Let $(X_n)_{n\ge 0}$ be a process for which $X_0$ is uniform on $(0,1)$ and, for $0<\alpha <1$ and $n\ge 0$, we have:
$X_{n+1} = \alpha X_n + 1 -\alpha$ with probability $X_n$
$X_{n+1} = \alpha X_n$ with probability $1-X_n$.
What are the limiting properties of $X_n$ as $n\to \infty$?
It is easy to see that $0<X_n<1$ and also that $X_n$ is a martingale, and so by the Martingale Convergence Theorem, we must have $\lim_{n\to\infty} X_n =X$ where $X$ is a random variable that is almost surely finite. But I cannot deduce anything else about $X$. Can we even say anything else, and if so, how?
Assuming that the choice of values of $X_{n+1}$ is made independently of $\mathcal{F}_n = \sigma(X_0, \cdots, X_n)$, notice that
$$\frac{X_{n+1} - \alpha X_n}{1-\alpha}$$
is a $\{0,1\}$-valued random variable, i.e., a Bernoulli random variable that converges almost surely to $X$. So $X$ is also a Bernoulli random variable. Then the law of $X$ is determined by the parameter
$$p = \Bbb{P}(X = 1) = \Bbb{E}[X]$$
which is easily computed from the martingale nature of $(X_n)$.
(This is essentially what @Did is explaining in the comment.)