Thus you wouldn't get closer and closer to 100% accuracy, but rather, you'd get the randomness response every single attempt
1. 'F1' which represents any possible function which does not invoke the H(F1) function internally.
2. 'F2' which represents any possible function which invokes the H(F2) function internally.
Turing's proof relies entirely on the existence of 'F2' functions and the fact that these kinds of functions always lead to a an infinite loop deadlock condition. Turing's proof relies on these 'F2' functions as a counter example to prove why H cannot exist. Turing's proof does not apply for functions of the 'F1' kind.
So my thinking is that assuming that there existed a probabilistic analyzer function `H` which worked with 'F1' functions and which could be modified to also work with 'F2' functions (I.e. by adding a probabilistic error % as I suggested), then that would invalidate Turing's counter-example. It would not give any answers about the halting problem itself though; just move the problem back to a state of uncertainty.
My idea aims to prove that if you have an 'H' which works for 'F1' then it can be easily modified to also work for 'F2' if you're prepared to accept an error probability each time you run the function.
A key part of my argument is that any F2 function can use an error % as an escape hatch to protect itself from infinite recursion and thus gives those functions the ability to decide their own output.
For these kinds of F2 functions, whatever output H(F2) returns becomes self-fulfilling after the recursion has finished unwinding and so it does not matter if H(F2) is true or false.