Thinking about it more, this isn’t a serious problem for the dilemma. While P(you can offer me k utility) goes to zero as k goes to infinity but there’s no reason to suppose it goes faster then 1/n does.
This means you can still set a similar dilemma, with a probability of you being able to offer me 2^n utility eventually becoming greater than (1/2)^n for sufficiently large n, satisfying the conditions for a St Petersburg Lottery.
Thinking about it more, this isn’t a serious problem for the dilemma. While P(you can offer me k utility) goes to zero as k goes to infinity but there’s no reason to suppose it goes faster then 1/n does.
Thinking about it more, this isn’t a serious problem for the dilemma. While P(you can offer me k utility) goes to zero as k goes to infinity but there’s no reason to suppose it goes faster then 1/n does.
This means you can still set a similar dilemma, with a probability of you being able to offer me 2^n utility eventually becoming greater than (1/2)^n for sufficiently large n, satisfying the conditions for a St Petersburg Lottery.
That’s just Pascal’s mugging, though; the problem that “the utility of a Turing machine can grow much faster than its prior probability shrinks”.