Question #380

What is the probability that in a room of n people, at least two have the same birthday?

Expert's answer

It is easier to first calculate the probability P’(n) that all n birthdays are different. It is clear that P(n) is zero when n>365. When n≤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 – 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 – 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 – (n–1)/365). Multiplying all these probabilities, we obtain the probability that all birthdays in the group will be different: P’(n) = 1*(1 – 1/365)*(1 – 2/365)*...*(1 – (n–1)/365)=(365*364*...*(365–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 – P’(n) = 1 – 365!/(365n*(365-n)!)

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 – 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 – 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 – (n–1)/365). Multiplying all these probabilities, we obtain the probability that all birthdays in the group will be different: P’(n) = 1*(1 – 1/365)*(1 – 2/365)*...*(1 – (n–1)/365)=(365*364*...*(365–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 – P’(n) = 1 – 365!/(365n*(365-n)!)

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