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Date: Tue, 9 Jan 1996 09:36:15 -0500 (EST)
From: "Jonathan M. Bresler" <jmb@FreeBSD.ORG>
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Subject: Re: Prognosis of 2.2-960107-SNAP
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On Mon, 8 Jan 1996, Paul Richards wrote:

> Isn't there some mathematical puzzle that says something like, you
> only need 27 people in a room for one of them to have a birthday the
> same day as yours? Given the numbers we have, we've probably got
> every day covered.

	nah.  to get a duplicate of a given birthday (say, yours) requires
from 2 to 366 (ignoring leap years) people in the room.  assuming that
birthdays are uniformly distributed, odds are (50/50) that you need 365/2
people in the room beside yourself. 

	now, if you will accept ANY person's birthday (as opposed your a
specific person, yours in the example above) then the number drops
considerably, again assuming that birthdays are uniformly distributed
thoughout the calendar year. 

	sometimes it is easier to calculate the odds of an event NOT 
happening and then substract from 1.0 to get the odds of the event 
occurring.  this is one of those cases.  the odds of one other person not 
having the same birthday are 1 - 364/365 (which equals 1/365).  for three 
people the odds are 1 - (364/365)(363/365) (which is messy when expressed 
as a sum instead of a product).  the 50/50 point is 23 people in the room.

caution: math follows (horrors!)

	here is a bc function to calculate the odds of ANY two people in 
a group having the same birthday

define day(cnt) {
        auto prod, i;
        prod = 1;
        for ( i = 1; i < cnt; i++) {
                prod *= (365 -i)/365
        }
        return (1 - prod)
}

	the 50/50 odds point is 23 (day(23) = .507310)

jmb


Jonathan M. Bresler        FreeBSD Postmaster         jmb@FreeBSD.ORG
play go. ride bike. hack FreeBSD.--ah the good life 
i am moving to a new job.                 PLEASE USE: jmb@FreeBSD.ORG