In Happy Birthday
, the question was if there are N people in a room, what is the probability that there are at least two people in the room who share a birthday?
What if instead exactly two was required? If there are N people in a room, what is the probability that there are exactly two people in the room who share a birthday?
(Note: Assume leap year doesn't exist, and the birthdays are randomly distributed throughout the year.)
(In reply to re: thoughts
by Robby Goetschalckx)
Okay, I did catch one error in my formula, but still didn't fully understand your comment. I'm probably just being muddle-headed but I can't see where I didn't take into account the first two people. The proverbial nth person can have the same birthday as any of the other n-1 people.
num of n-1 poss BDs for
n diff BDs nth person Probability
2 365 1 365/365^2
3 365*364 2 365*364*2/365^3
4 365*364*363 3 365*364*363*3/365^4
There are 365!/(366-n)! ways for n-1 people to NOT share a birthday.
There are then n-1 ways for an nth person to share a birthday with one of the others. Order is irrelevant.
This is out of a total possible 365^n possible combinations of birthdays in a group of n people.
P = (365!/(366-n)!)*(n-1)/365^n