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A red die, a blue die, and a yellow die (all six sided) are rolled. Given that no two of the dice land on the same number, what is the conditional probability that blue is less than yellow which is less than red?

The Answer is a sixth. I have absolutely no idea how to do this though.

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Imagine recording the outcomes as $(a,b,c)$, where $a$ is the number on the red, $b$ the number on the blue, and $c$ the number on the yellow. Fix a particular set of distinct numbers, such as $\{1,4,5\}$. All ordered triples (permutations) of these three numbers are equally likely. There are $3!$ such permutations. In exactly $1$ of these permutations, we have $\text{blue}\lt \text{yellow}\lt \text{red}$. Thus the required probability is $\frac{1}{3!}$.

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Since the three die rolls are unequal, every set of possible rolls $\{x,y,z\}$, with $x < y < z$, can be obtained in exactly six ways: $$ \begin{array}{|c|c|c|} \hline \text{blue} & \text{yellow} & \text{red} \\ \hline x & y & z \\ y & z & x \\ z & x & y \\ x & z & y \\ y & x & z \\ z & y & x \\ \hline \end{array} $$

Note that out of these six equally likely possibilities, the only one where blue < yellow < red is $x, y, z$ (since we assumed $x < y < z$).

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