You seem to be saying that after picking an envelope I have to go from saying that there's a probability of 0.5 that I will pick the smaller envelope to saying that the probability is unknown that I have picked the smaller envelope. — Michael

If there exists some bounded and normalized (meaning that the probabilities add up to 1) prior probability distribution that represents your expectation for the possible distributions of envelope pairs then, in that case, your average raised expectation for an always-switching strategy is zero, for reasons that many have expounded in this thread. However, any such prior bounded probability distribution which might represent your expectation is inconsistent with your being able to apply the principle of indifference to whatever case of X that you might observe in the first envelope. Your knowledge (or assumption) of this prior distribution rather allows you to calculate exactly the exact expectation conditionally on any X, and this is generally different from 1.25X.

If, on the other hand, you take the uniform and unbounded distribution to represent your prior expectation, then, in that case, you can apply the principle of indifference whatever X it is that you might observe in the first envelope. Under this wild assumption of a uniform expectation, for any number M, however large, your prior expectation was that it was infinitely more likely that X would turn out to be larger than M rather than smaller or equal to M. So, it is no surprise that you are expecting a gain from switching. (You can refer back to my Hilbert Grand Hotel example, earlier in this thread, for another illustration of the consequences that follow from assuming such a wild unbounded and uniform 'uninformed' probability distribution.)

In the case where you are confident that there is some bounded, and therefore non-uniform, prior distribution of envelope pairs, but you don't have a clue how to go about estimating what it might look like, the mere assumption that there exists such an unknown prior distribution is enough to rule out the wild degenerate case described in the prior paragraph. When faced with some determinate amount X, you will possibly not have a clue what the expectation from switching might be. But that is no reason for assuming it to be zero. Hence, there is no reason either for assuming (or inferring) that the expectation from switching is exactly 1.25X. What you can reasonably assume, rather, is that the higher the value of X is, the more risky it is to make the switch, and this knowledge that the risk of a loss increases roughly in proportion with the value of X is inconsistent with the unconditional application of the principle of indifference.

But what action does your answer entail? Switching or sticking? If you say it doesn't matter, and so you're being indifferent, isn't that the same as treating it as equally likely that the other envelope contains the larger amount as the smaller amount? And if you're treating them as equally likely then isn't it rational to switch?

Yes, it is rational to switch if you are justified in treating them as equally likely. But if it is axiomatic that they are equally likely, as most statements of the two-envelopes paradox seem to make it, then you must also infer that the prior distribution is uniform and unbounded with all the weirdness that such an ill-defined probability distribution entails.

If you are using the principle of indifference then criticizing people for using the principle of indifference, that is hypocritical. Either accept that as a standard starting point or don't — Jeremiah

What I criticized merely was a failure to draw a logical inference from one particular application of the principle of indifference. The logical inference that must be drawn from the assumption that the principle of indifference can be applied unconditionally on X (the value of the seen envelope) is that the prior distribution must therefore be assumed to be unbounded and uniform. If the player correctly draws this inference, then she can still apply the principle of indifference and expect to gain 0.25X (on average) from switching from her envelope (containing $X) to the other one and this expected gain doesn't yield a paradox since, in the case of such an unbounded distribution, however large X might be, it was infinitely unlikely that it be so small and, also, her average expectation from an always-switching strategy isn't any larger than her average expectation from an always-sticking strategy since both are infinite.

So you think you always have a 1.25 expected gain in every case? — Jeremiah

Not at all. I have rather argued that there is an 1.25X expected gain from switching in one specific case of a known distribution {{5,10},{10,20}} where, by your own argument, switching ought to be no better than sticking since we are ignorant of the case and, according to this argument, the cases therefore can only be treated separately and don't justify the 1.25*$10 expectation.

As far as I am concerned I already found the flaw. Take it or leave it, that is your choice. — Jeremiah

I have chosen a third option, which is to point out the logical flaw in your purported identification of "the flaw". As I suggested earlier, your own resolution of the paradox relies on an argument that proves too much, since it leads to wrong inferences about expectations in specific cases.

Hey, if you feel lucky then switch, if you think you are close to the cap don't, feel this one out, but you are not going to be able to quantify a positive gain based on the information we have. — Jeremiah

This is something I have never disputed. I have never purported to offer an optimal strategy or suggested that there is any way to come up with one. The two-envelopes paradox is, precisely, a paradox because under some widespread interpretations of "don't know" (regarding the prior distribution, and the probability that the open envelope is the smallest one) there appears to be two equally valid arguments that purport to conclude that switching yields a positive expectation or that it yields a null expectation. Since those two conclusions are inconsistent, the resolution of the paradox calls into finding the flaw in (at least) one of the two arguments. Considerations of well defined strategies only are meant (by me) to illustrating flaws in the arguments that purport to lead to two inconsistent conclusions on the basis of a common set of assumptions regarding the possible initial distributions. A few other participants in this thread (such as JeffJo, fdrake and both Andrews) have offered diagnoses similar to mines of the most common mistakes that lead one to the erroneous and paradoxical conclusions.

It is also rational to want ice-cream on a hot day. You still don't know anything about the distribution. You are speculating then trying to model your speculations. — Jeremiah

When I say that this can be inferred from that, or that it is fallacious to infer this from that, then I am either right or wrong about it; and in the case where you think I am wrong, an argument is forthcoming. None of my claims purport to be empirical or speculative (except when I explicitly hedged some as conjectures earlier in the thread).

I take the two-envelopes paradox to be a puzzle about probability theory and there is little point speculating rather than arguing logically about it. The use of models is perfectly fine for illustrative purposes, for conveying a concept across, or for supplying proofs of existence.

Statistics is a data science and uses repeated random events to make inference about an unknown distribution. We don't have repeated random events, we have one event. Seems like a clear divide to me. You can't learn much of anything about an unknown distribution with just one event. — Jeremiah

None of my arguments relied on being able to lean "much" about a distribution from one single observation. My arguments rather relied entirely on logical relations between definite claims about possible distributions and conditional probabilities.

You don't know the distribution, you don't know the limits and you only get once chance to switch. — Jeremiah

It is rational to want to maximize your expectation even when you only get one single chance to play, and it is irrational to dismiss your expectation merely on the ground that just one of the possible outcomes will be realized.

Suppose you are forced to play Russian roulette once. There are two revolvers, one with five bullets in the cylinder (and one empty chamber) and the other one with one bullet (and five empty chambers). The revolvers are truthfully labelled accordingly. You are free to pick any one. Are you arguing that since you only are going to play once, it's irrelevant which revolver you choose?

You are no longer talking about just probability anymore, since you can now sample the distribution you are now engaged in statistics, which is outside the scope of the OP — Jeremiah

It is rather difficult to divorce discussion of probabilities from discussion of statistics. You can't really build an insulating wall between those two disciplines. It doesn't make much sense to talk about probabilities of events that are singular, unique, occurrences not belonging to any sort of distribution. Probability distributions do have statistical properties. You just seem to want to outlaw, by fiat, arguments that make trouble for your case.

Actually only one case is true, while the other one does not exist. So they can't both be possible outcomes, not objectively. You are modeling your assumption of what you think is possible. However, just because you can think of something that doesn't mean it is objectively a possible outcome. — Jeremiah

Of course "only one case is true" at each iteration of the game. Still, the player doesn't know which one is true at each iteration of the game when he finds $10 in his envelope. But the player does know what the distribution is and, therefore, that, in the long run, each one of the two cases will be realized an approximately equal number of times. This is what allows her to calculate that her average gain from playing the game repeatedly will be $2.5 if she adopts the strategy of switching whenever her envelope contains $10. This is also what makes it rational for her to switch when the game is played only once and she is willing to risk losing $5 for an equal chance of winning $10 since she doesn't know which "one case is true" but knows them to be equally distributed.

The 1.25X come from considering expected gains over both cases, the larger and smaller. However when one case is true the other cannot be true, so it makes no sense to consider expected gains in this fashion. They should to be considered separately. — Jeremiah

There is a sort a move in philosophy that is called "proving too much". An argument proves to much when it succeeds in proving the thesis that one purported to demonstrate but, unfortunately, it also proves some corollary that this quite embarrassing.

The trouble with your argument is that, while it indeed (purportedly) concludes that you can't expect to have a positive gain from implementing an always-switching strategy, it also ought to yield this conclusion when the initial distribution is bounded, known, and X is also known.

Suppose for instance the initial distribution simply is {{5,10},{10,20}} with each pair equally probable. Suppose also you open your envelope and find $10. Should you switch? In this case, you should. Your raised expectation from switching is $2.5. This is what you tend to gain on average when you plays the game several times. But your own argument would lead us to conclude that the expectation from switching is the very same as the expectation from sticking. You are arguing that the two cases must be considered separately rather than being weighted in accordance with their posterior probabilities. Hence, in the case where the envelope contents are {5,10} your gain from switching is $5 when you have $5 and -$5 when you have $10, or, overall zero. And then you likewise would consider the {10,20} case "separately" and conclude that the switching strategy yields no increased expected gain in that case either (you either win or lose $10 from switching, in that case). The trouble is that this way to frame the problem offers you no guidance at all regarding what to do when you know that your envelope contains $10 and you don't know which of the two case {{5,10},{10,20}} is actual. It precludes you from making use of your knowledge that both of those two possible envelope distributions still are equiprobable conditionally on your having found $10 in the first envelope.

Do you the the point or not? — Jeremiah

No. You seem to be reaching for an argument that purports to show that your raised expectation from switching, conditionally on having found out that there is some determinate amount X in the first envelope, is zero regardless of the initial envelope pair distribution and regardless of X. I can't see how this argument can work just on the basis that you don't know the initial distribution. While it's true that, on the mere assumption that the initial distribution is bounded, the overall raised expectation of the always-switching strategy is zero, it's not generally true that the expectation, conditional on seeing the amount X in the first envelope, is zero. Although it may not be possible to know, or calculate, what this conditional expectation might be, there is no reason to conclude that it is zero.

Never said any thing about both being actual at once. — Jeremiah

You just said "The envelopes cannot be in both cases at once", followed with the word "therefore...". So it looked like you were making an issue of the fact that they can't be "in both cases" at once. But that is quite uncontroversial.

That is what I just did. The envelopes cannot be in both cases at once, therefore it makes no sense to hedge your expections that both cases are possible. — Jeremiah

The argument that purports to show that the expectation from switching is 1.25X doesn't rely on both possible cases (possible consistently with the information that is available to you, that is) being actual at once. It only relies on them being equiprobable; or both equally likely to be true, consistently with everything that you know.

And I am saying that doesn't really matter because it will always be amount A and amount B. — Jeremiah

It doesn't really matter for what? It does matter for invalidating the fallacious argument that purports to show that your expected gain from switching, conditionally on having initially opened an envelope with the determinate amount X in it, is 1.25X.

See that was easy. — Jeremiah

Sure, but that's not what the equiprobability assumption is. What I have been referring to as the equiprobability assumption is the assumption that your credence in having picked the smallest envelope, which is 1/2 before you open it, remains 1/2 conditionally on there being the determinate amount X in it for any X. This is an assumption that can only be reasonably held (if at all) if the distribution of possible envelope contents is assumed to be unbounded and uniform.

I have two envelopes, one with amount A and one with amount B. I flip a fair coin to choose one. What is my chance of getting B? — Jeremiah

1/2

I think we are safe, I doubt anything will blow up. — Jeremiah

Things have blown up long ago. It is precisely the endemic oversight of the logical dependency at issue that is the source of the apparent paradox being presented in the OP. There is an illicit move from an assumption of equiprobability regarding the conditional probability of one having picked the smallest envelope (conditional on X, whatever X one might pick) to the assumption that there might be a bounded distribution of possible envelope pairs that is merely unknown. Those two assumptions are logically inconsistent. Either the unknown distribution isn't (as it indeed can't be, in realistic cases) uniform and unbounded or the equiprobability assumption is true. But if the equiprobability assumption is true, then the initial distribution for possible contents of the smallest envelope in each possible pair must be unbounded and uniform.

It absolutely can be ignored. — Jeremiah

To ignore logical dependencies between claims in rational arguments is a recipe for disaster.

The filling of the envelopes and the selecting of the envelopes are two separate events. — Jeremiah

I know that. But assumptions regarding the method for filling up the possible envelope pairs (and hence their distribution) entail logical consequences for the conditional(*) expectations of one having picked either the smallest or the largest one within one given pair. This dependency relation between the two successive events can't be ignored.

(*) Conditional on the observed value of the first envelope, that is.

Never said it was. — Jeremiah

OK. Once one is reminded that such an 'uninformed' prior, regarding the initial possible contents of the envelopes, isn't reasonable, then, it follows that an uninformed prior regarding the conditional probability that one has picked the smallest envelope, conditional on its value being X, is likewise unreasonable for at least some value of X. And that's because an unconditional uninformed prior of 50% (valid for any observed X value) entails a uniform and unbounded distribution for the possible envelope contents.

If a loaded coin flips H 9 out 10 times, without that knowledge, an uninformative of 50/50 prior is completely justified. — Jeremiah

Yes, for sure, but if someone hands me, as a gift, an envelope containing some unknown amount of dollars for me to keep, an 'uninformative' prior that is a uniform distribution (or continuous probability density function) from zero to infinity isn't reasonable for me to use for almost any purpose.

In this case, it's the fact that the Hotel has countably infinitely many rooms that enables the assumption of equiprobability to hold. — Pierre-Normand

I was rather careless in this post where I had made use of Hilbert's Grand Hotel for illustrative purpose. In my thought experiment the equiprobability assumption didn't actually hold because, after the guests have moved to their new rooms, they can often deduce from the room number they end up in where it is that they came from. Initially, the guests are equally distributed in the rooms numbered 2,4,6, ... Then, after flipping a coin and moving, accordingly, from room X to either room Y = X/2 or 2*X, there are three cases to consider. They may end up (case A) in a room within the range (1,3,5,...); (case B) within the range (2,6,10,...); or, (case C) within the range (4,8,12,...).

In case A, the players can deduce from the number Y of the room where they moved being odd that they moved there from room 2*Y and hence will want to move back if given the opportunity.

In case B also, they can deduce the same thing since, although Y/2 is an integer, it is an odd number and hence not in the initial range. They will also want to move back if given the opportunity.

Only in case C will they not be able to deduce where it is that they came from. About half of them, on average, will have moved to room Y from room Y/2 and the other half from room 2*Y. Their expectation for moving back to the room where they came from, conditional on them being in room Y, is 1.25Y (whereas, in cases A and B, the guaranteed outcome of moving back was 2Y)

So, even though the expectation that any guest who starts up in room X, from the initial range (2,4,8,...), is 1.25X after 'switching', the conditional expectations for moving back from Y to the initial room is either 2Y or 1.25Y. Rational (and greedy) guests should always accept the option to move, and then, wherever they end up, provided only that they don't know where they came from, they also should always accept the option to move back. This only makes sense in Hilbert's Grand Hotel, though, and not in any hotel that has a merely finite number of rooms.

I have a feeling though that Michael will still think that absent knowledge of the distribution, he can turn back to 50% as an assumption. — Srap Tasmaner

The best way to counter that assumption, it seems to me, just is to remind oneself that even though one may not know what the distribution is, assuming only that it is not uniform and unbounded, then, whatever this unknown distribution might be, the raised expectation from switching always is zero. So, it doesn't matter that the actual distribution is unknown. It's still known that it must be such as to make the raised expectation from switching zero, for the reason illustrated by @Andrew M in one specific case.

On edit: "the raised expectation from switching" is meant to refer to the average expectation of the always-switching strategy. The specific expectation of switching, conditionally on having found some determinate amount X in the first envelope, can only be guessed and can't be assumed to be zero.

I recall they did. Then there was a very strong rumour that Snr. helped draft a statement about the meeting after the news of it broke, whilst on Air Force One. I think that is one of the subjects of the 'obstruction of justice' part of the investigation. — Wayfarer

If you are talking about Don Jr's statement about the meeting (being about adoption, etc.), then it's more than a strong rumor that Donald J. Trump helped draft it. Trump's legal team acknowledged that he dictated it to Don Jr. This admission came after repeated denials that Trump Sr had anything to do about it.

I think I'm just reluctant to see the simple situation of choosing between two envelopes of different values in terms of the strange behavior of infinity. — Srap Tasmaner

Yes, it is indeed the strange behavior of infinity that generates the paradox.

(Edit: Opps, there are some mistakes below that I'll correct later today)
((Edit again: I'll correct them in a separate reply to this post))

Maybe I can get some more mileage from my analogy with Cantor's Hotel(*). Suppose you are hosted in this Hotel in some random even numbered room X. (The rooms are labeled 1,2,3,...). You are then offered to toss a coin and move to room X/2 if you flip tails, and move to room 2X if you flip heads. You can also choose not to toss the coin and stay in your room. Whatever you do, when you leave the hotel you are being awarded your room's number worth of dollars. Suppose everyone who occupies an even numbered room gets offered the same deal and everyone chooses to flip the coin. After the move, some rooms will be empty and some rooms will have one or (at most) two occupants. But that's immaterial. The point is that guests can expect to increase their rewards by 1.25 on average. It is also true that if they had chosen to flip the coin without looking at their room number, and only know the number Y of the room where they relocated as dictated by the result of the coin toss, they would still expect to increase their reward expectation to 1.25Y if offered to move back to whatever initial room they came from (assuming they had forgotten what it is that they flipped and hence couldn't deduce where it is that they came from before choosing whether to move back).

In this case, it's the fact that the Hotel has countably infinitely many rooms that enables the assumption of equiprobability to hold. Before the coin flip, each occupant is equally likely to double her reward than she is to halve it and hence is warranted to flip. But conversely, after she moved, she is equally likely to have moved there from a rooms that was twice as valuable than from a room that was half as valuable and hence is warranted to move back if given the opportunity, assuming only that she has forgotten where she came from!

(*) On edit: What I referred to as Cantor's Hotel is better known as Hilbert's Grand Hotel, or Hilbert's Infinite Hotel.

I'm still confused. This makes it sound like the switching argument isn't fallacious -- it just makes an unwarranted assumption. So if every value of X were equally probable, then it would be true that you can expect a .25 gain from switching? I see how the math works, but if that's true, isn't it true whether you know the value of your envelope or not? And if that's true, isn't it also true for the other envelope as well? — Srap Tasmaner

Yes, yes and yes, but the assumption isn't merely unwarranted, it is impossible that it be true in any real world instantiation of the problem where the available amounts that can figure in the distribution all are smaller than some finite upper bound. It follows from the equiprobability assumption not only that every possible amount in the distribution are equiprobable but also that the distribution is unbounded(*). In that case, your prior expectation, before you open the envelope, is infinite. The probability that you would be dealt an amount X that is lower than some finite upper bound M is vanishingly small however big M might be. But if you do wind up with some finite amount X in your envelope, then your conditional expectation upon switching is 1.25X. That doesn't make the unconditional expectation from the always-switching strategy any larger than the unconditional expectation of the 'always-sticking' strategy since 1.25 times aleph-zero is aleph-zero. The situation is rather akin to Cantor's Hotel where all the (countably infinitely many) rooms are filled up and there nevertheless still is 'room' for accommodating twice as many guests.

(*) Some distributions, such as the Gauss or Poisson distributions, are unbounded and well behaved (i.e. yielding finite expectations) but they are not uniform and so don't satisfy the equiprobability assumption.

Yes. But I think the OP is asking for a general solution for one run with no special assumptions about the context (such as whole dollar amounts or million dollar limits). — Andrew M

For sure, I agree, since any non-uniform prior would violate the equiprobability condition that alone grounds the derivation of the unconditional 1.25X expectation from switching.

(...) To remove that option, I recast the problem with the envelopes containing IOUs rather than cash, for an amount that is a real number of cents, with an arbitrary but large number of decimal places shown. The amount is only rounded to the nearest cent (or dollar) when the IOU is cashed in. — andrewk

That's clever.

Yes. You learn something about the distribution when you open an envelope (namely, that it had an envelope with that seen amount). But not enough to calculate anything useful. It's like getting a bicycle with one wheel. You might wonder whether you could get somewhere with it, but you probably can't. — Andrew M

I disagree. Suppose you were to engage in one million iterations of that game and find that the seen envelope contents converge on a specific and roughly uniform distribution of amounts all belonging in the discrete range ($1,$2,$4,$8,$16), with deviations from 1/5 frequencies that aren't very statistically significant. (I am assuming that the game only allows for amounts in whole dollars, for simplicity). Wouldn't that information be useful? I would argue that the useful knowledge that you thus gain is being accrued progressively, one little bit at a time, and could be represented by the successive updating from an allegedly very tentative long tailed initial prior distribution that represents your initial expectation that the dealer likely doesn't have access to an amount of money in excess of one million dollars, say, with the peak of the distribution somewhere around $50, say. This initial prior could be very wrong, but through (Bayesian) updating it after each iteration of the game, will lead to a prior that converges towards the 'real' distribution. Hence, each envelope that you look at provides some useful information since it is more likely than not to lead you, at each step, to updating your initial prior in the direction of a more reliable one.

I did wonder -- maybe a week ago? it's somewhere in the thread -- if there isn't an inherent bias in the problem toward switching because of the space being bounded to the left, where the potential losses are also getting smaller and smaller, but unbounded to the right, where the potential gains keep getting bigger and bigger. — Srap Tasmaner

Yes, you can have a uniform discrete distribution that is bounded between 0 and M such that the possible values are M, M/2, M/4, ... In that limit case also, if the player takes M as the cutoff for not-switching (that is, she only switches when she sees a value lower than M) her expectation is 1.25X whenever she switches and her expectation is X=M when she is dealt M (and therefore doesn't switch). Her overall expectation if she were rather to always switch would only be X. The limit case where M tends towards infinity also yields an always switching strategy (with M=infinity being the cutoff) with an expectation that is both X (=infinity) and 1.25X (=infinity).

The argument is sound, so I probably won't spend any more time trying to figure out how to simulate knowing nothing about the sample space and its PDF. — Srap Tasmaner

One way to represent "knowing nothing" might be to take the limit of an initial distribution that is uniform between 0 and M, where M tends towards infinity. In that case, the cutoff value that you can rely on to maximise your gain also tends towards infinity. In practice, that means always switching; and the expected added value from switching tends towards 0.25X, where X is the value of the envelope that you were initially dealt. This still appears to give rise to the same paradox whereby switching increases your expectation by 1.25 even though switching doesn't change anything to the probabilistic distribution of the value of the envelope that you end up holding. But the paradox only is apparent since your expectation from the non-switching strategy tends towards infinity and 1.25 times infinity (or aleph-zero) still is infinity. It is thus both true that your expectation from switching remains the same and is increased by a 1.25 factor. In this limiting case, though, it is infinitely unlikely that you will be dealt an envelope containing an amount smaller than M, however large M might be. This also explains why, in this limiting case, you should always switch.

But it is still false that you have an expected gain of 1/4 the value of your envelope. You really don't. All these justifications for assigning 50% to more possibilities than two envelopes can hold are mistaken. You picked from one pair of envelopes. This is the only pair that matters. You either have the bigger or the smaller. Trading the bigger is a loss, trading the smaller is a gain, and it's the same amount each way. — Srap Tasmaner

I agree that in the context of any real world instantiation of this problem, what you say is true (because there is no real world instantiation of this problem, with finite stocks of money, that satisfies the condition of equiprobability tacitly assumed in the original formulation). The challenge that @Michael would have to answer is this: Why is it, on his view, that it isn't rationally mandatory for him to switch his choice even before he has looked into his envelope? And if it is rationally mandatory for him to switch without even looking -- because whatever the content X of his envelope might be, the expected gain from switching is 0.25X -- then why isn't it also rationally mandatory for him to switch yet again, if given the opportunity, without even looking?

Sorry, I'm a bit confused by your response. Did you read me as saying "this isn't the same as"? — Michael

Indeed! I misread you precisely in this way. We agree then.

... Isn't this the same as: ... — Michael

Why are they not the same? In the case where the unseen envelope M is the smaller one, its content is indeed S = a/2.

If you right-click on a TeX formula and select 'Show Math As...' then 'TeX commands', then you can copy and paste the code for that in between ... — andrewk

Thanks so much. I'll do this from now on.

So far so good. But we cannot do this:

$\small p=\cfrac12\to E(N\mid M=a)=\cfrac{5a}{4}$ — Srap Tasmaner

(How should I proceed in order to quote your formulas correctly?)

Why couldn't you do that? If the initial set up calls for randomly assigning values for the two envelopes in the finite range ((1,2),(2,4),(4,8)) for instance, then, in that case, assuming the player knows this to be the initial set up (and hence uses it as his prior) then the posterior probability conditionally on observing any value of M that isn't either 1 or 8 (that is, conditionally on values 2 or 4 being observed) p will indeed be 1/2.

I was arguing that it can't be 1/2 regardless of the value being observed in the first envelope unless the prior being assumed is an infinite and uniform (and hence non-normalisable) distribution.

Random selection, which means equal probability, mitigates observational bias by treating each n in a population the same. — Jeremiah

The equiprobability that I am talking about is the posterior equiprobability between the two possible contents of the second envelope: either X/2 or 2*X. This posterior equiprobability only is guaranteed by an unbounded prior distribution. If one rather assumes a prior distribution that assigns the same probability to a finite population of possible envelope contents, then such a prior isn't uniform since it assigns a zero probability (or zero probability density) to all the conceivable envelope contents that fall outside of this finite discrete (or continuous albeit bounded) range. (One might also consider the case of unbounded albeit convergent, and hence normalisable, probability density functions. And those aren't uniform either).

1. If the player does not know the amount in the chosen envelope then the expected gain from switching is zero. — Andrew M

You might not find that everyone agrees on this first claim since, under conditions of equiprobability, the paradox arises whereby (1) acquiring knowledge of the content X of the first envelope yields and expected value of 1.25*X for switching and (2) merely acquiring knowledge of the content of the first envelope ought not to change anything to the already determined expectation of switching. This is what makes the assumption of equiprobability so problematical (since (2) can be inferred from it).

On edit: I looked at earlier posts of yours in this thread, such as this one, and I see that we are on the same page.

By this do you just mean that if we know that the value of X is to be chosen from a distribution of 1 - 100 then if we open our envelope to find 150 then we know not to switch? — Michael

That's one particular case of a prior probability distribution (bounded, in this case) such that the posterior probability distribution (after one envelope was opened) doesn't satisfy the (posterior) equiprobability condition on the basis of which you had derived the positive expected value of the switching strategy. But I would conjecture that any non-uniform or bounded (prior) probability distribution whatsoever would likewise yield the violation of this equiprobability condition.