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Happy Birthday (3) (Posted on 2006-06-09) Difficulty: 3 of 5
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.)

No Solution Yet Submitted by Sir Percivale    
Rating: 4.0000 (3 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re(2): thoughts | Comment 4 of 12 |
(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


  Posted by Bob Smith on 2006-06-09 13:43:27
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