12. An ant is at a vertex of a tetrahedron. Every minute, it randomly moves to one of the adjacent vertices. What is the probability that after one hour, it stops at the original starting point? $\qquad$ .
12. $\frac{3^{59}+1}{4 \cdot 3^{59}}$ Let $P_{n}$ be the probability that the ant is at its initial position after $n$ minutes, then $P_{0}=1$.

Then, the probability that the ant is at one of the other three vertices after $n$ minutes is: $\frac{1}{3}\left(1-P_{n}\right)$.
Notice that, the ant can only reach the initial vertex from one of the other three vertices, and the probability of reaching the initial vertex from one of these three points is $\frac{1}{3}$. Thus, we have $P_{n+1}=\frac{1}{3}\left(1-P_{n}\right)$, which means $P_{n}-\frac{1}{4}=-\frac{1}{3}\left(P_{n-1}-\frac{1}{4}\right)$, hence $P_{n}=\frac{1}{4}+\frac{3}{4} \cdot\left(-\frac{1}{3}\right)^{n}$. Specifically, $P_{60}=\frac{1}{4}+\frac{3}{4} \times \frac{1}{3^{60}}=\frac{3^{59}+1}{4 \times 3^{59}}$