Box A has $1,000,000. Box B is opaque and its contents depend on a machine that predicts the future... (Title has not enough space space, Question is in text below.)

https://lemmy.dbzer0.com/post/125970

Box A has $1,000,000. Box B is opaque and its contents depend on a machine that predicts the future... (Title has not enough space, Question is in text below.) - Divisions by zero

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 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 [https://en.wikipedia.org/wiki/Newcomb%27s_paradox]), but I increased the money to make it more interesting.

The best case result is 1.001.000.000 (A+B) vs 1.000.000.000 (B) only. Worst case is I have 1.000.000 only.

I go with B only because the difference feels tiny / irrelevant.

Maybe I actually have free will and this is not determism kicking in, but who knows. I‘m not in for the odds with such a tiny benefit.

Worst case is I have 1.000.000 only.

Except that's not the worst case. If the machine predicted you would pick A&B, then B contains nothing, so if you then only picked B (i.e. the machine's prediction was wrong), then you get zero. THAT'S the worst case. The question doesn't assume the machine's predictions are correct.

Good point. Actually I was assuming that the machine was never wrong. That’s also what is defined in the Newcomb’s Paradox wiki page.

If that‘s not a 100% given, you are definitely right.