I want to know about the density of automatic sets where the DFA recognizing it has a particular nice form. I'm going to start with a simple version of the question and then add complications until we get to the case that led to the question.  I'm aware of some results about the density of automatic sets but I'm not sure if any of them do what I want.

So, **simplest version of the problem**: Suppose $S$ is a $b$-automatic set, and suppose that it's recognized, *when read from the little end* (right-to-left), by a DFA where every state has a path to a sink state (a state from which there is no escape).

Under this assumption, we can meaningfully speak of the probability of the machine accepting if we plug in a random infinite string on $\{0, ..., b-1\}$ (or, if you like, a random $b$-adic integer) -- treating the machine as a Markov chain, essentially -- because with probability $1$ we will eventually end up in a sink state, at which point the rest of the string is irrelevant and we can determine acceptance or rejection.

**Question**: Under these assumptions, is the density of $S$ equal to this probability?  (Natural density if possible, logarithmic density if not.)

Note that if we were to instead read from the big end, left to right, then certainly the natural density version of this question would have a negative answer (e.g.: consider numbers that start with a $1$ when written in base $3$).  But I'm specifically reading from the little end, right to left.

Also worth noting, if we add the additional condition that (other than the sink self-loops) the machine has no cycles, then the answer to the question is positive, even with natural density, because then you only need finite additivity to determine the probability.  However, the more general version requires countable additivity, which (natural/logarithmic) density doesn't satisfy.

**Complication #1**: What if instead $S$ is a $k$-dimensional $b$-automatic set, with the same conditions?  So we would be plugging in $k$ random infinite strings on $\{0,...,b-1\}$.  Does the probability equal the density here?

(Is logarithmic density a thing in this setting?  I guess you would weight $(n_1,...,n_k)$ by $\prod_i \frac{1}{n_i}$?  I don't know if there's a standard version.)

**Complication #2**: Let's go back to the one-dimensional case.  What if now, instead of being $b$-automatic, the set is Zeckendorf-automatic?

In this case, of course, we'll need to adjust our probabilities accordingly.  We're plugging in a random infinite string on $\{0,1\}$, but it's no longer uniformly random.  Rather, the first digit is a $0$ with probability $\frac{1}{\varphi}$ and a $1$ with probability $\frac{1}{\varphi^2}$, and we use this same distribution after a $0$, but after a $1$ we always put a $0$.  (So it's not actually a Markov chain anymore, because it's no longer memoryless.  Well, you could construe it as one, but that won't work for the next case.)

Does the probability equal the density in this case?  With whatever notion of density we can get.

(Note that once again here we get that the answer is yes, even for natural density, if we add the no-cycles condition.)

**Complication #3**: Combine complications #1 and #2; the set is both multidimensional and Zeckendorf-automatic.  You can fill in the details.

(This is the case that led to the question -- the set I'm looking at is a two-dimensional Zeckendorf-automatic set of the above form.)

Thanks all!  I would consider this the obvious way to compute the density of such a set, and it's kind of annoying to me that I can't quickly find anything to indicate that it's a valid method.  But I'd also be interested to hear about answers to the simplified versions of the problem, because I think it's quite interesting on its own.