Because then you can just number your possible outcomes by integers n and set p(n) to 1/U(n) * 1/2^n, which seems too easy to have been missed.
The reason why this wouldn’t work is that sometimes what you’re calling “U(n)” would fail to be well defined (because some computation doesn’t halt) whereas p(n) must always return something.
The reason why this wouldn’t work is that sometimes what you’re calling “U(n)” would fail to be well defined (because some computation doesn’t halt) whereas p(n) must always return something.
No; the utility function is stipulated to be computable.
What Manfred is calling U(n) here corresponds to what the paper would call U(phi_n(k)).
The utility function is defined as being computable over all possible input.
phi_n(k) may not halt.