We could introduce a bet. After making your choice you can pay $1,000 to double the contents of the opaque box.
Should one-boxers make this bet? Should two-boxers? If one-boxers should but two-boxers shouldn’t then does that not suggest that one-boxing is the more rational choice? Leaving $1,000 on the table might superficially seem less rational, but it’s a short-sighted assessment of the problem.
The word “risk” implies chance.
There is a probability involved, even in the transparent scenario. I’m a one-boxer, and there’s a chance the predictor is going to put $0 in the “mystery” box.
But that doesn’t matter, what matters is that two-boxers are going to get $0 almost certainly. In either standard Newcomb’s or transparent.
They can ignore the probability all they want, but it’s still there.
I didn’t see that before but it’s interesting. It’s a bit sloppy (or I don’t understand).
So for newcomb, C1, C2 is whether I pick one box or two box. If I get one-box, the expected is
If I pick two-boxes, it’s:
Now for your bet thing, it’s pretty different. If I don’t bet, then I get simply 0. If I bet heads then it’s:
And it’s the opposite if I bet on tails. So it’s easy to see that here if p > 0.5, it’s better to bet on heads. And you told me that out of the past 100000 results, 99.9% or something was heads so I assume p is high. I am not using some weird correlation.
Your use of ‘is going to’ implies the transparent box isn’t already filled. It is. Let’s say you already see the million in there. How do you argue that there’s still risk or chance about it? Suppose you see the box empty. Same question. What justifies you leaving the 1000 in either case?
Calling it a risk is kind of like betting that you were born as a boy. Risk implies you don’t yet know the outcome, but you apparently do in the transparent box scenario.
One arguably can still choose one’s ‘pre-existing’ preference at choice time, after the boxes have been filled. Just saying that there’s still arguably ‘causality’ going on at this point in time. It’s a thin argument since it’s not direct, but it’s a counter to say @Suny suggesting that he’s screwed by the predictor before he even became aware of the situation. That’s the same faulty argument as is used to attempt to deflect responsibility for choices due to determinism. Choice, even if not free, is still choice, and one can choose one’s disposition.
No. I said to switch it so betting gets you 1M. We can add the 1K noise to make it completely symmetrical, it doesn’t make a difference.
You bet heads and you win, you get 1M. You bet tails and you win, you get 1M plus 1K whether you win or lose (1,001,000 if you win, 1000 if you lose).
Now they are completely symmetrical. It doesn’t change anything.
Betting heads and winning 1M is a correlation. Just like choosing one-box and winning 1M.
There is no difference. In both p is high.
The risk already materialized, but it was there.
Let’s modify my example, instead of one button there’s now two. The first button locks in the result, but it remains hidden. The second button shows the result.
If I ask you to place your bet after locking the result with the first button, is the risk gone just because the result is already determined? No, for you it makes no difference.
You are muddying the waters with the transparent Newcomb’s. There’s a reason why the box was opaque in the first place: the risk is carried after you enter the room.
Let’s make the tables
| Prediction OB | Prediction TB | |
|---|---|---|
| OB | $1,000,000 | $0 |
| TB | $1,001,000 | $1,000 |
| result heads | result tails | |
|---|---|---|
| bet on heads | $1,000,000 | $0 |
| bet on tails | $1,000 | $1,001,000 |
In both cases, we have the condition about the actions not influencing (causally) the states, so that’s good. We now see immediately that in (only) one of the tables, there is a dominant strategy. So really, the Newcomb argument cannot apply to the bet problem.
But if you add probabilities to the states, then yes, you will see that if p is high, you should bet on heads. Just like you could add probabilities to the Newcomb states and see that you should two-box, whatever the probabilities are.
There is a probability in Newcomb’s problem. The probability that you will lose $1,000,000 if you choose two-box is above 99.99%.
It is the exact same problem, and you haven’t explained why you completely ignore the probability in one, but not in the other.
The only thing you can lose is the $1,000 you won’t take by one-boxing.
The tables aren’t even the same, so I am not sure how you justify this claim.
Anyway, I showed you the same kind of analysis for both problems, and it yields different answers, so I don’t know what more you need.
I see it as clarifying the waters. The risk is entirely gone if the answer is known (first button not only locks the answer, but displays it). The ‘bet’ (which isn’t one) is placed after that occurs.
My point is, choosing OB (even in the transparent scenario) is done for reasons other than risk management and betting. It clarifies the real reason one one-boxes, which is to be the kind of person that would do that. A cardboard image of a person with a word balloon attached saying “one box” would fare better than any thinking human, who runs the risk of thinking too hard. The cardboard guy makes his choice without knowing he did it, and without any knowledge of the box contents. Why does cardboard-man do so consistently well? Because he’s the kind of guy that would do that, bolstered by his total lack of choice in the matter.
All that risk and bettinging stuff is superfluous, which is illustrated by the transparent box case, and also by Suny’s ‘rational friend’ who sees the box contents (for what purpose??) and advises ‘TB’ unconditionally, without revealing what’s actually in the boxes.
The tables are exactly the same. You constructed them wrong.
| C1 | C2 | |
|---|---|---|
| Simple | 1M \cdot p | 1M \cdot (1-p) + 1K |
| Newcomb | 1M \cdot p | 1M \cdot (1-p) + 1K |
No. You have not explained why you consider the probability in my simple problem, but you completely ignore the probability in Newcomb’s problem.
Which cells are wrong?
You are ignoring the probability.
Nothing is being ignored. It’s just a table of outcomes. If you bet on heads and the result is tails, you get $0; that’s all the table is saying. It doesn’t matter what the probability of the result being tails is.
You are 100% ignoring the probability.
Without their probability.
Every rational person disagrees. You yourself said you wouldn’t choose tails.
You have still not explained why you are completely ignoring the probability.
That’s precisely the problem. Let’s add the probabilities that you conveniently removed.
Newcomb’s:
| Prediction OB | Prediction TB | |
|---|---|---|
| OB | p | $1,000,000 | 1-p | $0 |
| TB | 1-p | $1,001,000 | p | $1,000 |
Simple:
| result heads | result tails | |
|---|---|---|
| bet on heads | p | $1,000,000 | 1-p | $0 |
| bet on tails | p | $1,000 | 1-p | $1,001,000 |
Now we can compare every result:
The probability in both is 1-p: extremely unlikely.
It’s unlikely the result is tails given that the bet is tails. In Newcomb’s it’s unlikely the prediction is OB given that the choice is TB.
The probability in both is p: extremely likely.
It’s likely the result is heads given that the bet is heads. In Newcomb’s it’s likely the prediction is OB given that the choice is OB.
The probability in both is p: extremely likely.
It’s likely the result is heads given that the bet is tails. In Newcomb’s it’s likely the prediction is TB given that the choice is TB.
The probability in both is 1-p: extremely unlikely.
It’s unlikely the result is tails given that the bet is heads. In Newcomb’s it’s unlikely the prediction is TB given that the choice is OB.
In short:
| outcome | probability |
|---|---|
| $1,001,000 | 1-p |
| $1,000,000 | p |
| $1,000 | p |
| $0 | 1-p |
As you can see the probability of each outcome is identical in both problems.
You claim you are not ignoring anything, but you are ignoring the probability in Newcomb’s problem. In my simple problem you choose heads, since clearly that’s how you win $1,000,000 almost certainly because the probability p is very high. But you avoid explaining why you ignore the probability in Newcomb’s problem, even though the likelihood of winning $1,000,000 if you choose OB is equally very high because the probably p is almost certain.
I think it’s very clear why you are avoiding an explanation: because there isn’t one.
It’s simply not rational to ignore the probability.
| Prediction OB | Prediction TB | |
|---|---|---|
| OB | p ~ $1,000,000 | 1-p ~ $0 |
| TB | p ~ $1,001,000 | 1-p ~ $1,000 |
There I fixed it.
Except that “p” is completely made up: it’s not the one the Newcomb’s problem explicitly tells you is “almost certain”.
Your “p” — which I’m going to call q — is deliberately chosen to completely obscure the real p, which we know.
So once again you are refusing to engage with Newcomb’s problem and avoiding an explanation as to why.
At this point it’s clear beyond a shadow of a doubt that you are engaging in bad faith. Refusing to answer a very straightforward question because it defies your entire position just made it more obvious.
I’m just going to state for the record what you are doing.
It’s an obvious case of circular reasoning: you claim the probability p doesn’t matter because two-boxers obtain more money anyway, except the probability p tells us they don’t, so you have to completely ignore it in order to conclude that.
I already proved it’s irrational to ignore the probability p with my simple example, in which you did not ignore the probability and fully utilized the correlation even though there was no causation.
You are still refusing to explain why you completely ignore the probability p in one problem, but not the other. And you are refusing to explain why correlation with no causation is useful in one problem, but not the other.
I also proved it’s irrational to ignore the correlation even though you see no causation with my sunscreen example. By completely ignoring the correlation you arrived at the wrong conclusion, which I showed by explaining a potential common cause. After understanding the potential common cause, you changed your answer.
But you never acknowledged you were wrong. You claimed the typical “I was wrong but I was right” (given the available information).
You never addressed the point raised by many others — not just me — that correlation is evidence of causation (in fact, the only reason we suspect causation in the first place). According to you, the correlation magically becomes meaningful once you are told the causal network of a potential common cause.
But here’s the nail in the coffin for your position: there is a common cause.
I already proved why a rational observer shouldn’t discount the correlation: it’s evidence of causation. Even if there’s no direct causal link, there could be an indirect causal network, for example a common cause. We don’t need to know any specific common cause, a rational agent would consider the possibility of it existing, and that’s all that is needed.
But in fact there is a common cause in Newcomb’s problem: it’s you. Your beliefs and your character is what causes the prediction to be two-box, and also causes your choice.
Your mistake is thinking you and I are interchangeable, but we are not. I cannot choose to have your brain while making my choice, I will choose with my brain, and you will choose with your brain, both are predictable, and both will be the cause of the prediction to be two-box for you, and one-box for me.
Just because you don’t see a common cause doesn’t mean there is no causal network between choice and prediction.
But you already tipped your hand that you are never going to change your mind, no matter the arguments, no matter the evidence. I already showed you probability shouldn’t be ignored just because there’s no causation, didn’t matter. I already showed you just because there’s no direct causal link you can see doesn’t mean there’s no causation involved, didn’t matter. I already showed you there is a common cause in Newcomb’s problem, it won’t matter.
Personally I don’t see the point in engaging in a debate if there’s absolutely no possibility in changing your mind. An intellectually honest interlocutor must accept the possibility they might be wrong, even hypothetically. I call it “suspending belief”. Just like I know The Lord of the Rings is not realistic but I can suspend my disbelief and pretend it is for a second, so can you even though you “know” the correct answer is two-box, you could suspend your belief and pretend it isn’t for a second. But you cannot even do that.
So there’s really no point. I already showed your position has no grounds whatsoever.
How is that FTL messaging? You verifying the state of the box is still at sublight speeds. What am I missing? I’m imagining it as follows:
The quantum entanglement mind reading machine causes, at a high level of accuracy, one of a pair of particles to collapse into 1/2 up spin if you choose 1B. It’s entangled particle is in the box and according to current theory instantaneously collapses in 1/2 down spin (if I understand it correctly). This effect is regardless of distance. The rule is that if you measure down spin on the particle in the box, you get 1,000,000.
You open the box, measure it, get 1,000,000. No FTL involved I think?
QM doesn’t work that way. There cannot be a way to force any superposition state to collapse to a state of your choice. It is by definition unpredictable.
Aliens predicting behaviour is totally normal. 
What about Stern-Gerlach filtering then?