I’m not convinced it’s quite the same. If you owe the mafia $1001000 and they’re coming to collect the money this afternoon, you’re best off if you toss a coin to decide whether to choose two boxes. Omega, if I remember the formulation correctly, doesn’t stand for such tricks.
I could change the rules and decide not to stand for such tricks (mixed strategies) either. EDIT: No, I couldn’t.
And on the other hand, Omega could deal with mixed strategies perfectly well, and I don’t really understand why people make it so that he explicitly doesn’t tolerate mixed strategies in their problems. For example, in Newcomb’s Problem, if you one-box with probability p, Omega can just fill box B with probability p—for example if p=0.5 your expected winnings in Newcomb’s Problem are $500,500.
In the traditional formulation of Newcomb’s Problem (at least here on Less Wrong), if Omega predicts you’ll use a randomizer, it will leave box B empty.
That’s weird. Assuming human decision making is caused by neural processes, which aren’t perfectly reliable, there’d be no way for a human to not use a randomizer.
Omega could deal with mixed strategies perfectly well, and I don’t really understand why people make it so that he explicitly doesn’t tolerate mixed strategies in their problems.
Use of a mixed strategy might tarnish Omega’s reputation.
No, I can get $1001,000. If I randomly choose to take one box the first time, then both boxes will contain money the second time, where I might randomly choose to take both.
(Unless randomising devices are all somehow forced to come up with the same result both times)
1 box then 2 box = $1,001,000
1 box then 1 box = $1,000,000
2 box then 2 box = $1,000
2 box then 1 box = $0
$2,002,000 divided by 4 is $500,500. Effectively you’re betting a million dollars on two coinflips, the first to get your money back (1-box on the first day) and the second to get $1000 (2-box on the second day). Omega could just use a randomizer if it thinks you will, in which case people would say “Omega always guesses right, unless you use a randomizer. But it’s stupid to use one anyway.”
Where p is the probability of 1 boxing, $E = p^2 $1,000,000 + p(1-p $1,001,000 + (1-p)^2 * $1,000 = $999,000 p + $1000.
So the smart thing to do is clearly always one-box, unless showing up Omega who thinks he’s so big is worth $499,500 to you.
I completely agree that to maximise your expected gain you should one-box every time. I was thinking of the specific case where you really, really need $1001,000 and are willing to reduce your expected gain to maximise the chance of getting it.
I’m not convinced it’s quite the same. If you owe the mafia $1001000 and they’re coming to collect the money this afternoon, you’re best off if you toss a coin to decide whether to choose two boxes. Omega, if I remember the formulation correctly, doesn’t stand for such tricks.
I could change the rules and decide not to stand for such tricks (mixed strategies) either. EDIT: No, I couldn’t.
And on the other hand, Omega could deal with mixed strategies perfectly well, and I don’t really understand why people make it so that he explicitly doesn’t tolerate mixed strategies in their problems. For example, in Newcomb’s Problem, if you one-box with probability p, Omega can just fill box B with probability p—for example if p=0.5 your expected winnings in Newcomb’s Problem are $500,500.
That sounds tricky—unless you are a mind-reading superintelligence!
Yeah, you’re right. I can’t decide to not stand for mixed strategies, only Omega can.
In the traditional formulation of Newcomb’s Problem (at least here on Less Wrong), if Omega predicts you’ll use a randomizer, it will leave box B empty.
That’s weird. Assuming human decision making is caused by neural processes, which aren’t perfectly reliable, there’d be no way for a human to not use a randomizer.
We assume that Omega is powerful enough to simulate your brain and the environment precisely, and that quantumness is negligible.
In that case, you could still say that there’s no way not to use a randomizer, but Omega would be using the same randomizer with the same seed.
If you use flipping a coin as a randomizer, Omega could simulate that too. But traditionally using coins doesn’t fly while using brains is okay.
Use of a mixed strategy might tarnish Omega’s reputation.
The first time you enter the room, the boxes are both empty, so you can’t ever get more than $1,000,000. But you’re otherwise correct.
No, I can get $1001,000. If I randomly choose to take one box the first time, then both boxes will contain money the second time, where I might randomly choose to take both.
(Unless randomising devices are all somehow forced to come up with the same result both times)
Sorry, my mistake. I misread the OP.
Hang on a minute though
1 box then 2 box = $1,001,000 1 box then 1 box = $1,000,000 2 box then 2 box = $1,000 2 box then 1 box = $0
$2,002,000 divided by 4 is $500,500. Effectively you’re betting a million dollars on two coinflips, the first to get your money back (1-box on the first day) and the second to get $1000 (2-box on the second day). Omega could just use a randomizer if it thinks you will, in which case people would say “Omega always guesses right, unless you use a randomizer. But it’s stupid to use one anyway.”
Where p is the probability of 1 boxing, $E = p^2 $1,000,000 + p(1-p $1,001,000 + (1-p)^2 * $1,000 = $999,000 p + $1000. So the smart thing to do is clearly always one-box, unless showing up Omega who thinks he’s so big is worth $499,500 to you.
I completely agree that to maximise your expected gain you should one-box every time. I was thinking of the specific case where you really, really need $1001,000 and are willing to reduce your expected gain to maximise the chance of getting it.