If the machine predicts that you will take both Boxes A and B, Box B will be empty. But if the machine predicts that you will take Box B only, then Box B will contain $1,000,000,000. The machine has already done it’s prediction and the contents of box B has already been set. Which box/boxes do you take?

To reiterate, you choices are:

-Box A and B

-Box B only

(“Box A only” is not an option because no one is that stupid lol)

Please explain your reasoning.

My answer is:

spoiler

I mean I’d choose Box B only, I’d just gamble on the machine being right. If the machine is wrong, I’ll break that thing.

This is based on Newcomb’s Paradox (https://en.wikipedia.org/wiki/Newcomb’s_paradox), but I increased the money to make it more interesting.

  • Lvxferre@lemmy.mlEnglish
    15·
    1 year ago

    I’ll abstract the problem a tiny bit:

    • a = the prize in box A
    • ka = the potential prize in box B; i.e. “k times larger than a”
    • p = the odds of a false positive. That is, the odds that you pick box B only and it got nothing, because dumb machine assumed that you’d pick A too.
    • n = the odds of a false negative. That is, the odds that you pick A+B and you get the prize in B, because the machine thought that you wouldn’t pick A.

    So the output table for all your choices would be:

    1. pick nothing: 0
    2. pick A: a
    3. pick B: (1-p)ka
    4. pick A+B: a + nka

    Alternative 4 supersedes 1 and 2, so the only real choice is between 3 (pick B) or 4 (pick A+B).

    You should pick A+B if a + nka > (1-p)ka. This is a bit messy, so let’s say that the odds of a false positive are the same as the odds of a false negative; that is, n=p. So we can simplify the inequation into

    • a + nka > (1-n)ka // subbing “p” with “n”
    • 1 + nk > (1-n)k // divided everything by a
    • 1 + nk - (1-n)k > 0 // changed sides of a term
    • 1 + 2nk -k > 0 // some cleaning
    • n > (k-1)/2k // isolating the junk constant

    In OP’s example, k=1000, so n > (1000-1)/(2*1000) → n > 999/2000 → n > 49.95%.

    So you should always pick B. And additionally, pick A if the odds that the machine is wrong are higher than 49.95%; otherwise just B.

    Note that 49.95% is really close to 50% (a coin toss), so we’re actually dealing with a machine that can actually predict the future somewhat reliably, n should be way lower, so you’re probably better off picking B and ignoring A.