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How many people do you have to assemble in a room (people picked at random from...

How many people do you have to assemble in a room (people picked at random from the general population) before there is at least an even chance (P = 0.5) that at least two have the same birthdate (e.g., February 4 or September 9)? Assume that the year has 365 days; that is, don’t worry about leap year. Hint: Check your answer by using the same method to calculate the probability that two people will have the same birthday if there are 366 people in the room.

Math   Statistics-And-Probability   
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Answer #1

The answer is 23 people and the reasoning (ignoging leap years) goes :- Start with one person in a room. Add a second person and the probability that he has a different birthday is 364/365. Add a third and the probability that he has a different birthday to the first two is 363/365. The probability that there are now no duplicate birthdays is (364/365)*(363/365). Add a fourth and the probability that he has a different birthday to the first three is 362/365. The probability that there are now no duplicate birthdays is (364/365)*(363/365)*(362/365). etc.

So, if 'n' people are selected then probability that there are no duplicate birthdays is -

The probability that there are no duplicate birthdays falls below 50% when the 23rd person arrives.

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