A bug crawls along a triangular iron ring. At each vertex, it has an equal chance of crawling to one of the other two vertices. Find the probability that it returns to the starting point after 10 crawls.
Analyze the number of ways that satisfy the conditions using a recursive method.
Let $P_{n}$ be the number of ways to return to the starting point after $n$ moves, then $P_{n}=2^{n-1}-P_{n-1}$.
$$
\begin{aligned}
P_{10} & =2^{9}-P_{9}=2^{9}-2^{8}+P_{8}=\cdots \\
& =2^{9}-2^{8}+2^{7}-2^{6}+2^{5}-2^{4}+2^{3}-2^{2}+2^{1}-2+P_{1} \\
& =342 .
\end{aligned}
$$

Therefore,
$$
P=\frac{342}{2^{10}}=\frac{171}{2^{9}}
$$
!Note: By using the recursive relationship, the number of ways that satisfy the conditions is cleverly obtained!