The post you’re commenting on argues that Pascal’s mugging is already solved by merely letting the utility function be bounded by Kolmogorov complexity. Obviously, having it be uniformly bounded also solves the problem, but why resort to something so drastic if you don’t need to?
The OP is not living in the least convenient possible world. In particular, let X be the worst thing that could happen. Suppose that at the end of the day you have calculated that X will occur with probability 10^(-100) if you don’t pay the mugger $5. Assuming that you wouldn’t pay the mugger, then by definition of the utility function it follows that u($5) > 10^(-100) u(X). So u(X) < 10^(100) u($5) and is therefore bounded. Since u(X) is the worst thing that could happen, this means that your entire utility function is bounded.
See also my reply to wedrifid where this argument is slightly expanded.
The post you’re commenting on argues that Pascal’s mugging is already solved by merely letting the utility function be bounded by Kolmogorov complexity. Obviously, having it be uniformly bounded also solves the problem, but why resort to something so drastic if you don’t need to?
The OP is not living in the least convenient possible world. In particular, let X be the worst thing that could happen. Suppose that at the end of the day you have calculated that X will occur with probability 10^(-100) if you don’t pay the mugger $5. Assuming that you wouldn’t pay the mugger, then by definition of the utility function it follows that u($5) > 10^(-100) u(X). So u(X) < 10^(100) u($5) and is therefore bounded. Since u(X) is the worst thing that could happen, this means that your entire utility function is bounded.
See also my reply to wedrifid where this argument is slightly expanded.
If your utility function is not bounded (below), then there is no “worst thing that could happen.”
See my reply to komponisto in the comment above.