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.
I’ll abstract the problem a tiny bit:
So the output table for all your choices would be:
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
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.
I’ll abtract the problem…
Proceeds to teach calculus
BRB, finding a way to insert derivation by parts into that. :^)
Damn dude.