Let $g$ be the golden number (or another algebraic integer in $(0,1)$ that fullfills an equation with coefficients $\pm 1$). Consider the random walk on $\mathbb{R}$ starting with $0$ and walking $g^{n}$ with probability $1/2$ left or right in step $n+1$. What is the probability to return to $0$ exactly $m$-times. Here $m$ maybe $0$ and $\infty$ as well. If $g$ ist not as described the walk will of course not return at all. 



Random walks with exponential decreasing steps
In the case of the golden number $g = (\sqrt{5}-1)/2$, we have
$g^n = (-1)^n (F\_{n-1} - F\_n g)$ where $F\_n$ is the $n$'th Fibonacci number.
Let $a\_i, i=0,1,2,\ldots$ be $+1$ if the $i+1$'th step is to the right, $-1$ if it is to the left. Then we return to the origin after $n$ steps iff
 $\sum\_{i=0}^{n-1} a\_i F\_i = 0$ and $\sum\_{i=0}^{n-1} a\_i F\_{i-1} = 0$.


Note that $\sum\_{i=0}^{n-1} a\_i F\_i \equiv 1 \mod 2$ if $n \equiv 2 \mod 3$
and $\sum\_{i=0}^{n-1} a\_i F\_{i-1} \equiv 1 \mod 2$ if $n \equiv 1 \mod 3$, so 
the only possibilities for return are with $n$ divisible by $3$. Moreover, if
 $a\_{n-3}, a\_{n-2},a\_{n-1}$ are not either $[1,1,-1]$ or $[-1,-1,1]$ then we have
$$ \eqalign{\left| \sum\_{i=0}^{n-1} a\_i F\_i \right| &\ge F\_{n-1} - F\_{n-2} + F\_{n-3} - \sum\_{i=0}^{n-4} F\_i \cr &= -1-\dfrac{ \sqrt {5}-1 }{2} \left( \dfrac{1-\sqrt {5}}{2}
 \right) ^{n-3}+ \dfrac{ \sqrt {5}+1}{2} \left( \dfrac{\sqrt {5
}+1}{2} \right) ^{n-3}\cr
&> 0 \ \text{for}\ n \ge 4}$$
So the only ways to ever return to the origin are to return to the origin after every $3$ steps, each three steps being either $+,-,-$ or $-,+,+$.
The probability of having exactly $m$ returns to the origin is then the probability that the first $3m$ steps follow this pattern but the next $3$ do not, thus $3 \times 4^{-m-1}$. 


EDIT: A simpler argument in the case of the golden number:


Note that if $(a\_0, a\_1, a\_2)$ is not $(1,-1,-1)$ or $(-1,1,1)$,
$$|a\_0 + a\_1 g + a\_2 g^2| \ge 3 - \sqrt{5} > g = \sum\_{n=3}^\infty g^n$$
So if you don't return to the origin after the first $3$ steps, you are too
far away to ever return. Similarly for $(a\_{3n}, a\_{3n+1}, a\_{3n+2})$ where
$\sum\_{i=0}^{3n-1} a\_i g^i = 0$... 


A similar argument works for the next case, where $g$ is the real root
(approximately $0.5436890127$) of $X^3 + X^2 + X - 1$. If 
$(a\_0,a\_1,a\_2,a\_3)$ is not $(1,-1,-1,-1)$ or $(-1,1,1,1)$, 
$$|a\_0 + a\_1 g + a\_2 g^2 + a\_3 g^3| \ge 1 - g - g^2 + g^3 
> \dfrac{g^4}{1-g} = \sum\_{i=4}^\infty g^i$$
so the only way to ever return to the origin is to return every $4$ steps.


However, this argument won't work for the positive root (approximately $0.8483748957$) of $X^4 + X^3 + X^2 - X - 1$. For this case, is it possible to return to the origin without returning after the first $5$ steps?


