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I was surprised to find that we did not have a question on this Stack Exchange on this particular problem in decision theory. So I have chosen to add a question on Newcomb's Paradox, though I would like to limit the scope of it to its economic considerations more than its philosophical considerations.

For those unfamiliar with the problem, here is the usual formulation of the problem.


You are presented with two boxes.

  • Box A is clear and has \$1,000 in it.
  • Box B is opaque and either has \$0 or \$1,000,000 in it, which is decided by another player, let's say a machine here.

You can choose to take just Box B, or both Box A and B, but before you choose, a machine makes a decision about what is in Box B, based on what it thinks you will do. The machine has played this game many times before and empirically is almost always correct in predicting what the player will do.

  • If it thinks you are going to choose both boxes, the machine will place \$0 in Box B.
  • If it thinks you are going to choose just Box B, the machine will place \$1,000,000 in Box B.

Empirically, people are split on whether to play the one box or two box strategy. The intuition for both cases goes likes this.

For those who pick one box, the intuition is that if the machine is almost always right, then the expected value/utility of picking just Box B will be much much higher than picking both boxes. So obviously one should pick just Box B and not be greedy.

For those who pick two boxes, the intuition is that picking both boxes seems to be a dominant strategy. If the machine has already picked putting nothing in Box B, it doesn't hurt, and can only help, if you pick both boxes. If the machine has already picked putting \$1,000,000 in Box B, then...well you should still want both boxes!


So what's the rub? Philosophers like getting into the weeds with this question in discussing causality or free will. But perhaps one criticism of this thought experiment is that the scenario described here is not rigorously defined within the realm of game theory. The machine's only goal is to predict the player's move correctly, but it does not actually have a defined payoff for making the correct prediction. One can map the set of choices and player payoffs as a tree, where the player cannot observe the choice of the machine, but the information set needs more elaboration than this. How does the machine make its prediction exactly so that it is "almost always right"? Why does the machine get to completely preclude mixed strategies and just...know what the player will do most of the time? Essentially, the whole game, strategies and payoffs and information, seems all very hazy. Should it even be considered a game? Or is this just a weird decision theory question?

With paradoxes, trying to add in extra assumptions, or showing why an existing assumption is ill-defined, this can go a long way to providing a possible solution to a paradox. So my question(s) are:

  • How could one alter the formulation of Newcomb's Paradox to make it decisively analyzable with game theory or decision theory, and what strategy/decision should the player make in that case?
  • If you would choose not to alter the formulation of the paradox, what is the correct analysis of the problem?

Two ending points.

$^1$ I hope I have formulated my question so that there could be multiple correct answers that are objectively correct, and not just opinion based.

$^2$ If you think Newcomb's Paradox is just so incoherent that one can't really alter it without really altering what makes the paradox interesting, you can talk about that too. I am not asking that as a question because I think that would be too opinion based.

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In order to start making this tractable, I'd start by stating in which order the two events – the player making their choice, and the machine deciding how much money to put into box B – happen. There's two options:

  1. First, the machine deposits money into box B (based on a guess as to how the player will choose); the player chooses afterward.
  2. First, the player chooses; the machine deposits money into box B afterward.

(Technically, there's a third option: both actions being performed at the same time. I'd consider this to be equivalent to the first, in that the machine cannot observe the player's actual choice, and the machine's choice is a given for the player.)

The first option is not interesting as a one-off game, as choosing both boxes is clearly a dominant strategy. However, it would be interesting as an iterated game where the player's goal is to maximize their total payout, while the machine's goal is to minimize it. Obviously the player's optimal strategy would depend on the strategy (programming) of the machine, and vice versa. In order to identify overall good strategies, one could perhaps come up with different strategies for both player and machine, pit them against each other over a certain number of rounds, and see which strategy is robust against a wide variety of adversary strategies, the way that Axelrod did for prisoner's dilemma strategies in the early 1980s.

For the second option, the machine's decision-making process needs to be clarified; what does it mean for the machine to be "almost always correct"? I'd implement this as follows: the machine has access to an oracle that will tell it the player's choice, but which, with probability $p > 0$, will give an incorrect answer. The machine always follows the oracle and puts \$1,000,000 into box B iff the oracle says the player did not choose both boxes. This scenario could then be analyzed using standard tools, and the player's optimal strategy determined for both one-off and iterated games, depending on the value of $p$.

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    $\begingroup$ The downvote is rather harsh. There are no “options” as you describe, the game goes as described in the question: “you can choose to take just box B, it both box A and B, but before you choose, a machine makes a decision about what is in box B, based on what it thinks you will do.” The order of the events is what describes the game, the options are the choices with respect to boxes. $\endgroup$ – Brennan Aug 25 at 2:53

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