Question 331587
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For the probability that 13 people have different birthdays, you need the probability that the 2nd person has a different birthday than the first, namely *[tex \Large \frac{364}{365}], times the probability that the 3rd person has a different birthday than the first two, namely *[tex \Large \frac{363}{365}], and so on:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P(13,0)\ =\ \frac{365!}{352!\,\cdot\,365^{13}}]


Then for the probability that at least two share a birthday,


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P(13,\geq2)\ =\ 1\ -\ P(13,0)]


In general for *[tex \Large n] people,


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P(n,\geq2)\ =\ 1\ -\ \frac{365!}{(365\,-\,n)!\,\cdot\,365^{n}}]


As a sanity check on your arithmetic work, 23 people is roughly the break-even point, that is where the probability is very close to 50/50.  If you have 50 people it is nearly certain that at least 2 will share a birthday. Your answer for 13 people should be somewhere in the 20% range. 


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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