Random suborbits of a rotation Let $u_n = x + n\alpha \pmod 1$ with $\alpha$ irrational. We know that $(u_n)_{n \geq 0}$ is dense in $\mathbb{R}/\mathbb{Z}$ (equivalently $(u_n)_{n \geq 0}$ visits every open interval infinitely many times).
Now, let $(\epsilon_n)_{n \geq 0}$ be a sequence of independent uniform Bernoulli random variables on $\{0,1\}$, and define the random integers $K_n=\sum_{i = 0}^n \epsilon_i2^i$. Is it true that the random sequence $(u_{K_n})_{n \geq 0}$ almost surely visits every open interval infinitely many times ? 
As a conditional and secondary question, if it is true, I would also like to know:


*

*is it more generally true for random integers $K_n$ defined in the same way but using another representation of the integer numbers instead of the binary one ? 

*is it more generally true for the orbits of a minimal homeomorphism ? (after replacing intervals with open sets)
 A: The answer to the last question is negative.  Let $T:\{0,1\}\to\{0,1\}$ be the map that takes $0$ to $1$ and $1$ to $0$.  Clearly, $T$ is a minimal homeomorphism of $\{0,1\}$ with the discrete topology.  However, with probability $1/2$, all the integers $K_n$ are even, and therefore $\left(T^{K_n}(0)\right)_{n\geq 0}$ is not dense. 
A: For your initial question I think I have an answer. WLOG assume $x=0$.
Assume that $\alpha$ has a universal binary expansion, i.e., the one which contains every finite 0-1 word as a factor. (This is a generic property in $\alpha$.) Then the sequence $2^n\alpha\bmod1$ is clearly dense in $(0,1)$. (Alternatively, you may use Weyl's criterion, as Vaughn suggested, but my condition is an explicit one.)
Now, we have $K_n=K_{n-1}+\epsilon_n2^n$, whence we have a map
$$
x\mapsto \begin{cases} x+2^n\alpha\bmod1,& 
\\ x,\end{cases}
$$
with equal probabilities. 
Fix $(a,b)\subset(0,1)$. Since $(\epsilon_n)$ is generic, for any $x=K_{n-1}\alpha\bmod1$ we will have $\epsilon_n=\epsilon_{n+1}=\dots=\epsilon_{n+k-1}=0, \epsilon_{n+k}=1$ with a positive probability, where $n+k$ is such that $x+2^{n+k}\bmod1\in(a,b)$. And this will happen infinitely many times by some general probability nonsense. ;) 
So, the claim appears to be true for any such $\alpha$. 
I think if a counterexample exists, one might want to look at a Liouville $\alpha$ with very sparse 1s in its binary expansion. 
EDIT. Come to think about it, there's no reason why we should stick to $2^n\alpha$. We have $K_{n+k}=K_n+2^nN$, where $0\le N\le 2^{k-1}$, and all of these have positive probability (where $k$ is chosen large enough to satisfy $b-a>>2^{-k}$). So, since the set $\{ N2^n\alpha\bmod1 \mid 0\le N\le 2^k-1\}$ intersects every sufficiently large interval, with probability 1 we will have $K_n\alpha\in (a,b)$. 
So, the answer is YES for all irrational $\alpha$.
