Isn't all of this true whichever of a and b is larger, and whatever their ratio? — Srap Tasmaner

Yes, this is true of the unconditional expected values of sticking or switching. But those are not the same as the conditional values of sticking or switching (conditional, that is, on the specific value of one of the two envelopes). In the case where the prior distribution isn't uniform and unbounded, then, the unconditional values of sticking and switching are equal to each other, and they are finite. In the case where the prior distribution is uniform and unbounded, the unconditional values still are equal to each other since they are both infinite. And also, the conditional values of switching or sticking both are equal to 1.25v, conditional on v being the value of any one of the two envelopes. (The trick, of course, is to see why this doesn't lead to a contradiction in the case where the prior distribution is uniform and unbounded. It's the same reason why the guests of the Hilbert Rational Hotel are rationally justified to blindly move to new rooms, and, if they have already moved, are rationally justified to blindly move back to their previous rooms).

But it isn't logically consistent. With anything. That's what I keep trying to say over and over.

1.25v is based on the demonstrably-false assumption that Pr(X=v/2)=Pr(X=v) regardless of what v is. It's like saying that the hypotenuse of every right triangle is 5 because, if the legs were 3 and 4, the hypotenuse would be 5. — JeffJo

What you are saying is correct in any case (most cases?) where the prior probability distribution of the envelope values isn't unbounded and uniform. In the case where it is, then there is no inconsistency between the expected value of the unseen envelope being 1.25v conditionally on the value of the seen envelope being v, and this being the case regardless of which one of the two envelopes has been seen.

Exp(other) = (v/2)*Pr(picked higher) + (2v)*Pr(picked lower) is a mathematically incorrect formula, because it uses the probabilities of the wrong events.

Actually, the formula is correct in the special case where the prior distribution is uniform and unbounded, since, in that special case, the conditional probabilities Pr(picked higher|V=v) and Pr(picked lower|V=v) remain exactly 1/2 whatever v might be. In the more general case, the formula should rather be written:

Exp(other) = (v/2)*Pr(picked higher|V=v) + (2v)*Pr(picked lower|V=v)

Exp(other) = (v/2)*Pr(V=v|picked higher) + (2v)*Pr(V=v|picked lower) is the mathematically correct formula, because it uses the probabilities of the correct events.

Are you sure you didn't rather mean to write "Exp(other) = (v/2)*Pr(picked higher|V=v) + (2v)*Pr(picked lower|V=v)"?

A pedant would insist you need to include one probability from the probability distribution of whichever you choose. But it divides out so it isn't necessary in practice. — JeffJo

Edited: I had posted an objection that doesn't apply to what you said since I overlooked that you were only here considering the case where both envelopes remain sealed. I agree with your post.

And since the OP does not include information relating to this, it does not reside in this "real world." — JeffJo

That's fine with me. In that case, one must be open to embracing both horns of the dilemma, and realize that there being an unconditional expectation 1.25v for switching, whatever value v one might find in the first envelope, isn't logically inconsistent with there being an unconditional expectation 1.25w for sticking, whatever value w one might be shown to be in the other envelope (as @Srap Tasmaner had suggested, by way of reductio of the unconditional switching strategy). The situation would therefore be analogous to the one that I illustrated with my Hilbert's Infinite Hotel thought experiment.

I'm not sure what "real world" has to do with anything. But... — JeffJo

It has to do with the sorts of inferences that are warranted on the ground of the assumption that the player still "doesn't know" whether her envelope is or isn't the largest one whatever value she finds in it. And this, in turn, depends on how "doesn't know" is meant to be interpreted. Is that meant to imply that the the player is entitled to apply the principle of indifference and therefore assign a 50% probability (exactly) to each one of the two possibilities irrespective of the value that she finds in her envelope? This is what I would take to entail that the prior distribution is uniform, and not to be consistent with a prior belief that the amount of money in the universe is finite.

When I say that the prior distribution is uniform, I mean this to represent the player's prior expectation (or credence) that whatever real positive value v it is that she will find in her envelope, it remains equally likely, from her point of view, that the other envelope might contain v/2 or 2*v. This would also entail that the prior expectation that the player would find a value v in her first envelope that is lower than some upper bound M, however large M might be, is infinitesimal. That such uniform and unbounded prior expectations don't apply to rational agents being faced with "real world" problems is what I meant.

It'll be fine once we use a MCMC and get the HDI. — Jeremiah

Yes, everything will be fine and dandy. How did I not think of that...

A normal distribution does not have to have a mid of 0, nor do they need negative values. — Jeremiah

I did not say that it has to be centered on zero. Normal distributions are unbounded on both sides, however. They assign positive probability densities to all real values.

A normal prior would actually make more sense, as empirical investigations have shown it robust against possible skewness. — Jeremiah

I am not sure how you would define a normal prior for this problem since it is being assumed, is it not, that the amount of money that can be found in the envelopes is positive? If negative values are allowed, and the player can incur a debt, then, of course, a normal prior centered on zero yields no expected gain from switching. Maybe a Poisson distribution would be a better fit for the player's prior credence in a real world situation where negative amounts are precluded but no more information is explicitly given. But such a prior credence would also fail to satisfy the principle of indifference as applied to the 'v is lowest' and 'v is highest' possibilities, conditionally on most values v being observed in the first envelope.

A random variable is defined by a real world function — Jeremiah

That's a bit like saying that a geometrical circle is defined by a real world cheese wheel.

(...) That distribution is an unknown function F1(x). After picking high/low with 50:50 probability, the value in our envelope is a new random variable V. Its distribution is another unknown function F2(v), but we do know something about it. Probability theory tells us that F2(v) = [F1(v)+F1(2v)]/2. But it also tells us that the distribution of the "other" envelope, random variable Y, is F3(y) = [F1(y)+F1(2y)]/2. Y is, of course, not independent of V. The point is that it isn't F3(v/2)=F3(v)=1/2, either. — JeffJo

I agree with this, and with much of what you wrote before. I command you for the clarity and rigor of your explanation.

Looking in the envelope does change our role from that of the game player, to the gamemaster. Just like seeing the color of your die does not. Simply "knowing" v (and I use quotes because "treat it as an unknown" really means "treat it as if you know the value is v, where v can be any *single* value in the range of V") does not change increase our knowledge in any way.

I am not so sure about that. The game master does have prior exact knowledge of the function F1(x), which I possibly misleadingly earlier called the "initial distribution". According to the OP specification, the player doesn't necessarily know what this function is (although, under one interpretation of the problem, it must be assumed to be uniform and unbounded). When the player opens her envelope her epistemic position remains distinct from that of the gamemaster since she still is ignorant of F1(x).

My earlier point, which is a concession to some of the things @Jeremiah and @Michael have said, is that the nature of the "system" whereby some probability distribution is being generated, and which F1(x) represents, only is relevant to a player who plays this specific game, either once or several times. But what it is that I may not have conveyed sufficiently clearly, is that when the player not only plays this game only once, but also, isn't seeking to maximize her expected value with respect to the specific probability space defined by this "game" of "system", but rather with respect to the more general probability space being generated by the distribution of all the possible games that satisfy the OP general specification, the specific F1(x) that is being known to the game master is irrelevant to the player at any step. The mutual dependences of F1, F2 and F3, though, as you correctly point out, are indeed relevant to the player's decision. They are relevant to constraining the player's relevant probability space. This is the point that @Michael may have missed.

Claiming this case is "ideal" is an entirely subjective standard pumped full of observational bias. — Jeremiah

The OP doesn't specify that the thought experiment must have physically possible instantiations. The only ideal, here, consists in pursuing strictly the mathematical/logical consequences of the principle of indifference even when it is interpreted such as to entails that the relevant distribution of possible pairs of envelope contents must be assumed by the player to be uniform and unbounded.

I suggested at one point in this thread that if told the value of the other envelope instead of your own, then you would want not to switch; I found this conclusion absurd but my interlocutor did not. Go figure. — Srap Tasmaner

That was a very pointed challenge, and I think it has force in the case where the analysis is assumed to apply to a game that can be instantiated in the real physical world and that can be played by finite creatures such as us. But in the ideal case where the prior distribution is assumed to be uniform and unbounded, then, the conclusion, although seemingly absurd, would be warranted.

(...) This is the fallacy. You reason from the fact that, given the criterion of success, you would have a 1 in 2 chance of picking the envelope that meets that criterion, to a 1 to 2 chance that the unknown criterion of success is the one your chosen envelope meets. (...) — Srap Tasmaner

This is a very neat analysis and I quite agree. One way to solve a paradox, of course, is to disclose the hidden fallacy in the argument that yields one of the two inconsistent conclusions. Your explanation indeed shows that, in the case where the criterion of success is unknown, and we don't have any reason to judge that the two possible success criteria are equiprobable, then it is fallacious to infer that the expected value of switching is 1.25v (where v is the value observed in the first envelope).

This is consistent with what I (and some others) have been arguing although I have also highlighted another source of the paradox whereby the equiprobability assumption (regarding the two possible albeit unknown success criteria) would appear to be warranted, without any reliance on the fallacy that you have diagnosed, and that is what occurs in the case where the prior distribution of possible envelope pairs which the player is entitled to judge to be possible is assumed to be uniform and unbounded. This is the ideal case, physically impossible to realize in the real world, that I have illustrated by means of my Hilbert Rational Hotel analogy.

No, it isn't fair to say that. No more than saying that the probability of heads is different for a single flip of a random coin, than for the flips of 100 random coins — JeffJo

Well, that's missing the point ;-) My purpose was to highlight a point that @Srap Tasmaner, yourself and I now seem to be agreeing on. Suppose, again, that a die throwing game will be played only once (i.e., there will be only one throw) with a die chosen at random between two oppositely biased dice as earlier described. The point was that, in the special case where the die that has been picked at random is biased towards 6, and hence the "initial distribution" is likewise biased, this fact is irrelevant to the "problem space" that the player is entitled to rely on. The initial coin picking occurred within the black box of the player's ignorance, as it were, such that the whole situation can still be treated (by her) as the single throw of a fair die.

This is what I took @Srap Tasmaner to mean when he said that "the player" doesn't "interact" with the "sample space", which is a statement that I glossed as the player's "problem space" (maybe not the best expression) not being (fully) constrained by the "initial distribution" of envelope pairs, which must be treated as being unknown but, as you've pointed out repeatedly, still belongs to a range that is subject to definite mathematical constraints that can't be ignored.

Hilbert's Grand Hotel Revisited

Here is an improvement on my earlier Hilbert Grand Hotel analogy to the two-envelope paradox. The present modification makes it into a much closer analogy.

Rather than considering an Infinite Hotel where the countably infinitely many rooms are numbered with the natural numbers, this time, they will be numbered with the (also countably infinitely many) strictly positive rational numbers. Hence, for instance, room 3/2 will be located midway between room 1 and room 2, and will be small enough to fit between them. Guests of the Hilbert Rational Hotel who want to go into rooms corresponding to rational numbers that have very large denominators will have to swallow very many magic shrinking pills, let us assume (and may have to leave their luggage behind). Initially, infinitely many guests are being randomly distributed such that there are, on average, about one hundred guests in each room. Every morning, each guest is awarded $Q, where Q is their rational room number. Every night, each guest is being given the opportunity to flip a coin to determine if she will move from room Q to either room Q/2 (if she lands tails up) or to room 2*Q (if she lands heads up). Let us suppose that all the guests are greedy and rational and, hence, all choose to flip the coin (rather than stay) in order to increase their expected earnings to 1.25*Q on the next morning. (If we take a random sample of guests within any given bounded segment of Hilbert's Rational Hotel, it is clear that they fare better, on average, than they would if they had all chosen to stay. On average, they fare 1.25 times better.)

The first thing to notice is that, after the daily reshuffling of guests, the average population of the rooms stays the same since each room is accessible from exactly two other rooms and as many guests, on average, move from room Q to room 2Q, for any Q, than the reverse. So, although any finite random sample of guests improves its fare 1.25 times, on average, the average room occupancy doesn't change and the overall population density (measured as guests per room) on any segment of the hotel doesn't vary at all.

Now, suppose that on some particular nights -- on Sunday nights, say -- all the guests swallow a pill that makes them forget what room it is that they had moved from on the previous night. They are then offered the opportunity to blindly move back to whatever room it is that they previously occupied. If they are rational and greedy, they ought to choose to move back since, if they now are in room Q, it is equally likely that they came from room Q/2 than it is that they came from room 2*Q. Hence, on average, the guests who move back to the room where they came from fare 1.25 times better than they would if they stayed.

This explains why, likewise, in the two-envelope game, with an unbounded and uniform distribution, (if such a thing is intelligible at all), it appears rational to switch envelopes in order to increase one's expected value by 1.25 and, nevertheless, after one has switched blindly, it would appear to be equally rational to switch back in order to increase one's expected value by 1.25. Such distributions, though, can no more be instantiated in real games than Hilbert's Grand Hotel can be built.

Why would your ignorance preclude you from facing a choice and making a decision? In the OP, you make at least two choices: which envelope to claim, and whether to keep it or trade it for the other. Whether you end up with the most or the least you could get depends on those two decisions and nothing else. What the amounts are depends on someone else. — Srap Tasmaner

Your ignorance doesn't preclude you from facing a choice and making a decision. What it precludes you from doing is basing your decision to switch (if you decide to) on a determinate expected gain, since the value of such an expected gain (conditionally on your having seen v in the first envelope) is unknown (and, in particular, is not known to be zero).

Only in the case where the distribution is uniform and unbounded can you know what the expected gain from switching is, and know it to be precisely 1.25v. (And that's because, in the unbounded case, it is precluded that you might ever hit the top of the distribution(*) and hence lose in one fell swoop all the expected gains potentially accrued from the other cases in the average time that it takes for this large loss to randomly occur). In that unbounded case, you would know that switching will earn you either v/2 or 2*v with equal probability. It merely appears paradoxical, in that unbounded case, that you are always entitled to switch but that, nevertheless, if you were to switch blindly the first time, and only open the second envelope to find some value w, then, you would still expect 1.25w from switching back. It is this apparent paradox, occurring with infinite and uniform distributions, that my Hilbert Grand Hotel analogy was meant to illustrate.

(*) I am only considering uniform probability distributions over elements of a single discrete doubling sequence, for simplicity. We can also assume it to be truncated below, at $1, say.

Conversely, the expected gain that the player calculates will still be the unconditional gain of zero since she doesn't know the initial distribution or both amounts in the selected envelope pair. — Andrew M

On the assumption, of course, that the player takes this initial distribution to be bounded above by M for some (possibly uncknown) M; or that, if unbounded, its sum or integral converges.

There are values in envelopes. How they got there can be discussed, and that can be interesting when the player has, say, partial knowledge of that process, but it is not the source of the paradox, in my opinion. — Srap Tasmaner

Whereas, on my view, it is the source of the paradox ;-)

This makes no sense to me. Initial distribution of what? If these are pairs of envelopes from which will be chosen the pair that the player confronts, then not only is this sample space unknown to the player, she never interacts with it. She will face the pair chosen and no other. — Srap Tasmaner

Yes, it is tempting to say that if the game is only being played once then the shape of the initial distribution(*) isn't relevant to the definition of the player's known problem space. That's a fair point, and I may not have been sufficiently sensitive to it. This is broadly the argument @Michael and @Jeremiah are making, although they are reaching opposite conclusions regarding the expected gain from switching. (And they're, in two different ways, both right; hence the paradox). The player doesn't know what this distribution is, and, since she only is playing once, and hence only is being dealt one single envelope pair, how is it relevant what the other unrealized envelope pairs (and their probabilistic frequencies) were within this specific distribution? But it is actually relevant what the distribution might look like since there are logical constraints on the shape of that distribution that can be inferred from the assumptions that ground either strategies (i.e. switching or being indifferent to switching).

The advocate of the switching strategy is actually right in saying that if, even after she has seen that her envelope contains v, she still has no reason to assign an at least twice higher probability to the other envelope being v/2 rather than 2v, then she is justified in switching. The switching strategy can only be justified since, precisely, the prior probability distribution of this game, which is only being played once, doesn't interact with this player's problem space. When she gets to play such a game again, if ever, the prior distribution might be different.

What rather defines the problem space of this player, in accordance with the general specification of the problem that is provided to her (as described in the OP), is the range of possible envelope pairs (and their probabilistic weighs) that is being merely consistent with the general (and abstract) specification of the problem. There are however (roughly) two ways to generate such a range: bounded or unbounded. If it's unbounded, then the switching strategy is justified since the expected gain from switching is 1,25v conditionally on any v being found in the first envelope. If the range is bounded, because the player assumes that there is some finite amount of money M in the universe, however big M might be, then her problem space is such that the average expected gain from switching is zero. This is because her problem space is built up from some arbitrary weighing of all the possible bounded prior distributions, and all of those individually yield an average expected gain from switching that is zero, and so is any weighted sum of them.

(*) I am defining this initial distribution with reference to the method, unknown to the player, by means of which the initial contents of the two envelopes is effectively being determined, by means of a pseudo-random number generator, quantum device, or whatever.

I'm not sure which of JeffJo's examples you're referring to. — Srap Tasmaner

I was not referring to a specific example but rather to his general resolution of the apparent problem. It both justifies the zero expected gain for the unconditional switching strategy and explains why the indifference principle can't be applied for inferring that the expected utility of switching is 1.25v, in the case where your envelope contains v.

As for my "tree" and what it predicts -- You face a choice at the beginning between two values, and the same choice at the end between those same two values. If you flip a coin each time, then your expectation is the average of those two values both times and it is unchanged.

I am not sure why you are saying that I am facing a choice rather than saying that I simply don't know whether my envelope is smallest or largest (within the pair that was picked). I am not facing a choice between two values. I am holding v, and I am offered to trade v for another envelope which, for all I know, might contain v/2 or 2v. One wrong inference that one might make is that just because I don't know whether the other envelope contains v/2 or 2v, and don't have any determinate means to estimate it, therefore the indifference principle applies and I can assign 1/2 to the probability of either cases. That would indeed yield an expected value of 1.25v for the second envelope. But that is a wrong inference from the fact that I don't know either P(v, v/2) or P(v, 2v). I only know that those two probabilities add up to one and their exact values are dependent on the prior distribution of possible envelope pairs. The only way for such an initial distribution to be such as to guarantee that P(v, v/2) = P(v, 2v) = 1/2 for any v would be for the initial distribution to be uniform and unbounded. For any other feasible way to randomly determine amounts of value pairs in accordance with some well behaved probability distribution, what the player can infer only is that, on average, after repeated trials, the expected value for switching tends towards zero.

Opening an envelope changes things somewhat, but only somewhat. It gives more substance to the word "switch", because having opened one envelope you will never be allowed to open another. You are now choosing not between two envelopes that can be treated as having equal values inside, although they do not, but between one that has a known value and another that cannot be treated as equal in value.

But it is, for all that, exactly the same choice over again, and however many steps there are between beginning and end, your expected take is the average of the values of the two envelopes. If there's an example in which that is not true, I would be surprised.

I disagree. Suppose the initial distribution is, unbeknownst to you, ((5,10), (10,20), (20,40)). In that case, if you are being dealt 5, the expected value of sticking is 5. You don't know what the expected gain of switching is. But it's not the case that it is therefore zero. That would only be zero if you knew for a fact that (5, 10) is half as likely as (5, 2.5) in the prior distribution.

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.