In a game of croquet, a ball begins at Wicket A. On a given move, a ball struck from Wicket A has a $\dfrac{1}{2}$ chance of staying at Wicket A and a $\dfrac{1}{2}$ chance of going to Wicket B. On a given move, a ball struck from Wicket B has a $\dfrac{1}{2}$ chance of going back to Wicket A, and a $\dfrac{1}{2}$ chance of hitting the goal post. What is the probability the ball hits the post the $n$th time the ball is hit?
$\textbf{What I've Tried:}$
I tried using Markov State chains to solve this problem, but it didn't really help me. The only useful observation I was really able to make is that we are really solving for $\dfrac{1}{4}$ multiplied by the probability that the ball is at Wicket A after $n-2$ hits.
$\textbf{Where I need help:}$
How would I calculate the probability that the ball is at Wicket A after $n-2$ hits? I was able to notice that this can be defined recursively, but solving for the characteristic polynomial was particularly nasty. Particularly, there should be a nice answer in terms of the Fibonacci sequence.
For each nonnegative integer $n$, let $a_n$ be the probability of being at position $\text{A}$ after $n$ steps, assuming $\text{A}$ as the starting position.
For $n\ge 2$, a return to position $\text{A}$ after $n$ steps happens if and only if the final two states are $\text{AA}$ or the final three states are $\text{ABA}$.
Hence for $n\ge 2$ we have the recursion $$ a_n=\frac{1}{2}a_{n-1}+\frac{1}{4}a_{n-2} $$ together with the initial values $a_0=1$ and $a_1={\Large{\frac{1}{2}}}$.
Applying the recursion by hand, we get \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|} \hline n&0&1&2&3&4&5&6&7&8&9&10\\ \hline 2^na_n&1&1&2&3&5&8&13&21&34&55&89\\ \hline \end{array} which suggests $$ a_n=\frac{F_{n+1}}{2^n} $$ a result which is easily proved by induction.
It follows that for $n\ge 2$, the probability of hitting the goal post at step $n$, assuming $\text{A}$ as the starting position, is $$ \frac{1}{4}a_{n-2}=\frac{F_{n-1}}{2^n} $$