涂色立方体：看见 5 个无色面时有 1 个涂色面的概率

“恰好 1 个涂色面”的小立方体共有 6 个（每个大面中心）。设事件 $B$ 为“可见 5 面无色”，事件 $C$ 为“恰好 1 个涂色面”。

• $P(C)=6/27=2/9$。
• $P(B\mid C)=1/6$（唯一涂色面必须朝下）。
• $P(B)$ 来自两类：中心块（概率 $1/27$，且必满足）与 1 涂色面块（概率 $2/9$，且以 $1/6$ 满足），所以

$$P(B)=\frac{1}{27}+\frac{2}{9}\cdot\frac{1}{6}=\frac{2}{27}.$$

因此

$$P(C\mid B)=\frac{P(B\mid C)P(C)}{P(B)}=\frac{\frac{1}{6}\cdot\frac{2}{9}}{\frac{2}{27}}=\frac{1}{2}.$$

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Original Explanation

We can update the probability the chosen cube has a colored side given that we know $5$ sides are not painted. Using Bayes theorem, Let $B$ be the event that the five sides of the cube shown are blank and let $C$ be the event of the cube having exactly one colored side. Then we have

$$\mathbb{P}[C \mid B] = \dfrac{\mathbb{P}[B \mid C]\cdot\mathbb{P}[C]}{\mathbb{P}[B]}$$

If the cube has a colored side, there's a $\frac{1}{6}$ chance that the painted side is not showing. Thus $\mathbb{P}[B \mid C] = \frac{1}{6}$. Also note that there are $6$ cubes that have only one painted face out of $27$. Thus $\mathbb{P}[C] = \frac{6}{27} = \frac{2}{9}$. Finally, the probability we have a cube that shows five unpainted sides is $\frac{1}{6}\cdot\frac{2}{9}$ from the cubes with one painted side and $1\cdot\frac{1}{27}$ from the center cube with no painted sides. Therefore, considering everything together we get

$$\mathbb{P}[C \mid B] = \dfrac{\mathbb{P}[B \mid C]\cdot\mathbb{P}[C]}{\mathbb{P}[B]} = \dfrac{\frac{1}{6}\cdot\frac{2}{9}}{\frac{1}{6}\cdot\frac{2}{9}+1\cdot\frac{1}{27}}=\dfrac{1}{2}$$