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Probability of the Same Birthday within a Group

Date: Tue, 22 Aug 95 16:54 EST
From: "Kyra K. Bromley"
Subject: Probability Question

What is the probability that, in a room of 30 people, two people will have
the same birthday?

The answer would be nice but I'd really love to know how to solve it, i.e.,
what equation(s) would you use?

Date: 809138109
From: Doctor Ken
Subject: Re: Probability Question

Hello there!

The easiest way to approach this problem is to figure out the probability
that, with a certain number of people "n" in the room, _none_ of them will
have the same birthday.  Once this probability dips down below 1/2, that's 
the critical number of people.  So let's do it.

To do probability, you always go back to the basic principle: take the number
of favorable outcomes and divide it by the total number of outcomes of any
sort.  In this case, "favorable" means the total number of ways we can have
n people with none of them having the same birthday.  There's a formula
called the "pick" formula (which is different from the "choose" formula)
which tells you the number of ways of picking a certain number of items, when
order matters.  We're picking days of the year from the calendar, and we
don't want any repeats, and we're going to let different orders count.  The
formula for 365 pick n is 365!/(365-n)!.  So that's our numerator.

The denominator is much simpler.  It's just the total number of ways that
we can have n people have birthdays.  Since there are 365 choices for the 
first, 365 choices for the second, and so on, there are 365^n different

So our total probability that no two of n people in a room have the same 
birthday is [365!/(365-n)!]/365^n, better known as 365!/[(365-n)!365^n].

It turns out that if you plug in values of n and see when this fraction
becomes 50.  Well, it's somewhere between 22 and 23.  The fraction for
22 is 0.524305, and for 23 is 0.492703.  So it's closest to 50% at 23.

For n=30, the fraction is 0.293684, a 29% chance that no two people will
have the same birthday, i.e. a 71% chance that some two people will.

-Doctor Ken,  The Geometry Forum
Associated Topics:
High School Probability

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