Never said it was.

Why don't you consider it from a categorical perceptive then. The OP never called for a solution based on expected gains.

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.

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

The OP never called for a solution based on expected gains — Jeremiah

Sure. But I'm not trying to figure out whether I should switch. I'm trying to figure out where the fallacy in the 5/4 argument is, and that's an expected gain argument.

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.

It is a subjective modeling based on expections when viewing Y. It confounds the objective with the subjective. The random event that determined the contents of the envelopes has already occurred and passed. That was the objective event. The selection is a separate event and it is subjective, which means these expectations are half-truths and everyone is just modeling their subjectivity. I am purposefully avoiding expected calculations for that reason. The envelopes are not in a possible state of two cases, they are in one case.

It absolutely can be ignored.

It absolutely can be ignored. — Jeremiah

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

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

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.

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?

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

See that was easy.

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.

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

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.

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. You need to either consider each case separately or just go off the fact you have two envelopes in front of you as the end results is the same.

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.

Never said anything about both being actual at once. Never meant that at all. Clearly that can't happen so I am not sure how you got that. You have to consider each case on its own and not together. We had this same issue at the start of the thread.

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.

Do you get the point or not?

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.

I don't really care about things you can't know.

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 since the chance event for the envelope has already been decided it makes no sense to consider expected gains in this fashion. They should to be considered separately.

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.

Actually only one case is true, while the other one does not exist. So they can't both be possible outcomes, not objectively. Remember the envelopes are already decided. 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.

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.

Already commented on "strategy of switching". 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

You only get one instance of the game. You don't get to open a 100+ envelopes.

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