Many solutions are available via google search: probability birthday same day
These "at least" kind of questions are worked out by calculating:
1 - P (none)
since the sum of probabilities of an event occuring and its not occuring is 1.
That is, P (event occurs) + P (event doesn't occur) = 1
P (at least two, i.e. two or more, people share their birthdate) = 1 - P (nobody has the same birthdate)
Now, if the numerator stands for the no. of favourable cases and the denominator represents the total no. of possibilities, then
The first person can have their birthday on any day of the year, i.e. 365 cases favourable out of 365.
For the second person to have a DIFFERENT birthdate, all days would be favourable EXCEPT the birthdate of the first person, i.e. 364 days out of 365.
For the third person to have a birthdate different from the first two, there would be (365-2) favourable days out of 365
and so on
Till the 23rd person, who needs to be born on the remaining (365-22) days [b'days of the previous 22 persons] to have a birthdate different from them all.
So the possibility that NOBODY shares the same b'day is:
365/365 * (365-1)/365 * (365-2)/365 * ...... * (365-22)/365
which, using permutation symbology, can be written as:
(365)_P_(23) / (365)^23
[Since the denominator gets repeated 23 times]
AND THE ANSWER IS:
1 - P (nobody shares the same birthday)
= 1 - (365)_P_(23) / (365)^23