 111.^ Brink 2012, Table 3, Conjecture 112.^ Z. E. Schnabel (1938) The Estimation of the Total Fish Population of a Lake, American Mathematical Monthly 45, 348–352.13.^ M. C. Wendl (2003) Collision Probability Between Sets of Random Variables (http://dx.doi.org/10.1016/S0167-7152(03)00168-8) , Statistics and ProbabilityLetters 64(3), 249–254.

14.

^ a b M. Abramson and W. O. J. Moser (1970) More Birthday Surprises, American Mathematical Monthly 77, 856–85815.^ D. E. Knuth; The Art of Computer Programming. Vol. 3, Sorting and Searching (Addison-Wesley, Reading, Massachusetts, 1973)16.^ P. Flajolet, P. J. Grabner, P. Kirschenhofer, H. Prodinger (1995), On Ramanujan's Q-Function, Journal of Computational and Applied Mathematics 58, 103–11617.^ They are Spencer Tracy and Gregory Peck (April 5), Rod Steiger and Adrien Brody (April 14), Paul Lukas and John Wayne (May 26), Emil Jannings and PhilipSeymour Hoffman (July 23), Robert De Niro and Sean Penn (August 17) and Ben Kingsley and Anthony Hopkins (December 31).

18.

^ They are Jane Wyman and Diane Keaton (January 5), Joanne Woodward and Elizabeth Taylor (February 27) and Barbra Streisand and Shirley MacLaine (April24).

19.

^ They are Norman Taurog and Victor Fleming (February 23), William Wyler and Sydney Pollack (July 1), Robert Redford and Roman Polanski (August 18),William Friedkin and Richard Attenborough (August 29) and George Stevens and Steven Spielberg (December 18).

20.

^ They are John Major and The Earl of Derby (March 29) and Spencer Perceval and The Viscount Goderich (November 1).21.^ C. Borgs, J. Chayes, and B. Pittel (2001) Phase Transition and Finite Size Scaling in the Integer Partition Problem, Random Structures and Algorithms19(3–4), 247–288.

22.

ReferencesJohn G. Kemeny, J. Laurie Snell, and Gerald Thompson Introduction to Finite Mathematics . The first edition, 1957E. 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, 856–858D. 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 (http://www.rsscse-edu.org.uk/tsj/wp-content/uploads/2011/03/matthews.pdf) Experimental test of the BirthdayParadox and other coincidenceshttp://www.efgh.com/math/birthday.htmhttp://planetmath.org/encyclopedia/BirthdayProblem.htmlWeisstein, Eric W., "Birthday Problem (http://mathworld.wolfram.com/BirthdayProblem.html) " from MathWorld.A humorous article explaining the paradox (http://www.damninteresting.com/?p=402)SOCR EduMaterials activities birthday experiment (http://wiki.stat.ucla.edu/socr/index.php/SOCR_EduMaterials_Activities_BirthdayExperiment)Understanding the Birthday Problem (Better Explained) (http://betterexplained.com/articles/understanding-the-birthday-paradox/)Eurobirthdays 2012. A birthday problem. (http://www.matifutbol.com/docs/units/eurobirthdays.html) A practical football example of the birthdayparadox.

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