I think the major unanswered question is how reliable do we think the machine is? 50%? 100%? I think the most interesting scenario is one where we are convinced that the machine actually predicts the future and always predicts correctly, so I’ll continue with that assumption in mind.

From one point of view, we have no reason not to take both boxes, since we can’t alter the machine’s prediction now, it’s already happened. I think however that this undermines my premise. Choosing both boxes only makes sense if we don’t actually believe the machine predicts the future.

One would be tempted to say "alright, then I will choose only box B, as the machine will have predicted that and I will get lots of money. If I were to choose both boxes, the machine would have predicted that too, and I would get much less money.

My argument is that both answers are wrong in a sneaky way: assuming an actual perfect predictor, my answer is box B only. However, the important part here is that this will not be, in fact, a choice. The result was already determined ahead of time, so I really only had that one option.

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.

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.

Box A only because ain’t nobody got time for any “Paradox” BS & daddy’s got bills to pay. Trick someone else with all that Time Travel nonsense!