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”.
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”.