This question has nothing to do with Turing machines, it's more about Measure Theory. It's well-known that it's not possible to define uniform finite measure on infinite countable sets such as integers. There are 2 common ways around it:

First method: define a set of measures on a filter and take a limit. In your example fix N, assume probability 1/N for each value. Then calculate whatever probability you want to calculate. Then take limit of your result for N->infinity.

Second method: define non-uniform measure that sums up to a finite number. In your example for each value n assume probability C/A^n, for some fixed value A, and C is a fixed normalizing coefficient that would make the whole series converge to 1. Then compute whatever probability you need in this non-uniform measure. Then take limit A->1.

This question has nothing to do with Turing machines, it's more about Measure Theory. It's well-known that it's not possible to define uniform finite measure on infinite countable sets such as integers. There are 2 common ways around it:

First method: define a set of measures on a filter and take a limit. In your example fix N, assume probability 1/N for each value. Then calculate whatever probability you want to calculate. Then take limit of your result for N->infinity.

Second method: define non-uniform measure that sums up to a finite number. In your example for each value n assume probability C/A^n, for some fixed value A, and C is a fixed normalizing coefficient that would make the whole series converge to 1. Then compute whatever probability you need in this non-uniform measure. Then take limit A->1.

EDIT: A small (not-too-rigorous, but essentially right) demonstration how the second method of dealing with probabilities over infinite countable sets works.

A question: let N be a "random" positive integer. What is the "probability" that N has no multiple prime factors? That is, what's the probability that N does not divide a complete square other than 1?

The difficulty: the set of integers is an infinite countable set, therefore one cannot assign equal measure to all integers.

The idea: assign to every integer N measure 1/N^s for some parameter s. Then compute the "probability" in that measure. Then take a limit s->0. Since 1/N^0=1 for any N that would correspond in the limit to the uniform measure that cannot be assigned directly.

A sketch of the core of the solution: the sum 1/N^s is of course the Riemann zeta-function \zeta(s). Rewrite the expression according to Euler's product formula. Write down Euler product with single prime factors. Divide the 2nd expression by the 1st one. Get the answer 1/\zeta(2s).

The method in question at work: take the limit s->0. Arrive to the answer 1/\zeta(2). This is possible for technical reasons, due to the singularity at \zeta(1).

The answer: 1/\zeta(2), which happens to be 6/\pi^2.

The confirmation: take first 1000000 integers and count how many of them don't divide a complete square. Observe that the proportion is approximately 6/\pi^2.

The conclusion: assigning non-uniform probabilities and taking the limit works rather well.