Update: in response to comment below, I'm not sure anymore the probability in question was 0.

Let's try the measure that gives an equal weight for any true statement of fixed length $N$ (written in mathematical English).

Then we have statements of the form "S or 1+2=3" which form more then $1/10^{20}$th of all true statements of a given length. On the other hand, the statements "S and Z" (where Z is an undecidable problem) form some positive (bounded below) measure subset in the statements of given length as well.

So, yes, for the measure described above the measure of provable statements is within some positive bounds $[a, b]$ for any $N$ greater than some $N_0$.

But that measure is proportional, for a fixed $N$, to the characteristic function of the set of all true statements, so, no, my answer doesn't give a "natural" measure.

I think the probability of "a random true statement is provable" is 0 in any good formalization of your problem.

Here's the reasoning: consider a true undecidable statement S. Now I would say that in any reasonable definition of probability the long statement will contain S with probabilitiy that goes to 1 as its length increases. One can't prove it until there's no definition of this probability, but the related fact for strings is straightforward:

Consider any string S. Then the probability $P(n) := \{$the random string of length $n$ contains S$\}$ goes to 1 as $n\to \infty$.

The proof can be done by considering the random strings of the form (chunk 1)(chunk 2)...(chunk N) where all chunks are of the same length as S.

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I think the probability of "a random true statement is provable" is 0 in any good formalization of your problem.

Here's the reasoning: consider a true undecidable statement S. Now I would say that in any reasonable definition of probability the long statement will contain S with probabilitiy that goes to 1 as its length increases. One can't prove it until there's no definition of this probability, but the related fact for strings is straightforward:

Consider any string S. Then the probability $P(n) := \{$the random string of length $n$ contains S$\}$ goes to 1 as $n\to \infty$.

The proof can be done by considering the random strings of the form (chunk 1)(chunk 2)...(chunk N) where all chunks are of the same length as S.