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 configurations. 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
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