三个骰子严格递增的概率

先算三次都不相同的概率：

$$1\cdot\frac{5}{6}\cdot\frac{4}{6}=\frac{5}{9}.$$

在互异条件下，三次结果的 6 种相对顺序等可能，严格递增占 1 种，因此条件概率为 $1/6$。

所以总概率为

$$\frac{5}{9}\cdot\frac{1}{6}=\frac{5}{54}.$$

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Original Explanation

First, the probability that the three dice are all different is $$1 \times \frac{5}{6} \times \frac{4}{6} = \frac{5}{9}.$$ Then, given they are distinct, the chance that they are in strictly increasing order is $\frac{1}{6}$. Therefore, $$\frac{5}{9} \times \frac{1}{6} = \frac{5}{54}.$$