You could say, that the strategy “always take the demons bet” has an expected value of 0 QUALYs
This is not true. The expected value of this strategy is undefined. [Edit: commenters reasonably disagree.]
So maybe we want to keep our normal expected-utility-maximizing behavior for some nicely-behaved prospects, where “nicely-behaved” includes conditions like “expected values are defined,” and accept that we might have to throw up our hands otherwise.
That said, I agree that thought experiments like this give us at least some reason to disfavor simple expected-utility-maximizing, but I caution against the jump to expected-utility-maximizing is wrong since (in my opinion) other (non-nihilistic) theories are even worse. It’s easy to criticize a particular theory; if you offered a particular alternative, there would be strong reasons against that theory.
Regardless, note that you can have a bounded utility function and be a longtermist, depending on the specifics of the theory. And it’s prima facie the case that (say) x-risk reduction is very good even if you only care about (say) the next billion years.
If you follow the strategy, won’t you eventually get three sixes? Which means that any run through this procedure will end up with a value of zero, no?
Does the undefined come from the chance that you will never get three sixes, combined with the absolutely enormous utility of that extremely improbable eventuality?
Medium answer: observe that the value after winning n bets is at least 10×2^n, and the limit of (215/216)^n × 10×2^n as n→∞ is undefined (infinite). Or let ω be an infinite number and observe that the infinitesimal probability (215/216)^ω times the infinite value 10×2^ω is 10×(2×215/216)^ω which is infinite. (Note also that the EVs of the strategy “bet with probability p” goes to ∞ as p→1.)
Edit: hmm, in response to comments, I’d rephrase as follows.
Yes, the “always bet” strategy has value 0 with probability 1. But if a random variable is 0 with probability measure 1 and is undefined with probability measure 0, we can’t just say it’s identical to the zero random variable or that it has expected value zero (I think, happy to be corrected with a link to a math source). And while ‘infinite bets’ doesn’t really make sense, if we have to think of ‘the state after infinite bets’ I think we could describe its value with the aforementioned random variable.
While you will never actually create a world with this strategy, I don’t think the expected value is defined because ‘after infinite bets’ you could still be talking to the demon (with probability 0, but still possibly, and talking-with-the-demon-after-winning-infinitely-many-bets seems significant even with probability 0).
But if a random variable is 0 with probability measure 1 and is undefined with probability measure 0, we can’t just say it’s identical to the zero random variable or that it has expected value zero (I think, happy to be corrected with a link to a math source).
The definition of expected value is ∫xP[x]. If the set of discontinuities of a function has measure zero, then it is still Riemann integrable. So the integral exists despite not being identical to the zero random variable, and the value is zero. In the general case you have to use measure theory, but I don’t think it’s needed here.
Also, there’s no reason our intuitions about the goodness of the infinite sequence of bets has to match the expected value.
Saying that the expected value of this strategy is undefined seems like underselling it. The expected value is positive infinity since the cumulative reward is increasing strictly faster than the cumulative probability of getting nothing.
This might be a disagreement about whether or not it’s appropriate to use “infinity” as a number (i.e. a value). Mathematically, if a function approaches infinity as the input approaches infinity, I think typically you’re supposed say the limit is “undefined”, as opposed to saying the limit is “infinity”. So whether this is (a) underselling it or (b) just writing accurately depends on the audience.
I agree with you that the limit of the EV of “bet until you win n times” is infinite as n→∞. But I agree with Guy Raveh that we probably can’t just take this limit and call it the EV of “always bet.” Maybe it depends on what precise question we’re asking...
It looks to me like there’s some confusion in the other comments regarding this. The expected value is, in fact, defined, and it is zero. The problem is that if you look at a sequence of n bets and take n to infinity, that expected value does go to positive infinity. So thinking in terms of adding one bet each time is actually deceiving.
In general, a sequence of pointwise converging random variables does not converge in expected value to the expected value of the limit variable. That requires uniform convergence.
Infinities sometimes break our intuitions. Luckily, our lives and the universe’s “life” are both finite.
The expected value is, in fact, defined, and it is zero.
Is the random variable you’re thinking of, whose expectation is zero, just the random variable that’s uniformly zero? That doesn’t seem to me to be the right way to describe the “bet” strategy; I would prefer to say the random variable is undefined. (But calling it zero certainly doesn’t seem to be a crazy convention.)
It’s zero on the event “three sixes are rolled at some point” and infinity on the event that they’re never rolled. The probability of that second event is zero, though. So the expected value is zero.
Wouldn’t the EV be 0? My reasoning: - The condition for the universe to be created is when you stop betting. - Under this strategy, the only way where you stop betting is when you get three sixes.
So there is no plausible scenario where any amount of QALY is being created. Am I missing something?
Well, what value do we assign a scenario in which you’re still talking with the demon? If those get value 0, then sure, the EV is 0. But calling those scenarios 0 seems problematic, I think.
Some academic papers including A paradox for tiny probabilities and enormous values would be of interest to you.
This is not true. The expected value of this strategy is undefined. [Edit: commenters reasonably disagree.]
So maybe we want to keep our normal expected-utility-maximizing behavior for some nicely-behaved prospects, where “nicely-behaved” includes conditions like “expected values are defined,” and accept that we might have to throw up our hands otherwise.
That said, I agree that thought experiments like this give us at least some reason to disfavor simple expected-utility-maximizing, but I caution against the jump to expected-utility-maximizing is wrong since (in my opinion) other (non-nihilistic) theories are even worse. It’s easy to criticize a particular theory; if you offered a particular alternative, there would be strong reasons against that theory.
Regardless, note that you can have a bounded utility function and be a longtermist, depending on the specifics of the theory. And it’s prima facie the case that (say) x-risk reduction is very good even if you only care about (say) the next billion years.
(QUALY should be QALY)
If you follow the strategy, won’t you eventually get three sixes? Which means that any run through this procedure will end up with a value of zero, no?
Does the undefined come from the chance that you will never get three sixes, combined with the absolutely enormous utility of that extremely improbable eventuality?
Short answer: yes.
Medium answer: observe that the value after winning n bets is at least 10×2^n, and the limit of (215/216)^n × 10×2^n as n→∞ is undefined (infinite). Or let ω be an infinite number and observe that the infinitesimal probability (215/216)^ω times the infinite value 10×2^ω is 10×(2×215/216)^ω which is infinite. (Note also that the EVs of the strategy “bet with probability p” goes to ∞ as p→1.)
Edit: hmm, in response to comments, I’d rephrase as follows.
Yes, the “always bet” strategy has value 0 with probability 1. But if a random variable is 0 with probability measure 1 and is undefined with probability measure 0, we can’t just say it’s identical to the zero random variable or that it has expected value zero (I think, happy to be corrected with a link to a math source). And while ‘infinite bets’ doesn’t really make sense, if we have to think of ‘the state after infinite bets’ I think we could describe its value with the aforementioned random variable.
While you will never actually create a world with this strategy, I don’t think the expected value is defined because ‘after infinite bets’ you could still be talking to the demon (with probability 0, but still possibly, and talking-with-the-demon-after-winning-infinitely-many-bets seems significant even with probability 0).
The definition of expected value is ∫xP[x]. If the set of discontinuities of a function has measure zero, then it is still Riemann integrable. So the integral exists despite not being identical to the zero random variable, and the value is zero. In the general case you have to use measure theory, but I don’t think it’s needed here.
Also, there’s no reason our intuitions about the goodness of the infinite sequence of bets has to match the expected value.
Saying that the expected value of this strategy is undefined seems like underselling it. The expected value is positive infinity since the cumulative reward is increasing strictly faster than the cumulative probability of getting nothing.
This might be a disagreement about whether or not it’s appropriate to use “infinity” as a number (i.e. a value). Mathematically, if a function approaches infinity as the input approaches infinity, I think typically you’re supposed say the limit is “undefined”, as opposed to saying the limit is “infinity”. So whether this is (a) underselling it or (b) just writing accurately depends on the audience.
I agree with you that the limit of the EV of “bet until you win n times” is infinite as n→∞. But I agree with Guy Raveh that we probably can’t just take this limit and call it the EV of “always bet.” Maybe it depends on what precise question we’re asking...
(parent comment edited)
It looks to me like there’s some confusion in the other comments regarding this. The expected value is, in fact, defined, and it is zero. The problem is that if you look at a sequence of n bets and take n to infinity, that expected value does go to positive infinity. So thinking in terms of adding one bet each time is actually deceiving.
In general, a sequence of pointwise converging random variables does not converge in expected value to the expected value of the limit variable. That requires uniform convergence.
Infinities sometimes break our intuitions. Luckily, our lives and the universe’s “life” are both finite.
Is the random variable you’re thinking of, whose expectation is zero, just the random variable that’s uniformly zero? That doesn’t seem to me to be the right way to describe the “bet” strategy; I would prefer to say the random variable is undefined. (But calling it zero certainly doesn’t seem to be a crazy convention.)
It’s zero on the event “three sixes are rolled at some point” and infinity on the event that they’re never rolled. The probability of that second event is zero, though. So the expected value is zero.
Wouldn’t the EV be 0?
My reasoning:
- The condition for the universe to be created is when you stop betting.
- Under this strategy, the only way where you stop betting is when you get three sixes.
So there is no plausible scenario where any amount of QALY is being created. Am I missing something?
Well, what value do we assign a scenario in which you’re still talking with the demon? If those get value 0, then sure, the EV is 0. But calling those scenarios 0 seems problematic, I think.