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.