wedrifid comments on A Thought on Pascal's Mugging - Less Wrong

12 Post author: komponisto 10 December 2010 06:08AM

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Comment author: wedrifid 13 December 2010 02:12:33AM 0 points [-]

As an aside, if you wouldn't pay him then the definition of utility implies that u($5) > 10^(-100) u(X), which implies that u(X), and therefore the entire utility function, is bounded.

This doesn't actually imply that the entire utility function is bounded. It is still possible that u(Y) is infinite, where Y is something that is valued positively.

As an aside we can now consider the possibility of Pascal's Samaritan.

Assume a utility function such that u(Y) is infinite (and neutral with respect to risk). Further assume that you predict that $5 would increase your chance of achieving Y by 1/3^^^3. A Pascal Samaritan can offer to pay you $5 for the opportunity to give you a 90% chance of sending the entire universe into the hell state X. Do you take the $5?

Comment author: jsteinhardt 13 December 2010 02:35:16AM *  0 points [-]

From my reply to komponisto (incidentally, both you and he seem to be making the same objections in parallel, which suggests that I'm not doing a very good job of explaining myself, sorry):

Suppose that there is nothing so bad that you would pay $5 to stop it from happening with probability 10^(-100). Let X be a state of the universe. Then u(-$5) < 10^(-100) u(X), so u(X) > 10^(100) u(-$5). Since u(X) > 10^(100) u(-$5) for all X, u is bounded below.

Similarly, suppose that there is nothing so good that you would pay $5 to have a 10^(-100) chance of it happening. Then u($5) > 10^(100) u(X) for all X, so u(X) < 10^(100) u($5), hence u is also bounded above.

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