Are "almost all" decidable languages not in P?

Question

There's a famous classical circuit complexity result by Shannon that says almost all languages require exponential circuits [[1]], proven by comparing the number of distinct circuits of $n$ variables versus the number of boolean functions of $n$ variables.

Similarly, there's a result that "almost all" languages are undecidable, because the set of TMs is countable while the set of languages is uncountable.

Question: is there a result of similar flavor about complexity classes? E.g. something like "almost all" decidable languages are not in P?

The question seems interesting to me because it's much less clear how to even define "almost all". Both sets here are countably infinite so cardinality arguments don't work. We could maybe do something like $$\lim_{n \to \infty} \frac{\text{# of polytime TMs with n states}}{\text{# of total TMs with n states}}$$ ? But that's obviously tremendously complicated and out of reach. Are there simpler definitions that would capture a similar meaning?

Answer 1

This answer relates to the number of decidable languages (and not the number of TMs that accept decidable languages).

So my intuition would say no, there sre no more decidable problems not in P than the problems in $P$, since you can reduce a problem $L$ of running time $O(f(n))$ into a problem $L’=\{w0^{f(|w|)}|w\in L\}$ in $P$, i.e. we add many zeros to each instance, to make the size of the input large enough, that the original algorithm has linear running time on the resulting instance.

Note that this is not a complete proof, but I would assume you can use this fact to build a proof.

Note also that such a reduction is not really interesting (we can’t really use it to make any complexity assumptions about membership of $P$) since the reduction itself is not computable in log space and it is not even in FTIME(g(n)) for any computable g, since you can always choose a harder problem to begin with.