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Topic: A probability question...
Replies: 7   Last Post: Nov 18, 1999 9:09 AM

 Messages: [ Previous | Next ]
 Pat Ballew Posts: 356 Registered: 12/3/04
RE: A probability question...
Posted: Nov 18, 1999 9:09 AM

A resampleing approache to finding the probability of "exactly two" with the
same birthday, using the resampleing power of Fathom, Data Desk, Minitab or
Excel (who did I leave out that will hate me). Put the numbers 1-365 (forget
leap year, I won't play) and pull a sample of 35 without replacement. Count how
many duplicates of each number there are (this is harder in some programs than
others, but I usually create a seperate variable with a count of how many equal
this x value). Then if the max is 2 you have a winner, if the max is other than
2 something else happened. Create a measure for this (one for success, zero for
failure), do it 1000 times and total the successes.

Pat Ballew,
Misawa, Jp

"Statistics means never having to say you're certain."

Math Words & Other Words
http://www.geocities.com/paris/rue/1861/etyindex.html

The Mathboy's page http://www.geocities.com/paris/rue/1861

-----Original Message-----
From: Tim Erickson <tim@eeps.com> at EDU-INTERNET
Sent: Wednesday, November 17, 1999 2:46 PM
To: MRowe14@aol.com at EDU-INTERNET; apstat-l@cln.etc.bc.ca at
EDU-INTERNET
Subject: RE: A probability question...

>The question is: In a
>class of 35 students, what is the probability of 2 students having the same
>birthday.

This is one of those questions best answered by looking at the
complement: what's the probability that, in a class of 35 students, all
have different birthdays?

We look at them one at a time: The probability that the first one is
different is 1.
The probability that the second is different from the first is 364/365.
That the third is different from the first two is 363/365.

and so forth, so that the 35th's probability of being different from all
others is
(365-35+1)/365 = 331/365.

To get the probability that all of these are true together, we multiply,
to get

P(all different) = (365 * 364 * .... * 331) / (365)^35
= 365!/( (365-35)! 365^35)

which means that the probability that at least two have the same (note
this is different from your "exactly two," but I suspect that this is
what you mean) is 1 - P, or about 0.814.

This is a famously surprising result. By the way, my usual ways of
computing these things resulted (not surprisingly) in
your-number-is-too-big errors (since 365! is larger than the number of
particles in the universe), so I computed it as if it were a spreadsheet,
one fraction at a time. That let me also find the interesting tidbit that
the probability of a match crosses 50% when you have 23 people.

Tim Erickson

Author's plug: since I'm the author of _Data in Depth_, the book that
comes with Fathom (the new stats software from Key Curriculum Press), I
should point out that a related problem appears on p 251 of that book:

Suppose people come into a room one at a time. Whenever a new person
enters the room, he or she announces his or her birthday. If it matches
that of anyone else in the room, the door is locked and no one else is
let in.

On the average, how many people will get into the room? What's the
distribution of this number?

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Date Subject Author
10/28/97 LARRY HOLLOMAN