MATLAB Answers

Does MATLAB have a Birthday Problem?

Asked by Derek O'Connor on 5 Feb 2011

The Birthday Paradox or problem asks for the probability that in a room of n people, 2 or more have the same birthday (not date), assuming all years have N = 365 days. It is called a paradox because most people are surprised by the answer when there are (say) 30 people in the room.

Many applications require long sequences (or large vectors) of random numbers. In MATLAB these are supplied by the simple statement r = rand(n,1). This statement fills the vector r(n,1) with double precision random numbers uniformly distributed on (0,1).

Here is my question:

What is the probability that duplicates occur in r = rand(n,1), as n gets large?

I have an answer to this question but I would like to see what others come up with so that I can check the validity of my answer.


Oleg Komarov
on 5 Feb 2011

I vote this question for best Title :D

Malcolm Lidierth
on 8 Feb 2011

In the UK, when the BBC first moved its football highlights show ('Match of the Day') from its traditional Saturday night slot, the birth rate tripled 9 months later. The birthday paradox, makes a false assumption: birthdays are not random - they vary seasonally and are influenced by all sorts of things.

Derek O'Connor on 9 Feb 2011


Thank you for sharing that with us. However, it seems to have escaped your notice that we are not talking about actual births or birthdays here.



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5 Answers

Answer by Derek O'Connor on 5 Feb 2011
 Accepted answer

Reply from Derek O'Connor:

I must apologize for the omission of the RNG used. It was in my original draft but somehow got left out. I'm using the default Matlab rand(n,1) Here is what the docs (R2008a) say:

"Use the Mersenne Twister algorithm by Nishimura and Matsumoto (the default in MATLAB® Versions 7.4 and later). This method generates double-precision values in the closed interval [2^(-53), 1-2^(-53)], with a period of (2^19937-1)/2."

My answer is here, which is based on the RNG information above:

A summary of my analysis is that the probability of duplicates rises significantly after n = 10^8 and has converged to 1 at n = 10^9.

Here are the results of testing, with 5 runs for each value of n.

n = 1*10^8, Prob = 0.426..., Duplicates = (1, 0, 1, 0, 1). Avg = 0.6

n = 5*10^8, Prob = 0.999..., Duplicates = (18, 18, 18, 16, 20). Avg = 16.0

n = 1*10^9, Prob = 1.000..., Duplicates = (66, 64,43, 48, 62). Avg = 56.6

I am not an expert on the mathematics or implementation of RNGs and tend to accept what the docs tell me until I'm told otherwise.

This is not "just a mathematics" question. I say this at the end of my notes above which starts with the real question behind my original question:

What are the implications of getting duplicates in long vectors? This question cannot be answered in general because the answer depends on how these random vectors are used. If they are being used to test a new sorting algorithm, for example, then it is difficult to see how a few duplicates would matter.

I would not be so sanguine about the use of such vectors in medical research. A lot of DNA research seems to involve large amounts of computation which may use large vectors.\footnote{I don't really know what is done, so this is purely speculative.} This is particularly troubling, given the appalling record that bio-medical researchers have in misusing and abusing probability, statistics, and software.

See the case of Anil Potti:

I would certainly be uneasy if I found out that the Boeing or Airbus I was flying in used long random vectors as part of their avionics software. Hence my often-repeated question to students:

Would you fly in a plane whose software was written by you? "


Derek O'Connor.


I'm replying to comments here because I can't use that tiny comment window.

|Dear Bruno, Walter, et al.

I realize that I have not been clear in my answers to your questions and that my statement that rand picks "integer multiples of 2^(-53)" is just plain wrong. Before I get into the floating point aspects of the problem (ugh!), let me state clearly what probability model the analysis is based on:

I assumed that r = rand(n,1) picks r1, r2,...,rn from a bag of N = 2^(53)-1 'objects', WITH replacement, independently, and with equal probability Pr(ri) = 1/N. In other words its the standard "random sample of size n from a population of size N, with replacement" model. The standard textbook analysis and results follow, giving 1-P(N,n) = Pr(1 or more duplicates).

I think these are reasonable assumptions to make about any good random number generator, in any programming language. Hence the analysis and probability calculations are not specific to Matlab, and carry over to any programming language. Of course, for each situation we need to know N, but that is the only parameter needed.

But what about the 'objects' in the bag? In this case they are double precision floating point numbers. But who cares? The Matlab program that counts the number of duplicates merely tests for the equality of two objects. These objects could be integers, singles, doubles, or even matrices.

But you sorted these objects before you tested for equality, and this implies a '<' on the set of objects. Yes, the '<' was used to reduce the search for duplicates to O(nlogn). If '<' does not apply to the bag of objects then we must use an O(n^2) search algorithm that uses '=' only.

And now to the floating point aspects of the problem, where I'll meet my Waterloo, more than likely.

An IEEE 64-bit double precision floating point number is represented as

fl(x) = +/- s x 2^e where s (significand) is a 52-bit integer and e (exponent) is an 11-bit integer and the sign is 1 bit. The precision is p = 53 bits, and machine epsilon emach = 2^(1-p) = 2^(-52) is the distance between fl(1)=1 and the next higher fpn. = 1+emach = 1+2^(-52). A floating point number is normalized if the first bit of s is 1. In base 2 this can be a hidden or implied bit which is not actually stored.

Distribution Properties:

  1. There are 2^(p-1) = 2^(52) normalized dpfpns in [1, 1-2^(-52)]
  2. These dpfpns are uniformly distributed with spacing emach = 2^(-52).
  3. These dpfpns have the form x = 1 + k*emach, k = 0,1,2,...,2^(52)-1.

These numbers have the following bit patterns, with fixed exponent 2^0:

k = 0          1.000...000 x 2^0
k = 1          1.000...001 x 2^0
k = 2^(52)-3   1.111...101 x 2^0
k = 2^(52)-2   1.111...110 x 2^0
k = 2^(52)-1   1.111...111 x 2^0

If we wanted uniform numbers in [1, 1-2^(-52)] we could randomly pick an integer k between 0 and 2^(52)-1 and return r = 1+k*emach. Alternatively, we may view a random number as one of the 2^(52) possible bit patterns (000...000) ... (111...111).

Matlab's rand however, picks dpfpns from [2^(-53), 1-2^(-53)]. If we assume that all significand bit patterns are generated then using 2 and 3 above we get

x = 2^(-53) + k*2^(-53)*emach = 2^(-53) + k*2^(-105), k = 0,1, 2^(52)-1.

But this does not give the correct result. To get numbers in the range [2^(-53), 1-2^(-53)] we need to change the exponent. As Walter pointed out 53 different exponents are used. And as Bruno pointed out this means non-uniformity.

Right now I'm stuck. I'm missing something obvious but I can't see what it is. Floating point Aghhhh!

Does this invalidate the Birthday problem analysis, calculations and Matlab test results? No, because I'm not assuming anything about the values of the random numbers generated. I test only for equality. However, I'm not sure: should N = 2^(52) or 2^(53)?





Mark Shore
on 6 Feb 2011

Rearranging a formula in the Wikipedia reference noted by Walter gives, for n draws from a possible 2^53 distinct values, the probability p of at least one duplicated value of

p = 1 - exp(-n^2/(2*2^53)).

This is similar to your formula 1.4, if rearranged.

Derek O'Connor on 6 Feb 2011


Yes. It is a further simplification (n-1) ~ n for large n.

Remember, all I have done is to apply the standard Birthday Problem analysis to Matlab's rand(n,1), assuming N = 2^(53)-1.

Derek O'Connor

Walter Roberson
on 6 Feb 2011


>> num2hex(2^-53)
ans =
>> num2hex(1-2^(-53))
ans =

That is 53 different exponents that are used over the range of random numbers you have documented.

The only integral multiple of 2^(-53) that has an exponent of 3ca is 2^(-53) itself, with an all-0 mantissa. 2*2^(-53) has an exponent of 3cb and an all-0 mantissa. 3*2^(-53) has an exponent of 3cb and a mantissa whose first bit is set. 4*2^(-53) has an exponent of 3cc and a mantissa of 0.

If we follow this further than it becomes clear from the pigeon-hole principle that not all of the values starting with 3cf can be represented.


>> num2hex(hex2num('3fe0ffffffffffff') / 2^(-53))
ans =

It follows that 3fe0ffffffffffff is not an exact integral multiple of 2^(-53) and thus will not be randomly generated. (it is the representable number just below 0.53125 exactly)

Any argument made on the basis of equi-probable bit patterns after a constant exponent are suspect. One _can_, however, make arguments about taking integer multiples of 2^(-53) and then normalizing the numbers.

IEEE 754 does not actually store 53 bits of mantissa: it uses 52 bits of mantissa plus the "hidden bit" implied by the exponent.

Answer by Richard Willey
on 7 Feb 2011

Hi Derek

Your question about the birthday problem is best viewed as a special case of a more general topic: How should measure the statistical properties of a random number generator?

Rather than using this formulation of the birthday problem to reinvent the wheel, I recommend looking at established literature on this topic. A lot of very qualified experts have spent a lot of time and effort developing criteria to measure the quality of random number generators. The “Diehard” test suite is one such test (Diehard uses a reformulated version of the birthday paradox as one part of the test suite). Alternatively, there is another test suite that uses a combination of

  1. A monobit test (testing the distribution of ones and zeros in a sequence)
  2. A “Poker” test (a modified chi-square test)
  3. A Longruns test (checking whether there are runs of length 34 or greater within a 20,000 bit sequence)
  4. An autocorrelation test

Most of MATLAB’s code can be inspected by users. If you have questions about the specific implementation of a given algorithm – for example, what version of Twister does MathWorks use – you typically have the option to inspect the function and see how the code was written.




Walter Roberson
on 7 Feb 2011

The @Randstream.list documentation does indicate which MT algorithm is used and gives a link to a site that has implementation.

However, the description of the code indicate only 32 bit precision for the (0,1) algorithm, and we have demonstrated at least 53 bit precision, which is documented as occurring for [0,1) . Possibly Mathworks wrapped the 53 bit version with a rejection of an exact 0 result; such an implementation is quite plausible, but we don't *know* since the documentation that is available to us does not specify.

The release notes introducing RandStream might have documented the precision and range of MT, but I am unable to dig the information out of the current documentation. I think dropping that table of producible values was a mis-service to the users.

Derek O'Connor on 7 Feb 2011

Dear Richard,

You have completely misconstrued my question, which is a simple one:

What is the probability that rand(n,1) has 1 or more duplicates as n gets large.

This is a general question that applies to any programming language that has the equivalent of rand(n,1).

I must state most emphatically that I did not set out to test the statistical properties of your random number generator. Indeed, I assume in the analysis that rand is perfect, in the sense that it returns each of the 53-bit strings with equal probability. That is, it randomly samples, with replacement, from a bag of N = 2^(53)-1

Perhaps the use of the word "Problem" in the title is the problem. The word is used in the sense "Homework Problem". Perhaps you
inferred, mistakenly, that I thought Matlab's rand had a problem. My assumption that rand(n,1) is perfect refutes that inference. In fact I state in the Conclusion to my note


"This is not a deficiency in Matlab's implementation of the Mersenne Twister random number generator, but the inevitable
result of using any 53-bit generator to generate long vectors (and has nothing to do with the period)."

I have noticed that you are the second MathWorks-er that has misconstrued my question. Jiro Doke, in missing the point of my question, and dismissing it as "mathematical", seemed to think that
I was complaining about Matlab's rand:

"This is more of a mathematics question, rather than a MATLAB question. You should extend the question, state what you have
discovered (in regards to "rand"), and ask specific MATLAB programming questions to help you get the final answer."


Derek O'Connor

Richard Willey
on 9 Feb 2011

Hi Derek
Sorry that I misconstrued your original question. Here’s one last quick comment:
Random number generators are designed to sample from specific distributions.
A random number generator that generates normally distributed random numbers is not identical to a random number generator that generates normally distributed random numbers (but excludes duplicate values)
Suppose you were using an RNG to generate random integers from a Poisson distribution with lambda = 10. Would you be happy if the RNG was “gimmicked” to exclude duplicate values? I’d argue that this would be a very flawed RNG because the samples being generated aren’t actually coming from a Poisson distribution.
Let’s move back to the example using “rand”. We’re dramatically changing the granularity of the problem, but the basic principle remains the same…
At the end of the day, I don’t think it’s reasonable evaluate an RNG based on a design criteria which is (arguably) contrary to the way in one would expect an RNG to behave (in this case, deliberately excluding duplicate values)
If you have an application that needs RNGs that are generated from “some distribution that looks very much like a normal distribution but is guaranteed never to produce a duplicate value” then you should probably find an RNG designed to do precisely that.
From my own perspective, I’d worry about implementations that were this persnicetky…



Answer by Walter Roberson
on 5 Feb 2011

Upper bound: 1/2+(1/2)*sqrt(1+8*S*ln(2)) where S is the number of distinct random numbers that can be generated.

Unfortunately Mathworks does not appear to document which Twister version they are using. Traditionally their random number generators were (0,1); mt19937ar over (0,1) is 32 bits precision, but over [0,1) is 53 bits precision.

If we speculate that Mathworks used the (0,1) version with 32 bits precision, then according to Wikipedia the 50% probability is at 7.7 × 10^4 numbers.

  1 Comment

Peter Perkins
on 8 Feb 2011

Walter, the User Guide includes this: "mt19937ar: The Mersenne Twister, as described in [9], with Mersenne prime 2^19937-1. This is the generator documented at It has a period of 2^19937-1. Each U(0,1) value is created using two 32-bit integers from the generator; the possible values are all multiples of 2^-53 strictly within the interval (0,1)."

Answer by Mark Shore
on 5 Feb 2011

Using the formula Walter gives leads to the following numbers for a 50% probability:

32-bit 7.716e4

53-bit 1.112e8

57-bit 4.470e8

64-bit 5.056e9

80-bit 1.294e12

A brute-force check of 1e10 random numbers

for n=1:1000 mindiff(n) = min(diff(sort(rand(1e7,1))))/eps(1); end

found 11 identical values, or a ~50% probability for 4.44e8 numbers, suggesting that the equivalent of a 57-bit precision RNG is in use.


Mark Shore
on 6 Feb 2011

@ Jan, to avoid any misunderstanding, my 57-bit hypothesis was based on faulty analysis and is incorrect.

Bruno Luong
on 6 Feb 2011

@Jan: Walter is right by stating the fact that Matlab rand() command returns only integer-multiple of 2^(-53).

Walter Roberson
on 6 Feb 2011

Jan, the testing I did was unable to find a non-zero difference smaller than 2^(-53) as well. But it could just be quite rare. I don't think I have time to generate 2^53-ish values to check!

Answer by Walter Roberson
on 8 Feb 2011

The Matlab content of this question is:

  • How many distinct random numbers can be generated in Matlab. (We still don't know for sure, but 2^53-1 appears likely.)
  • Are the numbers generated with equal probability. (The algorithm would have to be examined in detail to be sure. If there is a bias, it is minute.)
  • Is the period of the generator greater than the number of different possible values, or is it equal to the number of possible values?

Anything beyond that is a straight-forward application of the Birthday Paradox and is thus a mathematical question rather than a Matlab question. This forum exists to serve Matlab questions, not to serve mathematical questions and especially not to discuss the risks of fly-by-wire or to discuss deliberate fraud by Matlab users.

One only has to look at the period of MT and compare that to the largest number directly representable in Matlab in order to see that rand() will eventually generate a duplicate. Software that expects otherwise is broken.

It would be fair game to ask Mathworks to create a random number generator with no repetitions for those people who need one; it would even be fair game to ask Mathworks to rename rand() to make it clear that duplicates were possible. Those would be Matlab issues. Considering the amount of legacy code that uses rand(), I think Mathworks would want some strong evidence that "rand with replacement" was widely enough misunderstood to be worth renaming; I personally don't think speculation about possible trouble in genome software would qualify.


Bruno Luong
on 9 Feb 2011

Personally I'll believe on the doc and also what Peter wrote earlier:
[ So, mt19937ar in MATLAB produces integer multiples of 2^-53 strictly between 0 and 1,]

There is no indication of the twister on FEX submission is the exact code of Matlab RAND().

Peter Perkins
on 10 Feb 2011

While the code on the FEX is indeed a wrapper around N&M's original mt19937ar code, the wrapper differs slightly from the implementation that the RAND function has used for the MT since that generator was officially supported circa R2007b. You have noticed two things: the range, and the speed. There is also the treatment of a zero seed. I can't recall if there are other differences, but in any case, RAND is RAND, and the FEX submission is the FEX submission.

The doc for RAND and my earlier comments are correct, and in particular RAND (internally) rejects an exact (double precision) zero in the very infrequent event that it happens to generate one.

Hoping this clears it all up. I almost hesitate to even mention it, but as described in the RandStream doc, you can also set the FullPrecision property of an mt19937ar stream to false, and get values that are multiples of 2^-32.

Bruno Luong
on 10 Feb 2011

Thanks again Peter. That settles the question.

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