View Poll Results: Which boxes will you take?

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Thread: Newcomb's problem

  1. #1
    Senior Member Dim7's Avatar
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    Default Newcomb's problem

    There's two boxes. Box A is transparent and definitely contains 1000 euros, dollars, or the equivalent amount of money in whatever currency you identify with. Box B is opaque. You know that it either has 1 000 000 units of your preferred form of money or nothing at all. You can choose which box to take... or take both!

    Now that shouldn't be much of a choice. Whatever is in Box B, if you take both boxes you get 1000 more units of money than if you just take box B. And by choosing box B only, you get nothing in case it turns out to be empty. What a pointless thought experiment.

    Except here's the thing: Whether box B is empty or not depends on the prediction of a predictor - let's say a supercomputer that scanned your brain before. If the predictor predicted that you'll take both boxes, box B will be empty. Only if it predicted that you'll only take box B will it contain any money. However now as you're pondering whether to take both or just box B it has been already decided - the money is either already in there or the box is empty and nothing will be done about that anymore.

    The predictor has played this "game" with thousands of people and has never failed in its predictions. It would be very strange if you happened to be an exception, so if you take box B only it will almost certainly contain the large amount of money and if you take both boxes it will almost certainly be empty. However it's also too late to do anything about the content of box B, and whether it is empty or not you'll win more money by taking both boxes.

    What will you choose?
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  3. #2
    Senior Member Dim7's Avatar
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    How did it take so much text for me to explain? Here's the tl;dr version

    Box A contains 1000 dollars for sure, box B may contain either 1 000 000 dollars or nothing. Box B contains the money if the predictor predicted you will only choose box B, if the predictor predicted you'll take both it's empty. The contents of the box won't change by now as you're pondering the choice, it's either already there or it's empty. However the predictor has played the game with thousands of people and has never been wrong.
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  4. #3
    Senior Member Art Rock's Avatar
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    Well, whatever you do will not change the money on the table under the boxes, so just take both.
    I treat my music like I treat my pets. It’s something to own, care about and curate with attention to detail. From a blog by hjr.

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  6. #4
    Senior Member EdwardBast's Avatar
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    Quote Originally Posted by Dim7 View Post
    How did it take so much text for me to explain? Here's the tl;dr version

    Box A contains 1000 dollars for sure, box B may contain either 1 000 000 dollars or nothing. Box B contains the money if the predictor predicted you will only choose box B, if the predictor predicted you'll take both it's empty. The contents of the box won't change by now as you're pondering the choice, it's either already there or it's empty. However the predictor has played the game with thousands of people and has never been wrong.
    As you've described the situation, the chooser is aware of the above information. If this is so, then there is no problem. One chooses B and the infallible predictor gives you the big payout every time. Or am I missing something?

    What greater comfort does time afford than the objects of terror re-encountered and their fraudulence exposed in the flash of reason?
    — William Gaddis, The Recognitions

    Originality is a device untalented people use to impress other untalented people and to protect themselves from talented people.
    Basil Valentine

  7. #5
    Senior Member Dim7's Avatar
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    Quote Originally Posted by EdwardBast View Post
    As you've described the situation, the chooser is aware of the above information. If this is so, then there is no problem. One chooses B and the infallible predictor gives you the big payout every time. Or am I missing something?
    I would refrain from saying the predictor is literally infallible, since that can never be known under any hypothetical scenario. Just that the predictor hasn't failed so far, despite thousands of predictions.

    The problem with "one-boxing", although I personally think it's the rational choice, is of course that it's like choosing with the assumption one's choice can determine retro-causally what happened in the past. One can imagine ever weaker predictors, that have very occasionally been wrong. At some point if the predictor is fallible enough, taking B only starts to seem very silly, even if statistically people who only take box B win more money.
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  8. #6
    Senior Member KenOC's Avatar
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    I'd go with the predictor. After all, it seems far more dependable in its forecasts (verifiable forecasts, please note) than any of those deities we're so happy to place our trust in.

    Anyway, say what you want about Box B, most likely until opened it contains only Schrodinger's cat.
    Last edited by KenOC; May-23-2020 at 20:53.


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  10. #7
    Senior Member Ethereality's Avatar
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    I don't know if this is a valid solution, but it's what I would recommend trying. The predictor will most certainly predict my choice if I force myself to choose with no information. Therefore, if I'm unable to flip a coin, I would generate artificial numbers via mass complication. It will be much harder to predict what that number is going to come up from my causation than it is predicting what box I will choose, as you will see.

    But first, if I choose Box B and Box B+A as my targets of generation, the predictor knows the same likelihood of me choosing Box A as it does Box B and Box B+A. In other words, while it may not make sense to include Box A as a target of generation, because if I decided to include it within the randomization the predictor would predict that I would want to include it as well as be more likely give me less chance to earn money ie. it would block one of the $million options, giving me a 66% chance of earning $600,000 avg instead of a 50% chance of earning $1,000,000, I could complexly randomize what my targets would be first, to try to fool the predictor more. The predictor might already know which option I'm going with, 3 options or 2 options, so I could randomize what I'm actually choosing from.

    But to actually randomize a number, let's skip past this to the final stage. I put Box B as 'odd' and Box B+A as 'even,' and then complicate numbers to a decided degree, so that how many times I complicate the numbers won't be biased later when I know if the final number is even or odd and then decide to stop. Knowing the end number will give me a bias. So, first I would decide beforehand how many steps to take. I go with 3 steps. Then based on the time I begin thinking, I might think of different things to do in that moment than I would a minute prior, ie. complicating the issue somewhat for the predictor. For step 1, for instance, one minute I might think of a random subject, like anatomy, and then think hard about how many body parts from anatomy I remember. If the predictor guesses this, then oh well. Then for step 2 I might think of a number 1-10, and then count a certain type of object in vision in the room, then when done counting the type of object (like screws in sockets, corners, total rivets on something), I would hold up 1 finger out of the {1-10} and continue to count more sets of objects until the targeted (1-10) sets, adding all the sets to the total each time. If the predictor predicts this, then by gosh, oh well. Then for part 3, I might think of another number 1-10 and then think of that many random colors, and then add all the letters of all the colors to the total. If the predictor guesses this, then oh well. Based on if the final number is odd or even, I will know which final choice to make, and hopefully have a 50% chance of getting away with some dough.
    Last edited by Ethereality; Today at 03:47.

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