There are finitely many $n$-state and $k$-symbol Turing machines. Call the number $N(n,k)= (2k(n + 1))^{kn}$ (according to Wikipedia https://en.wikipedia.org/wiki/Busy_beaver). Each of them either halts or does not halt.
Thus, we can speak of a (uniformly) random $n$-state and $k$-symbol Turing machine that halts. Let $A(n,k)$ denote the average average number of steps such a machine runs before halting. Is $A(n,k)$ computable?
The maximum number of steps $M(n,k)$ that such a machine takes is uncomputable and larger than any computable function.
In particular, is the following argument true: It holds that $A(n,k) \geq \frac{M(n,k)}{N(n,k)}$ and if $A(n,k)$ was computable then $A(n,k)N(n,k)$ would be a computable function larger than $M(n,k)$ which is a contradiction.