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.
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.
No information regarding the machine's accuracy is provided, but the fact that you are asked to make a choice implies that it is not perfect. The question explicitly specifies that the prediction has already been made and the contents of box B have already been set. You can't retroactively change the past and make the money appear or disappear by making a decision, so if your choice must match the prediction, then it's not your choice at all. You lack free will, and the decision has already been made for you by the machine. In that case the entire question is meaningless.
You never know the shenanigans of a machine, and one million is largely enough for me until I die, or if science gives us the option to live forever I bet machines will do the work for us :-)
Edit: as I believe the machine can be wrong, I’d probably take A + B
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.
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.
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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