I also agree with Dagon’s first paragraph. Then, since I don’t know which game Omega is playing except that either is possible, I will assign 0.5 probability to each game, calculate expected utilities (reject → $5000, accept → $550) and reject.
For general form I will reject if k > 1/k + 1, which is the same as k*k—k − 1 > 0 or k > (1+sqrt(5))/2. Otherwise i will accept.
It seems like I’m missing something, though, because it’s not clear why you chose these payoffs and not the ones that give some kind of nice answer.
Either is possible and no mention is made of how it’s chosen (in fact, it’s explicitly stated that the probability is not known), so why would you assign 50% rather than 0% to the chance of game A? If Omega mentioned a few irrelevant options (games C through K) which favored reject, but which it NEVER used (but you don’t know that), would you change your acceptance?
There’s no good reason for assigning 50% probability to game A but neither is there a good reason to assign any other probability. I guess I can say that i’m using something like “fair Omega prior” that assumes that Omega is not trying to trick me.
You and Gurkenglas seem to assume that Omega would try to minimize your reward. What is the reason for that?
You and Gurkenglas seem to assume that Omega would try to minimize your reward
Base rate pessimism and TANSTAAFL. Offers of free money are almost always tricks, so my prior is that the next offer is also a trick. I expect not to be paid at all, so choosing the option that’s clearly a violation if I’m not paid is a much clearer cheat than choosing the one where Omega can claim to play by the rules and not pay me.
If you state that I don’t know a probability, I have to use other assumptions. 50⁄50 is a lazy assumption.
Note: this boils down to “where do you get your priors?”, which is unsolved in Bayesean rationality.
What can I say, your prior does make sense in the real world. Mine was based on the other problems featuring Omega (Newcomb’s problem and Counterfactual mugging) where apart from messing with your intuitions Omega was not playing any dirty tricks.
You could also make a version where you don’t know what X is. In this case always reject strategy doesn’t work since you would reject k*X in real life after the simulation rejected X. It seems like if you must precommit to one choice, you would have to accept (and get (X+X/k)/2 on average) but if you have a source of randomness, you could try to reject your cake and eat it too. If you accept with probability p and reject with probability 1 - p, your expected utility would be (p*X + (1-p)*p*k*X + p*p*X/k)/2. If you know the value of k, you can calculate the best p and see if random strategy is better than always-accept. I’m still not sure where this is going though.
I also agree with Dagon’s first paragraph. Then, since I don’t know which game Omega is playing except that either is possible, I will assign 0.5 probability to each game, calculate expected utilities (reject → $5000, accept → $550) and reject.
For general form I will reject if k > 1/k + 1, which is the same as k*k—k − 1 > 0 or k > (1+sqrt(5))/2. Otherwise i will accept.
It seems like I’m missing something, though, because it’s not clear why you chose these payoffs and not the ones that give some kind of nice answer.
Either is possible and no mention is made of how it’s chosen (in fact, it’s explicitly stated that the probability is not known), so why would you assign 50% rather than 0% to the chance of game A? If Omega mentioned a few irrelevant options (games C through K) which favored reject, but which it NEVER used (but you don’t know that), would you change your acceptance?
There’s no good reason for assigning 50% probability to game A but neither is there a good reason to assign any other probability. I guess I can say that i’m using something like “fair Omega prior” that assumes that Omega is not trying to trick me.
You and Gurkenglas seem to assume that Omega would try to minimize your reward. What is the reason for that?
Base rate pessimism and TANSTAAFL. Offers of free money are almost always tricks, so my prior is that the next offer is also a trick. I expect not to be paid at all, so choosing the option that’s clearly a violation if I’m not paid is a much clearer cheat than choosing the one where Omega can claim to play by the rules and not pay me.
If you state that I don’t know a probability, I have to use other assumptions. 50⁄50 is a lazy assumption.
Note: this boils down to “where do you get your priors?”, which is unsolved in Bayesean rationality.
What can I say, your prior does make sense in the real world. Mine was based on the other problems featuring Omega (Newcomb’s problem and Counterfactual mugging) where apart from messing with your intuitions Omega was not playing any dirty tricks.
I think this is a different guy named Omega. No mention of prediction or causality tricks, which are the hallmarks of Newcomb’s problem.
You could also make a version where you don’t know what X is. In this case always reject strategy doesn’t work since you would reject
k*X
in real life after the simulation rejected X. It seems like if you must precommit to one choice, you would have to accept (and get(X+X/k)/2
on average) but if you have a source of randomness, you could try to reject your cake and eat it too. If you accept with probability p and reject with probability1 - p
, your expected utility would be(p*X + (1-p)*p*k*X + p*p*X/k)/2
. If you know the value of k, you can calculate the best p and see if random strategy is better than always-accept. I’m still not sure where this is going though.