递增骰序

为了解决这个问题，我们可以使用条件概率。回想一下 $P(A \cap B) = P(A) \times P(B \mid A)$。让我们定义以下事件：

$A$ - 我们滚动三个不同数字的事件

$B$ - 我们滚动 3 个严格递增数字的事件

我们的目标是找到：

$P(\textrm{Three unique numbers rolled in strictly increasing order}) = P(A \cap B)$ $P(A \cap B) = P(A) \times P( B \mid A )$

第一项只是绘图中的 $\frac{6}{6} \times \frac{5}{6} \times \frac{4}{6}$，无需替换。第二项是 $\frac{1}{3!} = \frac{1}{6}$，因为只有一种方法可以将三个不同的数字排列为严格递增。因此我们的答案是$\frac{6}{6} \times \frac{5}{6} \times \frac{4}{6} \times \frac{1}{6} = \frac{5}{54}$

---

Original Explanation

To solve this problem, we can use conditional probability. Recall that $P(A \cap B) = P(A) \times P(B \mid A)$. Let us define the following events:

$A$ - the event that we roll three different numbers

$B$ - the event that we roll 3 strictly increasing numbers

Our goal is to find:

$P(\textrm{Three unique numbers rolled in strictly increasing order}) = P(A \cap B)$ $P(A \cap B) = P(A) \times P( B \mid A )$

The first term is simply $\frac{6}{6} \times \frac{5}{6} \times \frac{4}{6}$ from drawing without replacement. The second term is $\frac{1}{3!} = \frac{1}{6}$, since there is only one way of permuting three distinct numbers to be strictly increasing. Therefore our answer is $\frac{6}{6} \times \frac{5}{6} \times \frac{4}{6} \times \frac{1}{6} = \frac{5}{54}$