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A red die, a blue die, and a yellow die (all six-sided) are rolled. We are interested in the probability that the number appearing on the blue die is less than that appearing on the yellow die which is less than that appearing on the red die $P(B < Y < R)$.

I know one way to calculate this is:

E = event that no two dice land on the same number:

$$P(E) = \frac{6 \cdot 5 \cdot 4}{6 \cdot 6 \cdot 6} = \frac{5}{9}$$

F = event that B < Y < R and no two dice land on the same number

$$P(F) = P(E) \cdot P(B < Y < R|E) = \frac{5}{9} \cdot \frac{1}{6} = \frac{5}{54}$$

Is there another way to find the number of outcomes where $B < Y < R$ using the formula $N(E)/N(S)$, where $N(S) = 6 \cdot 6 \cdot 6 = 216$?

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There are $\binom{6}{3}$ ways to obtain three different numbers when three dice are thrown. Given such as outcome, there is only one way to arrange these numbers such that $B < Y < R$. Hence, the probability that the number appearing on the blue die is less than the number appearing on the yellow die and the number appearing on the yellow die is less than the number appearing on the red die is $$\frac{\binom{6}{3}}{6^3} = \frac{20}{216} = \frac{5}{54}$$

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    $\begingroup$ Ooh, I like that, nice! $\endgroup$ – Bram28 Sep 27 '17 at 20:43
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Your configuration is of size $216$. You can get the same value ($aaa$) on all $3$ dice in $6$ ways. You can get two equal value and the third different ($aab$) in $6 \times 5 \times 3=90$ ways (the dice taking the value $b$ can be chosen in $3$ different ways). You can get all $3$ different ($abc$) in $6 \times 5 \times 4=120$ ways. Quick sanity check $6+90+120=216$.

Now if you have $3$ different values these can be oredered in $6$ different ways, so for the specic order you require that is $20$ ways. So the answer is $\frac{20}{216}=\color{blue}{\frac{5}{54}}$.

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