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I think I solved this problem but I would like to know if I am right or wrong, I am not quite sure.

We assume that the year has 365 days and the birthdays are uniformly distributed. We want to find out what's the probability that at least 2 of n people share the same birthday. So my sample space would be: $$\Omega = \left \{ (w_{1},... ,w_{n}) | n \geq 2, w_{i}\in{1,... ,365} \right \}$$ $$|\Omega| = 365^{n}$$

Let $A$ be the event that at least 2 people share the same birthday. Let $A_{0}$ be the event that no people share the same birthday/n people were born on different days. $\Rightarrow |A_{0}| = \binom{365}{n}$ This implies: $$P(A) = P(A_{0}^{c}) = 1- P(A_{0}) = 1-\frac{\binom{365}{n}}{365^{n}}$$ Is this true? Or where did I go wrong? The thing that I am really not sure about is the cardinality of $\Omega$.

Thanks for your help

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    $\begingroup$ Looks good to me. Please note that probability-theory is about more advanced topics. I removed it. $\endgroup$
    – AlexR
    Commented Apr 22, 2015 at 19:55
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    $\begingroup$ I think that your counting does not take order into account, and $|A_0| = 365!/(365-n)!$ instead. $\endgroup$
    – Rolf Hoyer
    Commented Apr 22, 2015 at 20:33

1 Answer 1

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This is the famous birthday problem, for which you should be able to find many explanations online (some on this site).

Your answer is headed in the right direction, but the denominator of the fraction counts ordered sequences of possible birthdays, whereas the numerator counts unordered sequences. (So the comment that suggest your answer 'looks good' should be reconsidered.)

If you change the numerator to the permutations of 365 days taken n at a time, it should be OK. (But the comment that suggests the numerator should be a factorial should also be reconsidered.)

If $n = 23$, the probability will be a bit larger than 1/2. In R, this is computed as follows:

  1-prod(365:(365-22))/365^23
  ## 0.5072972

Note: A very common beginners' error in using combinatorics to do probability problems is to mix unordered and ordered counts in the same expression. It seems everyone needs to make this mistake once. Perhaps in your case it will be only this once.

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