Probability Puzzle Paradox: Which Boxes To Take?

We did Newcomb’s Paradox (so called) years ago, and a twist in premises after that. But I thought it would be fun to revisit, because I had some new ideas about less-than-perfect predictors.

The set up is simple. Two boxes in front of you, A and B. A has \$1,000 no matter what, B has a million or nothing. A Guesser loads the boxes. You get to pick (and keep) B, or both A and B. But if Guesser guesses you’ll take A+B, and so hope for the max money, he puts \$0 in B. If you decide to pick B, but not both, Guesser puts a million bucks in B.

Which box or boxes do you take? Think about it first.

The paradox is first framed by saying the Guesser is perfect. In the link above, I show all the scenarios lead to you picking box B, since that gives the most money. There is no way to fool a perfect Guesser. Any way you come to the choice A+B is guessed by perfect Guesser, and he leaves B empty. This includes the strategy whereby you employ some kind of “randomizer”, to make your choice unpredictable even to you until you have to make the choice, because Guesser will know you are trying to cheat him and he will leave B empty.

It’s supposed to be a paradox because there are two decision or game theory “optimal” choices, one being maximizing something called expected utility, which says pick B, and the other “strategic dominance”, which says pick A+B because no matter what you get a grand. Wokepedia has a good explanation.

But that only shows you that theoretical “optimums” may have nothing to do with Reality. If you choose the strategy “Gimme the most”, then choosing B is optimal.

There are concerns, some say, about free will. If Guesser can know what you’re going to do with certainty, then you don’t have free will. This doesn’t follow. You know your kid is going to take a cookie if you leave them on the counter to cool, but that doesn’t mean it wasn’t your kid’s free choice to take a cookie. Your prediction is not causative of him taking the cookie.

You can immediately see that conclusion has application in theology. So it seems to follow that because God is omniscient, it does not follow that we do not have free will. And indeed, this is the solution advocated by William Lane Craig in his excellent The Only Wise God: The Compatibility of Divine Foreknowledge and Human Freedom, which I encourage you to read.

Notice, too, that we never (well, or rarely) question whether our physical predictive models are causative. We can predict the sun will rise in the east, and do an excellent job, but few now believe because we predict this we are causing it. Though some make similar (confusing) arguments in quantum mechanics.

However, let’s stick closer to fallibility. Robert Nozick was apparently the first philosopher to look into Newcomb, and he (says Wokepedia) wanted to skirt around problems like divine foreknowledge, and so supposed Guesser was not necessarily perfect. But he didn’t say how imperfect, and implied something near to perfection.

Martin Gardner apparently wrote about this decades ago, so I know I risk making an ass out of myself, because if it was already done by Gardner, then there is little sense in trying outdo him. Nevertheless, I go on glibly because I haven’t been able to find Gardner’s solution.

As I’ve said many times, sometimes notation helps sometimes it hinders. Here it helps. Let Y be the Guesser’s guess of what you’ll decide, A+B or B. Let X be your guess, A+B or B. We’ll let R be the rules of the game, which contain all other knowledge we have regarding the setup, including your own choices. We want to build imperfection into Guesser, since if he’s perfect, we know the answer (choose B). We have:

$$\Pr(Y = A+B | X = A+B,R) = p_{A+B},$$

$$\Pr(Y = B | X = B,R) = p_{B},$$

where these read the probability Guesser guesses correctly when you pick A+B, and when you pick B, both assuming R. These two chances need not be equal. Since the guess of Guesser drives the money, his guess is equivalent to what’s in the boxes.

Box B has a million either when Guesser guesses correctly you’ll pick B, or when he fails to guess you’ll pick A+B. Box B is empty when he fails to guess you’ll pick B or when he guesses correctly you’ll pick A+B. In other words:

B has a million = “Y = B & X = A+B or Y = B & X = B”

And that has probability

$$\Pr(Y = B | X = A+B,R) \times\Pr(X=A+B|R) + \Pr(Y = B | X = B,R) \times\Pr(X=B|R), $$

or

$$(1-p_{A+B}) \times\Pr(X=A+B|R) + p_B \times (1-\Pr(X=A+B|R)). $$

If Guesser is perfect, then $p_{A+B} = p_{B} = 1$, and if you choose A+B, i.e. Pr(X=A+B|R) = 1—remembering probability is not causative, but you are!—, then the probability B has a million is 0. If again Guesser is perfect and you choose B alone, then Pr(X=A+B|R) =0 and the probability B has a million is 1.

And indeed, when Guesser is perfect the equation simplifies to (1-Pr(X=A+B|R)) = Pr(X=B|R). So it all works, and shows even if you pick some mechanism to “randomize”, if $p_{A+B} = p_{B} = 1$ then the best strategy is Pr(X=B|R) = 1, i.e. pick box B.

We want to know what happens if the Guesser is imperfect, and either $p_{A+B} < 1$ or $p_{B} < 1$. Obviously, we want to know if we can beat the million and come home with that extra thousand. These are situations when Pr(X=A+B|R) = 1, i.e. you pick A+B, and Guesser guesses wrong, which he does with probability (1-p_{A+B}). The extreme case is that Guesser always gets it wrong when you decide A+B, i.e. p_{A+B} = 0, so the chance B has million is certain.

At what probability (1-p_{A+B}) is it best for you to pick A+B? Don’t forget Guesser might now goof, and might get it wrong if you pick B, and so leave B empty. Thus, if you pick A+B the chance is (1-p_{A+B}) B has the million and you take home an extra grand; if you pick B the chance is p_{B} and nothing extra, so you’d pick A+B if

$$1-p_{A+B} > p_B$$,

else pick B.

That ignores the nicety of the extra grand, so you might adopt a game theory “optimal” strategy, like expected value. It doesn’t follow that what is optimal in some theory is optimal for you. Speaking for myself, if I could get the million I wouldn’t care about the extra thousand. However, for the sake of completeness, pick A+B if

$$(1-p_{A+B})\times \$1,001,000 > p_B\times \$1,000,000,$$

else pick B.

How we learn of Guesser’s prowess is anyone’s guess (great pun!).

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