A cricket randomly hops between $4$ leaves, on each turn hopping to one of the other $3$ leaves with equal probability. After $4$ hops what is the probability that the cricket has returned to the leaf where it started?

[2022 AMC 8 Problem 25 Picture.jpg](https://artofproblemsolving.com/wiki/index.php/File:2022_AMC_8_Problem_25_Picture.jpg)
$\textbf{(A) }\frac{2}{9}\qquad\textbf{(B) }\frac{19}{80}\qquad\textbf{(C) }\frac{20}{81}\qquad\textbf{(D) }\frac{1}{4}\qquad\textbf{(E) }\frac{7}{27}$
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 $\frac13.$
If the cricket is at $B,$ then the probability that it hops to $B$ next is $\frac23.$
We apply casework to the possible paths of the cricket:


$A \rightarrow B \rightarrow A \rightarrow B \rightarrow A$ 
The probability for this case is $1\cdot\frac13\cdot1\cdot\frac13=\frac19.$
$A \rightarrow B \rightarrow B \rightarrow B \rightarrow A$ 
The probability for this case is $1\cdot\frac23\cdot\frac23\cdot\frac13=\frac{4}{27}.$

Together, the probability that the cricket returns to $A$ after $4$ hops is $\frac19+\frac{4}{27}=\boxed{\textbf{(E) }\frac{7}{27}}.$
~MRENTHUSIASM