I am trying to calculate the probability that at least 2 people having birthday the same day from a group that consists of: a) 20 or b)45 people.

I am also very interested in learning the mathematical notion, how to write it up correctly and not just calculate it.

I saw that other posts are talking about leap years and more advanced applications, so i just want to stick with the basics to better understand it.

So to calculate the probability of at least 2 people having a birthday in the same day is the same as if we were calculating the complement of "nobody has a birthday in the same day."

So $S=20$, then $P$(At least 2 people have a birth on the same day)$=1-P$(nobody has a birthday in the same day) = $1 - \frac{365!}{(365-20)! \cdot365^{20}}$. is this correct?

If you can, please explain to me how to write it correctly mathematically, not just how to calculate it.

Thank you very much for your assistance.

EDIT: if someone can help to show how it can be done using binomial notation it might be very interesting for me and i assume that for others as well, to select some out off the total etc... if you can offer different ways to do this, it will aid me a lot by learning how to approach a problem using various techniques and think outside of the box.

thank you very much again


1 Answer 1


Let's compute the probability of the opposite, i.e. all having different birthdays. Then, in a group of $n$ people, you have $365$ choices for the first one and $365-1=364$ for the second one (since the only birthday not allowed is the one you just picked for the first one), $363$ for the third one, etc.

Without constraints, you would have $365^n$ possibilities, so the probability for $n$ people is $$ \begin{split} p &= \frac{365}{365} \times \frac{365-1}{365} \times \frac{365-2}{365} \times \ldots \times \frac{365-n+1}{365} \\ &= \frac{365 \times \ldots \times (365-n+1)}{365^n} \frac{1 \times \ldots \times (365-n)}{1 \times \ldots \times (365-n)} \\ &= \frac{365!}{(365-n)! 365^n} \end{split} $$

  • $\begingroup$ understood, thank you very much $\endgroup$
    – mathnoobie
    Jan 6, 2020 at 21:58
  • $\begingroup$ @mathnoobie If you don´t have any further questions, please mark the answer as accepted. $\endgroup$ Jan 6, 2020 at 22:06

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