BirthdaysProbability puzzles require you to weigh all the possibilities and pick the most likely outcome.
How many people do you need to enter a room before the probability of any 2 or more people sharing a birthday (day and month only, not year) is greater than 50%?
Assume for the sake of the puzzle that birthdays in the population at large are equally spread over a 365 day year.
HintIt is easier to calculate the probability that all birthdays in the group are unique.
AnswerWe get the solution by calculating the solution to the opposite case: How many people are necessary before the chance of them all having unique birthdays is less than 50%.
Let Pn be the probability of n people having unique birthdays.
Obviously, with 1 person in the room, P1 = 1.0
If there are n people in the room, the probability of the (n+1)th person also having a unique birthday is (365-n)/365.
Hence P(n+1) given Pn is (365-n)/365.
So P(n+1) = Pn * (365-n)/365.
If we start with n=1 and count upwards, calculating Pn, we find that P22 = 0.52 and P23 = 0.49.
Hence, with 23 people in the room, the probability that 2 or more people will share a birthday is 51%.
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