Suppose I am generating random 4-digit numbers. Obviously there are 10,000 possible numbers, but the chances are I will get a duplicate long before I generate that many.

Can anyone explain how I would work out how many numbers I would be able to generate and still be (say) 90% sure that there weren't any duplicates?

To be clear, I am treating these numbers as strings, so all will have four digits, even if there are leading zeroes. Also, I'm using a random number generator, so am assuming that any 4-digit number is as likely to be generated as any other.

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    $\begingroup$ en.wikipedia.org/wiki/Birthday_problem $\endgroup$
    – Tim
    Jun 6, 2023 at 19:13
  • $\begingroup$ @Tim Thanks for the link. If I understand correctly, using the last method shown, the probability of at least 2 people out of n sharing a birthday is (365Pn)/(365^n), then applying it to my question, the probability of any two numbers out of 10,000 generated being the same would be (10000P4)/(10000^4), which is around 0.04. Is that correct? Thanks again. $\endgroup$ Jun 6, 2023 at 19:27
  • $\begingroup$ $(365Pn)/(365^n)$ is the probability that no two people have the same birthday $\endgroup$
    – Taladris
    Jun 8, 2023 at 9:59

2 Answers 2


The approximation given in the Wikipedia article (also mentioned in the paper by Diaconis & Mosteller (1989)) works well here. Suppose $N$ 4-digit numbers are drawn with replacement from a pool of $c$ possible numbers, $N\leq c$. According to Diaconis & Mosteller (1989), if $c$ is large and $N$ is small compared to $c^{2/3}$, the following approximation is useful. An approximation for the probability of no duplicates is $$ \exp(-N^2/2c) $$ You want this probability to be $\geq90\%$. In your case, $c = 10^4$. Solving the equation (there are two solutions but only one is positive) for $N$ so that $\exp(-N^2/2c)\geq0.9$ gives $N\leq \sqrt{-\log(0.9)\times2\times10^4}= 45.9044$. So you can draw around $46$ at most and more than that will result in a greater than $10\%$ probability of duplicates. In general, replace $0.9$ with the desired probability of no duplicates. This formula is also given in the Wikipedia article.

Thanks to @Henry, a better approximation to the probability of no duplicates is $$ \exp(-N(N-1)/2c) $$ resulting in $N\leq\sqrt{1/4 - 2c\log{(p)}} + 1/2$. Using your situation, we get $N\approx 46.4071$ with a probability of no duplicates of $0.9017$ which is extremely close to the true probability of $0.9015$.

R provides the functions pbirthday and qbirthday that calculate the probabilities of no duplicates exactly.

Here is a plot comparing the two approximations with the exact probabilities in the relevant range:


The approximation by Henry is very close to the exact value, i.e. the black and green dots are basically identical.

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    $\begingroup$ As a quick check, R comes with p and q functions for this. qbirthday(.1,10000) gives 47 while pbirthday(47,10000) gives 0.1026129 and pbirthday(46,10000) gives 0.09846583 Yes, it looks like you're right, the approximation seems to work quite well. $\endgroup$
    – Glen_b
    Jun 7, 2023 at 2:54
  • $\begingroup$ @Glen_b Thanks, I already forgot about the birthday functions! $\endgroup$ Jun 7, 2023 at 5:37
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    $\begingroup$ You can get slightly closer than the approximations $p \approx \exp\left(-\frac{N^2}{2c}\right)$ and $N \approx \sqrt{-2c\log(p)}$ with $p \approx \exp\left(-\frac{N(N-1)}{2c}\right)$ and $N \approx \sqrt{\frac14-2c\log(p)}+\frac12$. So with $c=10^4$ and $N=46$ you get about $0.9017$ instead of $0.8996$ and closer to the true $0.9015$; with $p=0.9$ you get less than or equal to about $46.407$ instead of $45.904$ (using the gamma function would suggest about $46.372$ is the best you can do) $\endgroup$
    – Henry
    Jun 7, 2023 at 21:53
  • $\begingroup$ @Henry Thanks for the improvement, I added it to the answer. Do you have a source on this improved approximation? $\endgroup$ Jun 8, 2023 at 10:00
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    $\begingroup$ @COOLSerdash The Wikipedia article you link to derives $\bar p(n) \approx e^{-\frac{n(n-1)}{2 \times 365}}$ before saying $\bar p(n) \approx e^{-\frac{n^2}{2 \times 365}}$ is an "even coarser approximation". My approximate expression for $N$ just solves the quadratic. $\endgroup$
    – Henry
    Jun 8, 2023 at 10:19

The other answer uses an approximation. Here is an answer with an exact formula.

The probability that there are no duplicates when drawing $k$ numbers among $N=10000$ is : $$P_{N,k} = \frac{\text{number of draws without duplicates}}{\text{total number of draws}} = \frac{\frac{N!}{(N-k)!}}{N^k} = \frac{N!}{{N^k}(N-k)!}$$

This is a strictly decreasing function of $k$. You want to know the largest value of $k$ such that this probability is still higher than $90%$.

We can find it with a quick binary search computer program: $$P_{10000, 46} = 90.15\%$$ $$P_{10000, 47} = 89.74\%$$

So the final answer is: you can draw 46 numbers.

This is matched very closely by the value of 45.9 found by the approximation formula!

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    $\begingroup$ Thanks for your very clear answer. It was hard to choose which one to mark, so please don't be offended that I chose COOLSerdash's $\endgroup$ Jun 8, 2023 at 17:29

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