Therefore p(n) is equal to 1 minus the above probability. The event that at least one pair has the same birthday is complementary to the event that all n birthdays are different. When n is less or equal to 365, we haveīelow is the graph of probabililty of no match against n: Of course, if n is larger than 365, by the pigeonhole priciple, there must be two birthdays on the same day, so the probability is 0. In computing the probability p(n) that in a room of n people, there exists at least a pair that has the same birthday, we ignore the variation in distribution (in reality, not all the dates are equally likely) and assume the distribution of birthdays are uniform around a year of 365 days.It is easier first to calculate the probability that all n birthdays are different. Below is a graph showing the probability against size of the group. And of course, the probability reaches 100% if there are 367 or more people. For 57 or more people, the probability reaches more than 99%. Do you know whether there are two students in your class having the same birthday? (This question is different from is there any student in your class who has the same birthday as you.) The answer in probability is quite surprising: in a group of at least 23 randomly chosen people, the probability that some pair of them having the same birthday is more than 50%. As an application of the Poisson approximation to Binomial, we consider the Birthday problem, which is quite interesting.
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