Exercises EG.3 and EG.4 in Probability with martingales, by David Williams 
Let $G$ be the free group with two generators $a$, $b$. Start at time 0 with unit element 1, the empty word. At each second multiple the current word on the right by one of the four elements $a$, $a^{-1}$, $b$, $b^{-1}$ choosing each with probability 1/4 independently of previous choices. For instance the choices 
  $a,a,b,a^{-1},a,b^{-1},a^{-1},a,b$ yield the reduced word of length 3, $aab$ at time 9. 
EG.3 Prove that the probability that the reduced word $1$ ever occurs at a positive time is $1/3$. 
EG.4 Now suppose elements $a$, $a^{-1}$, $b$, $b^{-1}$ are chosen with probability $\alpha$, $\alpha$, $\beta$, $\beta$ respectively, where $2\alpha+2\beta=1$. Prove that the conditional probability that the reduced word $1$ ever occurs at a positive time given that $a$ is chosen at time $1$ is the unique root in $(0,1)$  of the polynomial $3x^3 + (3-4\alpha^{-1})x^2+x+1$. 

I know little about Markov Chains, and about limiting distributions but the chain is not of the form that a limiting distribution exists, so I am not able to make use of that facts. Chain I have in mind is of length of reduced word, states are $\{0,1,2,\dots,\}$. Now $p_{k\to k+1}=3/4,\ p_{k\to k-1}=1/4$ for $k\geq1$ and for $k=0,\ p_{0\to1}=1$. 
 A: I have only been able to solve this through guidance of Did.
$\def\P{\mathbb{P}}$
$\def\Ex{\mathbb{E}}$
$\def\Z{\text{\ensuremath{\mathbb{Z}}}}$
$\def\I{\textrm{I}}$
$\def\as{\textrm{a.s.}}$
EG.3 Use the Markov Chain theory developed in Markov chains, by Norris,
to calculate the hitting time to reach sate $0$ from state of length
$1$. Specifically, we model the length of reduced word as Markov Chain with length as it states. 
With $p:=\P\left(\left\{ \text{hitting }0\text{ given length is }1\right\} \right)$
we have 
\begin{align*}
p= & \frac{1}{4}+\frac{3}{4}\P\left(\left\{ \text{hitting }0\text{ given length is }2\right\} \right)\\
= & \frac{1}{4}+\frac{3}{4}\P\left(\left\{ \text{hitting }0\text{ given length is }1,\text{hitting 1 given length is 2}\right\} \right)\\
\overset{\left(1\right)}{=} & \frac{1}{4}+\frac{3}{4}\P\left(\left\{ \text{hitting }0\text{ given length is }1\right\} \right)\P\left(\left\{ \text{hitting }1\text{given length is }2\right\} \right)\\
\overset{\left(2\right)}{=} & \frac{1}{4}+\frac{3}{4}p^{2}.
\end{align*}
where in $\left(1\right):$ Strong Markov Property; Markov chains, by Norris
is used and in $\left(2\right):$ we exploit symmetry of setup that
\begin{align*}
\P\left(\left\{ \text{hitting }0\text{ given length is 1}\right\} \right)=\P\left(\left\{ \text{hitting }1\text{ given length is }2\right\} \right).
\end{align*}
Solving $p=\frac{1}{4}+\frac{3}{4}p^{2}$ we get $p=\frac{1}{3},1$
where the latter is discarded by observing the Markov Chain (higher
probability to visit states down the chain). 
EG.4
Similar to previous exercise let $p_{a}$ be the probability of interest;
i.e. 
\begin{align*}
p_{a}=\P\left(\left\{ \text{hitting $0$ given state is }a\right\} \right).
\end{align*}
Likewise generalize the definitions above and the argument of Strong
Markov Property above to get 
\begin{align*}
p_{aa}= & \P\left(\left\{ \text{hitting $0$ given state is }aa\right\} \right)=p_{a}^{2}\\
p_{ab}= & p_{a}p_{b}\\
p_{ab^{-1}}= & p_{a}p_{b}^{-1}=p_{a}p_{b}
\end{align*}
where the last equality is true by the symmetry of setup. Now note
that hitting time
\begin{align*}
p_{a}= & \alpha+\alpha p_{aa}+2\beta p_{ab}\\
= & \alpha+p_{a}^{2}\alpha+2\beta p_{a}p_{b}.
\end{align*}
Similarly, we find $p_{b}=\beta+\beta p_{b}^{2}+2\alpha p_{a}p_{b}$
but from equation above we have 
\begin{align*}
p_{b}=\frac{p_{a}-\alpha\left(1+p_{a}^{2}\right)}{2\beta p_{a}}.
\end{align*}
Substiting this into $p_{b}=\beta+\beta p_{b}^{2}+2\alpha p_{a}p_{b}$
we get after simplification
\begin{equation}
p_{a}^{2}\left(2a^{2}-4b^{2}+1\right)+3a^{2}p_{a}^{4}-a^{2}-4ap_{a}^{3}=0.\label{eq:EG4}
\end{equation}
Now setting $p_{a}=1$ above we get 
\begin{align*}
\left(2a^{2}-4b^{2}+1\right)+3a^{2}-a^{2}-4a=4\left(a^{2}-b^{2}-a\right)+1=4\left(\frac{1}{2}\left(a-b\right)-a\right)+1=0.
\end{align*}
Thus dividing original polynomial by $\alpha^{2}\left(p_{a}-1\right)$ (i.e. rejecting
this solution choice) we get $3p_{a}^{3}+\left(3-4\alpha^{-1}\right)p_{a}^{2}+p_{a}+1=0$.
This indeed has a unique solution in $\left(0,1\right)$ as can be
checked easily.  
A: Limiting distributions won't help you here, anyway. The long term probability of hitting a given state doesn't necessarily tell you anything about the probability of hitting it ever.
This is basically a combinatorics exercise. What is the probability of the $n$-th state being $1$? This is easier than it might seem at first. What strings of length $n$ would reduce to $1$? Let $a_n$ be the number of strings of length $n$ that reduce to $1$. I'm sure you can compute $a_n$ via some sort of recurrence relation.
