A $3 \times 3 \times 3$ cube grid has a mouse placed in one of its corner cubes, and a piece of cheese in the center cube. The mouse wanders in search of the cheese: at each step, it randomly moves to one of the adjacent cubes. On average, in how many steps will the mouse find the cheese?
I. solution. Let's call the corner cubes of the $3 \times 3$ cube $A$, the cubes located in the middle of the edges $B$, and the cubes located at the center of the faces $C$ cubes.

If the mouse is in an $A$ cube, it will definitely move to a $B$ cube in the next step. From a $B$ cube, it moves to an $A$ cube with a probability of $1 / 2$, and to a $C$ cube with a probability of $1 / 2$. From a $C$ cube, it moves to a $B$ cube with a probability of $4 / 5$, and finds the cheese with a probability of $1 / 5$.

For the mouse to reach a $C$ cube, an even number of steps is definitely required, since it can only move from a $B$ cube, and after the first step, the mouse is in a $B$ cube, and from then on (until it finds the cheese), it is in a $B$ cube every 2 steps. The center of the $3 \times 3$ cube can be reached from a $C$ cube, so an odd number of steps, and at least 3 steps, are required. After the first step, the mouse is in a $B$ cube. Let's examine the probability that the mouse will find the cheese after 2 steps. It must first move to $C$, then to the center of the cube, and according to the above, the probability of this is

$$
\frac{1}{2} \cdot \frac{1}{5}=\frac{1}{10}
$$

If the mouse does not find the cheese, it will be in a $B$ cube after 2 steps, so the probability of this is

$$
1-\frac{1}{10}=\frac{9}{10}
$$

Based on this, the expected number of steps is:

$$
\begin{aligned}
& 3 \cdot \frac{1}{10}+5 \cdot \frac{9}{10} \cdot \frac{1}{10}+7 \cdot\left(\frac{9}{10}\right)^{2} \cdot \frac{1}{10}+9 \cdot\left(\frac{9}{10}\right)^{3} \cdot \frac{1}{10}+\ldots= \\
= & \frac{1}{10}\left(3+5 \cdot \frac{9}{10}+7 \cdot\left(\frac{9}{10}\right)^{2}+\ldots\right)= \\
= & \frac{1}{10}\left[3\left(1+\frac{9}{10}+\left(\frac{9}{10}\right)^{2}+\ldots\right)+2\left(\frac{9}{10}+\left(\frac{9}{10}\right)^{2}+\ldots\right)+\right. \\
& \left.+2\left(\left(\frac{9}{10}\right)^{2}+\left(\frac{9}{10}\right)^{3}+\ldots\right)+2\left(\left(\frac{9}{10}\right)^{3}+\left(\frac{9}{10}\right)^{4}+\ldots\right)+\ldots\right]
\end{aligned}
$$

The sum formula for an infinite geometric series with common ratio $q$ is: $\frac{a_{1}}{1-q}$. Thus,

$$
1+\frac{9}{10}+\left(\frac{9}{10}\right)^{2}+\ldots=\frac{1}{1-\frac{9}{10}}=10
$$

Thus, the sought expected value is:

$$
\begin{aligned}
& \frac{1}{10}\left[3 \cdot 10+2 \cdot\left(\frac{9}{10}+\left(\frac{9}{10}\right)^{2}+\left(\frac{9}{10}\right)^{3}+\ldots\right)\left(1+\frac{9}{10}+\left(\frac{9}{10}\right)^{2}+\ldots\right)\right]= \\
= & \frac{1}{10}\left[3 \cdot 10+2 \cdot\left(\frac{9}{10}+\left(\frac{9}{10}\right)^{2}+\ldots\right) \cdot 10\right]= \\
= & 3+2\left(1+\frac{9}{10}+\left(\frac{9}{10}\right)^{2}+\ldots\right) \cdot \frac{9}{10}=3+2 \cdot 10 \cdot \frac{9}{10}=21
\end{aligned}
$$

II. solution. Let $k$ be the average number of steps required to reach the central unit cube from a corner cube; $l$, from an edge center cube; and $m$, from a face center cube - assuming these average values actually exist.

Since the mouse starts from a corner cube, the problem is to determine $k$.

We consider two cubes to be adjacent if they share a face. Starting from a corner cube, after one step, we will definitely reach an edge center cube, so:

$$
k=1+l
$$

From an edge center cube, with a probability of $1 / 2$, we move to another corner cube, and with a probability of $1 / 2$, to a face center cube:

$$
l=1+\frac{1}{2} k+\frac{1}{2} m
$$

If we reach a face center cube, from there, with a probability of $4 / 5$, we can move to an edge center cube:

$$
m=1+\frac{4}{5} l
$$

Substituting the expressions for $k$ from (1) and $m$ from (3) into (2), we get:

$$
l=1+\frac{1}{2}(1+l)+\frac{1}{2}\left(1+\frac{4}{5} l\right)
$$

from which $l=20$ and thus $k=21$.