The distribution of $x$ can be approximated with a logit-normal distribution for a large number of steps. The distribution will concentrate at 0 or 1, depending on $\mu$:

$$\mu=(1-\lambda)\ln\bigg(\frac{1-\pi_H}{1-\pi_L}\bigg)+\lambda\ln\bigg(\frac{\pi_H}{\pi_L}\bigg)$$

• $\mu<0$: $x\rightarrow1$
• $\mu>0$: $x\rightarrow0$
• $\mu=0$: $P(x\rightarrow0)=0.5$; $P(x\rightarrow1)=0.5$

I'll assume $\lambda_{right}+\lambda_{left}=1$ and use $\lambda\equiv\lambda_{right}$.

Let $x_{a,b}$ denote the position of a particle that has moved a total of $a$ times to the left and $b$ times to the right. First, notice that the sequence of the $n=a+b$ moves does not affect the value of $x_{a,b}$ (e.g., left-left-right-right results in the same position as left-right-right-left).

$$x_{a,b}=\bigg[1+\bigg(\frac{1-\pi_L}{1-\pi_H}\bigg)^a\bigg(\frac{\pi_L}{\pi_H}\bigg)^b\bigg]^{-1}$$

A logit transformation on $x_{a,b}$ results in:

$$y_{a,b}=a\ln\bigg(\frac{1-\pi_H}{1-\pi_L}\bigg)+b\ln\bigg(\frac{\pi_H}{\pi_L}\bigg)$$

Possible values of $y_{a,b}$ are equally spaced from $n\ln\Big(\frac{1-\pi_H}{1-\pi_L}\Big)$ to $n\ln\Big(\frac{\pi_H}{\pi_L}\Big)$ with intervals:

$$d=\ln\Big(\frac{\pi_H}{\pi_L}\Big)-\ln\Big(\frac{1-\pi_H}{1-\pi_L}\Big)$$

$b\sim\text{Bin}(n,\lambda)$, so $y_{a,b}$ follows a scaled and shifted binomial distribution:

$$b=\frac{y_{a,b}-n\ln\Big(\frac{1-\pi_H}{1-\pi_L}\Big)}{n\cdot{d}}\sim\text{Bin}(n,\lambda)$$

We can use the normal approximation to the binomial as $n$ gets large:

$$y_{a,b}\sim\mathcal{N}(n\mu,d^2n\lambda(1-\lambda))$$

Equivalently, $x_{a,b}$ can be approximated with the logit-normal distribution.

As $n$ increases, the magnitude of the ratio of the mean to the standard deviation increases if $\mu\neq0$, so $x$ will tend to either 0 or 1. For $\mu=0$, the mean of $x$ remains at 0.5, and, although the most probable positions are near 0.5, they make up an increasingly smaller proportion of the total probability.