You control the posterior probability of 1033 being composite, conditional on the event of playing the game where both numbers are 1033. Seen from outside that event (i.e. without conditioning on it), what you are controlling is the probability of Omega’s number being composite, but not the probability of Lottery number being composite. The expected value of composite 1033 comes from the event of Lottery number being composite, but you don’t control the probability of this event. Instead, you control the conditional probability of this event given another event (the game with both 1033). This conditional probability is therefore misleading for the purposes of overall expected utility maximization, where outcomes are weighed by their absolute (prior) probability, not conditional probability on arbitrary sub-events.
So, I can definitely see why this applies to the Ultimate Newcomb’s Problem. As a contrast to help me understand it, I’ve adjusted this problem so that P(playing the game, both numbers 1033|playing the game) = ~1. See my response to Manfred here.
It is, of course, possible that your algorithm results in you not playing the game at all—but if Omega does this every year, say, then the winners will be the ones who make the most when the numbers are the same, since no other option exists.
(If Omega can choose whether to let you play the game, which is the only game available to you, and the game has the rule that the numbers must be equal, then you should two-box to improve your chances of being allowed to play when the Lottery number is composite, and thus capture more of the composite Lottery outcomes. This works not because you are increasing the conditional probability of the number you get being composite (though you do), but because you are increasing the prior probability of playing the game with a composite Lottery number.)
Okay. I feel much more confident in my answer, then :P.
Double-checking: What if the lottery picks primes and composites with equal frequency?
… then, on average, you’ll get into half the games and make twice the money, so you should still two-box. I think.
So, assuming you/Manfred don’t poke holes into this, I’ll edit the original post. Thanks—having a clear, alternative two-box problem makes understanding the original much easier.
You control the posterior probability of 1033 being composite, conditional on the event of playing the game where both numbers are 1033. Seen from outside that event (i.e. without conditioning on it), what you are controlling is the probability of Omega’s number being composite, but not the probability of Lottery number being composite. The expected value of composite 1033 comes from the event of Lottery number being composite, but you don’t control the probability of this event. Instead, you control the conditional probability of this event given another event (the game with both 1033). This conditional probability is therefore misleading for the purposes of overall expected utility maximization, where outcomes are weighed by their absolute (prior) probability, not conditional probability on arbitrary sub-events.
Ah, I see. (Thanks! Some rigour helps a lot.)
So, I can definitely see why this applies to the Ultimate Newcomb’s Problem. As a contrast to help me understand it, I’ve adjusted this problem so that P(playing the game, both numbers 1033|playing the game) = ~1. See my response to Manfred here.
It is, of course, possible that your algorithm results in you not playing the game at all—but if Omega does this every year, say, then the winners will be the ones who make the most when the numbers are the same, since no other option exists.
(If Omega can choose whether to let you play the game, which is the only game available to you, and the game has the rule that the numbers must be equal, then you should two-box to improve your chances of being allowed to play when the Lottery number is composite, and thus capture more of the composite Lottery outcomes. This works not because you are increasing the conditional probability of the number you get being composite (though you do), but because you are increasing the prior probability of playing the game with a composite Lottery number.)
(Responded before your edit, so doubly-curious about your answer.)
Okay. I feel much more confident in my answer, then :P.
Double-checking: What if the lottery picks primes and composites with equal frequency?
… then, on average, you’ll get into half the games and make twice the money, so you should still two-box. I think.
So, assuming you/Manfred don’t poke holes into this, I’ll edit the original post. Thanks—having a clear, alternative two-box problem makes understanding the original much easier.