Let’s take a break from the insanity and wade into cool waters to start the weekend. I’ll take it that you’ve watched this video, which purports to give a solution to Newcomb’s paradox, or that you otherwise know about both.
The idea is simple. There are two boxes. A variable one has either a million or nothing, and the other always had a thousand. You can (A) choose the first variable box or (B) pick both. But there is an entity that can, with some unspecified degree of accuracy, predict what you will do, choose A or B. If the entity thinks you’ll pick A, he’ll stick the million in first box; if he thinks you’ll pick B (both boxes), he’ll leave the first box empty.
Obviously, you’d like to max out your purse. So, which choice should you make?
Here is the twist: all probability, really all information, is conditional. Keep it in mind. Not so easy to do; I know I always didn’t.
Wildberger’s solution uses “expected values”, of which I am not a universal fan. Not that they are without value, but there are deep suspicions about them in odd situations.
Skip that for now, and accept the idea of expected value (which is explained in the video). Wildberger says that, somehow, we know
p_A = Pr( Entity guess you pick A | you pick A, I),
p_B = Pr( Entity guess you pick B | you pick B, I).
The “I” is the information that lets you deduce the numbers. We do not (and few do) discuss where this I comes from. But see below.
Note that some versions of this paradox have p_A = p_B = 1; i.e. the entity is an infallible oracle.
Lastly, it’s convenient to let m = money in the fixed box, and mK = money in the variable box. Above, m = 1,000 and K = 1,000. Then we can do the expected value calculation.
E_A = p_A * (mK) + (1-p_A) * (0),
E_B = p_B * (m) + (1-p_B) * (m + mK).
If you choose A, only the variable box, you get mK if the entity guesses correctly, which has probability (we say) of p_A; if the entity, however, guesses you’ll pick B (even though you pick A), you get nada, with probability (1-p_A). The sum of both possibilities is the expected value when you pick A.
A similar argument holds for picking B, as you see.
Expected value logic says pick the strategy with the larger value, E_A or E_B. For example, you should pick A when E_A ≥ E_B, or when, as Wildberger shows us,
p_A + p_B ≥ 1 + 1/K.
Assuming the entity is an oracle, i.e. p_A + p_B = 1, you’d choose A when K ≥ 1. If K < 1, you’d pick B (both boxes). If either p_A or p_B is less than 1, which happens when the entity can make mistakes, you’d have to plug in what’s what and solve the simple calculation. So says expected value calculus.
Now the twist, which involves all that stuff about quantum mechanics and free will and whatnot Wildberger was referencing.
For ease, let’s stick with the entity-as-oracle, so that p_A = p_B = 1. Fix a K, which is known, both by you and the entity. Doesn’t matter which K you pick; it only matters if K ≥ 1 or K < 1. Above we said that pick A (only variable box) if K ≥ 1, else pick B (both boxes).
Since the entity is an oracle, this means the entity is never wrong about what you’d pick—no matter when you pick (saying the oracle never guesses wrong makes it irrelevant when he guesses; and even though somebody can predict what you will choose, you still have free will). That means the entity knows you know about expected value. Since we all know K ≥ 1, the entity knows you’d pick A. Which means he’d put the mK in the variable box. So far, it’s seems there is no problem and no twist. (Obviously as K goes to infinity, there is less incentive to choose B anyway.)
But, since you know the entity can’t guess wrong, because you started with expected value, and expected value says pick only the variable box K ≥ 1, why not pull a fast one and pick B (both boxes), since the entity assumes you’ll only pick the variable box?
Yet the entity can’t guess wrong! He’ll know you’ll try a fast one, and so leave zippo in the variable box. You only get m. But, we just figured out the entity would know this “second iteration”, as it were, and that we’d have to pick A, so we have the opportunity for another sly move. Yet the entity would have figured this, too, since he always guesses right. Meaning if you try to get clever, you end up with only m.
That works if you go to infinity in the chain of reasoning, too. If K ≥ 1, pick A and win mK. Any tricks brings you only m.
Now if K < 1, expected value says to pick B, both boxes. But the entity knows you’d do this, which means he’d—again—leave the variable box empty, meaning you ought not to pick both boxes, since you’d only get m.
Hold up. Why would you pick only the variable box, which only has mK < m in it? You’re better off picking both boxes (option B)! True, the entity would know this, and he’d leave the variable box empty, but you’d at least come away with m, and not less than m if you chose A.
That’s the twist. If you rely on expected value, the entity would know this. If K > 1, pick A and come away with mK > m; if K = 1, pick A or B, or if K < 1, choose B, and in both cases you come away with m. So it seems the size of the reward drives the decision. And there is no way to reward your greed. At least with oracles, expected value is not needed.
This is for the case of an omniscient entity. Playing this game instead only with a merely intelligent one (and finding a new “I”) and the answer can change. If there is enough interest, I’ll write up that solution, too. The theological aspects of this one are, I think, more or less obvious.
I couldn’t wait and did it anyway, since it’s too juicy to leave sit. Suppose first of all, for ease, that the intelligent entity guesses right about the same amount, i.e. p_A = p_B = p (this is your new I). Then you should choose A (only the variable box) when
p ≥ 1/2 + 1/(2K),
else choose B (both). This can be easily pictured:
For any “large” K, choose the variable box as long as you think the entity is at least as good as just guessing. For “small” K, or for inaccurate entities (when you think you have him fooled), choose B.
Here’s a blowup of the K < 1.
When you’re dealing with a dummy, be greedy. The case where the oracle always guesses wrong, i.e. p_A = p_B = 0, regardless of the value of K, says grab both (B).
Why is this fun? Because the p is deduced from your I, the information you are using to guess the guessing ability of your opponent. Your opponent does not “have” a probability of guessing your guess correctly. He has an ability, perhaps imperfect, perhaps flawed, but no probability.
This becomes a bit more obvious if you imagine playing this game iteratively, switching sides about who takes the pot. I’m doing a dice version of this to further annoy my friends and relatives.
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