Average performance of universal computation
By $T(A)$ I will mean the runtime of $A$ on the empty input, and will define $T(A) = \infty$ if $A$ doesn't halt. By $|A|$ I mean the length of $A$ in bits.

What you are trying to calculate in part (1) is:

$$T = \sum_{A | T(A) &lt; \infty} \frac{1}{2^{|A|}}T(A)$$

Where the fraction is the Chaitin probability of generating program $A$. For each length $n$ there is (at least) one machine $A_n$ such that for all halting $A$ with $|A| = n$ we have $T(A) \leq T(A_n) = BB(n)$. Here $BB$ is the relevant busy-beaver function, which grows faster than any computable function, in particular $BB(n) &gt; 2^n$. From this we have:

$$T \geq  \sum_{n = 1}^{\infty} \frac{1}{2^n}BB(n) &gt; \sum_{n = 1}^{\infty} 1$$

Thus your sum does not converge, and $T$ is not finite. I guess that answers the rest of your questions, too. If you want to achieve your goal, then you need to pick a better density function on programs, maybe something where probability of $A$ is proportional to say $\frac{1}{2^{BB(|A|)}}$. However, for learning more about this I recommend CS.SE.