Omega can be replaced by amnesia

Bongo

Let's play a game. Two times, I will give you an amnesia drug and let you enter a room with two boxes inside. Because of the drug, you won't know whether this is the first time you've entered the room. On the first time, both boxes will be empty. On the second time, box A contains $1000, and Box B contains $1,000,000 iff this is the second time and you took only box B the first time. You're in the room, do take both boxes or only box B?

This is equivalent to Newcomb's Problem in the sense that any strategy does equally well on both, where by "strategy" I mean a mapping from info to (probability distributions over) actions.

I suspect that any problem with Omega can be transformed into an equivalent problem with amnesia instead of Omega.

Does CDT return the winning answer in such transformed problems?

Discuss.

Let's play a game. Two times, I will give you an amnesia drug and let you enter a room with two boxes inside. Because of the drug, you won't know whether this is the first time you've entered the room. On the first time, both boxes will be empty. On the second time, box A contains $1000, and Box B contains $1,000,000 iff this is the second time and you took only box B the first time. You're in the room, do take both boxes or only box B?

This is equivalent to Newcomb's Problem in the sense that any strategy does equally well on both, where by "strategy" I mean a mapping from info to (probability distributions over) actions.

I suspect that any problem with Omega can be transformed into an equivalent problem with amnesia instead of Omega.

Does CDT return the winning answer in such transformed problems?

Discuss.

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.

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Omega can be replaced by amnesia

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Omega can be replaced by amnesia

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