(3) Michael and @andrewk have part of the right analysis, but something more is needed. — Srap Tasmaner

I disagree that the 50% chance can be assumed. If the sample space is finite, then there is a non-zero chance that the $10 amount is the maximum amount. — Andrew M

I agree with both of these. The point about the maximum possible amount has been made before, but has been lost in the length of the thread. I mentioned it somewhere early on, and in the interests of simplicity, declined to probe further because (1) I thought digging into it would have required some fancy notation like integral signs, and I couldn't remember how to invoke latex on this site (Edit: I see from the above that Michael has mastered it) and (2) I was pretty confident it would not make a material difference.

But given the continuing uncertainty I'll have a go. If nothing else, it should be able to bring out the intrinsically Bayesian nature of the calculation, which I think is camouflaged in the simplified versions discussed so far. Explicitly recognising a maximum possible value of X is part of recognising a prior distribution for X, that is then used in a Bayesian calculation.

I'll revert if I can either get latex to work, or manage to write the full-Bayesian calc without it.

Dormez bien, mon ami.

It's not about the games mistress's sample space. She doesn't have one. She has two amounts that she has always intended to put into the envelopes. It's about my sample space, which consists of the events that I cannot rule out based on my knowledge to date.

That knowledge - based on seeing 10 in my envelope - allows me to rule out every other event (an event being a particular sum being in the other envelope) except for 5 and 20. So that's my sample space - just two events.

To me, to say 'B is possible' means 'I would not be astonished if I discovered B to be the case'.

If I have 10 and somebody opens the other envelope, I will be astonished neither if it contains 5 nor if it contains 20. I would however be astonished - or think the games mistress has deceived me - if it contained 100, or any other number except 5 or 20.

I would say yes I do. I feel a discussion over the meaning of 'possible' may be approaching, but let's see where this goes.

Sorry, still not following you. One of what?

What does that mean? You know one of them must be in the distribution of X, but you don't know which — Srap Tasmaner

I mean that I know my envelope has 10 and that the other envelope has either 5 or 20.

I don't know what you mean by 'one of them must be in the distribution of X'.

The first question is about the prior distribution of X. My understanding is that most posters are assuming this is unknown to the player, and I have gone along with that. So the probability cannot be estimated.

The second question is about conditional probabilities. Given I know I have 10, the two possibilities that are open to me are 5, 10 and 10, 20, and I have no basis on which to distinguish between the two. So, being a Bayesian, I make the probabilities equal at 0.5. That approach is embedded in line 0 of the proof:

Proof: We use conditional probabilities to calculate the gain G from switching as follows, where the envelope we looked in contained £10:

E[G] = 0.5 * E[G | X= 5] + 0.5 * E[G | X= 10] ] . . . . . . . . . . . (0)
= 0.5 * E[-X | X= 5] + 0.5 * E[+X | X= 10] . . . . . . . . . . . (1)
= 0.5 * E[-5 | X= 5] + 0.5 * E[+10 | X= 10] . . . . . . . . . . . (2)
= 0.5 * (-5) + 0.5 * (+10) . . . . . . . . . . . (3)
= £2.50 . . . . . . . . . . . . (4)

I have no idea what a Frequentist would do.

If I have the larger envelope, then the other has 5 and the average value of an envelope is 7.5.
If I have the smaller, then the other has 20 and the average value of an envelope is 15.

(It is prima facie absurd that the average value of an envelope changes depending on whether you have the larger or the smaller of the two.) — Srap Tasmaner

Not if it is in two different games, which is what is happening here.

In the first case, the game has envelopes with 5 and 10 in them, average 7.5, and I got the larger envelope.
In the second case, the game has envelopes with 10 and 20 in them, average 15, and I got the smaller envelope.

What has changed is not just whether I have the larger envelope, but also what the games-mistress has put in the envelopes. So of course the average is different.

I regard all computer simulations as irrelevant to the question at hand. The question is one of mathematics, and must be addressed by mathematics, not programming.

I'm afraid the table(s) is(are) still not showing. It may be a conflict between my system and the web-site.

Perhaps you could enter the table as text rather than an embedded image.

Indeed. You have written a program with repeated plays that shows it is best to switch, and Srap appears to have written one that is only slightly different and shows that it makes no difference. That corroborates my contention that if one wishes to make an argument based on repeated plays, the devil will be in the detail of the set-up of the system of repeated plays, and changing the tiniest detail can change the result.

Your error, BTW, is having the two situations are that X = 5 or that X = 10 — Snakes Alive

Good, so you are challenging line 0 of my proof.

Why do you think it is an error? Do you not agree that, if I open the envelope and see 10, the possibilities are that X=5 or that X=10, and that I have no reason to favour one over the other?

The fact that one can think of it another way is neither here nor there. If you want to challenge this line of analysis, you need to find a flaw in it, not just say 'the correct way to think about it is this other way...'

BTW I tried to read your first post a day or so ago, but the embedded images were corrupted, so it was not comprehensible. If that has been fixed, by all means post a link (I don't know where in this long thread it is now) and I'll have another look.

That is a coherent response. Whether you realised it or not, you have sided with the Frequentist approach, and a Frequentist will indeed reject my calculations, saying that there are no probabilities for values of X.

I am a Bayesian, so the calculation is valid for me.

I would love to explore the Frequentist alternative, but nobody has so far (to my knowledge) offered a formal presentation of it - the Frequentist counterpart to my Bayesian calculation.

My current expectation is that there is no Frequentist alternative, because one cannot use any probability distributions at all. But I would be delighted to be shown to be wrong. I suggest you (implore you to?) have a go at writing a formal argument from the Frequentist perspective, in which none of the items are treated as random variables.

We could get into a discussion of what's wrong with it, but we'd be discussing the wrong thing. Computer programs - with very few exceptions like dedicated theorem-proving programs, which is a very niche area - can only ever be used to do calculations and generate intuitions. They cannot prove anything. To do that we need to use mathematics, which is what my proof does. If we stray away from mathematics, we just end up trading word salads, which is what most of this thread is.

If the computer program has any valid insights, it should be able to be expressed as a mathematical proof, that is presented as a series of numbered formal propositions, each being either a premise or else having a justification as to how it can be deduced from one or more previous propositions in the sequence. If you can produce such a proof, I promise to engage with it.

And here is mine. I haven't written out the justifications for the steps yet because most of them are obvious, so I'll wait until somebody engages it and challenges a line, before taking the trouble to write out the justification for it.

We use conditional probabilities to calculate the gain G from switching as follows, where the envelope we looked in contained £10:

E[G] = 0.5 * E[G | X= 5] + 0.5 * E[G | X= 10] ] . . . . . . . . . . . (0 - Premise)
= 0.5 * E[-X | X= 5] + 0.5 * E[+X | X= 10] . . . . . . . . . . . (1)
= 0.5 * E[-5 | X= 5] + 0.5 * E[+10 | X= 10] . . . . . . . . . . . (2)
= 0.5 * (-5) + 0.5 * (+10) . . . . . . . . . . . (3)
= £2.50 . . . . . . . . . . . . (4)

If you could prove that always switching is the best strategy over the long term, doesn't that amount to proving that you are more likely to have chosen the smaller envelope? Why doesn't that bother you? — Srap Tasmaner

There is no long term in this setup. There is a single offering. If we want to consider a long series of such activities, it needs to be set up precisely, and that has not been done. A number of crucial conditions will need to be specified, such as whether X is always the same and if not, how it varies from one play to the next. The best strategy will vary according to how those conditions are set up.

In any case, a setup with repeated plays is a completely different probability space, and does not entail anything for this one, or vice versa.

I'm going to reprise my proof, because three pages have been added and so far nobody seems to have attempted to point out an invalid step in it:

we use conditional probabilities to calculate the gain G from switching as follows, where the envelope we looked in contained £10:

E[G] = 0.5 * E[G | X= 5] + 0.5 * E[G | X= 10] ] . . . . . . . . . . . (0)

= 0.5 * E[-X | X= 5] + 0.5 * E[+X | X= 10] . . . . . . . . . . . (1)
= 0.5 * E[-5 | X= 5] + 0.5 * E[+10 | X= 10] . . . . . . . . . . . (2)
= 0.5 * (-5) + 0.5 * (+10) . . . . . . . . . . . (3)
= £2.50 . . . . . . . . . . . . (4) — andrewk

I agree with everything in that post. I too would love to see the world attain that state of having a single, trusted, global police force, and I feel uncomfortable at the celebration of national armed forces. Perhaps I'm just more pessimistic than you about the chance of ever getting there from here. I am a gloomy type more often than not, but am quite ready to believe that my pessimism about life is unwarranted. If only I could convert that intellectual belief (that optimism can be justifiable) to a belief of the heart.

On choosing an envelope and learning the amount in it, there is still no reason to prefer one envelope to the other. So there is no reason to switch. — Andrew M

That claim is not consistent with the following formal calculations from my earlier post:

we use conditional probabilities to calculate the expected value of the gain G from switching [after seeing 10 pounds in the first envelope] as follows:

E[G] = 0.5 * E[G | X= 5] + 0.5 * E[G | X= 10] ] . . . . . . . . . . . (0)

= 0.5 * E[-X | X= 5] + 0.5 * E[+X | X= 10] . . . . . . . . . . . (1)
= 0.5 * E[-5 | X= 5] + 0.5 * E[+10 | X= 10] . . . . . . . . . . . (2)
= 0.5 * (-5) + 0.5 * (+10) . . . . . . . . . . . (3)
= £2.50 . . . . . . . . . . . . (4) — andrewk

which say that, after opening the first envelope and finding 10 pounds in it, the expected gain from switching is 2 pounds fifty.

I have numbered the steps from 0 to 4. Which step(s) do you not believe?

Are you saying that you have reservations about my corrected version set out in that post? It is a fairly straightforward calculation based on conditional expectations, for the two possibilities. Is there a particular part of it that concerns you?

I don't think you understand the concept of a probability distribution and, in relation to statistics, I don't think you know what those density curves are for — Jeremiah

You've found me out. Please don't tell my employers that, or I'll lose my job.

[Jeremiah's] reasoning is:

I have £10. If X is 5 then I lose £X by switching. If X is 10 then I gain £X by switching.

So it's either -£X or +£X. This is symmetrical. — Michael

The conclusion of symmetry is incorrect, because X is £5 in the first case and £10 in the second case. So we can't say that there's no difference because we lose X in one case and gain X in the other.
That would be like saying that if I have two £1 coins in my pocket and a big hole in the pocket it makes no difference whether I take one of the coins out and put it in another, non-holy pocket, because in both cases I lose everything in the holy pocket.

Doing this approach correctly, we use conditional probabilities to calculate the gain G from switching as follows:

E[G] = 0.5 * E[G | X= 5] + 0.5 * E[G | X= 10] ] . . . . . . . . . . . (0)

= 0.5 * E[-X | X= 5] + 0.5 * E[+X | X= 10] . . . . . . . . . . . (1)
= 0.5 * E[-5 | X= 5] + 0.5 * E[+10 | X= 10] . . . . . . . . . . . (2)
= 0.5 * (-5) + 0.5 * (+10) . . . . . . . . . . . (3)
= £2.50 . . . . . . . . . . . . (4)

So this approach, when done correctly, reaches the same conclusion that you did - that the expected winnings are increased by switching.

Well technically this is a probability distribution: 1/2(X) + 1/2(2X). 50% is distributed to X and 50% is distributed to 2X.

That distribution is (1) unknown, as X is unknown and (2) not appropriate for estimating expected gains from switching, since it does not use all the available information after the envelope has been opened. The expected gains from switching, in the absence of any knowledge of how X was selected, is as @Michael calculated it earlier. But the question in the OP does not explicitly ask for expected gains from switching. It just asks 'should you switch?'. What is supposed to be the metric used to determine whether to switch?

This has nothing at all to do with distributions. — Jeremiah

That is not consistent with your use of the term 'expected value' in the following:

In both cases our expected value then is:

1/2(X) + 1/2(2X) = 1/2X+X.

Which is the mid point between X and 2X. — Jeremiah

'Expected value' only has meaning in the context of a probability distribution.

There are two possible amounts for the contents of the two envelopes either X or 2X. You don't know which is which — Jeremiah

This is not well-defined. It needs re-stating to make it unambiguous. Here are some options.

(1) A number X is drawn from a distribution f. A second number N is drawn from the Bernoulli distribution with parameter 0.5 (think of a coin flip). If N=1 (2) the amount X is placed in envelope 1 (2) and 2X is placed in envelope 2 (1). None of f, N or X are known to the player.

Note that this includes as a special case the setup where X is a fixed number known to the game operators, but not to the player. In that case the distribution f is just the uniform distribution on interval [X,X].

(2) Same as (1) but the player knows f.

In case 1, the player needs to assume a prior distribution as her guess of f. From then on, the calculation is the same, and it will use the known (case 2) or assumed (case 1) distribution f. The answer of whether expected winnings are increased by switching will depend on the slope and curvature of the PDF of f at the value that is observed when the envelope is opened.

From what I can see by attempts above, the assumption has been made implicitly that f is a uniform continuous distribution on the interval from 0 to some very high number such as the total of all banknotes in the world. That means the slope and curvature of the PDF at the value observed is zero.

Note the distribution must be continuous, which means that rather than banknotes, there must be an IOU specifying a real number that will be paid. If it was discrete, which is necessary if the envelope contains banknotes, then if the value was an odd multiple of the smallest currency unit, we'd know we have the smaller-valued envelope.

I suspect that even with that sparse, uniform distribution assumption, one might be able to make some inference. For instance, if the amount seen is very small, I suspect the Bayesian probability that it is the lower of the two amounts may be more than 0.5. But I've not thought about that much and it could be wrong.

I picked this version to generate more discussion. I think this version of the problem forces more conflict of ideas, which in turn generates more discussion. — Jeremiah

It was a good choice. In the version in wikipedia it is easy, and uncontroversial, to conclude that there is no reason to switch. Not so in this case.

Not true. — Jeremiah

You said this in reply to my pointing out that arguments based on expected values prior to opening an envelope are wrong because we don't even know if there is an expected value. To rebut that, you need to show that there is an expected value, eg that we can be confident the lower amount X is not drawn from a Cauchy distribution.

Yet in your answer you have not addressed this at all.

What exactly is it then, that you think is Not True?

I think you are trying to justify violence and pretend it is peace. — unenlightened

I am seeking to justify something - armed deterrence - that you call violence and that I do not. The state that results from that, I call peace and you do not.

Even though I think your terminology is very different from common usage, I am happy to adopt it for the purpose of this conversation, and call armed deterrence 'violence', except I insist it must be qualified as non-aggressive violence, to distinguish it from violent acts that I initiate - aggressive violence. With that terminology I am in favour of non-aggressive violence at the nation-state level. I am not in favour of it at the personal level, because I support gun control.

Your analogy of putting a gun to somebody's head does not fit, because that is aggressive violence. In that analogy, non-aggressive violence would be simply carrying a gun, or a big stick, to deter people from attacking me. But even then, as per the previous paragraph, the analogy doesn't work, because I do not support unrestricted gun ownership. What is suitable for nation-states is not necessarily suitable for individuals.

I have merely pointed out and lamented that the world that we live in is founded on violence. — unenlightened

As far as this goes, I am in passionate agreement. Your OP was poetic and moving. Poetry can evoke powerful emotions, in this case, for me, emotions of sadness that we, like all other animals, have violence indelibly written in our genes.

If all you are saying is 'It's a pity that we have armies', I agree. But if you want to advocate for the UK decommissioning its armed forces completely, I disagree. Perhaps you are not advocating that, in which case I misinterpreted your subsequent posts and I apologise.

It occurs to me that, on my understanding of the two main interpretations of probability, and assuming no knowledge of the prior distribution of the lower amount, a Bayesian interpretation can take the approach that, on seeing $10 in the envelope, the expected winnings from switching envelopes is $2.50, but a Frequentist interpretation does not provide any approach at all - it denies the correctness of a probabilistic calculation of the winnings: the winnings are either $10 or -$5, and whichever one it is, is certain - no partial probabilities.

Sounds like a good reason to be a Bayesian.

It appears we have different ideas of what constitutes violence and what constitutes peace. I doubt that difference can be bridged.

It's peaceful as long as the violence is only a threat.

If violence actually happens then it depends on why it happened. If it happens in the name of colonialism or other projections of power overseas, as has so often happened in the history of European countries and the US, then it increases the violence, and I abhor it. On the other hand if the violence is defensive, then it has only replaced other violence that was going to happen anyway when the aggressor invaded our country. If the military is well-trained and well-equipped, the violence will probably be less than what would have happened if the aggressor had invaded without military opposition.

I am in favour of having a defensive military service, but not an aggressive one. The Japanese have the right idea, calling it a defence service rather than an army.

I tend to agree with Hanover, and with Hobbes, that it is the threat of violence on our behalf by the armed forces and police that allows people like you and I that have been granted the rare privilege of a peaceful life, to continue to have that privilege. It keeps the Huns, Goths, Visigoths, Mongols, Romans, Angles, Saxons, Jutes, Vikings and Normans at bay.

I fully recognise the tragedy that that privilege is not extended to people growing up surrounded by violence in the Jasmine Allen Estate or such like, and I wish I knew how to fix that, but I don't.

Interesting. The problem is different. Unlike in the OP, in the Wiki case, the envelope has not been opened before the option to swap. So the player has no new information in the wiki case. I think their analysis in the Simple Case is wrong, because it assumes the existence of an expected value that may not exist, but I think the conclusion that there is no reason to switch may in spite of that error be correct in that case, but not in this one.

I am trained in mathematics, and use it in my work, but never studied formal logic until several years ago. I self-studied using what I could find on the internet, motivated by a desire to fully understand Godel's incompleteness theorems.

I think it has helped my thinking a great deal. Whereas previously I could tell what was a valid proof, just by years of working on proofs in mathematics, I did not have the vocabulary and concepts to explain exactly what was wrong with a flawed proof, or to defend the correctness of a valid one. Studying logic gave me those tools. I have found it also very useful in debate, and in critically assessing tendentious statements by politicians or lobbyists. It equips one to spot the loaded question, the hidden assumption, and such like. Whenever somebody claims A therefore B, I find myself thinking 'what rule of inference did you use to get from A to B, and if you didn't use one, what's the basis for your assertion?'

I am enthusiastic enough about it to even suggest that maybe some basic logic should be introduced to the curriculum in late High School.

It looks like an error to me. A set of truth assignments cannot satisfy both ¬φ[τ] and ∀x.φ[x] because the latter entails φ[τ].

Do you have a link? It could just be an error. Universities tend to have fairly high error rates in their materials these days, because all the fee income from undergrad students is funnelled into research to push the uni up the rankings list, rather than into quality control on student learning materials.

It is hard to argue that armed forces are not necessary to a peaceful life, as Hanover pointed out. I think of them as a necessary evil. I admire those who join the armed forces, or the police for that matter, for the right reasons - a desire to serve and protect. I fear that many join both for wrong and anti-social reasons, but that's humans for you. We still need the services.

What we don't need though, is a celebration of the weapons of destruction, like fighter jets, warships, missiles and tanks. Parading those is not just carrying a big stick. It's waving it about.

The trouble is though - human nature again - little boys just love military toys. I know this because I once was a little boy, and I loved them then. So if we want people to occasionally get enthused about the armed forces - and that is to some extent necessary in countries like ours that have no compulsory military service - it's a lot easier if we trot out the big toys for the kids to ooh and aah over, overlooking the sad truth that their purpose is to kill people and mangle flesh.

As Jeremiah points out, your code doesn't reflect the problem in the OP. Before an envelope is picked, the expected value of each envelope is the same. — Andrew M

It's necessary to distinguish between two cases, which is whether we know the distribution of X, ie the distribution for the lower of the two values.

If we don't know the distribution then we don't even know whether there is an expected value of the amount in either envelope. For instance, if the amount put in the envelope is a random draw from the Cauchy distribution, there is no expected value (the relevant integral for the Cauchy distribution does not converge).

If we do know the distribution, and it has an expected value, then it is correct that our expected value is the same for both envelopes. But that changes when we see the amount in the envelope, because that tells us where we are in the distribution. Say we know the distribution of the smaller amount X is uniform on [1,2]. Then if the amount we see is in [1,2] we know it is the smaller amount and we should switch. On the other hand if the amount is in [2,4] we know it is the larger amount and should not switch.

In the absence of knowing the distribution of X, any calculations based on expected values prior to opening the envelope are meaningless and wrong. Since the claimed paradox relies on such calculations, it dissolves.

If Γ ⊨ ¬φ[τ] for some ground term τ, then Γ ⊭ ∀x.φ[x]; true or false?

Isn't it true? — Rayan

That sounds correct.

Am I missing something? Because when I check the answers, it turns out that it is false... — Rayan

It sounds like you feel you've come up against something that is contrary to the above. I can't see it. What do you see that concerns you?

But isn't above a binary relation constant? It can't be is a block. — Rayan

Yes, that was a mistake. 'is a block' is unary. Instead let it be 'is the same shape as', which is binary.

The expected value of each envelope is $(X + 2X)/2 = $3X/2 = $1.5X. You should be indifferent between switching or not. — Andrew M

It looks like you're assuming the player's utility function is the identity function, which would be unusual and unrealistic. Even if we assume that, the quoted calculation doesn't take account of all the available information. There is new information, which is the known dollar value in the opened envelope. That changes the utility calculation.