Question 573882
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There are exactly two possibilities:  Either at least 2 have the same sandwich or everyone has a different sandwich.  Hence the probability that at least 2 have the same sandwich is equal to 1 minus the probability that they all have different sandwiches.  So what, you say?  Calculating the probability that they all have different sandwiches is a much simpler calculation, that's so what.


The probability that the first guy will order a different sandwich than everyone else is certainty.  There are eight different sandwiches and he can choose any of the eight without duplicating anyone.


The second guy only has 7 sandwiches to choose because one of the possible 8 was already used by the first guy, hence his probability of not duplicating is *[tex \LARGE \frac{7}{8}]


Repeating the pattern through all six guys, the probability of no duplication is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{8}{8}\ \times\ \frac{7}{8}\ \times\ \frac{6}{8}\ \times\ \frac{5}{8}\ \times\ \frac{4}{8}\ \times\ \frac{3}{8}\ =\ \frac{8!}{(8\,-\,6)!8^6}]


So the probability of at least one duplication is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 1\ -\ \frac{8!}{(8\,-\,6)!8^6}]


You can do the arithmetic.


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