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If you are given 3 standard 6-sided dice, and are asked to pick the order of the numbers that will appear; what is the probability that you will win, given that order DOES matter?

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What do you mean by "pick the order of the numbers"? –  joriki Oct 14 '12 at 17:38
    
You choose what numbers are going to be rolled, and the order they are going to show up in. –  jasouth Oct 14 '12 at 17:58
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3 Answers 3

I guess, it's still 1/216. Similar to choosing a number between 000 and 999 with odds 1/1000.

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How many different orderings are there? clearly $3!$:

a>b>c
a>c>b
b>a>c
b>c>a
c>a>b
c>b>a

where a,b,c are the first, second and third throw.

Is some order more likely than others? No, because of symmetry.

So the answer would be $\frac{1}{6}$.

The only thing that you did not clarify is what you want to do with ties... (I ignore them in this case)

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There are $3!=6$ different orders of three different numbers, corresponding to their $3!$ different permutations. –  joriki Oct 14 '12 at 17:41
    
@joriki I corrected it before you commented, but I think the OP might want to clarify what he meant by picking the order... I gave my own interpretation - that he is trying to guess the rank. –  Bitwise Oct 14 '12 at 17:44
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The size of the probability space is $6^{3}$. The event that you guess the right sequence of numbers has size 1. Thus, the probability is $\frac{1}{216}$, in fact only 1 of the possible sequence is the right one (given that order matters).

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