6. Let the edge length of the regular tetrahedron $ABCD$ be 1 cm. An ant starts from point $A$ and moves according to the following rule: at each vertex, it randomly chooses one of the three edges connected to that vertex and crawls along this edge to another vertex. What is the probability that it crawls 6 cm and returns exactly to vertex $A$? $\qquad$
6. $\frac{61}{243}$.

Let the probability that the ant returns to vertex $A$ after crawling $n(n \in \mathbf{N})$ centimeters be $P_{n}$. Then
$$
P_{1}=0, P_{n}=\frac{1}{3}\left(1-P_{n-1}\right)(n \geqslant 2) .
$$

When $n \geqslant 2$,
$$
P_{n}-\frac{1}{4}=-\frac{1}{3}\left(P_{n-1}-\frac{1}{4}\right) \text {, }
$$

and $\quad P_{1}-\frac{1}{4}=-\frac{1}{4} \neq 0$.
Thus, $\left\{P_{n}-\frac{1}{4}\right\}$ is a geometric sequence with the first term $-\frac{1}{4}$ and common ratio $-\frac{1}{3}$.
Hence, $P_{n}-\frac{1}{4}=-\frac{1}{4}\left(-\frac{1}{3}\right)^{n-1}$
$$
\Rightarrow P_{n}=\frac{1}{4}-\frac{1}{4}\left(-\frac{1}{3}\right)^{n-1} \quad(n \in \mathbf{N}) \text {. }
$$

Therefore, $P_{6}=\frac{1}{4}+\frac{1}{4} \times \frac{1}{243}=\frac{61}{243}$.