Probability That Someone Shares Your BirthdayDate: 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/ |
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