probablitity

Let us consider the case EXACTLY TWO PEOPLE
Sanity check says that the probability for exactly two out of 100 students to have the same birthday should be very small.

Total outcomes: there are 100 students; each may have a birthday in one of 365 days. = 365^100

Favorable... Hmm...
any 2 of 100 = 100C2=4950
have 365 days to be born=365*4590
in order to provide a condition that EXACTLY two
the third person has 364 days
the fourth has 363
....................
the hundredth 267

Finally, 4950*(365*364*...*267)/365^100= 5.7E-6

Okay.

This would be more interesting if the number of people were smaller, say 20 to 25 people. But here goes...

In order to find the probability of ANY TWO (which to me means AT LEAST TWO), we best find the probability that NOBODY has the same birthday (let's call this P), then subtract from one.

Say we pick any one person, The probability that that next person will not have the same birthday is 364/365, then next 363/365, and so on until we get to 266/365.

So P = (1)(364/365)(363/365)....(266/365) = (364!/265!)/(365^99)

You would need a computer to calculate this, but if you think about it, this number will end up being very small. (we are multiplying a bunch of numbers less than 1 together 99 times)

So my answer at an interview would be: "It is almost certain that at least two people will have the same birthday in a group of 100"

Numerical answer using computer is P = 3 x 10^-7
and 1 - P = .9999997

Interesting result: If there were only 23 people, the answer would be

1 - P
= 1 - (364!/342!) / (365^22)
= 1 - .4927
= .5073

which means that there is a slightly more than 50-50 chance that there will be at least 2 people with the same birthday in a group of 23 people.

You could make money on this if you could get someone to give you odds.