The OP asks, "What should you do?"

Think of the problem as being in the same family as Pascal's Wager, involving decision theory and epistemology. — Andrew M

Indeed. And just as is the case with Newcomb's problem, with the two-envelope paradox also, the dominance principle and the (maximum) expected utility principle appear to recommend inconsistent strategies when carelessly applied. Newcomb's problem is more controversial, even, than the two-envelope paradox. It is also quite rich in philosophical implications.

What area of philosophy do you think the significance would obtain? — Janus

Probability is a big philosophical topic. It is quite tightly enmeshed with both metaphysics and epistemology. Michael Ayers wrote a lovely book, The Refutation of Determinism, which explores some of the philosophical problems associated with the concepts of probability, necessity and possibility, and also pursues some implications for the problem of free will and determinism. There is also a close connection with epistemology, Gettier examples, and some of the most puzzling paradoxes of Barn Facade County, which arise, it seems to me, from assumptions regarding the grounding of knowledge that are closely related to some of the assumption that give rise to the two-envelope paradox. Maybe I'll create a new topic about this when time permits.

A first step, which is being pursued in this thread, is to get clear on (what should be) the uncontroversial steps in the mathematical reasoning.

Sorry, I'm not following this. This sounds like you think I said your expected gain when you have the smaller envelope is zero, which is insane. — Srap Tasmaner

No, that's not what I was saying. I was rather suggesting that, assuming there is some determinate albeit unknown probability distribution of possible envelope pairs, then, conditional on some one specific pair having been selected from this initial range, and consistently with $5 being one of the two amounts within this pair (because $5 is the amount that you have seen in your envelope, say), then, the expected gain of switching appears to be zero, according to your decision tree analysis. According to @JeffJo, it could be $2.5, $6.25, $10, or something else, and not necessarily zero. What it actually is, is unknown to the player. What is known to the player only is the average gain from the unconditional switching strategy. And that is zero.

It's truly remarkable that a question which is of no philosophical significance or interest could generate so many responses on a philosophy forum! — Janus

It does have some philosophical implications. Some of @andrewk's replies raised good philosophical points regarding the status and significance of probability distributions, which are being involved in the analysis of this apparent paradox.

Here's my decision tree again (...) — Srap Tasmaner

Yours isn't really a decision tree that the player must make use of since there is no decision for the player to make at the first node. Imagine a game where there are two dice, one of which is biased towards six (and hence against one) and the other die is equally biased towards one (and hence against six). None of the dice are biased against any of the other possible results, 2,3,4 and 5, which therefore still have a 1/6 chance of occurring. Suppose the game involves two steps. In the first step, the player is dealt one of the two dice randomly. In the second step the player rolls this die and is awarded the result in dollars. What is the player's expected reward? It is $3.5, of course, and upon playing the game increasingly many times, the player can expect the exact same uniform random distribution of rewards ($1,$2,$3,$4,$5,$6) as she would expect from repeatedly throwing one single unbiased die. Indeed, so long as the two dice look the same, and thus can't be reidentified from one iteration of the game to the next, she would have no way to know that the dice aren't unbiased. There just isn't any point in distinguishing two steps of the "decision" procedure, since the first "step" isn't acted upon, yields no information to the player, and can thus be included into a black box, as it were. Either this game, played with two dice biased in opposite directions, or the same game played with one single unbiased die, can be simulated with the very same pseudo-random number generator. Those two games really only are two different implementations of the very same game, and both call for the exact same strategies being implemented for achieving a given goal.

In the case of the two envelopes paradox, the case is similar. The player never has the opportunity to chose which branch to take at the first node. So, the player must treat this bifurcation as occurring within a black box, as it were, and assign each branch some probability. But, unlike my example with two equally biased dice, those probabilities are unknown. @JeffJo indeed treats them as unknown, but he demonstrates that, whatever they are, over the whole range of possible dealings of two envelopes that may occur at the first step of the game, they simply divide out in the calculation of the expected gain of the switching strategy, which is zero in all cases of possible (bounded, or, at least, convergent) initial distributions. Where @JeffJo's approach seems to me to be superior to yours is that it doesn't yield an incorrect verdict for the specific cases where the prior distribution is such as to yield envelope pairs where, conditionally on being dealt either the smaller or the larger amount from this pair, the expected gain from switching isn't zero. Your own approach seems to yield an incorrect result, in that case, it seems to me.

This still looks like you're considering what would happen if we always stick or always switch over a number of repeated games. I'm just talking about playing one game. There's £10 in my envelope. If it's the lower bound then I'm guaranteed to gain £10 by switching. If it's the upper bound then I'm guaranteed to lose £5 by switching. If it's in the middle then there's an expected gain of £2.50 for switching. I don't know the distribution and so I treat each case as equally likely, as per the principle of indifference. There's an expected gain of £2.50 for switching, and so it is rational to switch. — Michael

This works if you are treating all the possible lower and upper bounds of the initial distribution as being equally likely, which is effectively the same as assuming a game where the distribution is uniform and unbounded. In that case, your expected value for switching is indeed 1.25 * v, conditionally on whatever value v you have found in your envelope, because there is no upper bound to the distribution. The paradox arises.

If we then play repeated games then I can use the information from each subsequent game and switch conditionally, as per this strategy (or in R), to realize the .25 gain.

If there is an upper bound M to the distribution, and you are allowed to play the game repeatedly, then you will eventually realize that the losses incurred whenever you switch after being initially dealt the maximum value M tend to wipe out your cumulative gains from the other situations. If you play the game x times, then your cumulative gain (which will tends towards zero) will tend, as x grows larger, towards being x times the average expected value of the gains for switching while playing the game only once. This average expected value will therefore also be zero. To repeat, conditionally on where v is situated in the bounded distribution, the expected value of switching could be either one of 2*v, 1.25*v or 0.5*v. On average, it will be v.

And some of the time it will be 2v, because it could also be the lower bound. So given that v = 10, the expected value is one of 20, 12.5, or 5. We can be indifferent about this too, in which case we have 1/3 * 20 + 1/3 * 12.5 + 1/3 * 5 = 12.5. — Michael

No. The conditional expected values of switching conditional on v = 10, and conditional on 10 being either at the top, bottom or middle of the distribution aren't merely such that we are indifferent between the three of them. They have known dependency relations between them. Although we don't know what those three possible conditional expected values are, for some given v (such as v = 10), we nevertheless know that, on average, for all the possible value of v in the bounded distribution, the weighted sum of the three of them is v rather than 1.25v.

But I don't know if my envelope contains the upper bound. Why would I play as if it is, if I have no reason to believe so? — Michael

Which is why you have no reason to switch, or not to switch. It may be that your value v is at the top of the distribution, or that it isn't. The only thing that you can deduce for certain, provided only that the distribution is bounded, if you are entirely ignorant of the probability that v might be at the top of the distribution, is that, whatever this probability might be, it always is such that the average expected values of switching, conditional on having been dealt some random envelope from the distribution, is the same as the average expected value of sticking. Sure, most of the time, the conditional expected value will be 1.25v. But some of the times it will be 0.5v.

No, because the expected gain is $333,334, which is more than my 333,334X. — Michael

But then, if you agree not to disregard the amount of the improbable jackpot while calculating the expected value of the lottery ticket purchase, then, likewise, you can't disregard the improbable loss incurred in the case where v is it a the top of the bounded distribution while calculating the expected value of the switching decision.

But we're just talking about a single game, so whether or not there is a cumulative expected gain for this strategy is irrelevant. — Michael

Unless we are considering that the player's preference is being accurately modeled by some non-linear utility curve, as @andrewk earlier discussed, then the task simply is to choose between either sticking or switching as a means to maximizing one's expected value. It's irrelevant that the game is being played once rather than twice, or ten, or infinitely many times. If a lottery ticket costs more than the sum total of the possible winnings weighed by their probabilities, then it's not worth buying such a ticket even only once.

If it's more likely that the expected gain for my single game is > X than < X then it is rational to switch.

By that argument, it would be irrational to purchase a $1 lottery ticket that gives you a 1/3 chance to win one million dollars since it is more likely that you will lose $1 (two chances in three) than it is that you will gain $999999. Or, maybe, you believe that it is only irrational to buy such a ticket in the case where you can only play this game once?

Or if I have no reason to believe that it's more likely that the other envelope contains the smaller amount then it is rational to switch, as I am effectively treating a gain (of X) as at least as likely as a loss (of 0.5X).

Sure, and in some cases, when your value v=X happens to be at the top of the distribution, you are making an improbable mistake that will lead you to incur a big loss. You are arguing that you can "effectively" disregard the size of this improbable potential loss on the ground that you only are playing the game once; just like, in the lottery case, presumably, you should entirely disregards the size of the improbable jackpot if your argument were sound.

The other way to phrase the difference is that my solution uses the same value for the chosen envelope (10) and your solution uses different values for the chosen envelope (sometimes 10 and sometimes 20 (or 5)). — Michael

There is another difference between the two methods @JeffJo presented, as both of them would be applied to the determinate value that is being observed in your envelope. The first one only is valid when we dismiss the known fact, in the case where the distribution is merely known to be bounded above (even though the upper bound is unknown), that the loss incurred from switching when the value v happens to be at the top of the distribution cancels out the expected gains from switching when the value v isn't at the top of the distribution. The second method doesn't make this false assumption and hence is valid for all bounded distributions.

then I am effectively treating both cases as being equally likely, and if I am treating both cases as being equally likely then it is rational to switch. — Michael

Yes, you are justified in treating the "cases" (namely, the cases of being dealt the largest or smallest envelope within the pair) as equally likely but you aren't justified in inferring that, just because the conditional expected gain most often is 1.25X, and occasionally 2X, when X is at the bottom of the range, therefore it is rational to switch rather than stick. That would only be rationally justified if there weren't unlikely cases (namely, being dealt the value at the top of the distribution) where the conditional loss from switching (0.5X) is so large as to nullify the cumulative expected gains from all the other cases.

There's also the possibility that £10 is the bottom of the distribution, in which case the expected value for switching is £20. — Michael

Sure. We can consider a distribution that is bounded on both sides, such as (£10,£20,£40,£80), with envelope pairs equally distributed between ((£10,£20),(£20,£40),(£40,£80)).

The issue is this: are you prepared to apply the principle of indifference, and hence to rely on an expected value of 1.25X, for switching for any value in the (£10,£20,£40,£80) range? In the case where you switch from £10, you will have underestimated your conditional expected gain by half. In the cases where you switch from either £20 or £40, your expectations will match reality. In the case where you switch from £80 then your expected loss will be twice as large as the gain that you expected. The average of the expected gains for all the possible cases still will be zero.

Assuming your goal merely is to maximise your expected value, you have not reason to favor switching over sticking.
— Pierre-Normand

Which, as I said before, is equivalent to treating it as equally likely that the other envelope contains the smaller amount as the larger amount, and so it is rational to switch.

It is not rational since you are ignoring the errors that you are making when the envelope contents are situated at both ends of the distribution, and the expected loss at the top wipes out all of the expected gains from the bottom and the middle of the distribution. When this is accounted for, the principle of indifference only tells you that while it is most likely that your expected gain is 1.25X, and it might occasionally be 2X, in the cases where it is 0.5X, the loss is so large that, on average, you expected gain from switching still remains exactly X.

If there's £10 in my envelope then the expected value for switching is £12.50, and the expected value for switching back is £10. — Michael

That's only true if £10 isn't at the top of the distribution. When the bounded and uniform distribution for single envelope contents is for instance (£10, £5, £2.5, £1.25 ...) then the expected value for switching from £10 (which is also a unique outcome) is minus £5 and, when it occurs, it tends to wipe out all of the gains that you made when you switched from smaller amounts. Even if you play the game only once, this mere possibility also nullifies your expected gain. Assuming your goal merely is to maximise your expected value, you have not reason to favor switching over sticking.

I have no way of knowing that my value is "average". Perhaps the 10^100 in my envelope is a puny value because the upper bound is Graham's number. — Michael

Sure, you will never know for sure that the value that you get is close to the top of the distribution. But the main point is that you will have no reason to apply the principle of indifference to justify your switching decision on the basis an expected gain of 1.25X. If you were so justified, then you would be equally justified to switch back, on the ground of the very same argument, before you even looked into the second envelope. That's a reductio of the claim that the expectation of switching is 1.25X.

My argument is that given how arbitrarily large the numbers in the envelopes can be (using points rather than money), there isn't really a point at which one would consider it more likely that your envelope has the larger value. If my envelope is 10 then it's rational to switch. If it's 1,000 then it's rational too switch. If it's 10100 then it's rational to switch. — Michael

If the distribution is somewhat uniform with an unfathomably large (albeit finite) upper bound, and you know this, then you can't generally expect to get such puny values. If you do, conditionally on that, then for sure, you ought to switch, and your expectation will be close to or exactly 1.25X. But what are you rationally to do when you get "average" values from the distribution, which are unfathomably large? Is it still rational to switch? On what ground? There will be no reason then to ground your decision of an (average) expectation of 1.25X. The average expectation from switching still will be zero.

Suppose the expectation is (close to) 1.25X for purpose of reductio. You are being dealt some unfathomably large amount X which is typical from the actual distribution. We suppose that you are generally warranted to switch on the basis of the principle of indifference, and thus the expectation that switching yields the expected value of (roughly) 1.25X. After you've switched, but before you are permitted to look at the content of the second envelope, you are being given to opportunity to switch back. Is it rational for you to switch back? By the very same argument that justified your initial switch, you should deduce that the expectation for switching back is roughly 1.25Y, where Y is the content of the second envelope. But that's an inconsistency.

Sure, the practical limitations of real life play a role, but I wonder if such limitations go against the spirit of the problem. What if instead of money it's points, and the goal of the game is to earn the most points? There isn't really a limit, except as to what can be written on paper, but with such things as Knuth's up-arrow notation, unfathomably large numbers like Graham's number aren't a problem. — Michael

The practical limitations indeed go against the spirit of the idealized two-envelopes problem. That's because if the prior distribution is bounded, albeit unknown, then the paradox doesn't arise. The average raised expectation for the unconditional always-switch strategy, always is zero. In order that the expectation be exactly 1.25X, whatever X, and therefore also, the always-switch strategy superior to the always-stick strategy, then there ought to be no bound to the prior distribution of possible envelope values, not just an unfathomably large albeit finite bound such a Graham's number. If the upper bound is Graham's number, then the average raised expectation from the always-switch strategy still is exactly zero.

So my takeaway is that if it isn't rational to stick then it's rational to switch. — Michael

Sure, but it is one thing to say that it is rational (or isn't irrational) to switch when you find X in your envelope ($10,000 say) and it is another to say that it is rational to behave, whatever your personal utility curve might be, as if the expected value(*) from switching always is exactly $12,500. That is rationally unjustified.

(*) I mean expected value in the technical sense, as the long term expected average of the individual pay offs.

But this just seems to be saying that there's no reason to believe that it's more likely that the other envelope contains the smaller amount and no reason to believe that it's more likely that the other envelope contains the larger amount and so you're effectively treating each case as equally likely, in which case it would be rational to switch. — Michael

Yes, that is roughly true for some mid-range values of X. See the second paragraph of my edited post for more discussion about real cases.

So what's the rational decision if you know that the prior distribution is isn't uniform and unbounded? There's £10 in your envelope. Should you stick or switch? — Michael

I am fine with acknowledging that there isn't any such thing as the rational decision to make in the vaguely specified case where you merely have a reasonable expectation that the amount of money can't be infinitely large but you don't have any precise idea how very high the distribution might be tailing off.

Suppose for instance that tomorrow morning you are being called to play this game with real money in the context of some scientific experiment conducted by the psychology department of your local university. You are to play only once and keep the money. Suppose you open your envelope and find $96 in it. Is it rational to expect 1.25*$96 from switching? I am not committed to saying this. What I am committed to say merely is that it is increasingly irrational to expect 1.25*X (or more) from switching as the value that you find in your envelope increases to ever lumpier sums.

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.