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Two people are said to have the same birthday if they were born in the same month and on the same day of the month (the year of birth is not relevant). If there is are 40 people in a room, compute the probability that at least two people have the same birthday.

I calculated the probability as follows (40*39)/2 = 780. (364/365)^780 = 11.77%. This seems very low. Can someone please provide me with the correct equation or let me know if what I have is correct? Thanks so much.

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    What is the probability that everyone has different birthdays? Lets label the people $a_1,a_2,a_3,\dots,a_{40}$... The probability $a_2$'s birthday is different than $a_1$ is $\frac{364}{365}$. Given that $a_2$ and $a_1$ have different birthdays, the probability that $a_3$'s birthday is different than both $a_2$ and $a_1$ is $\frac{363}{365}$. Given that $a_1,a_2,a_3$ all have different birthdays the probability of $a_4$ having a different birthday than all of $a_3,a_2,a_1$ is $\frac{362}{365}$... multiplying these we get... – 2017-02-21
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    As an aside, this is the classic birthday problem, which has been discussed at length on this site hundreds of time already. Pick at random from the related bar, search in the top search bar, or just [visit the wikipedia page for the problem](https://en.wikipedia.org/wiki/Birthday_problem) for more information. Your solution is incorrect because these are in fact dependent events, not independent events. If $a_1$ has the same birthday as $a_2$ and $a_2$ has the same birthday as $a_3$ then obviously $a_1$ and $a_3$ have the same birthday. – 2017-02-21
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    Actually your $11.77\%$ is slightly too high – 2017-02-21
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    @Henry $11.77\%$ is slightly too high for probability that noone shares a birthday. It is horrifically low for the probability that at least two people share a birthday. – 2017-02-21
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    Would it not just be $2 \times \frac{1}{365}$? $\frac{1}{365}$ chance of same bday? – 2017-02-22
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    @JMoravitz - yes - that was what I had intended to say - thank you. It is widely known in the classical birthday problem that the probability is less than $50\%$ that nobody shares a birthday if there are $23$ people, and that goes down as the number of people goes up – 2017-02-22
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    @Nick the chance of a specific pair of people sharing a birthday is (*assumed to be*) $\frac{1}{365}$ (*not $2\times \frac{1}{365}$*). That isn't particularly helpful to this problem however. If you were to read the wiki link I gave in my second comment (*or spend more than thirty seconds searching for an answer*), you would see the probability that at least one pair shares a birthday is [this](http://www.wolframalpha.com/input/?i=1-p(365,40)%2F365%5E40). – 2017-02-22

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