涂色 3×3×3 立方体：看见 5 个白面时底面为红的概

满足“可见 5 面都白”的只有两类小立方体：

• 内部中心块（1 个）：6 面都白，必满足。
• 面中心块（6 个）：恰好 1 面为红。要让可见 5 面都白，红面必须朝下，概率为 $1/6$。

因此权重：中心块 $1\times 1=1$，面中心块 $6\times (1/6)=1$。

归一化后所求概率为

$$\frac{1}{1+1}=\frac{1}{2}.$$

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Original Explanation

Let $R$ be the event that the last side not visible to you is red and $W$ be the event that the other sides visible are not red. We are looking for $\mathbb{P}[R \mid W]$, using conditional probability we have that:

$$\mathbb{P}[R \mid W] = \dfrac{\mathbb{P}[R \cap W]}{\mathbb{P}[W]}$$

The definition of $\mathbb{P}[R \cap W]$, occurs when we have $1$ red side that's facing away from us, and $5$ sides that are white which we can see. When considering how many cubes have exactly $1$ red side we have there are $6$ which are the cubes in the middle of each side, all other cubes have no red paint, or have $2$ or more sides painted red. Since the orientation is random, the probability the red side is facing away from us is $\dfrac{1}{6}$. Therefore:

$$\mathbb{P}[R \cap W] = \dfrac{1}{6} \times \dfrac{6}{27} = \dfrac{1}{27}$$

Then, in the denominator, we are looking for the probability that $5$ sides of a cube that are visible are white. We can calculate this by segmenting our cubes into groups based on how many red sides they have. The center cube is white on all sides so there is no need to orient it. Therefore, $\mathbb{P}[W] = \dfrac{1}{27} + \dfrac{1}{27} = \dfrac{2}{27}$. (There is 1 fully white cube, and 6 cubes with 1 red side that each have a probability of $\frac{1}{6}$ probability of showing all white)

Substituting in, we get $\mathbb{P}[R \mid W] = \dfrac{1}{2}$.