Okay, so it should be the probability that a TM halts rather than compiles.

Technically you're right about random number generation, but there are pseudo random number generators with over 10^80 length cycles, ISAAC for example.

That should be random enough to get a fair gauge whether a TM is random.

There's a quote, "if you go large enough, the primes seem to crop up randomly, but we're not quite sure what random means."

So, for example, you could write a TM that goes through numbers one at a time starting from some large value, and whether or not it's prime could be used to generate randomness.

My query is, can you definitively come up with a number that is non-computable?

Can you give me an example?

The computable reals are enumerable as are the set of computable infinite binary sequences. I choose to use binary sequences because there are two representations for each dyadic rational in binary but there is only one representation per infinite binary sequence.

To show the existence of some non-computable infinite binary sequence, just consider a listing of the computable binary sequences (note that such a listing generally requires the axiom schema of specification or some choice I believe because the halting problem essentially ensures that there is no explicit function for generating the listing despite the listing absolutely being countable.):

$t_1 = t_{1,1}, t_{1,2}, t_{1,3}, ...$

$t_2 = t_{2,1}, t_{2,2}, t_{2,3}, ...$

$t_3 = t_{3,1}, t_{3,2}, t_{3,3}, ...$

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.

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Using Cantor's diagonal method, create a new infinite binary sequence that isn't in the list. To do this, let $s(t_{i,j}) = 0$ if $t_{i,j} = 1$ and let $s(t_{i,j}) = 1$ if $t_{i,j} = 0$:

$a = s(t_{1,1}), s(t_{2,2}), s(t_{3,3}), ...$

We now know that $a$ is different than every member of the list. If the list includes all computable infinite binary sequences, then $a$ must further be a non-computable infinite binary sequence.

Hope that helps!