Do the powers of $2$ in ternary eventually all contain a $0$? It seems that the last power of $2$ that only has $1$'s and $2$'s in its ternary expansion is $2^{15} = 1122221122_3$. Empirically, this is true upto $2^{10^7}$. Is it true in general?
The context of this question is that some of us were looking at the operation of given a number taking the product of its digits. The base $10$ case seemed too complicated, and the first non-trivial case is base $3$, where every number immediately reduces to a power of $2$ (or $0$).
 A: As a probabilistic approximate argument, $2^n$ has slightly more than $0.63\, n$ digits when written in ternary, since $\log_3(2) \approx 0.63092975$.  Your example $2^{15}$ has ten ternary digits, slightly more than $9.45$  
The first and last of these digits will not be zero, but there is no obvious pattern for the others, so perhaps about a third of the other digits are zero. They are not random (this is number theory) but we might consider what would happen if they were  
That would suggest that the chance that at least one digit was zero might be about $1-\left(\frac23\right)^{0.63n-2}$ and so the chance there would be at least one case of no zeros from $2^n$ onwards might for large $n$ be around $$1 - \prod\limits_{m=n}^\infty \left(1-\left(\frac23\right)^{0.63m-2}\right) \approx \sum\limits_{m=n}^\infty \left(\frac23\right)^{0.63m-2} \approx \int\limits_{x=n}^\infty \left(\frac23\right)^{0.63x-2}\, dx \approx \frac{1}{-0.63 \log\left(\frac23\right)}\left(\frac23\right)^{0.63n-2}$$ 
which for $n=15$ is about $0.19$ so seeing an example there is not much of a surprise, while for $n=10^7$ would be smaller than $10^{-1000000}$.  Hence Henning Makholm's comment that this is ridiculously small 
