If you select 20 people at random. What is the probability that at least one pair of people share the same birthday? (Ignore leap years.)

I was thinking of the total number of possibilities so like (20)^20, for the denominator of the fraction. But I am not sure about this. Also am not too sure about how to do the numerator i.e. the choosing using factorials?

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    $\begingroup$ You better look at the complement problem: no pair has the same birthday. $\endgroup$ – joy May 19 '16 at 19:21
  • $\begingroup$ How many choices for a birthdate does the first person have? Does he have 20 choices? or does he have 365 choices? How many choices for the second person if we don't care about repetition? if we do? How many ways can 20 people have birthdays? $20^{20}$ or something else? Maybe something to do with $365$? As a general suggestion, it is easier to find the probability of the opposite event, in this case all people have different birthdays, and subtract that probability from one. $\endgroup$ – JMoravitz May 19 '16 at 19:21
  • $\begingroup$ The total number of possibilities for their birthdays is $365^{20}$. The total number of possibilities if the birthdays must be unique is $365! / (365-20)!$. $\endgroup$ – mjqxxxx May 19 '16 at 19:22
  • $\begingroup$ $1-\frac{\binom{365}{20}\cdot20!}{365^{20}}$ $\endgroup$ – barak manos May 19 '16 at 19:39
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    $\begingroup$ en.wikipedia.org/wiki/Birthday_problem <=== Birthday Problem in Wikipedia $\endgroup$ – Felix Marin May 19 '16 at 20:30

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