8. A small bug crawls on the surface of a regular octahedron,

each second it moves from one vertex to one of the four adjacent vertices with equal probability. Then the probability that the bug returns to the initial position at the 8th second is $\qquad$ .
8. $\frac{43}{256}$.

As shown in Figure 3, after crawling $n$ steps, let the probability of the insect being at point $A$ be $p_{n}$, the probability of being at one of points $B, C, D, E$ be $q_{n}$, and the probability of being at point $F$ be $r_{n}$.
$$
\begin{array}{l}
\text { Then } p_{n+1}=\frac{1}{4} q_{n}, \\
q_{n+1}=p_{n}+r_{n}+\frac{1}{2} q_{n}, \\
r_{n+1}=\frac{1}{4} q_{n} .
\end{array}
$$

Given $p_{1}=0, q_{1}=1, r_{1}=0$, by recursive calculation we get
$$
p_{8}=\frac{43}{256} \text {. }
$$