Strictly speaking, if you somehow knew in advance (time travel?) that you would “die in a plane crash”, then avoiding flying would indeed, presumably, result in a plane crash occurring as you walk down the street.
If you know your attempt will fail in advance, you don’t need to try very hard. If you don’t, then it is reasonable to avoid dangerous situations.
If you know your attempt will fail in advance, you don’t need to try very hard.
I actually don’t believe this is true, for most mechanisms of “mysterious future knowledge”, including most (philosophical) forms of time travel that don’t allow change. Unless I had some specific details about the mechanism of prediction that changed the situation I would go ahead and try very hard despite knowing it is futile. I know this is a total waste… it’s as if I am just leaving $10,000 on the ground or something! (ie. I assert that newcomblike reasoning applies.)
In Newcomb’s problem, Omega knows what you will do using their superintelligence. Since you know you cannot two-box successfully, you should one-box.
In this case you are trying (futilely) so that you, very crudely speaking, are less likely to be in the futile situation in the first place.
If Omega didn’t know what you would do with a fair degree of accuracy, two-boxing would work, obviously.
Yes, then it wouldn’t be Newcomb’s Problem. The important feature in the problem isn’t boxes with arbitrary amounts of money in them. It is about interacting with a powerful predictor whose prediction has already been made and acted upon. See, in particular, the Transparent Newcomb’s Problem (where you can ourtright see how much money is there). That makes the situation seem even more like this one.
Even closer would be the Transparent Newcomb’s Problem combined with an Omega that is only 99% accurate. You find yourself looking at an empty ‘big’ box. What do you do? I’m saying you still one box the empty box. That makes it far less likely that you will be in a situation where you see an empty box at all.
Being a person who avoids plane crashes makes it less likely that you will be told “you will die in a plane crash”, yes.
But probability is subjective—once you have the information that you will die in a car crash, your subjective estimate of this should vastly increase, regardless of the precautions you take.
But probability is subjective—once you have the information that you will die in a car crash, your subjective estimate of this should vastly increase, regardless of the precautions you take.
Absolutely. And I’m saying that you update that probability, perform a (naive) expected utility function calculation that says “don’t bother trying to prevent plane crashes” then go ahead and try to avoid plane crashes anyway. Because in this kind of situation maximising expected utility is actually a mistake.
(To those who consider this claim to be bizarre without seeing context, note that we are talking situations such as within time-loops.)
In the specific “infallible oracle says you’re going to die in a plane crash” scenario, you might live considerably longer by giving the cosmos fewer opportunities to throw plane crashes at you.
I was assuming a time was given. wedrifid was claiming that you should avoid plane-crash causing actions even if you know that the crash will occur regardless.
Yes, you are correct. Or at least it is true that I am not trying to make a “manipulate time of death” point. Let’s say we have been given a reliably predicted and literal “half life” that we know has already incorporated all our future actions.
So the odds of my receiving that message are the same as the odds of my death by plane, but having recieved it I can freely act to increase the odds of my plane-related death without repercussions. I think.
If you know the time, then that becomes even easier to deal with—there’s no particular need to avoid plane crash opportunities that do not take place at that time. In fact, it then becomes possible to try to avoid it by other means—for example, faking your own plane-crash-related demise and leaving the fake evidence there for the time traveller to find.
If you know the time of your death in advance, then the means become important only at or near that time.
Let’s take this a step further. (And for this reply I will neglect all acausal timey-wimey manipulation considerat ions.)
If you know the time of your death you have the chance to exploit your temporary immortality. Play Russian Roulette for cash. Contrive extreme scenarios that will either result in significant gain or certain death. The details of ensuring that it is hard to be seriously injured without outright death will take some arranging but there is a powerful “fixed point in time and space” to be exploited.
Problem with playing russian roulette under those circumstances is that you might suffer debilitating but technically nonfatal brain damage. It’s actually surprisingly difficult to arrange situations where there’s a chance of death but no chance of horrific incapacitation.
Problem with playing russian roulette under those circumstances is that you might suffer debilitating but technically nonfatal brain damage.
Yes, that was acknowledged as the limiting factor. However it is not a significant problem when playing a few of rounds of Russian Roulette. In fact, even assuming you play the Roulette to the death with two different people in sequence you still only create two bits of selection pressure towards the incapacitation. You can more than offset this comparatively trivial amount of increased injured-not-dead risk (relative to the average Russian Roulette player) by buying hollow point rounds for the gun and researching optimal form for suicide-by-handgun.
The point is, yes exploiting death-immunity for optimization other outcomes increases the risk of injury in the same proportion that the probability of the desired outcome is increased but this doesn’t become a significant factor for something as trivial as a moderate foray into Russian Roulette. It would definitely become a factor if you started trying to brute force 512 bit encryption with a death machine. That is, you would still end up with practically 0 chance of brute forcing the encryption and your expected outcome would come down to whether it is more likely for the machine to not work at all or for it to merely incapacitate you.
This is a situation where you really do have to shut up and multiply. If you try to push the anti-death too far you will just end up with (otherwise) low probability likely undesirable outcomes occurring. On the other hand if you completely ignore the influence of “death” outcomes being magically redacted from the set of possible outcomes you will definitely make incorrect expected utility calculations when deciding what is best to do. This is particularly the case given that there is a strict upper bound on how bad a “horrific incapacitation” can be. ie. It could hurt a bit for a few hours till your certain death.
This scenario is very different and far safer than many other “exploit the impossible physics” scenarios in as much as the possibility of bringing disaster upon yourself and others is comparatively low. (ie. In other scenarios it is comparatively simple/probable for the universe to just to throw a metaphorical meteor at you and kill everyone nearby as a way to stop your poorly calibrated munchkinism.)
It’s actually surprisingly difficult to arrange situations where there’s a chance of death but no chance of horrific incapacitation.
I shall assume you mean “sufficiently low chance of horrific incapacitation for your purposes”.
It isn’t especially difficult for the kind of person who can get time travelling prophets to give him advice to also have a collaborator with a gun.
Call P1 the probability that someone who plays Russian Roulette in a submarine will survive and suffer a debilitating injury. P1 is, I agree, negligible. Call P2 the probability that someone who plays Russian Roulette in a submarine and survives will suffer a debilitating injury. P2 is, it seems clear, significantly larger than P1.
What you seem to be saying is that if I know with certainty (somehow or other) that I will die in an airplane, then I can safely play Russian Roulette in a submarine, because there’s (we posit for simplicity) only two things to worry about: death (which now has P=0) or non-fatal debilitating injury (which has always had P=P1, which is negligible for my purposes).
But I’m not quite clear on why, once I become certain I won’t die, my probability of a non-fatal debilitating injury doesn’t immediately become P2.
The probability does become P2, but in many cases, we can argue that P2 is negligible as well.
In the submarine case, things are weird, because your chances of dying are quite high even if you win at Russian Roulette. So let’s consider the plain Russian Roulette: you and another person take turns trying to shoot yourself until one of you succeeds. For simplicity’s sake, suppose that each of you is equally likely to win.
Then P1 = Pr[injured & survive] and P2 = Pr[injured | survive] = P1 / Pr[survive]. But Pr[survive] is always at least 1/2: if your opponent shoots himself before you do, then you definitely survive. Therefore P2 is at most twice P1, and is negligible whenever P1 is negligible.
Then P1 = Pr[injured & survive] and P2 = Pr[injured | survive] = P1 / Pr[survive]. But Pr[survive] is always at least 1/2: if your opponent shoots himself before you do, then you definitely survive. Therefore P2 is at most twice P1, and is negligible whenever P1 is negligible.
I see how you calculated that, but I think you’re looking at the wrong pieces of evidence, and I agree with TheOtherDave.
You have an even split chance of getting the real bullet in play, so let’s put that down:
P[bullet] = 0.5
P[¬bullet] = 0.5
Then, given that you DO get the bullet, you have a very high chance of being dead if you don’t know how you will die:
P[die | bullet] = 0.99
P[¬die | bullet] = 0.01
Of course, this means that overall, P[die] = 0.495, and P[injury] = 0.005. However, if you already also know that P[¬die] = 1, then...
P[bullet & ¬die] = 0.5
P[¬bullet & ¬die] = 0.5
...because P[bullet] is computed before its causal effects (death or injury) can enter the picture, which means you’re left with P[injury] = 0.5 (a hundred times larger than P1!).
Thus, while the first chance of injury is negligible, the chance of injury once you already know that you won’t die is massively larger, given that P[injury XOR death | bullet] = 1 (which is implied in the problem statement, I would assume).
Edit: I realize that this makes the assumption that your chances of getting the bullet doesn’t correlate with knowing how you will die, but it most clearly exposes the difference between your calculation and other possible calculations. This is not the correct way to calculate the probabilities in real life, since it’s much more likely that non-death is achieved by not having the bullet in the first place (or by failing to play Russian Roulette at all), but there’s all kinds of parameters you can play with here. All I’m saying is that P2 isn’t necessarily at most twice P1, it all depends on the other implicit conditions and priors.
The chances of injury are then not quite double what they would have been without death-immunity. This is reasonably low, because the prior odds of survival at all are reasonably high (0.505) - had the experiment been riskier, such that there was only a 0.01% chance of survival overall, then the chance of injury in the death-immunity case would be correspondingly higher.
(We also have not yet taken into account the effect of the first player—in such a game of Russian Roulette, he who shoots first has a higher prior probability of death).
I agree that this is how it should be reasoned ideally, which I only realized after first posting the grandparent. See other comments and the edit for how I arrived at the 50⁄50 reasoning. If you know the answer for the bottom/last question in this comment, I’d be interested to know.
It depends on how exactly this time-travel-related knowledge of how you die will work. My calculation is correct if a random self-consistent time loop is chosen (which I think is reasonable) -- there are far more self-consistent time loops in which you survive because you don’t get the bullet, than ones in which you survive because a bullet failed to kill you.
Terrible things start happening if there’s some sort of “lazy pruning” of possibilities (which I think is what you’re suggesting). Then the probability you get shot is 0.5, and then if you do get shot the self-consistency condition eliminates the branches in which you die, so you are nearly guaranteed to survive in some horrible fashion.
I don’t like the second option because it requires thinking of branching possibilities as some sort of actual discrete things, which I find dubious. But it’s a bit silly to argue about interpretations of time travel, isn’t it?
This is a bit of what I was getting at with the edit in the grandparent: basically, it’s not very bayesian to stick to a 50⁄50 bullet chance when you know you will not die.
However, I was also considering the least convenient possible world: You already know that you won’t die, and since the worst you have to fear is debilitating permanent non-fatal injury (which you apparently don’t care about if it’s for Science!), you decide to repeatedly play Russian Roulette with tons of people, just to test probabilities or for fun or something.
Then what happens? Does it become more probable that you’ll just randomly end up always not getting the bullet with .98 probability (if the chance of surviving a bullet was 1%), which will to an outside view apparently defy all probabilities? Or does it instead stick to the 50⁄50 and on average you’ll get injured every second game (assuming you have a way of recovering from the injuries and playing again) without ever dying?
More importantly, which scenario should an ideal bayesian agent expect? This I have no idea, which is why I think it’s not trivial or obvious that P2 = 2*P1.
The calculation that P2=2*P1 obviously only applies to one game. If you play lots of games sequentially, then the probability increases stack. (Edit: I incorrectly said that the ratio doubles with every game, which is obviously false)
Another way of thinking about this: absent time travel, if you survive a bullet with 1% probability, then after N games your probability of surviving unscathed is 1/2^N, and your total probability of surviving is (101/200)^N. Therefore, given that you survive, your probability of surviving unscathed should be the ratio of these, or (100/101)^N.
(All of this is assuming the random self-consistent time loop interpretation of time travel.)
Hmm, interesting. That would imply that, to a third-party, there’s some random guy who wins 99% of the time at Russian Roulette. At this point, it should legally be considered murder.
Death sentence by plane crash sounds appropriate, in this case.
It is murder, but you’re going to have terrible trouble proving that (especially if he’s careful about documenting how fair the russian roulette is). To avoid murder charges, the hypothetical psychopathic death-immune person can go so far as to arrange a tournament, with 2^n entries for some integer n. In this arrangement, one person must survive every round, and thus it does not look suspicious afterwards that he did survive every round (plus he gets 2^n prizes for going through n rounds).
I’m pretty sure that the russian roulette itself is illegal just about everywhere, though. No matter how it’s done, it’s either murder or assisted suicide.
I’m pretty sure that the russian roulette itself is illegal just about everywhere, though. No matter how it’s done, it’s either murder or assisted suicide.
I would have expected a different label to apply. Neither of those seems accurate. In fact I didn’t think even assisted suicide got called “assisted suicide”.
Call P2 the probability that someone who plays Russian Roulette in a submarine and survives will suffer a debilitating injury. P2 is, it seems clear, significantly larger than P1.
To be precise (and absent any other interventions) P2 is larger than P1 by a factor of 2 (in two person to the death with randomized start case).
But I’m not quite clear on why, once I become certain I won’t die, my probability of a non-fatal debilitating injury doesn’t immediately become P2.
I thought I was fairly clear that that was exactly what I was arguing. Including the implication that it doesn’t immediately become more than P2. This (and other previously unlikely failure modes) have to be considered seriously but to overweight their importance is a mistake.
Ah! I see. Yeah, it seems I wasn’t thinking about Actual Russian Roulette, in which two players take turns and the most likely route to survival is my opponent blowing his brains out first, but rather Utterly Ridiculous Hypothetical Variation on Russian Roulette, in which I simply pull the trigger over and over while pointing at my own head, and the most likely route to survival is a nonfatal bullet wound.
Ah! I see. Yeah, it seems I wasn’t thinking about Actual Russian Roulette, in which two players take turns and the most likely route to survival is my opponent blowing his brains out first, but rather Utterly Ridiculous Hypothetical Variation on Russian Roulette, in which I simply pull the trigger over and over while pointing at my own head, and the most likely route to survival is a nonfatal bullet wound.
Ahh, yes. That seems to be an impractical course of action even with faux-immortality. It may be worthwhile if some strange person was willing to pay exorbitant amounts of cash per shot and you were also permitted to spin the cartridge after every shot (or five shots) in order to randomize it. Then the accumulated improbability (ie. magnified injury and ‘unknown black swan’ possibility) would ultimately injure or otherwise interrupt you but only after your legacy had been improved significantly.
It does become P2. But… aside from the submarine issue… if you play Russian Roulette against one other person, then with probability 1⁄2, the other person gets shot before you do. Assuming that the loser either dies or is incapacitated, we can write P1 as Pr[survive|shot] and P2 as Pr[shot|survive], and by a simple application of Bayes P2 is at most twice P1.
If you’re the sort of person who would take advantage of such knowledge to engage in dangerous activities, does that increase the probability that your reported time of death will be really soon?
On the other hand, if you’re the kind of person who (on discovering that you will die in a plane crash) takes care to avoid plane crashes, wouldn’t that increase your expected life span?
Moreover, these two attitudes—avoiding plane crashes and engaging in non-plane-related risky activities—are not mutually exclusive.
If you’re the sort of person who would take advantage of such knowledge to engage in dangerous activities, does that increase the probability that your reported time of death will be really soon?
Absolutely. Note the parenthetical. The grandparent adopted the policy of ignoring this kind of consideration for the purpose of exploring the implied tangent a little further. I actually think not actively avoiding death, particularly death by the means predicted, is a mistake.
You can do the same thing if you know only the means of your death and not the time in advance; merely set up your death-stunts to avoid that means of death. (For example, if you know with certainty that you will die in a plane crash but not when, you can play Russian Roulette for cash on a submarine).
An important safety tip becomes clear; if you’re involved in a time loop and know the means of your death, then keep a very close eye on scientific literature and engineering projects. Make sure that you hear the rumours of the aqua-plane before it is built and can thus plan accordingly...
So … I should do things that result in less expected utility … why?
I am happy to continue the conversation if you are interested. I am trying to unpack just where your intuitions diverge from mine. I’d like to know what your choice would be when faced with Newcomb’s Problem with transparent boxes and an imperfect predictor when you notice that the large box is empty. I take the empty large box, which isn’t a choice that maximises my expected utility and in fact gives me nothing, which is the worst possible outcome from that game. What do you do?
Two boxes, sitting there on the ground, unguarded, no traps, nobody else has a legal claim to the contents? Seriously? You can have the empty one if you’d like, I’ll take the one with the money. If you ask nicely I might even give you half.
I don’t understand what you’re gaining from this “rationality” that won’t let you accept a free lunch when an insane godlike being drops it in your lap.
I don’t understand what you’re gaining from this “rationality” that won’t let you accept a free lunch when an insane godlike being drops it in your lap.
No, you’re not. You’re getting an empty box, and hoping that by doing so you’ll convince Omega to put a million dollars in the next box, or in a box presented to you in some alternate universe.
And by this exact reasoning, which Omega has successfully predicted, you will one-box, and thus Omega has successfully predicted that you will one-box and made the correct decision to leave the box empty.
Remember to trace your causal arrows both ways if you want a winning CDT.
Remember also Omega is a superintelligence. The recursive prediction is exactly why it’s rational to “irrationally” one-box.
And by this exact reasoning, which Omega has successfully predicted, you will one-box, and thus Omega has successfully predicted that you will one-box and made the correct decision to leave the box empty.
Yes, that’s why I took the one box with more money in it.
Strictly speaking the scenario being discussed is one in which Omega left a transparent box of money and another transparent box which was empty in front of Wedrifid, then I came by, confirmed Wedrifid’s disinterest in the money, and left the scene marginally richer. I personally have never been offered money by Omega, don’t expect to be any time soon, and am comfortable with the possibility of not being able to outwit something that’s defined as being vastly smarter than me.
Remember also Omega is an insane superintelligence, with unlimited resources but no clear agenda beyond boredom. If appeasing such an entity was my best prospect for survival, I would develop whatever specialized cognitive structures were necessary; it’s not, so I don’t, and consider myself lucky.
Ah, then in that case, you win. With that scenario there’s really nothing you could do better than what you propose. I was under the impression you were discussing a standard transparent Newcomb.
The counterfactual mugging isn’t that strange if you think of it as a form of entrance fee for a positive-expected-utility bet—a bet you happened to lose in this instance, but it is good to have the decision theory that will allow you to enter it in the abstract.
The problem is that people aren’t that good in understanding that your specific decision isn’t separate from your decision theory under a specific context … DecisionTheory(Context)=Decision. To have your decision theory be a winning decision theory in general, you may have to eventually accept some individual ‘losing’ decisions: That’s the price to pay for having a winning decision theory overall.
If Parfit’s hitchhiker “updates” on the fact that he’s now reached the city and therefore doesn’t need to pay the driver, and furthermore if Parfit’s hitchhiker knows in advance that he’ll update on that fact in that manner, then he’ll die.
If right now we had mindscanners/simulators that could perform such counterfactual experiments on our minds, and if this sort of bet could therefore become part of everyday existence, being the sort of person that pays the counterfactual mugger would eventually be seen by all to be of positive-utility—because such people would eventually be offered the winning side of that bet (free money in the tenfold of your cost).
While the sort of person that wouldn’t be paying the counterfactual mugger would never be given such free money at all.
The likelihood of encountering the winning side of the bet is proportional to the likelihood of encountering its losing side. As such, whether you are likely to encounter the bet once in your lifetime, or to encounter it a hundred times, doesn’t seem to significantly affect the decision theory you ought possess in advance if you want to maximize your utility.
In addition to Omega asking you to give him 100$ because the coin came up tails, also imagine Omega coming to you and saying “Here’s 100,000$, because the coin came up heads and you’re the type of person that would have given me 100$ if it had come up tails.”
That scenario makes it obvious to me that being the person that would give Omega 100$ if it had come up heads is the winning type of person...
If the coin therein is defined as a quantum one then yes, without hesitation. If it is a logical coin then things get complicated.
All is explained.
This is more ambiguous than you realize. Sure, the dismissive part came through but it doesn’t quite give your answer. ie. Not all people would give the same response to counterfactual mugging as Transparent Probabilistic Newcomb’s and you may notice that even I had to provide multiple caveats to provide my own answer there despite for most part making the same kind of decision.
Let’s just assume your answer is “Two Box!”. In that case I wonder whether the problem is that you just outright two box on pure Newcomb’s Problem or whether you revert to CDT intuitions when the details get complicated. Assuming you win at Newcomb’s Problem but two box on the variant then I suppose that would indicate the problem is in one of:
Being able to see the money rather than being merely being aware of it through abstract thought switched you into a CDT based ‘near mode’ thought pattern.
You want to be the kind of person who two-boxes when unlucky even though this means that you may actually not have been unlucky at all but instead have manufactured your own undesirable circumstance. (Even more people stumble here, assuming they get this far.)
The most generous assumption would be that your problem comes at the final option—that one is actually damn confusing. However I note that your previous comments about always updating on the free money available and then following expected utility maximisation are only really compatible with the option “outright two box on simple Newcomb’s Problem”. In that case all the extra discussion here is kind of redundant!
I think we need a nice simple visual taxonomy of where people fall regarding decision theoretic bullet-biting. It would save so much time when this kind of thing. Then when a new situation comes up (like this one with dealing with time traveling prophets) we could skip straight to, for example, “Oh, you’re a Newcomb’s One-Boxer but a Transparent Two-Boxer. To be consistent with that kind of implied decision algorithm then yes, you would not bother with flight-risk avoidance.”
Not if you mistakenly believe, as CDTers do, in human free will in a predictable (by Omega) universe.
“Free will” isn’t incompatible with a predictable (by Omega) universe. I also doubt that all CDTers believe the same thing about human free will in said universe.
I think this is the kind of causal loop he has in mind. But a key feature of the hypothesis is that you can’t predict what’s meant to happen. In that case, he’s equally good at predicting any outcome, so it’s a perfectly uninformative hypothesis.
Strictly speaking, if you somehow knew in advance (time travel?) that you would “die in a plane crash”, then avoiding flying would indeed, presumably, result in a plane crash occurring as you walk down the street.
If you know your attempt will fail in advance, you don’t need to try very hard. If you don’t, then it is reasonable to avoid dangerous situations.
I actually don’t believe this is true, for most mechanisms of “mysterious future knowledge”, including most (philosophical) forms of time travel that don’t allow change. Unless I had some specific details about the mechanism of prediction that changed the situation I would go ahead and try very hard despite knowing it is futile. I know this is a total waste… it’s as if I am just leaving $10,000 on the ground or something! (ie. I assert that newcomblike reasoning applies.)
I don’t understand this.
In Newcomb’s problem, Omega knows what you will do using their superintelligence. Since you know you cannot two-box successfully, you should one-box.
If Omega didn’t know what you would do with a fair degree of accuracy, two-boxing would work, obviously.
In this case you are trying (futilely) so that you, very crudely speaking, are less likely to be in the futile situation in the first place.
Yes, then it wouldn’t be Newcomb’s Problem. The important feature in the problem isn’t boxes with arbitrary amounts of money in them. It is about interacting with a powerful predictor whose prediction has already been made and acted upon. See, in particular, the Transparent Newcomb’s Problem (where you can ourtright see how much money is there). That makes the situation seem even more like this one.
Even closer would be the Transparent Newcomb’s Problem combined with an Omega that is only 99% accurate. You find yourself looking at an empty ‘big’ box. What do you do? I’m saying you still one box the empty box. That makes it far less likely that you will be in a situation where you see an empty box at all.
Being a person who avoids plane crashes makes it less likely that you will be told “you will die in a plane crash”, yes.
But probability is subjective—once you have the information that you will die in a car crash, your subjective estimate of this should vastly increase, regardless of the precautions you take.
Absolutely. And I’m saying that you update that probability, perform a (naive) expected utility function calculation that says “don’t bother trying to prevent plane crashes” then go ahead and try to avoid plane crashes anyway. Because in this kind of situation maximising expected utility is actually a mistake.
(To those who consider this claim to be bizarre without seeing context, note that we are talking situations such as within time-loops.)
So … I should do things that result in less expected utility … why?
In the specific “infallible oracle says you’re going to die in a plane crash” scenario, you might live considerably longer by giving the cosmos fewer opportunities to throw plane crashes at you.
I was assuming a time was given. wedrifid was claiming that you should avoid plane-crash causing actions even if you know that the crash will occur regardless.
Yes, you are correct. Or at least it is true that I am not trying to make a “manipulate time of death” point. Let’s say we have been given a reliably predicted and literal “half life” that we know has already incorporated all our future actions.
OK.
So the odds of my receiving that message are the same as the odds of my death by plane, but having recieved it I can freely act to increase the odds of my plane-related death without repercussions. I think.
If you know the time, then that becomes even easier to deal with—there’s no particular need to avoid plane crash opportunities that do not take place at that time. In fact, it then becomes possible to try to avoid it by other means—for example, faking your own plane-crash-related demise and leaving the fake evidence there for the time traveller to find.
If you know the time of your death in advance, then the means become important only at or near that time.
Let’s take this a step further. (And for this reply I will neglect all acausal timey-wimey manipulation considerat ions.)
If you know the time of your death you have the chance to exploit your temporary immortality. Play Russian Roulette for cash. Contrive extreme scenarios that will either result in significant gain or certain death. The details of ensuring that it is hard to be seriously injured without outright death will take some arranging but there is a powerful “fixed point in time and space” to be exploited.
Problem with playing russian roulette under those circumstances is that you might suffer debilitating but technically nonfatal brain damage. It’s actually surprisingly difficult to arrange situations where there’s a chance of death but no chance of horrific incapacitation.
Yes, that was acknowledged as the limiting factor. However it is not a significant problem when playing a few of rounds of Russian Roulette. In fact, even assuming you play the Roulette to the death with two different people in sequence you still only create two bits of selection pressure towards the incapacitation. You can more than offset this comparatively trivial amount of increased injured-not-dead risk (relative to the average Russian Roulette player) by buying hollow point rounds for the gun and researching optimal form for suicide-by-handgun.
The point is, yes exploiting death-immunity for optimization other outcomes increases the risk of injury in the same proportion that the probability of the desired outcome is increased but this doesn’t become a significant factor for something as trivial as a moderate foray into Russian Roulette. It would definitely become a factor if you started trying to brute force 512 bit encryption with a death machine. That is, you would still end up with practically 0 chance of brute forcing the encryption and your expected outcome would come down to whether it is more likely for the machine to not work at all or for it to merely incapacitate you.
This is a situation where you really do have to shut up and multiply. If you try to push the anti-death too far you will just end up with (otherwise) low probability likely undesirable outcomes occurring. On the other hand if you completely ignore the influence of “death” outcomes being magically redacted from the set of possible outcomes you will definitely make incorrect expected utility calculations when deciding what is best to do. This is particularly the case given that there is a strict upper bound on how bad a “horrific incapacitation” can be. ie. It could hurt a bit for a few hours till your certain death.
This scenario is very different and far safer than many other “exploit the impossible physics” scenarios in as much as the possibility of bringing disaster upon yourself and others is comparatively low. (ie. In other scenarios it is comparatively simple/probable for the universe to just to throw a metaphorical meteor at you and kill everyone nearby as a way to stop your poorly calibrated munchkinism.)
I shall assume you mean “sufficiently low chance of horrific incapacitation for your purposes”.
It isn’t especially difficult for the kind of person who can get time travelling prophets to give him advice to also have a collaborator with a gun.
I have to admit, you’ve sort of lost me here.
Call P1 the probability that someone who plays Russian Roulette in a submarine will survive and suffer a debilitating injury. P1 is, I agree, negligible.
Call P2 the probability that someone who plays Russian Roulette in a submarine and survives will suffer a debilitating injury. P2 is, it seems clear, significantly larger than P1.
What you seem to be saying is that if I know with certainty (somehow or other) that I will die in an airplane, then I can safely play Russian Roulette in a submarine, because there’s (we posit for simplicity) only two things to worry about: death (which now has P=0) or non-fatal debilitating injury (which has always had P=P1, which is negligible for my purposes).
But I’m not quite clear on why, once I become certain I won’t die, my probability of a non-fatal debilitating injury doesn’t immediately become P2.
The probability does become P2, but in many cases, we can argue that P2 is negligible as well.
In the submarine case, things are weird, because your chances of dying are quite high even if you win at Russian Roulette. So let’s consider the plain Russian Roulette: you and another person take turns trying to shoot yourself until one of you succeeds. For simplicity’s sake, suppose that each of you is equally likely to win.
Then P1 = Pr[injured & survive] and P2 = Pr[injured | survive] = P1 / Pr[survive]. But Pr[survive] is always at least 1/2: if your opponent shoots himself before you do, then you definitely survive. Therefore P2 is at most twice P1, and is negligible whenever P1 is negligible.
I see how you calculated that, but I think you’re looking at the wrong pieces of evidence, and I agree with TheOtherDave.
You have an even split chance of getting the real bullet in play, so let’s put that down:
P[bullet] = 0.5 P[¬bullet] = 0.5
Then, given that you DO get the bullet, you have a very high chance of being dead if you don’t know how you will die:
P[die | bullet] = 0.99 P[¬die | bullet] = 0.01
Of course, this means that overall, P[die] = 0.495, and P[injury] = 0.005. However, if you already also know that P[¬die] = 1, then...
P[bullet & ¬die] = 0.5 P[¬bullet & ¬die] = 0.5
...because P[bullet] is computed before its causal effects (death or injury) can enter the picture, which means you’re left with P[injury] = 0.5 (a hundred times larger than P1!).
Thus, while the first chance of injury is negligible, the chance of injury once you already know that you won’t die is massively larger, given that P[injury XOR death | bullet] = 1 (which is implied in the problem statement, I would assume).
Edit: I realize that this makes the assumption that your chances of getting the bullet doesn’t correlate with knowing how you will die, but it most clearly exposes the difference between your calculation and other possible calculations. This is not the correct way to calculate the probabilities in real life, since it’s much more likely that non-death is achieved by not having the bullet in the first place (or by failing to play Russian Roulette at all), but there’s all kinds of parameters you can play with here. All I’m saying is that P2 isn’t necessarily at most twice P1, it all depends on the other implicit conditions and priors.
No, that’s not right. What we’re interested in here, is P[injury|¬die]. Using Bayes’ Theroem:
P[injury|¬die] = {P[¬die|injury]*P[injury]}/P[¬die]
Using the figures you assume, and recalling that “injury” refers only to non-fatal injury (hence P[¬die|injury]==1):
P[injury|¬die] = {1*0.005}/0.505 = 1⁄101 = approx. 0.00990099
The chances of injury are then not quite double what they would have been without death-immunity. This is reasonably low, because the prior odds of survival at all are reasonably high (0.505) - had the experiment been riskier, such that there was only a 0.01% chance of survival overall, then the chance of injury in the death-immunity case would be correspondingly higher.
(We also have not yet taken into account the effect of the first player—in such a game of Russian Roulette, he who shoots first has a higher prior probability of death).
Thanks for the full Bayes Theorem breakdown.
I agree that this is how it should be reasoned ideally, which I only realized after first posting the grandparent. See other comments and the edit for how I arrived at the 50⁄50 reasoning. If you know the answer for the bottom/last question in this comment, I’d be interested to know.
It depends on how exactly this time-travel-related knowledge of how you die will work. My calculation is correct if a random self-consistent time loop is chosen (which I think is reasonable) -- there are far more self-consistent time loops in which you survive because you don’t get the bullet, than ones in which you survive because a bullet failed to kill you.
Terrible things start happening if there’s some sort of “lazy pruning” of possibilities (which I think is what you’re suggesting). Then the probability you get shot is 0.5, and then if you do get shot the self-consistency condition eliminates the branches in which you die, so you are nearly guaranteed to survive in some horrible fashion.
I don’t like the second option because it requires thinking of branching possibilities as some sort of actual discrete things, which I find dubious. But it’s a bit silly to argue about interpretations of time travel, isn’t it?
This is a bit of what I was getting at with the edit in the grandparent: basically, it’s not very bayesian to stick to a 50⁄50 bullet chance when you know you will not die.
However, I was also considering the least convenient possible world: You already know that you won’t die, and since the worst you have to fear is debilitating permanent non-fatal injury (which you apparently don’t care about if it’s for Science!), you decide to repeatedly play Russian Roulette with tons of people, just to test probabilities or for fun or something.
Then what happens? Does it become more probable that you’ll just randomly end up always not getting the bullet with .98 probability (if the chance of surviving a bullet was 1%), which will to an outside view apparently defy all probabilities? Or does it instead stick to the 50⁄50 and on average you’ll get injured every second game (assuming you have a way of recovering from the injuries and playing again) without ever dying?
More importantly, which scenario should an ideal bayesian agent expect? This I have no idea, which is why I think it’s not trivial or obvious that P2 = 2*P1.
The calculation that P2=2*P1 obviously only applies to one game. If you play lots of games sequentially, then the probability increases stack. (Edit: I incorrectly said that the ratio doubles with every game, which is obviously false)
Another way of thinking about this: absent time travel, if you survive a bullet with 1% probability, then after N games your probability of surviving unscathed is 1/2^N, and your total probability of surviving is (101/200)^N. Therefore, given that you survive, your probability of surviving unscathed should be the ratio of these, or (100/101)^N.
(All of this is assuming the random self-consistent time loop interpretation of time travel.)
Hmm, interesting. That would imply that, to a third-party, there’s some random guy who wins 99% of the time at Russian Roulette. At this point, it should legally be considered murder.
Death sentence by plane crash sounds appropriate, in this case.
It is murder, but you’re going to have terrible trouble proving that (especially if he’s careful about documenting how fair the russian roulette is). To avoid murder charges, the hypothetical psychopathic death-immune person can go so far as to arrange a tournament, with 2^n entries for some integer n. In this arrangement, one person must survive every round, and thus it does not look suspicious afterwards that he did survive every round (plus he gets 2^n prizes for going through n rounds).
I’m pretty sure that the russian roulette itself is illegal just about everywhere, though. No matter how it’s done, it’s either murder or assisted suicide.
Or only enter other people’s tournaments and have them document their own procedure.
I would have expected a different label to apply. Neither of those seems accurate. In fact I didn’t think even assisted suicide got called “assisted suicide”.
To be precise (and absent any other interventions) P2 is larger than P1 by a factor of 2 (in two person to the death with randomized start case).
I thought I was fairly clear that that was exactly what I was arguing. Including the implication that it doesn’t immediately become more than P2. This (and other previously unlikely failure modes) have to be considered seriously but to overweight their importance is a mistake.
Ah! I see. Yeah, it seems I wasn’t thinking about Actual Russian Roulette, in which two players take turns and the most likely route to survival is my opponent blowing his brains out first, but rather Utterly Ridiculous Hypothetical Variation on Russian Roulette, in which I simply pull the trigger over and over while pointing at my own head, and the most likely route to survival is a nonfatal bullet wound.
Ahh, yes. That seems to be an impractical course of action even with faux-immortality. It may be worthwhile if some strange person was willing to pay exorbitant amounts of cash per shot and you were also permitted to spin the cartridge after every shot (or five shots) in order to randomize it. Then the accumulated improbability (ie. magnified injury and ‘unknown black swan’ possibility) would ultimately injure or otherwise interrupt you but only after your legacy had been improved significantly.
I’ve made the exact same mistake before. Maybe there should be (or is) a name for that.
It does become P2. But… aside from the submarine issue… if you play Russian Roulette against one other person, then with probability 1⁄2, the other person gets shot before you do. Assuming that the loser either dies or is incapacitated, we can write P1 as Pr[survive|shot] and P2 as Pr[shot|survive], and by a simple application of Bayes P2 is at most twice P1.
If you’re the sort of person who would take advantage of such knowledge to engage in dangerous activities, does that increase the probability that your reported time of death will be really soon?
On the other hand, if you’re the kind of person who (on discovering that you will die in a plane crash) takes care to avoid plane crashes, wouldn’t that increase your expected life span?
Moreover, these two attitudes—avoiding plane crashes and engaging in non-plane-related risky activities—are not mutually exclusive.
Absolutely. Note the parenthetical. The grandparent adopted the policy of ignoring this kind of consideration for the purpose of exploring the implied tangent a little further. I actually think not actively avoiding death, particularly death by the means predicted, is a mistake.
You can do the same thing if you know only the means of your death and not the time in advance; merely set up your death-stunts to avoid that means of death. (For example, if you know with certainty that you will die in a plane crash but not when, you can play Russian Roulette for cash on a submarine).
And then the experimental aqua-plane crashes into you.
An important safety tip becomes clear; if you’re involved in a time loop and know the means of your death, then keep a very close eye on scientific literature and engineering projects. Make sure that you hear the rumours of the aqua-plane before it is built and can thus plan accordingly...
… True.
But you could still be injured by a plane crash or other mishap at another time, at standard probabilities.
And you should still charter your own plane to avoid collateral damage.
I am happy to continue the conversation if you are interested. I am trying to unpack just where your intuitions diverge from mine. I’d like to know what your choice would be when faced with Newcomb’s Problem with transparent boxes and an imperfect predictor when you notice that the large box is empty. I take the empty large box, which isn’t a choice that maximises my expected utility and in fact gives me nothing, which is the worst possible outcome from that game. What do you do?
Two boxes, sitting there on the ground, unguarded, no traps, nobody else has a legal claim to the contents? Seriously? You can have the empty one if you’d like, I’ll take the one with the money. If you ask nicely I might even give you half.
I don’t understand what you’re gaining from this “rationality” that won’t let you accept a free lunch when an insane godlike being drops it in your lap.
A million dollars.
No, you’re not. You’re getting an empty box, and hoping that by doing so you’ll convince Omega to put a million dollars in the next box, or in a box presented to you in some alternate universe.
And by this exact reasoning, which Omega has successfully predicted, you will one-box, and thus Omega has successfully predicted that you will one-box and made the correct decision to leave the box empty.
Remember to trace your causal arrows both ways if you want a winning CDT.
Remember also Omega is a superintelligence. The recursive prediction is exactly why it’s rational to “irrationally” one-box.
Yes, that’s why I took the one box with more money in it.
Strictly speaking the scenario being discussed is one in which Omega left a transparent box of money and another transparent box which was empty in front of Wedrifid, then I came by, confirmed Wedrifid’s disinterest in the money, and left the scene marginally richer. I personally have never been offered money by Omega, don’t expect to be any time soon, and am comfortable with the possibility of not being able to outwit something that’s defined as being vastly smarter than me.
Remember also Omega is an insane superintelligence, with unlimited resources but no clear agenda beyond boredom. If appeasing such an entity was my best prospect for survival, I would develop whatever specialized cognitive structures were necessary; it’s not, so I don’t, and consider myself lucky.
Ah, then in that case, you win. With that scenario there’s really nothing you could do better than what you propose. I was under the impression you were discussing a standard transparent Newcomb.
Oh, so you pay counterfactual muggers?
All is explained.
The counterfactual mugging isn’t that strange if you think of it as a form of entrance fee for a positive-expected-utility bet—a bet you happened to lose in this instance, but it is good to have the decision theory that will allow you to enter it in the abstract.
The problem is that people aren’t that good in understanding that your specific decision isn’t separate from your decision theory under a specific context … DecisionTheory(Context)=Decision. To have your decision theory be a winning decision theory in general, you may have to eventually accept some individual ‘losing’ decisions: That’s the price to pay for having a winning decision theory overall.
I doubt that a decision theory that simply refuses to update on certain forms of evidence can win consistently.
If Parfit’s hitchhiker “updates” on the fact that he’s now reached the city and therefore doesn’t need to pay the driver, and furthermore if Parfit’s hitchhiker knows in advance that he’ll update on that fact in that manner, then he’ll die.
If right now we had mindscanners/simulators that could perform such counterfactual experiments on our minds, and if this sort of bet could therefore become part of everyday existence, being the sort of person that pays the counterfactual mugger would eventually be seen by all to be of positive-utility—because such people would eventually be offered the winning side of that bet (free money in the tenfold of your cost).
While the sort of person that wouldn’t be paying the counterfactual mugger would never be given such free money at all.
If, and only if, you regularly encounter such bets.
The likelihood of encountering the winning side of the bet is proportional to the likelihood of encountering its losing side. As such, whether you are likely to encounter the bet once in your lifetime, or to encounter it a hundred times, doesn’t seem to significantly affect the decision theory you ought possess in advance if you want to maximize your utility.
In addition to Omega asking you to give him 100$ because the coin came up tails, also imagine Omega coming to you and saying “Here’s 100,000$, because the coin came up heads and you’re the type of person that would have given me 100$ if it had come up tails.”
That scenario makes it obvious to me that being the person that would give Omega 100$ if it had come up heads is the winning type of person...
If the coin therein is defined as a quantum one then yes, without hesitation. If it is a logical coin then things get complicated.
This is more ambiguous than you realize. Sure, the dismissive part came through but it doesn’t quite give your answer. ie. Not all people would give the same response to counterfactual mugging as Transparent Probabilistic Newcomb’s and you may notice that even I had to provide multiple caveats to provide my own answer there despite for most part making the same kind of decision.
Let’s just assume your answer is “Two Box!”. In that case I wonder whether the problem is that you just outright two box on pure Newcomb’s Problem or whether you revert to CDT intuitions when the details get complicated. Assuming you win at Newcomb’s Problem but two box on the variant then I suppose that would indicate the problem is in one of:
Being able to see the money rather than being merely being aware of it through abstract thought switched you into a CDT based ‘near mode’ thought pattern.
Changing the problem from a simplified “assume a spherical cow of uniform density” problem to one that actually allows uncertainty changes things for you. (It does for some.)
You want to be the kind of person who two-boxes when unlucky even though this means that you may actually not have been unlucky at all but instead have manufactured your own undesirable circumstance. (Even more people stumble here, assuming they get this far.)
The most generous assumption would be that your problem comes at the final option—that one is actually damn confusing. However I note that your previous comments about always updating on the free money available and then following expected utility maximisation are only really compatible with the option “outright two box on simple Newcomb’s Problem”. In that case all the extra discussion here is kind of redundant!
I think we need a nice simple visual taxonomy of where people fall regarding decision theoretic bullet-biting. It would save so much time when this kind of thing. Then when a new situation comes up (like this one with dealing with time traveling prophets) we could skip straight to, for example, “Oh, you’re a Newcomb’s One-Boxer but a Transparent Two-Boxer. To be consistent with that kind of implied decision algorithm then yes, you would not bother with flight-risk avoidance.”
Not if you mistakenly believe, as CDTers do, in human free will in a predictable (by Omega) universe.
“Free will” isn’t incompatible with a predictable (by Omega) universe. I also doubt that all CDTers believe the same thing about human free will in said universe.
I think this is the kind of causal loop he has in mind. But a key feature of the hypothesis is that you can’t predict what’s meant to happen. In that case, he’s equally good at predicting any outcome, so it’s a perfectly uninformative hypothesis.
That was exactly my point. If he could make such a prediction, he would be correct. Since he can’t...