3 ms·

The classic solution is that $s are not equal to utility. People have some discount rate. But you can trivially rephrase the problem, so the $ increase at the s

The classic solution is that $s are not equal to utility. People have some discount rate. But you can trivially rephrase the problem, so the $ increase at the same rate of your discount rate, and you get the same problem. Only a bounded utility function is a solution - there must be some amount of money where literally even a trillion dollars more doesn't matter. That seems acceptable, but it still means this game is worth some amount to play, and that amount can still grow very large before reaching your bound. Also there must be some things which we can't bound. There is a very related problem called Pascal's Mugging: http://wiki.lesswrong.com/wiki/Pascal's_mugging http://wiki.lesswrong.com/wiki/Pascal's_mugging In Pascal's mugging, a mugger asks you to pay him $5, or he will kill 3↑↑↑3 people (an incomprehensibly huge number, that for all intents in purposes, might as well be infinity.) He says that he is the matrix lord and likes playing games with simulated people. This is of course, incredibly unlikely. But is the probability he is telling the truth greater than 1/3↑↑↑3? Is $5 worth more than a human life? If so you should pay him. This is a general problem with expected utility. EU only cares about the average utility. The utility of all the possible outcomes, weighted by their probability. A single outlier can throw the average case off a lot. EU is forced to trade away utility from the majority of probable outcomes to really weird unlikely outcomes, like the mugger, or winning an infinite series of coin flips. EU is optimal in most everyday problems, but it can fail in extreme cases.

arielby 11y ago
The problem with the original lottery is that most of its value is from high-EV tiny-probability events, e.g. the 2^{-50} probability of winning 2^50 dollars. The practical result of that event does not seem to be worth 2^50 utilons, to say the least. It is hard to think about events worth that many. However, many perturbations of this lottery can actually be good bets. For example, suppose you gain 3^n dollars with probability 2^{-n}. Then you have a 1/128 chance of winning $2187, a 1/256 change of winning $6561, and this game starts looking much nicer. The "Pascal's Mugging" divergence is a different problem, where Solomonoff-style priors imply negative-exponential probabilities of Busy-Beaverish payoffs. Ordinary priors don't really have this problem.
- Houshalter 11y ago
  It seems like the same class of problems, because they are both about high payoff, low probability bets. Solomonoff induction is just a formalization used to show the result is very general. Any reasonable prior should have similar cases. Unless you really believe the mugger being a matrix lord has 0 probability, or that God has 0 probability, etc, you are forced to act as if they are true. Which results in wasted effort in the vast majority of possible outcomes, in exchange for a massive payoff in incredibly rare outcomes. Assigning 0 probability is not something you should do lightly. It would mean you could wake up and find yourself outside of the matrix, and you still would not believe it had any chance of being true. It would mean God himself could come to and say "yeah it's all real." And you would be forced to believe there is still 0 probability he exists.

arielby 11y ago

The problem with the original lottery is that most of its value is from high-EV tiny-probability events, e.g. the 2^{-50} probability of winning 2^50 dollars. The practical result of that event does not seem to be worth 2^50 utilons, to say the least. It is hard to think about events worth that many. However, many perturbations of this lottery can actually be good bets. For example, suppose you gain 3^n dollars with probability 2^{-n}. Then you have a 1/128 chance of winning $2187, a 1/256 change of winning $6561, and this game starts looking much nicer. The "Pascal's Mugging" divergence is a different problem, where Solomonoff-style priors imply negative-exponential probabilities of Busy-Beaverish payoffs. Ordinary priors don't really have this problem.

Houshalter 11y ago
It seems like the same class of problems, because they are both about high payoff, low probability bets. Solomonoff induction is just a formalization used to show the result is very general. Any reasonable prior should have similar cases. Unless you really believe the mugger being a matrix lord has 0 probability, or that God has 0 probability, etc, you are forced to act as if they are true. Which results in wasted effort in the vast majority of possible outcomes, in exchange for a massive payoff in incredibly rare outcomes. Assigning 0 probability is not something you should do lightly. It would mean you could wake up and find yourself outside of the matrix, and you still would not believe it had any chance of being true. It would mean God himself could come to and say "yeah it's all real." And you would be forced to believe there is still 0 probability he exists.