Happy Birthday to Me! ... and him... and her...

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I was talking to a friend of mine recently who makes his living as an actuary. We were talking about birthdays and I asked him why it seems like every time I go out to dinner on my birthday, there's at least one other person there celebrating his or her birthday, too, effectively stealing my thunder.

My actuary friend explained that if you got 23 people together in a room, there's a 50-50 chance of at least one coincidental birthday.

Seeing as restaurants usually seat at least twice that number (well, not the hoity toity establishments some of you might frequent, but, for those of us still throwing their birthday parties at T.G.I.F.s"¦), the odds get even better.

After the jump, you'll find a complete breakdown for those who are curious to see the math involved.

To figure out the exact probability of finding two people with the same birthday in a given group, it turns out to be easier to ask the opposite question: what is the probability that NO two will share a birthday, i.e., that they will all have different birthdays? With just two people, the probability that they have different birthdays is 364/365, or about .997. If a third person joins them, the probability that this new person has a different birthday from those two (i.e., the probability that all three will have different birthdays) is (364/365) x (363/365), about .992. With a fourth person, the probability that all four have different birthdays is (364/365) x (363/365) x (362/365), which comes out at around .983. And so on. The answers to these multiplications get steadily smaller. When a twenty-third person enters the room, the final fraction that you multiply by is 343/365, and the answer you get drops below .5 for the first time, being approximately .493. This is the probability that all 23 people have a different birthday. So, the probability that at least two people share a birthday is 1 - .493 = .507, just greater than 1/2.

Statistics courtesy of Math Guy over at NPR.

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July 10, 2006 - 3:25am
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