Omega's Idiot Brother, Epsilon

by OrphanWilde 4y25th Nov 20151 min read7 comments

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Epsilon walks up to you with two boxes, A and b, labeled in rather childish-looking handwriting written in crayon.

"In box A," he intones, sounding like he's trying to be foreboding, which might work better when he hits puberty, "I may or may not have placed a million of your human dollars."  He pauses for a moment, then nods.  "Yes.  I may or may not have placed a million dollars in this box.  If I expect you to open Box B, the million dollars won't be there.  Box B will contain, regardless of what you do, one thousand dollars.  You may choose to take one box, or both; I will leave with any boxes you do not take."

You've been anticipating this.  He's appeared to around twelve thousand people so far.  Out of eight thousand people who accepted both boxes, eighty found the million dollars missing, and walked away with $1,000; the other seven thousand nine hundred and twenty people walked away with $1,001,000 dollars.  Out of the four thousand people who opened only box A, only four found it empty.

The agreement is unanimous: Epsilon is really quite bad at this.  So, do you one-box, or two-box?


There are some important differences here with the original problem.  First, Epsilon won't let you open either box until you've decided whether to open one or both, and will leave with the other box.  Second, while Epsilon's false positive rate on identifying two-boxers is quite impressive, making mistakes about one-boxers only .1% of the time, his false negative rate is quite unimpressive - he catches 1% of everybody who engages in it.  Whatever heuristic he's using, clearly, he prefers to let two-boxers slide than to accidentally punish one-boxers.

I'm curious to know whether anybody would two-box in this scenario and why, and particularly curious in the reasoning of anybody whose answer is different between the original Newcomb problem and this one.

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