Talk:Two envelopes problem/Arguments

Variant with no constrains

https://docs.google.com/file/d/0B7Z3B3DjCNt6VjJLT284TDVSZGs/edit?usp=sharing

Suppose you are allowed to open the envelope before you make the decision to swap.

1. Suppose the amount in the envelope you opened is $10
2. The other envelope may contain either $5 (meaning the one you chose is the higher value of the two with P=.5) or $20 (meaning the one you chose is the lower value of the two with P=.5).
3. So the expected value of the money in the other envelope is

$E(X)=\$5*1/2+\$20*1/2=\$12.5>\$10$

This is greater than the one you opened. So you gain on average by swapping. This applied to any amount of money you discover in the envelope and suggesting you should always swap no matter what you find. This implies that opening the envelope and finding the amount is not needed in deciding the action of swapping.

This is contradictory to the fact that the probability of choosing the envelope with more money is .5, and swapping the envelop will not increase that probability (1-.5=.5).

Solution:

The probability, P=.5 of you choosing the envelope with the higher (or lower) amount does not imply that the other envelope have 50% chance containing 1/2 of amount you see and 50% chance containing 2X of the amount. There’s only 2 possible cases in this game.

Case #1: The game organizer put a $5 and $10 in each envelope. You have 50% chance of getting the $10 and once you chose the envelope of $10. The amount that the other contains is $5 with 100% probability.

Case #2: The game organizer put a $20 and $10 in each envelope. You have 50% chance of getting the $20 and once you chose the envelope of $10. The amount that the other contains is $20 with 100% probability.

The probability of the game organizer doing Case #1 is P and Case #2 is Q. The expected value of the other envelope once you open the $10 envelope is:

$E(X|X_{0}=\$10)=\$5*P+\$20*Q$

However, from your perspective, P and Q is unknown. So the expected value of the other envelope cannot be determined.

The probability of you choosing the higher value in either case is 0.5. So the probability of winning this game (chooing the envelope with the higher amount) is

$.5*P+.5*Q=.5*1=.5$

So no swapping is needed.

Modification to the setup:

To satisfy the original conclusion of "always swapping", the following is needed:

You are given the following information before the game:

The game organizer flip a fair coin.

1. if heads, they will put $5 and $10 in each envelope
2. if tails, they will put $10 and $20 in each envelope

Once you open the envelope that contains $10, the expected value of the other envelope is indeed higher and hence you should swap.

More thoughts on classifying resolutions (and therefore problems)

Richard, I think you are absolutely right with your Anna Karenina point, there are many ways it can go wrong. I still think that the fundamental distinction should be between bounded and unbounded, even though you point out that mathematically speaking this is not an important difference.

The problem in most of its forms refers to money or the value of objects. I suspect that that is how it was intended by its original proposers and how it is understood by most (non-mathematician) readers.

For the bounded version it is easy describe the problem clearly and to give a simple resolution that is mathematically sound and can be put into words so that most people can understand it. I asked you to confirm that solution just to make sure that I had not missed anything but it is exactly as I thought. If the sum in your envelope should be more than half the upper bound it must be the larger sum so swapping will result in a certain loss, and a large one. No more is required to completely resolve the real-world version of the paradox.

Once you move to an unbounded sum, you need some mathematics. It makes no sense to say, 'Tell me in simple language what the problem is if there is no upper bound'; some understanding of mathematics and infinity is essential. In the powers of two version, which I still find the hardest to deal with, most people have no conception of what, no upper bound would mean in practice. With a lower bound only, not only would it be almost certain that the sums could not be fitted into the envelopes, but it would be almost certain that the sums could not even be described, even if the envelopes were the size of the known universe. This quickly takes us a very long way from two gentlemen comparing the values of their neckties. With no lower bound we almost certainly still could not describe the sums in the envelopes and (as you pointed out) expectations are meaningless. You will, to all intents or purposes, have nothing or an infinite amount, so swapping is pointless.

I am not suggesting that we do not discuss these purely mathematical versions just pointing out that, although P(B=2A | A = a) is not 1/2 for all a, applies to both bounded and unbounded versions there is still a big difference between the two cases. Martin Hogbin (talk) 13:43, 27 November 2014 (UTC)

Your pont of view is also common in the literature. Still I would like to point out that the Littlewood problem, which derived from Schrödinger and seems to be the original exchange paradox, was about an impoper prior, and improper priors have been at item of controversy since Laplace. So if you deal with the problem by treating the bounded case as if that is all there is, you are cutting off the problem at its roots. Richard Gill (talk) 07:32, 28 November 2014 (UTC)

Do you have any record of exactly what Schrödinger version of the problem was? I cannot find any clear reference to the problem in Littlewood either, although I do agree with one thing he says, 'Mathematics (by which I shall mean pure mathematics) has no grip on the real world ; if probability is to deal with the real world it must contain elements outside mathematics'. Kraitchik and Gardner both refer to real-world physical objects and money. The problem as stated in the article also refers to envelopes and money. It is contrived paradox based on a propsed line of reasoning that is mathematically ill-defined. In the real world the paradox is easily resolved, you should not always swap because the sum in your envelope might be above half the maximum sum allowed.

Once you try to get round this by having no upper bound you move unavoidably into the world of pure mathematics. I have never seen a mathematically well defined statement of this paradox (I hope I would know one if I saw it) and I suspect that one does not exist. So, once you get into the world of mathematics, the answer is 'What paradox?'. Martin Hogbin (talk) 19:47, 28 November 2014 (UTC)

See Littlewood, page 4: "Mathematics with minimal raw material" Example (4).

If you have never seen a mathematically well defined statement of this paradox then you haven't read my paper. Nor understood any of Boris Tsirelson's comments on the problem. Expectations are *not* meaningless without an upper bound. They are being used in this situation throughout science and needed in this situation througout science. And this is usually not problematic.

Dear Martin, whenever you use "real numbers" and theorems of calculus in the real world, you are also in the realm of pure mathematics and infinities. The calculus was invented because it is easier to compute integrals than sums. But in the real world everything is finite and discrete. Integrals and derivatives are pure mathematical fictions. You are short-sighted. Laplace wanted to solve practical problems. He introduced the principle of insufficient reason in order to solve practical problems. His solution breaks down in the two envelopes problem. So Laplace has a big practical problem even if you don't see any practical problem. If this particular situation shows that Laplace's principle has to be abandoned - or perhaps more accurately - sometimes needs to be abandoned - then we need to know how to distinguish problems where it works from problems where it doesn't work. That is a very very practical issue.

Every time a scientist uses a normal distribution and talks about its expectation value or standard deviation they are taking expectations of unbounded random variables. It seems you want to have 99.99% of modern scientific literature thrown in the rubbish bin. Richard Gill (talk) 12:18, 29 November 2014 (UTC)

If any amount a is a possible amount to be in Envelope A, then it should have positive probability. But moreover, if amount a is possible, then surely twice that amount is possible too? Even if much less likely. So there can be no definite bound to the amount which might be in Envelope A. (An unbounded distribution can still easily have a very small expectation value). In my opinion, the argument that the amount has to be bounded is facile. It fails to distinguish carefully between actual unknown amounts and our beliefs about them. Richard Gill (talk) 05:59, 30 November 2014 (UTC)

[edit conflict]I am not trying to make such strong claims as you seem to think. I have never said that expectations are meaningless without an upper bound.

One understanding of the argument for switching is based on the assumption that P(B=2A | A = a) is 1/2 for all a. The unbounded powers of two distribution achieves this assumption. Having an upper bound makes the resolution simple; having a lower bound does not resolve the paradox and makes the problem a bit more realistic. The problem with the distribution with only a lower bound is that the probability that either envelope contains a sum that could be put into it, or even represented in any way inside it, is zero. This is a far worse problem than unbounded expectations in normal distributions or infinities in calculus or Zeno's paradox. Unbounded numbers can be used in practical calculations in cases where they are almost certainly not goint to happen but in the TEP, right at the start of the problem, both envelopes contain a sum that is physically imposssible. The unbounded problem is a non-starter.

At least for non-mathematicians, and that is who most of our readers will be. For the mathematical reader I no not want to remove the mathematical details or dumb down the paradox in any way, but for the average reader, we should not start with an imposssible situation. My point is that to many people there is a natural division between bounded and unbounded, even if mathematically this is not so significant.

Regarding what I said about a mathematically well defined statement of this paradox I was talking about pure mathematics, which does not have money or envelopes, or, in my opinion and that of some others, probability. I do not think it is possible to create the paradox in pure mathematics although I have to admit that, if I saw one, I would probably not understand it. Martin Hogbin (talk) 12:19, 30 November 2014 (UTC)

In fact the mathematical issue here is that if the probability of twice as large an amount as a given amount rapidly declines, then expectation values are finite, and already in small samples, averages are close to expectation values. Moreover, truncating a distribution by some high cut-off hardly changes the mean value. So for practical purposes the bounded/unbounded distinction is *irrelevant*. The important thing is that the tail of the distribution falls off rather rapidly. This situation is often met in practice. But also we often meet in practice the opposite situation. So-called heavy tailed distributions. If you know that some value exceeds some large value, it becomes more and more likely that it exceeds it by more and more, as that value increases! Black swans. The financial markets. Hurricanes and global warming. So: unbounded distributions with infinite expectation values are actually very very appropriate is mathematical models for all kinds of significant real world phenomena! Of course, the height of an extreme storm-flood at the Netherlands North-Sea coast is bounded since it can't be larger than what it would be if all the water in the world were stacked in the North Sea. Yet from a prediction and mathematical modelling point of view a good model does *not* have an upper bound, but on the contrary, has a heavy tail ... Richard Gill (talk) 09:35, 30 November 2014 (UTC)

Yes but are these not different problems from the basic one? Martin Hogbin (talk) 12:19, 30 November 2014 (UTC)

All specific problems are different from one another. I am mentioning these other problems as counter-example to your argument, according to which all distributions with infinite support are banned from practical science and supposed only to be of interest within pure mathematics. Richard Gill (talk) 08:28, 7 December 2014 (UTC)

You are exaggerating my position somewhat. I am not trying to make any mathematical points or to ban the infinite from from practical science, I am trying to find a good way to explain the resolution of the TEP to the average reader. If you prefer, I am making an arbitrary distinction to aid readers' understanding. Many people do not accept or understand the infinite so the division into bounded and unbounded will be quite natural for them. No mathematics need be lost or corrupted. As an example, we are free split the proof of Pythagoras' theorem into two parts, one where the hypotenuse was over 7 cm long and the other where it was not, if we wish. That would, of course be completely pointless, but we could do it.

In the bounded TEP there is a very simple resolution that everyone can understand. I suspect that most readers will stop there and not worry about the unbounded case. Now I suspect that you would like to try and make them consider the unbounded case, because the resolution does not rely on the distribution being bounded. We cannot force them to do that, although we might try to encourage them. The first step, I suggest, would be simply to state immediately after that bounded resolution that, mathematically speaking, the resolution of the paradox does not rely on the boundedness and that there is a mathematically sound resolution that applies equally to the unbounded case. That statement alone might beenough for many readers but we could then go on to give the full resolution. Martin Hogbin (talk) 10:45, 16 December 2014 (UTC)

How about a very short general resolution first

If we start by saying:

For the proposed argument for switching to be valid, it is necessary that the probability that the unchosen envelope contains twice the sum in the chosen one must always be 1/2, regardless of what sum might be in the chosen envelope. It can be shown mathematically that, if this is the case, the expected sum in both envelopes must be infinite and switching is therefore pointless.

As an example, consider the case where the sum in each envelope must be a positive power of 2 (2, 4, 8, 16, 32 ...). If there is no upper limit on the sums that might be in the envelopes, then the expected sum in both envelopes is infinite. If there is an upper limit, for example, 512, then in the case that the chosen envelope contains 512, the other envelope must contain less, so the statement that the unchosen envelope always contains twice the sum in the chosen envelope is false and the argument for switching fails.

For all possible ways of filling the envelopes, it can be shown that if the probability that unchosen envelope contains twice the sum in the chosen one is always 1/2, regardless of what sum might be in the chosen envelope, the expected sum in both envelopes must be infinite. Martin Hogbin (talk) 09:59, 17 December 2014 (UTC)

Not quite right Martin. You did not understand the logic and/or the math. Yes, in order for the proposed argument for switching to be valid it is necessary that the sum in the chosen one must always be 1/2, regardless of what sum might be in the chosen envelope. No, this does not imply the expectation is infinite. It is far worse than that. In this case, if c is some possible amount in either of the envelopes, then the probability distribution of, say, the smaller amount x would have to give equal probability to each of the amounts ..., c/16, c/8, c/4, c/2, c, 2c, 4c, 8c, 16c, ... where the sequence is extended indefinitely on both sides. This is a so-called improper probability distribution - it cannot be normalised so that the probabilities add up to one, because if we did that, the probability of each separate amount would have to be zero. There is no uniform distribution an infinite discrete set. In fact, without normalising, we can say that there is infinitely more probability that the amount is both larger than any amount M, and infinitely more probability that the amount is smaller than any amount m, than the amount of probability that it lies between any two finite amounts m and M, with m < M. Richard Gill (talk) 08:14, 4 January 2015 (UTC)

I limited the sums that might be in the envelopes to positive powers of two, because this better represents the problem as stated. In that case is it not true that, if the probability that unchosen envelope contains twice the sum in the chosen one is always >=1/2 the expected sum in both envelopes must be infinite?

I do understand that if any power of 2 is permitted then the expectation in both envelopes is not well defined but I do not think that this situation is envisaged by the average reader. Could we not add that case as a third option. See below.

PS and things are not helped by going to continuous distributions. The continuous analogue of the improper discrete distribution I just mentioned, would be to let the logarithm of the smaller amount x be continuously uniformly distributed between - infinity and + infinity. In the Bayesian literature this is an often used and rather *natural* prior distribution, representing complete lack of knowledge, about an amount x>0, but it is an improper distribution, and that means that quite a lot of standard probability calculus breaks down. Richard Gill (talk) 08:19, 4 January 2015 (UTC)

One more question. If we accept that the (unconditional) expectation in both enveloped is undefined, how do we show that this makes switching pointless. What two quantities do we need to compare to show this? Do we, for example, need to compare the unconditional expectation in the first envelope with the (conditional) expectation in the second envelope given the expected sum in the first envelope?

Rather than saying that switching is pointless, is it better to say that the question is ill-posed. Martin Hogbin (talk) 09:45, 4 January 2015 (UTC)

To put this another way, if we allow any power of 2,, for any finite (neither infinite not infinitessimal) sum that we might find in the first envelope the paradoxical argument holds and we should switch. The problem is that we almost certainly will not find a finite sum in our first envelope. Martin Hogbin (talk) 09:55, 4 January 2015 (UTC)

How about a very short general resolution first - Version 2

If we start by saying:

For the proposed argument for switching to be valid, it is necessary that the probability that the unchosen envelope contains twice the sum in the chosen one must always be 1/2, regardless of what sum might be in the chosen envelope. It can be shown mathematically that, if this is the case, the expected sum in both envelopes is undefined and switching is therefore pointless.

As an example, consider the case where the sum in each envelope must be a positive power of 2 (2, 4, 8, 16, 32 ...). If there is no upper limit on the sums that might be in the envelopes, then the expected sum in both envelopes is infinite and switching is pointless. If there is an upper limit, for example, 512, then in the case that the chosen envelope contains 512, the other envelope must contain less, so the statement that the unchosen envelope always contains twice the sum in the chosen envelope is false and the argument for switching fails.

If we allow the possible sums in the envelopes to be any power of 2 (... 1/32, 1/16, 1/8. 1/4, 1/2, 1, 2, 4, 8, 16, 32 ...) then the expectation in both envelopes is undefined and switching is again pointless.

For all possible ways of filling the envelopes, it can be shown that if the probability that unchosen envelope contains twice the sum in the chosen one is always 1/2, regardless of what sum might be in the chosen envelope, the expected sum in both envelopes is undefined. Martin Hogbin (talk) 09:45, 4 January 2015 (UTC)

Discussion with Evgeniy E

I don't see why the conclusion that the sum in any envelope has no mathematical expectation — in other words, that the expectation is infinite — is destructive for the argument. The argument does not rely on existence of this expectation. I would like to point out that the argument for switching must fail whatsoever the distribution of the sums is. Reasons that make people not switch are merely logical, not ones of probability: we know that there is no difference between the envelopes, whichever envelope is initially chosen. All difference that may exist may follow not from the mental act of initial choice, but from some properties of the envelopes themselves; the formulation of the problem gives no such special properties, whatsoever the distribution of sums is. So, the distribution of sums in these envelopes is not even important; we can safely assume that the sums may take any real values from zero onwards, probabilities non-assigned. No? - Evgeniy E. (talk) 12:31, 17 December 2014 (UTC)

The argument for switching that is presented to us explicitly calculates an expectation as a reason to swap. Martin Hogbin (talk) 14:10, 17 December 2014 (UTC

You mean, implicitly… Well, the words "the expected value of the money" are indeed present, but no probabilities are distributed across the ranges of possible sums, which makes think that these values are not taken in money units. The probability variable we have to deal with is discrete, not continiuous. We are given an amount of money in the initially chosen envelope, but the value of this amount is not significant and is not subject to probability laws, existence of this amount should only aid imagination. That is my reading. As you see, it depends on how you read it. - Evgeniy E. (talk) 14:35, 17 December 2014 (UTC)

It is hard to see how the words "the expected value of the money" can be taken to have any meaning other than an expectation value, given the proposed calculation.

You say, 'no probabilities are distributed across the ranges of possible sums' and that is correct. Howver we are lead to believe that the probaility of the other envelope being double the chosen one is always 1/2, but this cannot be , unless the expected sums are both infinite. That is the trick Martin Hogbin (talk) 15:23, 17 December 2014 (UTC)

I am under impression that this probability is not derived from any calculation; otherwise, prerequisites for such calculation would be given and the calculation itself expressed. This probability is only assigned on the stage of formulating the mathematical face of the problem. The words "value of the money" do not specify the unit of assessing the amount of money; assuming, that this unit is something like dollars, pounds, rubles or tenges, makes the initial amount float across different experiments, which is contrary to the spirit of the paradox. (I. e. why we need it, why we are interested). So, I am led to see that this unit is supposed to be the amount in the initially chosen envelope.

If I understand you correctly, you say that in the conditions of the paradox as you see them, mathematical expectations do not exist ('infinity' is not a numeric value), and therefore their utilisation in any kind of argument makes no sense. - Evgeniy E. (talk) 11:36, 18 December 2014 (UTC)

This is not my argument it is that of many mathematical sources. You can find links to them on the article page.

Articles in WP must be based on reliable sources not our own private research or opinion. Martin Hogbin (talk) 10:23, 19 December 2014 (UTC)

Articles indeed, but I understand that this page is rather a "safety valve to let steam go", as we say in Russian, a method for directing unavoidable arguments about the problem somewhere where they provide no danger, than a way to discuss the article itself, at least the wording on the main talk page made me think so. Sorry for inconvenience, if any is provided. - Evgeniy E. (talk) 13:40, 19 December 2014 (UTC)

The problem with the paradox is that it is not clear exactly what the argument for switching is meant to be. We as solvers of the paradox are therefore forced to try to fomulate exactly what the proposer of the argument for switching is trying to say. Much of the literature on the subject falls into two categories in this respect.

One category assumes that the proposer of the switching argument is asking us to compare the (unconditional) expectation in the chosen envelope with (unconditional) expectation in the other one. I, and some others here, do not much like that interpretation of the bogus argument for switching because it do not seem to be in accord with a natural interpretation of the problem or with its history. Nevertheless we have included this interpretation, and its associated resolution, in the article.

The other interpreatation, which I prefer and which is the interpretataion being addressed above, is to take it that the proposer is trying to calculate the conditional expectation in the unchosen envelope, given a specific (but unknown) sum in the chosen envelope. This interpretation seems to me to match the wording of most versions of the problem better.

This argument for switching only works if the sum in the unchosen envelope always has a probability of 1/2 of being twice the sum in the chosen envelope. If this is true, the (unconditional) expectation in both envelopes is infinite/undefined. So, yes, the argument for switching does actually work in this case but, in the long run at least, swapping is pointless. Martin Hogbin (talk) 11:47, 19 December 2014 (UTC)

I think, the main conclusion we have to learn from this discussion is that no interpretation of this problem is natural, any interpretation depends on the personal history of the interpreter. From my side, I confess that I do not understand how an argument for swapping may both work and not work at the same time, but that may have something to do with me, and I leave it to you. :-) - Evgeniy E. (talk) 13:40, 19 December 2014 (UTC)

Yes I agree. I have said before that this paradox is a self-inflicted injury. In fact it is doubly so. Firstly the paradox has to be created from an obviously non-paradoxical situation by proposing a bogus argument for swapping, then the resolver of the paradox has to decide exactly what the bogus argument is supposed to be before the paradox can be resolved. But, as Richard Gill points out, it was devised to tease people. Martin Hogbin (talk) 15:15, 19 December 2014 (UTC)

The argument for swapping does not both work and not work at the same time. It does not work, finito. The problem is to explain why it doesn't work. Martin should not have said "the argument for swapping only works if ...". He should have said "the argument for swapping only appears to work if ...". In that case, one simply has to think a little bit further. The argument derails, a little bit later. Richard Gill (talk) 08:53, 21 December 2014 (UTC)

Richard, see my suggestion 'How about a very short general resolution first' above. This expresses the position more accurately, 'For the proposed argument for switching to be valid, it is necssary that the probability that the unchosen envelope contains twice the sum in the chosen one must always be 1/2'. Martin Hogbin (talk) 10:24, 21 December 2014 (UTC)

To make what my personal history suggests me more clear, I say that this injury is inflicted not by us, but by our minds. The situation in which a feeling seems to provide a reason for what is reasonless is as paradoxical as if we seemed to have a natural reason to believe that one cookie and one cookie make three cookies. Basics of probability theory, like basics of arithmetics, are a natural theory, i. e. one that a mind should be able to cope with intuitively, if provided with a little number of alternatives; at least, the mind must fail to find an intuitive solution, but not fall into a trap of incorrect handling; whenever the opposite happens, the reason must be found out. That is why I rejected any version that constructs a distribution of probabilities across al possible sums: it is evident that the mind, when falling into the trap, does not do it. What it does is rather this one: 1) formulate two discrete probability variables of two values each for the two options, 2) fill them with values as the wording of the problem suggests: 2 per ½ and ½ per ½ for the first option, and 1 per ½ and 1 per ½ for the second option, 3) compare the means, 4) see that the mean 'for the first option' is greater, 5) choose the envelope that corresponds to the first option. - Evgeniy E. (talk) 17:17, 21 December 2014 (UTC)

It is nothing to do with infinite expectations. It is all about improper distributions. Different (and worse) pathology. Richard Gill (talk) 08:21, 4 January 2015 (UTC)

Question

Why is it better to switch in nalebuff? — Preceding unsigned comment added by 2601:58B:4280:1180:2914:7D36:9319:2337 (talk) 04:24, 13 October 2021 (UTC)