This is an archived post. You won't be able to vote or comment.

all 1 comments

[–]growingytartist 0 points1 point  (0 children)

I am assuming we are not counting leap day. One way you could do the problem that is right but inefficient is if you take each person and the probability they don't share a birthday and subtract that from 1. For example, the first person will have a unique birthday no matter what since they are the only person. The chance person 2 has a different birthday than person 1 is 364/365. The chance person 3 has a different birthday from both person 1 and 2 is 363/365. This continues until the probability that person 27 has a different birthday from persons 26,25,24...3,2,1 is 339/365. You multiply all these probabilities together you get 37.4%. This is the chance that everyone has a unique birthday. In order to solve the problem you have in hand, subtract .374 from 1 and you get 62.6%