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...
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...
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