All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Happy Birthday (3) (Posted on 2006-06-09)
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.)
 thoughts | Comment 1 of 12

So the first n-1 people in the room have different birthdays

365*364*363.../365^(n-1)

and the nth person has the same as one of the prior n-1

(n-1)/365

Therefore the probability of exactly two having the same birthday is

P = (365!/(365-n)!)*(n-1)/365^n

 Posted by Bob Smith on 2006-06-09 07:38:11

 Search: Search body:
Forums (0)