My Understanding Of Three Puzzling Paradoxes

[Keep_Relentless]

Hi everyone :)

I had written this in correspondence with a friend recently and now I will paste it here for the interest of anybody who has seriously pondered these problems. If anybody is interested in a detailed discussion of a particular problem we might take it to another thread. Otherwise, I hope you find this post somewhat thought-provoking :)

"Newcomb's problem: You are presented with two boxes, box A and box B. Inside box A is $10,000 and inside box B is either $1,000,000 or nothing. You can choose to take either both boxes or just box B. A fortune-teller has predicted your choice, and if they predicted you would take only box B then they have placed $1,000,000 inside, and if they have predicted you would take both boxes, they have placed nothing in box B. The fortune-teller has a 100% accuracy rate and has done this many times. Yet the amount in box B is already fixed. So, do you take one box, or do you take both boxes?
Argh this is such a beautiful problem. And it divides people almost evenly, even professional philosophers.

The surprise hanging paradox: A judge tells a convicted prisoner that he will be hanged at noon next week (during the week), but that the hanging will be a surprise, he will not know the particular date until it happens. So the prisoner thinks to himself, well, it cannot be Friday then, because by Friday morning it would not then be a surprise. But if it's not going to be Friday, it also can't be Thursday, for the same reason, come Thursday morning, if it's not Friday, it won't be a surprise. But then it can't be Wednesday either. Or Tuesday. Or Monday. So the prisoner concludes it won't happen at all. But then it does happen, on Wednesday, and guess what? It's a surprise.
This is a lovely mess of a problem but I do believe it can be understood precisely.

The St. Petersburg paradox: This one is actually my favourite. So you are offered a game to play. Someone will flip a coin. If it comes up tails, you win $2, but if heads, your win is doubled to $4, and the coin is flipped again. If it comes up heads again, your win is doubled to $8, and the coin is flipped again. If it comes up heads again, your win is doubled to $16 and another flip is made. This happens again and again doubling as many times as it comes up heads, and the first time it hits tails, the game ends and you take home the win. So you have a 1/2 chance of getting $2, a 1/4 chance of getting $4, a 1/8 chance of getting $8, and so on.
To calculate how much this game is worth, you just add all possibilities together with their appropriate weight. So you get 1/2 times 2, plus a 1/4 times 4, plus a 1/8 times 8... So you get 1+1+1+1+1... forever, indicating this game is worth infinity.
The question is, why is this such an unrealistic result? You wouldn't sell your house to play this game. Why not? Where's the problem with the logic?"




"Firstly Newcomb's problem is so brilliant because either answer may be perfectly logical depending on your perspective. The people who say take one box (which is what I would do too, by the way) look at the fact that taking one box always nets a million dollars and taking two boxes always nets ten thousand. People who take both boxes, however, look at the fact that the money has already been placed in box B, so the choice is between the money in box B, and the money in box B plus an extra ten thousand.
I personally think there is something suspect in the second way of reasoning, in that if you do choose to take both boxes, that choice itself implies that there will be nothing in box B. Whereas if you choose to take one box, that choice implies there will be a million dollars inside.
I look at it by analogy of a golf swing. Just by analysing the motion of a golfer's swing after he hit the ball, you can infer whether it was a good hit or not - even though what happens with a golfer's swing after the hit has no causal effect on the path of the golf ball. So a good swing post-hit doesn't cause a good hit (which has already happened), but it does imply one. Similarly, a choice to take one box doesn't cause the million dollars to appear, but it does imply it.

I think I know what's going on with the surprise hanging, it's just quite confusing. Firstly, the judge says two things: 1. You will be hanged (on a particular day of the week), and 2. It will definitely come as a surprise to you. Note that after the prisoner followed his line of reasoning (which we'll get to), he decided to throw out claim #1, instead of claim #2. This was rather arbitrary of him. He could have concluded "Ok, I have a 20% chance of being hung on Friday, which means I have a 20% chance that it will not be a surprise, despite what the judge said". Anyway, he decided to cling onto claim #2, so he reasoned that it can't be Friday. Let's follow him and say hypothetically claim #2 must be correct. If it can't be Friday, and it must be a surprise, it also can't be Thursday. Do you see why? It's called (in game theory) iterative deletion: once you delete certain options you come up with a new situation, and then you may be able to delete more, and then more. The reason it couldn't be Thursday is because he would know 100% by Thursday morning that it was going to be Thursday (if it couldn't be Friday). Then it wouldn't be a surprise. So he eliminates Thursday, but now he has the same argument for Wednesday (come Wednesday morning, if he knew for sure it wouldn't be Thursday or Friday, he would know 100% that it must be Wednesday, and it would not be a surprise). And the same argument for Tuesday, and finally for Monday.
So where did he go wrong? In my estimation, by eliminating any of the days at all, and taking the judge's second claim too seriously. It's pretty neat that the fact that it was a surprise that it would happen at all was only an outcome of believing that it must be a surprise.

Shame you had no thoughts on the St. Petersburg paradox because it's quite neat. Basically this is a game where your chance of winning hundreds of thousands is one in hundreds of thousands, winning billions one in billions, and so on, and your average (median) win will be a few bucks, but the math says the game is worth infinity so you should be willing to bet literally any amount to play. And in terms of pure math, this is an unassailable conclusion, you should be willing to bet any amount to play if you are not risk-averse (which you probably should be).
But why in the real world wouldn't we judge this game to be worth all that much? Do you see why it doesn't seem worth that much? Double your money as many times as you get heads? You may get an astronomical amount of money, but your odds of that are astronomically small. And what if you do? Let's say you get mind-bogglingly lucky and you get 50 heads in a row for somewhere around a quadrillion dollars. What value does a quadrillion dollars have, over a billion? And who's going to pay you a quadrillion dolllars? So we see that for very high numbers the game completely loses its meaning. Would you rather win 50 nonillion dollars or 1,000 decillion dollars? With the mathematical assumption that money has a linear value, one should be WAY better than the other, but in reality that's not the case.
In fact money doesn't have a linear value to people, and that's the other assumption I want to attack. Studies show that money has a logarithmic value to people, that is, the more you have, the less valuable it is to have more. If we replace the calculation of 1/2 times 2 + 1/4 times 4 + 1/8 times 8 ... with 1/2 times log(2) + 1/4 times log(4) + 1/8 times log(8) etc... we get an infinite sum of 2log(2), taking the inverse log of that gives us an expected value in dollar terms of $4 for this game, which aligns much better with most people's intuition. Other variants of this game can be handled similarly, I calculated one to be realistically worth somewhere around $50,000."

[charon]

if they predicted you would take only box B then they have placed $1,000,000 inside, and if they have predicted you would take both boxes, they have placed nothing in box B.

You're saying the fortune teller can alter the content of the box depending on his prediction. This invalidates the problem. The fortune teller can't be in charge of the boxes.

(But if you choose both boxes you would still win $10,000).

if it's not going to be Friday, it also can't be Thursday

The only day which is certain is Friday if nothing has happened on the other days. You can't say anything in advance about Monday to Thursday until they are over.

You wouldn't sell your house to play this game. Why not?

Because you could lose the first time or at any time. It's unpredictable.

Sorry, I'm no fun at all, am I? :-)

[Keep_Relentless]

You're saying the fortune teller can change the content of the box after he has made his predicton. This invalidates the problem.

Yes I am saying that, but it doesn't invalidate the problem. First the fortune-teller makes a prediction, then the money is put in the box (or not), and then you are presented with the boxes and your choice.

The only day which is certain is Friday if nothing has happened on the other days. You can't say anything in advance about Monday to Thursday until they are over.

Yes. But note the reason for this. The reason (as far as I can see) is that the judge's claim #2, the hanging will definitely be a surprise, is untenable.

You wouldn't sell your house to play this game. Why not?

Because you could lose the first time or at any time. It's unpredictable.

Sorry, I'm no fun at all, am I? :-)

Thanks for playing :D

[charon]

First the fortune-teller makes a prediction, then the money is put in the box (or not)

By whom? You can't have the fortune teller in charge of the boxes! He wouldn't be telling a fortune, would he?

the judge's claim #2, the hanging will definitely be a surprise, is untenable.

It's only untenable on Friday because it's no longer a surprise.

Thanks for playing :D

I don't think I did much playing :-)

[Keep_Relentless]

By whom? You can't have the fortune teller in charge of the boxes! He wouldn't be telling a fortune, would he?

The fortune-teller is in charge of stacking the boxes. The player is in charge of choosing to take one or both. All you know is that every time someone has taken both, the fortune-teller has placed nothing in box B beforehand. And they have never been wrong. I'm quite sure it is a well-defined problem, but it's very confusing how to reason your way to a solution.

the judge's claim #2, the hanging will definitely be a surprise, is untenable.

It's only untenable on Friday because it's no longer a surprise.

Yes, that's right. More generally I think, for any given set of days, it will not be a surprise on the last day. So the most the judge could ever say is that it will probably be a surprise. By assuming it will definitely be a surprise, the prisoner wound up confused.

[Positor]

Keep_Relentless,

I basically agree with you about the surprise hanging paradox.

So where did he go wrong? In my estimation, by eliminating any of the days at all, and taking the judge's second claim too seriously. It's pretty neat that the fact that it was a surprise that it would happen at all was only an outcome of believing that it must be a surprise.

Here's an interesting point:

If the hanging had not taken place by Friday morning, and the prisoner were asked: "Would it surprise you if you were now told you were to be hanged today?", how would he answer? According to the conclusion of his argument, he would answer Yes (since he has concluded that he will not be hanged at all). But according to a premise of the same argument, he would answer No (since his argument begins: "It would not be a surprise on Friday"). So his argument is illogical. It cannot be valid if its conclusion contradicts one of its premises!


I find Newcomb's problem somewhat artificial, as it requires us to assume (albeit just for the sake of the argument) that an infallible fortune-teller can exist. This throws up difficult questions about free will, predestination, compatibilism etc. And since infallible fortune-tellers are counterfactual, any assumptions we make in this case about free will etc will not be grounded in reality.

[charon]

it requires us to assume (albeit just for the sake of the argument) that an infallible fortune-teller can exist. This throws up difficult questions about free will, predestination, compatibilism etc. And since infallible fortune-tellers are counterfactual, any assumptions we make in this case about free will etc will not be grounded in reality.

Quite, which leads me to wonder if Keep_Relentless has posed the problem correctly.

There are actually Wiki pages for each of these problems which involve extensive and intricate mathematical workings out.

If we're not being forced into such calculations (not that I can do them) it's possible the problems aren't being stated accurately.

If they are being accurately stated then poo :-)

[Keep_Relentless]

Here's an interesting point:

If the hanging had not taken place by Friday morning, and the prisoner were asked: "Would it surprise you if you were now told you were to be hanged today?", how would he answer? According to the conclusion of his argument, he would answer Yes (since he has concluded that he will not be hanged at all). But according to a premise of the same argument, he would answer No (since his argument begins: "It would not be a surprise on Friday"). So his argument is illogical. It cannot be valid if its conclusion contradicts one of its premises!

Absolutely right, so he cannot even do away with claim #1 that there will be a hanging, even after he has eliminated every possible day! So there must be something wrong with his premise which is claim #2, the hanging will (definitely) be a surprise.

I only wonder why this doesn't appear to be an agreed-upon resolution in the philosophical community. Are people busy quibbling about the nature of a "surprise"?

I find Newcomb's problem somewhat artificial, as it requires us to assume (albeit just for the sake of the argument) that an infallible fortune-teller can exist. This throws up difficult questions about free will, predestination, compatibilism etc. And since infallible fortune-tellers are counterfactual, any assumptions we make in this case about free will etc will not be grounded in reality.

Well yes this is an extra dimension of the problem and an important one. It doesn't really trouble me, I just imagine a superintelligent biotech system that can read our future choices from our bio data or something. Besides, I don't think that necessarily invalidates free will (as in the responsibility for making choices). But of course all this is much more contentious and difficult than a logic problem.

[Keep_Relentless]

I'm happy to learn different ways of approaching the problems, I have more or less exhausted those I am familiar with in the opening post

[charon]

I only wonder why this doesn't appear to be an agreed-upon resolution in the philosophical community. Are people busy quibbling about the nature of a "surprise"?

Maybe that's why I don't belong to the 'philosophical community', they'll mess anything up :-)

I suppose then we have to say the judge was just plain wrong because he hadn't reckoned on the logical skills of the prisoner.

On the other hand, if it were emphasised beyond doubt as an integral part and premise of the problem that the hanging day would always be a surprise even on the Friday, then it certainly would be a problem. But an insoluble one that cheats the puzzler.

[Positor]

I'm happy to learn different ways of approaching the problems, I have more or less exhausted those I am familiar with in the opening post

I would be interested in your views on the Sleeping Beauty problem. I have always found this an interesting puzzle.

[Keep_Relentless]

I would be interested in your views on the Sleeping Beauty problem. I have always found this an interesting puzzle.

Thank you I appreciate this. I haven't seen this one.

I readily noticed the tension between the so-called "thirder" and "halfer" positions, but I'll have to give this a good think to do it some justice.

Edit: After a few hours the thirder position makes more sense to me, but still can't say I'm sure what's going on.

[charon]

She would only know it was tails if she can remember two awakenings. But she won't remember so she can't answer the question.

[Keep_Relentless]

She would only know it was tails if she can remember two awakenings. But she won't remember so she can't answer the question.

No one disputes that she can't say for sure, what is in dispute is her degree of confidence (as a percentage or fraction).

[charon]

On my planet speculative answers are no answer, they remain speculations.

_{Plus I sleep well at night :-)}

[charon]

Sorry, have to say this.

her degree of confidence (as a percentage or fraction).

She has no memory therefore she has no basis for confidence one way or the other. Any answer she gives is a complete shot in the dark. So the answer is zero in terms of justifiable confidence and 50/50 whether she'll guess correctly.

Poor girl. I hope she got a nice trip to Disneyland and suffered no after-effects from the drug.

[Keep_Relentless]

She has no memory therefore she has no basis for confidence one way or the other. Any answer she gives is a complete shot in the dark.

It's true she has no memory of the day or the coin toss, but she doesn't have no information because she was told (for the sake of argument) exactly what the procedure she is subjected to would involve. So having been given some information, she ought to be able to reason her way to some kind of conclusion. There are two likely answers and the main difficulty is in deciding between them. I think you are expecting some of these problems to be much easier than they actually are.

I will put my own thoughts about sleeping beauty into writing later. For now I'll say it seems useful to consider a variant where if tails is flipped, she gets woken up a million times.

[charon]

she was told (for the sake of argument) exactly what the procedure she is subjected to would involve. So having been given some information, she ought to be able to reason her way to some kind of conclusion.

The one would not enlighten the other. Knowing what was going to happen in advance would not help her choose heads or tails later.

There are two likely answers and the main difficulty is in deciding between them.

But that's my point. Due to memory loss she has no information to decide with.

Sorry, we're going round and round. I'm afraid I'm a bit allergic to all that :-)

I think you are expecting some of these problems to be much easier than they actually are.

No, harder!

[Positor]

I will put my own thoughts about sleeping beauty into writing later. For now I'll say it seems useful to consider a variant where if tails is flipped, she gets woken up a million times.

Yes, I think that is useful. If 'Heads', 'Tails and it's Monday' and 'Tails and it's Tuesday' each have equal (1/3) probability, then, in the variant, 'Heads' and each one of the 'Tails' awakenings has equal (1/1,000,001) probability. So 'Heads' and 'Tails and it's the 624,147th awakening' would have equal probability. This seems implausible!

In both versions, it is true that there are more (very many more in the variant) permutations in which Sleeping Beauty loses if she bets Heads; but this is countered by the fact that each single Tails permutation has a lower (very much lower in the variant) possibility of arising than the sole Heads permutation. It seems that this myriad of remote possibilities should add up to 1/2. (Heads = 1/2, Tails = 1/2.)

Consequently, having changed my mind from the "halfer" to the "thirder" position, I am now leaning back to "halfer"!

[charon]

...

[Keep_Relentless]

I think some light on the sleeping beauty problem has been dawning finally if I'm not mistaken. Ultimately I think both the halfer and thirder positions are correct answers but to two different interpretations of the question.

The halfer is answering the question in terms of the outside probability that heads was flipped, the answer as it is at any objective point in time, the answer from the point of view of the trial.
The thirder is answering the question in terms of the proportion of heads situations there are relative to the pool of situations Sleeping Beauty finds herself in.
They are two different questions, and what's more, there is a third question. Note that by saying the probability of heads is 1/2, Sleeping Beauty is often wrong, and is more often right by saying 1/3. But this doesn't resolve the question because in fact she is most often right by saying it's always tails and p(heads) is 0.

So we have three questions, as far as I can see, with three different answers. What is the probability that heads was flipped for the experiment? 1/2. What are the odds of Sleeping Beauty being in a heads situation out of her given situations? 1/3. What should Sleeping Beauty answer to be most often correct? 0.

I can't say I feel entirely satisfied having reached this conclusion, however, because it doesn't explain why these distinct points of view arise or what they mean.