Probabilities : A red die, a blue die, and a yellow die (all six-sided) are rolled.

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 $$\frac{\binom{6}{3}}{6^3} = \frac{20}{216} = \frac{5}{54}$$

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 $\frac{20}{216}=\color{blue}{\frac{5}{54}}$.