# Answer to Question #380 in Statistics and Probability for Emre

Question #380
What is the probability that in a room of n people, at least two have the same birthday?
1
2010-07-07T05:18:45-0400
It is easier to first calculate the probability P&rsquo;(n) that all n birthdays are different. It is clear that P(n) is zero when n&gt;365. When n&le;365 we will think as follows.

Take one random person from the group and remember his birthday. Then take a second random person, the probability that his birthday does not coincide with the birthday of the first man is (1 &ndash; 1/365). Then take the third man, the probability that his birthday does not coincide with the birthdays of the first two men is: (1 &ndash; 2/365). Reasoning by analogy, we reach out to the last man, for whom the probability of discrepancy between his birthday and all previous will be: (1 &ndash; (n&ndash;1)/365). Multiplying all these probabilities, we obtain the probability that all birthdays in the group will be different: P&rsquo;(n) = 1*(1 &ndash; 1/365)*(1 &ndash; 2/365)*...*(1 &ndash; (n&ndash;1)/365)=(365*364*...*(365&ndash;n+1))/365n = 365!/(365n*(365-n)!).

The event of at least two of the n persons having the same birthday is complementary to all n birthdays being different. Therefore, its probability P(n) is: P(n) = 1 &ndash; P&rsquo;(n) = 1 &ndash; 365!/(365n*(365-n)!)

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