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While our country celebrates its birthday today, here’s an interesting problem from the field of mathematics.  I first learned of the Birthday Problem in a statistics class I took at NYU (at least I remembered something, Professor Melnick).

In summary, the Birthday Problem calculates the probability that at least two people from a randomly chosen group share a birthday. 

At the extremes this is easy to understand.  Excluding leap year, there are 365 potential birthdays.  If you have a randomly chosen group of 366, it follows that the probability is 100% that at least two people share a birthday.  The outcome is absolutely assured.  Conversely, if two people meet in the street, and compare their birthdays, if is very unlikely that they will be the same.

What makes the Problem particularly interesting is how small a group you need to assemble to make the probability higher than 50%.

Without reseaching the answer (you are on the honor code), how large would you expect the group to have to be to make the probability 50%?  How about 95%?  The answers may surprise you…

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Dan Caruso said, July 5th, 2008 at 12:31 pm

93 and 175 if I am wrong, let me know and i will double check how I did it.

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Michael Wilson said, July 6th, 2008 at 4:34 pm

A little high in your guess. The answer is posted on July 5th post Birthday Problem – solved. Mathematical equation is: P(n) = 1 – p(n) where: p(n) is the probability that all birthdays are unique from a sample set of n. This is calculated as: p(n) = 365! / 365^n(365-n)! Note that excel was unable to calculate the factorial of 365 for me without an add-on, so perhaps paper and pencil was an understatement.

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