The length of the reduced word can be thought of as Markov chain $$0\underset{1}{\overset{\tfrac{1}{4}}{\leftrightarrows}}1\underset{\tfrac{3}{4}}{\overset{\tfrac{1}{4}}{\leftrightarrows}}2\underset{\tfrac{3}{4}}{\overset{\tfrac{1}{4}}{\leftrightarrows}}\cdots.$$ Let $p$ be the probability of ever returning to state $0$ . Then, since after the first step the length is $1$ , $p$ must satisfy $$p=\tfrac{1}{4}+\tfrac{3}{4}p^{2},$$ or $$p=\tfrac{1}{3}\text{ or }1.$$

Let $p_{n}^{(k)}$ ($p_{n}$ ) be the probability of reaching state $0$ from state $n$ in at most $k$ (in any number of) steps. Let $A_{n}^{(k)}$ ($A_{n}$ ) be the corresponding events. Then, as it is easy to see, $A_{n}=\cup_{k\ge1}A_{n}^{(k)}$ , and, hence, $p_{n}=\lim_{k\to\infty}p_{n}^{(k)}$ . So, if we show that $p_{1}^{(k)}\le\tfrac{1}{3}$ for all $k\ge1$ , then we will show that $p=p_{1}\le\tfrac{1}{3}$ . For this, suppose that $x_{n}\ge0$ , $n\ge1$ , solve the system $$\begin{align*} x_{1} & =\tfrac{1}{4}+\tfrac{3}{4}x_{2},\\ x_{n} & =\tfrac{1}{4}x_{n-1}+\tfrac{3}{4}x_{n+1},\text{ }n\ge2. \end{align*}$$ Then, $p_{n}^{(k)}\le x_{n}$ for all $n\ge1$ . Indeed, $$\begin{align*} p_{1}^{(1)}=\tfrac{1}{4} & \le x_{1},\\ p_{n}^{(1)}=0 & \le x_{n},\text{ }n\ge2, \end{align*}$$ and $$\begin{align*} p_{1}^{(k+1)}=\tfrac{1}{4}+\tfrac{3}{4}p_{2}^{(k)} & \le\tfrac{1}{4}+\tfrac{3}{4}x_{2}=x_{1}\\ p_{n}^{(k+1)}=\tfrac{1}{4}p_{n-1}^{(k)}+\tfrac{3}{4}p_{n+1}^{(k)} & \le\tfrac{1}{4}x_{n-1}+\tfrac{3}{4}x_{n+1}=x_{n},\text{ }n\ge2. \end{align*}$$ It remains to note that $$x_{n}=(\tfrac{1}{3})^{n},\text{ }n\ge1,$$ indeed solves the system.

Intuitively the last statement of the problem is clear because at each step (unless when the current word is empty, which happens only finitely often with probability $1$ ) the length is increased by $1$ with probability $\tfrac{3}{4}$ , and decreased by $1$ with probability $\tfrac{1}{4}$ , so on average it increases by $\tfrac{1}{2}$ . A bit more formally, let $w_{n}$ be the word length after $n$ steps. Consider i.i.d. $X_{n}\in\{-1,+1\}$ where $\mathbb{P}(X_{n}=+1)=\tfrac{3}{4}$ , $n\ge1$ . Let $Y_{n}=2I(w_{n-1}=0,X_{n}=-1)$ . Then, we can interpret $w_{n}=\sum_{k\le n}X_{k}+\sum_{k\le n}Y_{k}$ . By the strong LLN, $\mathbb{P}(\tfrac{1}{n}\sum_{k\le n}X_{k}\to\mathbb{E}X_{n}=\tfrac{1}{2})=1$ . Further, if $q_{n}=\mathbb{P}(w_{n}=0)$ , then $$\begin{align*} q_{n} & \le\mathbb{P}(\sum_{k\le n}X_{k}\le0)\\ & \le\mathbb{P}(\sum_{k\le n}(X_{k}+1)/2\le3n/4\cdot2/3)\\ & \le e^{-\tfrac{n}{24}}, \end{align*}$$ and $\sum_{n}q_{n}<\infty$ . Therefore, with probability $1$ , only a finite number of $Y_{n}$ is positive (intuitively, we return to $0$ only a finite number of times with probability $1$ ), and, hence, $\tfrac{1}{n}\sum_{k\le n}Y_{k}\to0$ with probability $1$ .