Probability That Someone Shares Your Birthday

Date: 05/14/2002 at 16:40:38
From: Sam
Subject: Probabilty Someone Shares MY Birthday?

What is the minimum number of people we would need to assemble in a 
group such that the probability that at least one person in the group 
has the same birthday as you is greater than 50%?

Date: 05/14/2002 at 18:26:35
From: Doctor Ian
Subject: Re: Probabilty Some Shares MY Birthday?

Hi Sam,

Suppose you're in a room, and you roll a die, and you get a 2.  

Now someone else walks in, and rolls a die of his own.  What is 
the probability that he'll get something different than you?  
He's got 5 other numbers to choose from, so the probability is 
5/6.  

If a third person walks in and does the same thing, the 
probability that they _both_ get something different from you is 
(5/6)(5/6).  

If a fourth person does it, the probability is (5/6)(5/6)(5/6), 
or (5/6)^3.  Does this make sense?

Now, if the probability that NO ONE ELSE rolls a 2 is less than 
50%, then the probability that SOMEONE ELSE rolls a 2 must be 
more than 50%, right? 

So how many people do we have to cram into the room to get the 
probability that NO ONE ELSE rolls a 2 down to 50%?  We're 
looking for n such that

  (5/6)^n < 1/2

Well, 

  (5/6)^2 = 25/36

  (5/6)^3 = 125/216

  (5/6)^4 = 625/1296

which is just a little less than half.  So if you have four other 
people in the room, the probability that NO ONE ELSE rolled your 
number is less than half... which means that the probability that 
SOMEONE ELSE rolled your number is greater than half. 

Did you follow that?  You can find out by applying the same 
reasoning to your problem.  It's really the same thing, except 
instead of 6 sides, each die has 365 sides, one for each day of 
the year that might be someone's birthday.  (I'm ignoring leap 
years here, to keep things simple.) 

It turns out to be a larger than I would have thought!

Can you take it from here?

- Doctor Ian, The Math Forum
  http://mathforum.org/dr.math/