# Math Help - probability of same birthdays

1. ## probability of same birthdays

in a lecture we derived a formula for the probablilty of at least two people having the same birthday which was as follows,

(365^n -(365!/(365 - n)!))/365^n

why does this not give the probability as 1 for n = 365

2. Originally Posted by hmmmm
in a lecture we derived a formula for the probablilty of at least two people having the same birthday which was as follows,

(365^n -(365!/(365 - n)!))/365^n

why does this not give the probability as 1 for n = 365
Read this: Birthday problem - Wikipedia, the free encyclopedia

3. Originally Posted by hmmmm
in a lecture we derived a formula for the probablilty of at least two people having the same birthday which was as follows,

(365^n -(365!/(365 - n)!))/365^n

why does this not give the probability as 1 for n = 365
the problem assumes there are 365 days you can be born on (ignores leep years and junk)
if there are 365 people in a room, it is possible for each of them to have a different birthday since there are 365 days to have one on.
366 (or more), on the other hand, evaluates to 1.