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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(4): thoughts | Comment 10 of 12 |
(In reply to re(3): thoughts by Charlie)

I'm not arguing that I'm right, I'm trying to figure out where I went wrong.  I ran simulations too.

In any "exactly two" scenario, there will be n-1 birthdays that don't match and one that matches one of the other n-1 birthdays.  That's the reason I set up the formula the way I did.

I guess I'm starting to see where I may need to multiply my formula by ((n-1) or n or n/2) in order to account for the multiple potential occupants of that nth position, but that still doesn't fix the formula.

Can my approach be fixed or is Charlie's the only way to go?

Edited on June 12, 2006, 8:01 am

Edited on June 12, 2006, 9:02 am
  Posted by Bob Smith on 2006-06-12 07:59:24

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