jmc

Let $A$ denote the leaf where the cricket starts and $B$ denote one of the other $3$ leaves. Note that: If the cricket is at $A,$ then the probability that it hops to $B$ next is $1.$ If the cricket is at $B,$ then the probability that it hops to $A$ next is $\frac{1}{3}.$ * If the cricket is at $B,$ then the probability that it hops to $B$ next is $\frac{2}{3}.$ We apply casework to the possible paths of the cricket: 1. $A\to B\to A\to B\to A$ The probability for this case is $1\cdot \frac{1}{3}\cdot 1\cdot \frac{1}{3}=\frac{1}{9}.$ 2. $A\to B\to B\to B\to A$ The probability for this case is $1\cdot \frac{2}{3}\cdot \frac{2}{3}\cdot \frac{1}{3}=\frac{4}{27}.$ Together, the probability that the cricket returns to $A$ after $4$ hops is $\frac{1}{9}+\frac{4}{27}=\boxed{\frac{7}{27}}.$