For example, 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.

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For example, 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.
Analyzing, we use the method of recursion to find the number of ways that satisfy the conditions.
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,
$$
P=\frac{342}{2^{10}}=\frac{171}{2^{9}} .
$$