Question 191368
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You were right the first time.  You have three independent events, each with a probability of *[tex \LARGE {1 \over 6}], hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left({1 \over 6}\right)\left({1 \over 6}\right)\left({1 \over 6}\right) = {1 \over 216}]


If the order didn't matter, then you would have:


A 1 in 2 chance (i.e. 3 out of 6) ways to get a 1, a 2, or a 3, then


A 1 in 3 chance (i.e. 2 out of 6) ways to get one of the two remaining numbers, then


A 1 in 6 chance of getting the last remaining number, for:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left({1 \over 2}\right)\left({1 \over 3}\right)\left({1 \over 6}\right) = {1 \over 36}]


Now this all makes sense if you consider that the probability of getting 1, 2, and 3 in any other specific order is also *[tex \LARGE {1 \over 216}], and there are 6 different orders that you can have for three numbers (3! = 6), and then:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6 \times {1 \over 216} = {6 \over 216} = {1 \over 36}]


The exact same probability as if order didn't matter.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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