My thinking for this question is that it is just a slight variation of the standard Cantor set and will therefore have the same similarity dimension. My logic is that at each new step, the interval can be split into 2 copies of the previous one, each with a scaling factor of $\frac{1}{3}$ and therefore the similarity dimension is $\frac{\log 2}{\log 3}$.
Am I correct in my thinking here?
In fact this is not similar to the cantor set. At every timepoint you're reducing the size to $\frac{2}{3}$ of its previous size. This means we get $N=1$ copies of the previous one with a scaling factor of $\frac{2}{3}$.
That yields a dimension of $\frac{\log 1}{\log \frac{2}{3}} = 0$.
That's no surprise, as the set you described is not a fractal, it's just the limit of the interval $[\sum_{i=1}^n \frac{2^{i-1}}{3^i},1]$ which is $[\lim_{n\to\infty}\frac{1}{3}\sum_{i=0}^n \frac{2^i}{3^i},1] = \{1\}$