Yes, 0 is no more a probability than 1 is. You are correct that I do not assign 100% certainty to the idea that 100% certainty is impossible. The proposition is of precisely that form though, that it is impossible—I would expect to find that it was simply not true at all, rather than expect to see it almost always hold true but sometimes break down. In any case, yes, I would be willing to make many such bets. I would happily accept a bet of one penny, right now, against a source of effectively limitless resources, for one example.
As to what probability you assign; I do not find it in the slightest improbable that you claim 100% certainty in full honesty. I do question, though, whether you would make literally any bet offered to you. Would you take the other side of my bet; having limitless resources, or a FAI, or something, would you be willing to bet losing it in exchange for a value roughly equal to that of a penny right now? In fact, you ought to be willing to risk losing it for no gain—you’d be indifferent on the bet, and you get free signaling from it.
Would you take the other side of my bet; having limitless resources, or a FAI, or something, would you be willing to bet losing it in exchange for a value roughly equal to that of a penny right now? In fact, you ought to be willing to risk losing it for no gain—you’d be indifferent on the bet, and you get free signaling from it.
Indeed, I would bet the world (or many worlds) that (A→A) to win a penny, or even to win nothing but reinforced signaling. In fact, refusal to use 1 and 0 as probabilities can lead to being money-pumped (or at least exploited, I may be misusing the term “money-pump”). Let’s say you assign a 1/10^100 probability that your mind has a critical logic error of some sort, causing you to bound probabilities to the range of (1/10^100, 1-1/10^100) (should be brackets but formatting won’t allow it). You can now be pascal’s mugged if the payoff offered is greater than the amount asked for by a factor of at least 10^100. If you claim the probability is less than 10^100 due to a leverage penalty or any other reason, you are admitting that your brain is capable of being more certain than the aforementioned number (and such a scenario can be set up for any such number).
That’s not how decision theory works. The bounds on my probabilities don’t actually apply quite like that. When I’m making a decision, I can usefully talk about the expected utility of taking the bet, under the assumption that I have not made an error, and then multiply that by the odds of me not making an error, adding the final result to the expected utility of taking the bet given that I have made an error. This will give me the correct expected utility for taking the bet, and will not result in me taking stupid bets just because of the chance I’ve made a logic error; after all, given that my entire reasoning is wrong, I shouldn’t expect taking the bet to be any better or worse than not taking it. In shorter terms: EU(action) = EU(action & ¬error) + EU(action & error); also EU(action & error) = EU(anyOtherAction & error), meaning that when I compare any 2 actions I get EU(action) - EU(otherAction) = EU(action & ¬error) - EU(otherAction & ¬error). Even though my probability estimates are affected by the presence of an error factor, my decisions are not. On the surface this seems like an argument that the distinction is somehow trivial or pointless; however, the critical difference comes in the fact that while I cannot predict the nature of such an error ahead of time, I can potentially recover from it iff I assign >0 probability to it occurring. Otherwise I will never ever assign it anything other than 0, no matter how much evidence I see. In the incredibly improbable event that I am wrong, given extraordinary amounts of evidence I can be convinced of that fact. And that will cause all of my other probabilities to update, which will cause my decisions to change.
Your calculations aren’t quite right. You’re treating EU(action) as though it were a probability value (like P(action)). EU(action) would be more logically written E(utility | action), which itself is an integral over utility * P(utility | action) for utility∈(-∞,∞), which, due to linearity of * and integrals, does have all the normal identities, like
In this case, if you do expand that out, using p<<1 for the probability of an error, which is independent of your action, and assuming E(utility|action1,error) = E(utility|action2,error), you get E(utility | action) = E(utility | error) * p + E(utility | action, ¬error) * (1 - p). Or for the difference between two actions, EU1 - EU2 = (EU1' - EU2') * (1 - p) where EU1', EU2' are the expected utilities assuming no errors.
Anyway, this seems like a good model for “there’s a superintelligent demon messing with my head” kind of error scenarios, but not so much for the everyday kind of math errors. For example, if I work out in my head that 51 is a prime number, I would accept an even odds bet on “51 is prime”. But, if I knew I had made an error in the proof somewhere, it would be a better idea not to take the bet, since less than half of numbers near 50 are prime.
Right, I didn’t quite work all the math out precisely, but at least the conclusion was correct. This model is, as you say, exclusively for fatal logic errors; the sorts where the law of non-contradiction doesn’t hold, or something equally unthinkable, such that everything you thought you knew is invalidated. It does not apply in the case of normal math errors for less obvious conclusions (well, it does, but your expected utility given no errors of this class still has to account for errors of other classes, where you can still make other predictions).
In fact, refusal to use 1 and 0 as probabilities can lead to being money-pumped (or at least exploited, I may be misusing the term “money-pump”)
The usage of “money-pump” is correct.
(Do note, however, that using 1 and 0 as probabilities when you in fact do not have that much certainty also implies the possibility for exploitation, and unlike the money pump scenario you do not even have the opportunity to learn from the first exploitation and self correct.)
Yes, 0 is no more a probability than 1 is. You are correct that I do not assign 100% certainty to the idea that 100% certainty is impossible. The proposition is of precisely that form though, that it is impossible—I would expect to find that it was simply not true at all, rather than expect to see it almost always hold true but sometimes break down. In any case, yes, I would be willing to make many such bets. I would happily accept a bet of one penny, right now, against a source of effectively limitless resources, for one example.
As to what probability you assign; I do not find it in the slightest improbable that you claim 100% certainty in full honesty. I do question, though, whether you would make literally any bet offered to you. Would you take the other side of my bet; having limitless resources, or a FAI, or something, would you be willing to bet losing it in exchange for a value roughly equal to that of a penny right now? In fact, you ought to be willing to risk losing it for no gain—you’d be indifferent on the bet, and you get free signaling from it.
Indeed, I would bet the world (or many worlds) that (A→A) to win a penny, or even to win nothing but reinforced signaling. In fact, refusal to use 1 and 0 as probabilities can lead to being money-pumped (or at least exploited, I may be misusing the term “money-pump”). Let’s say you assign a 1/10^100 probability that your mind has a critical logic error of some sort, causing you to bound probabilities to the range of (1/10^100, 1-1/10^100) (should be brackets but formatting won’t allow it). You can now be pascal’s mugged if the payoff offered is greater than the amount asked for by a factor of at least 10^100. If you claim the probability is less than 10^100 due to a leverage penalty or any other reason, you are admitting that your brain is capable of being more certain than the aforementioned number (and such a scenario can be set up for any such number).
That’s not how decision theory works. The bounds on my probabilities don’t actually apply quite like that. When I’m making a decision, I can usefully talk about the expected utility of taking the bet, under the assumption that I have not made an error, and then multiply that by the odds of me not making an error, adding the final result to the expected utility of taking the bet given that I have made an error. This will give me the correct expected utility for taking the bet, and will not result in me taking stupid bets just because of the chance I’ve made a logic error; after all, given that my entire reasoning is wrong, I shouldn’t expect taking the bet to be any better or worse than not taking it. In shorter terms: EU(action) = EU(action & ¬error) + EU(action & error); also EU(action & error) = EU(anyOtherAction & error), meaning that when I compare any 2 actions I get EU(action) - EU(otherAction) = EU(action & ¬error) - EU(otherAction & ¬error). Even though my probability estimates are affected by the presence of an error factor, my decisions are not. On the surface this seems like an argument that the distinction is somehow trivial or pointless; however, the critical difference comes in the fact that while I cannot predict the nature of such an error ahead of time, I can potentially recover from it iff I assign >0 probability to it occurring. Otherwise I will never ever assign it anything other than 0, no matter how much evidence I see. In the incredibly improbable event that I am wrong, given extraordinary amounts of evidence I can be convinced of that fact. And that will cause all of my other probabilities to update, which will cause my decisions to change.
Your calculations aren’t quite right. You’re treating
EU(action)as though it were a probability value (likeP(action)).EU(action)would be more logically writtenE(utility | action), which itself is an integral overutility * P(utility | action)forutility∈(-∞,∞), which, due to linearity of*and integrals, does have all the normal identities, likeE(utility | action) = E(utility | action, e) * P(e | action) + E(utility | action, ¬e) * P(¬e | action).In this case, if you do expand that out, using
p<<1for the probability of an error, which is independent of your action, and assumingE(utility|action1,error) = E(utility|action2,error), you getE(utility | action) = E(utility | error) * p + E(utility | action, ¬error) * (1 - p). Or for the difference between two actions,EU1 - EU2 = (EU1' - EU2') * (1 - p)whereEU1', EU2'are the expected utilities assuming no errors.Anyway, this seems like a good model for “there’s a superintelligent demon messing with my head” kind of error scenarios, but not so much for the everyday kind of math errors. For example, if I work out in my head that 51 is a prime number, I would accept an even odds bet on “51 is prime”. But, if I knew I had made an error in the proof somewhere, it would be a better idea not to take the bet, since less than half of numbers near 50 are prime.
Right, I didn’t quite work all the math out precisely, but at least the conclusion was correct. This model is, as you say, exclusively for fatal logic errors; the sorts where the law of non-contradiction doesn’t hold, or something equally unthinkable, such that everything you thought you knew is invalidated. It does not apply in the case of normal math errors for less obvious conclusions (well, it does, but your expected utility given no errors of this class still has to account for errors of other classes, where you can still make other predictions).
The usage of “money-pump” is correct.
(Do note, however, that using 1 and 0 as probabilities when you in fact do not have that much certainty also implies the possibility for exploitation, and unlike the money pump scenario you do not even have the opportunity to learn from the first exploitation and self correct.)