27.12*. (Belgium, 78). A point moves along the edges of the cube $A B C D A^{\prime} B^{\prime} C^{\prime} D^{\prime}$. From any vertex, it can move to one of the three edges (emanating from this vertex) at random with equal probability, equal to $1 / 3$. The vertices $B^{\prime}$ and $C^{\prime}$ are special in that, once the point reaches either of them, it stops moving. If the point starts moving from vertex $A$, what is the probability that it: 1) stops at vertex $B^{\prime} ; 2$ ) stops at vertex $C^{\prime}$; 3) never reaches either vertex $B^{\prime}$ or vertex $C^{\prime}$?
27.12. Note that from any vertex of the cube, the point will land in one of the vertices $B^{\prime}, C^{\prime}$ with a probability not less than $1 / 9$ (since for any vertex of the cube, different from $B^{\prime}$ and $C^{\prime}$, there exists a route passing through no more than two edges and connecting it to $B^{\prime}$ or $C^{\prime}$). Therefore, the probability that the point does not land in either $B^{\prime}$ or $C^{\prime}$ in two moves is no more than $1-1 / 9=8 / 9$, and the probability that the point does not land in either $B^{\prime}$ or $C^{\prime}$ in $2 k$ moves is no more than $(8 / 9)^{k}$. Consequently, the probability that the point never lands in either $B^{\prime}$ or $C^{\prime}$ does not exceed $(8 / 9)^{k}$ for any $k \in N$, and thus is 0. Therefore, the point will land in $B^{\prime}$ or $C^{\prime}$ with probability 1. Next, let $p$ denote the probability that the point will land in $B^{\prime}$ from $A$ (and, therefore, with probability $1-p$ the point will land in $C^{\prime}$ from $A$), and let $q$ denote the probability that the point will land in $B^{\prime}$ from $B$ (and, therefore, with probability $1-q$ the point will land in $C^{\prime}$ from $B$). By symmetry, the probability that the point will land in $B^{\prime}$ from $D$ is $(1-p)$, from $A^{\prime}$ in $B^{\prime}$ is $q$, and from $C$ or $D^{\prime}$ in $B^{\prime}$ is $(1-q)$. If the point is at $A$, then with probability $1 / 3$ it will land in $A^{\prime}$, with probability $1 / 3$ it will land in $D$, and with probability $1 / 3$ it will land in $B$. Therefore, the probability $\rho$ that the point will land in $B^{\prime}$ from $A$ is

$$
\frac{1}{3}(1-p)+\frac{1}{3} q+\frac{1}{3} q
$$

i.e., $4 p-2 q=1$. Similarly, the probability $q$ that the point will land in $B^{\prime}$ from $B$ is $\frac{1}{3}+\frac{1}{3}(1-q)+\frac{1}{3} p$, i.e., $4 q-p=2$. Thus, $p=4 / 7$, and the desired probability that the point will stop at vertex $B^{\prime}$ is $4 / 7$, and at vertex $C^{\prime}$ is $3 / 7$.