A and B every time because if the machine "predicts" you take both, you get 1kk usd, if the machine "predicts" you take only B, you get 1.1kkk usd. It's a free million dollars at least. Buy a house and invest the rest.
Regardless of whether the machine is right, if you don't believe it can perfectly predict what you'll do then taking both boxes is always better than just one.
The machine has already done it's prediction and the contents of box B has already been set. Which box/boxes do you take?
If my choices don't matter and the boxes are predetermined, what point is there to only taking one box? The machine already made its choice and filled the boxes, so taking both boxes is always the correct answer. Either I get $1,000,000 if the machine thought I would take both, or I get $1,001,000,000 if it didn't. This is a false dilemma, there is never a reason to take just one box.
This isn't a false dillemma. Imagine if the way the machine predicts is by copying your brain and putting it in a simulated reality, then the copy of you gets asked to choose which boxes to take, the exact same way and be given the exact same information. Under this assumption, the machine could predict with 100% accuracy what the real you would've chosen.
How do you know you are even the real you. You could just be the machine's simulation of the real you.
There is a dilemma and the dilemma is about how much you want to trust the machine.
If you are a simulation, then your choice doesn't matter. You will never get any real benefit from the boxes. It's like saying, "there is also a finite possibility that the machine is lying and all the boxes are empty". In which case, the choice is again irrelevant.
Situations in which your choice doesn't matter are not worth considering. Only the remaining possibility, that you are not a simulation and the machine is not lying, is worth considering.
Both! Critically, the contents of box B depend on the machine's prediction, not on whether it was correct or not (i.e. not on your subsequent choice). So it's effectively a 50/50 coin toss and irrelevant to the decision-making process. Let's break down the possibilities:
Machine predicts I take B only, box B contains $1B:
I take B only - I get $1B.
I take both - I get $1.001B
Machine predicts I take both, box B is empty:
I take B only - I get nothing.
I take both - I get $1M.
Regardless of what the machine predicts, taking both boxes produces a better result than taking only B. The question can be restated as "Do you take $1M plus a chance to win $1B or would you prefer $0 plus the same chance to win $1B?", in which case the answer becomes intuitively obvious.
But if it's true that the machine can perfectly predict what you will choose, then by definition your choice will be the same its prediction. In which case, you should choose one box.
Though OP never actually stated that the machine can perfectly predict the future. If that’s the case, then yes, you should just take box B. But we’re not given any information about how it makes its prediction. If @Sordid is correct in assuming it’s a 50-50, then their strategy of taking both is best. It really depends on how the machine makes its prediction.
I meant, let’s imagine the machine predicted B and is wrong (because I take A+B). I would call that scenario „I have free will - no determinism.“ Then I will have 1.000.000.000 „only“. That’s a good result.
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.
If (Box A Taken = True)
{place ($0) in Box B}
else
{place ($1,000,000,000) in Box B}
Machine doesnt care if you also take Box B, it only cares if Box A is one of the boxes taken. If you take no boxes, Box B would still have a billion dollars, although thats kinda dumb choice from a gameshow host's perspective.
Box A and B as the prediction has already been made so the choice has no bearing on the contents at this point. You either get the guaranteed million or both.
Well what you choose may not direct affect what is inside Box B, but there is still a huge difference between the two choices.
Imagine the way that the machine did it's prediction was copying your brain and making this copied brain choose in a simulation. Assuming the copied brain is completely identical to your brain, the machine could predict with 100% accuracy what the real you would choose. In this sense, what you choose can affect what's inside Box B (or rather, what your copied brain chooses can affect whats inside Box B).
One more thing to think about: How do you know that you aren't the simulated brain that's been copied?
It depends. If you mean flip a coin, then you should know that no coin flip or dice roll is truely random, it is random to us only because we couldn't predict it with our current technology. This scenario assumes that there are machines in the world that can predict the future, we just don't know whether this particular machine is accurate or not.
Now if you are talking about quantum-based randomness, I mean... I think the machine could just put $0 in the second box just to fuck with you.
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:
pick nothing: 0
pick A: a
pick B: (1-p)ka
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.
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