Just read Bostrom’s Pascal’s Mugging; Can’t the problem be solved as follows?
I have a probability estimate E0 in my head for the mugger giving me X (X being a lot) utility if I give them my money. E0 is not a number, as my brain does not seem to work with our traditional floating point numbers. What data structure actually represents E0, is not clear to me, but I can say E0 is a feeling of “empirically next to impossible, game-theoretically inadvisable to act on it being true”. Now, what’s the probability of I getting X utility tomorrow without giving the mugger my money? Let’s call that E1. E1 is “empirically next to impossible.” So giving my money to the mugger does NOT increase my expected utility gain at all! In fact, it decreases it, as I process E0 as a lower probability than E1 (because E0 is game-theoretically negative while E1 is neutral).
Now, you might say this is not solving the problem but bypassing it. I don’t feel this is true. Anyone who has studied numerical computation knows that errors are important and we can never have precise numbers.
The problem is solved if the limit as x approaches infinity, of p(x)*x, where x is the utility the mugger offers, is 0. (This is the case if p(x) ⇐ x^2. If that’s an upper bound, however loose, then the problem is solved.)
Just read Bostrom’s Pascal’s Mugging; Can’t the problem be solved as follows?
I have a probability estimate E0 in my head for the mugger giving me X (X being a lot) utility if I give them my money. E0 is not a number, as my brain does not seem to work with our traditional floating point numbers. What data structure actually represents E0, is not clear to me, but I can say E0 is a feeling of “empirically next to impossible, game-theoretically inadvisable to act on it being true”. Now, what’s the probability of I getting X utility tomorrow without giving the mugger my money? Let’s call that E1. E1 is “empirically next to impossible.” So giving my money to the mugger does NOT increase my expected utility gain at all! In fact, it decreases it, as I process E0 as a lower probability than E1 (because E0 is game-theoretically negative while E1 is neutral).
Now, you might say this is not solving the problem but bypassing it. I don’t feel this is true. Anyone who has studied numerical computation knows that errors are important and we can never have precise numbers.
The problem is solved if the limit as x approaches infinity, of p(x)*x, where x is the utility the mugger offers, is 0. (This is the case if p(x) ⇐ x^2. If that’s an upper bound, however loose, then the problem is solved.)