The Expected Value Approach to Newcomb's Problem

From the list of standard problems in rationality that have been talked to death but still don't have a strong consensus, allow me to re-present Newcomb's Problem:

A superintelligence from another galaxy, whom we shall call Omega, comes to Earth and sets about playing a strange little game. In this game, Omega selects a human being, sets down two boxes in front of them, and flies away.

Box A is transparent and contains a thousand dollars.
Box B is opaque, and contains either a million dollars, or nothing.

You can take both boxes, or take only box B.

And the twist is that Omega has put a million dollars in box B iff Omega has predicted that you will take only box B.

Omega has been correct on each of 100 observed occasions so far - everyone who took both boxes has found box B empty and received only a thousand dollars; everyone who took only box B has found B containing a million dollars. (We assume that box A vanishes in a puff of smoke if you take only box B; no one else can take box A afterward.)

Before you make your choice, Omega has flown off and moved on to its next game. Box B is already empty or already full.

Omega drops two boxes on the ground in front of you and flies off.

Do you take both boxes, or only box B?

This problem is famous for not only the fact that the answer is extensively controversial, but also for people who think they should one-box (take only box B) or two-box (take both box A and box B) almost always are very certain of their answer. I'm one of those people -- I'm very certain I should one-box.

Since I missed out on earlier Newcomb's discussion, I'd like to explore my approach to the problem here.

In Newcomb's Problem, there are only four possible outcomes, and they end up in this tree:

Omega predicts you will one-box:
- You one box: +$1000000
- You two box: +$1001000
Omega predicts you will two-box:
- You one box: +$0
- You two box: +$1000

We can use this knowledge to create an expected value formula for all four options (where P is the chance Omega guessed your choice correctly):

EV(One-box) = P*$1000000 + (1-P)*$0

EV(Two-box) = P*$1000 + (1-P)*$1001000

If we set the two equations equal to each other, we can solve for the value for P that would make both strategies equally viable, expected value wise. That value is 50.05%. So as long as we are confident that Omega has a better than 50.05% chance at predicting our choice, the expected value formula says we can expect to earn more from one-boxing.

But what of the dominant strategy approach? Clearly if Omega predicted one-box, you earn more by two-boxing and if Omega predicted two-box, you also earn more by two-boxing. Does this not count as a viable approach, and a worthy defense of two-boxing?

However, following this logic is specifically betting that Omega will predict incorrectly ...and if P < 50%, this is a losing bet.

LESSWRONG
LW

LESSWRONG
LW

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The Expected Value Approach to Newcomb's Problem

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