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What's your choice in Newcomb's problem? (lemmy.ml)

submitted 1 day ago by Objection@lemmy.ml to c/asklemmy@lemmy.ml

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Newcomb's problem is a thought experiment where you're presented with two boxes, and the option to take one or both. One box is transparent and always contains $1000. The second is a mystery box.

Before making the choice, a supercomputer (or team of psychologists, etc) predicted whether you would take one box or both. If it predicted you would take both, the mystery box is empty. If it predicted you'd take just the mystery box, then it contains $1,000,000. The predictor rarely makes mistakes.

This problem tends to split people 50-50 with each side thinking the answer is obvious.

An argument for two-boxing is that, once the prediction has been made, your choice no longer influences the outcome. The mystery box already has whatever it has, so there's no reason to leave the $1000 sitting there.

An argument for one-boxing is that, statistically, one-boxers tend to walk away with more money than two-boxers. It's unlikely that the computer guessed wrong, so rather than hoping that you can be the rare case where it did, you should assume that whatever you choose is what it predicted.

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[–] Arrkk@lemmy.world 2 points 13 hours ago

You can flip the problem around and have it be mathematically the same. The predictor has some knowable accuracy, you can run the experiment many times to determine what it is. Let's also replace the predictor with an Oracle, guaranteed 100% always correct, and we'll manually impose some error by doing the opposite of its prediction with some probability. This is fully indistinguishable from our original predictor.

Now, instead of the predictor making a prediction, let's choose our box first, then decide what to put in the mystery box afterwards, with some probability of being "wrong" (not putting the money in for the 1 box taker, or putting the money in for the 2 box taker). This is identical to having an Oracle, we know exactly what boxes will be taken, but there is some error in the system.

Now we ask, should you take one box or two? Obviously it depends on what the probability is. There's no more "fooling" the predictor. So, you do the EV calculation and find that if the probability is more than 50% accurate (in other words, if the probability of error is less than 50%), you should always take 1 box