Denote $A=\{0\}, B=\{0,1\}$. Then any subset of $\Omega:=\{A,B\}^{\mathbb N}$ is a continuum provided the number of $B$'s is infinite. We treat these as binary expansions of numbers in $[0,1]$.


For instance, $(AB)^\omega$ is the set $\left\{\sum\_{n=1}^\infty a\_n4^{-n}\mid a\_n\in\{0,1\}\right\}$, which is a Cantor set of Hausdorff dimension $\frac12$. 
Similarly, it is easy to show that $(AB^N)^\infty$ has Hausdorff dimension $\frac{N}{N+1}$.


Now, suppose the sequence of $A$'s and $B$'s is random: i.i.d. with distribution $\bigl(\frac1{N+1}, \frac{N}{N+1}\bigr)$. It is obvious that a generic sequence will be a Cantor set.


**Question.** Is it true that a subset of $\Omega$ has Hausdorff dimension $\frac{N}{N+1}$ almost surely? A weaker version: is this a null set almost surely? 



Random Cantor sets on the unit interval
Like the classical Cantor set, your set has a natural representation $C=\bigcap C\_n$, where $C\_1\supseteq C\_2\supseteq C\_3\supseteq\ldots$, and each $C\_n$ is a disjoint union of $2^{n-X\_n}$ intervals of length $2^{-n}$, with $X\_n$ denoting the number of $A$'s in your sequence up to position $n$.


So, covering by intervals of that length, we have that $H\_{2^{-n}}^d(C\_n)=2^{n(1-d)-X\_n}$. If we write $p=P(A)$, then $X\_n/n\to p$ almost surely, so $\dim C\le 1-p$ almost surely.


For the lower bound, we observe that $C$ will contain at least one point from each of the $2^{n-X\_n}$ intervals of length $2^{-n}$ whose union is $C\_n$. This means that we cannot find a $2^{-n}$ covering of $C$ that decreases the number of intervals needed by more than a fixed factor, compared to the one of $C\_n$ from above.


So $\dim C=1-p$ almost surely, as claimed.


