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.

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).

Or only enter other people’s tournaments and have them document their own procedure.

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...

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

timeof your death in advance, then themeansbecome 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

nota 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

verydifferent 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 surviveswill 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

beforeits causal effects (death or injury) can enter the picture, which means you’re left with P[injury] = 0.5 (ahundredtimes larger than P1!).Thus, while the first chance of injury is negligible, the chance of injury

once you already know that you won’t dieis 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

necessarilyat 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. 0.00990099The 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 allare 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

repeatedlyplay Russian Roulette withtonsof people, just to test probabilities or for fun or something.Then what happens? Does it become more probable that you’ll just

randomlyend 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

mustsurvive 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’stournaments 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 thanP2. 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

maybe 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

exactsame 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

actuallythink 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

beforeit 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.