The other big picture issue that has gotten short shrift in this thread, by focusing on the open envelope, is the paradox of trading an unopened envelope (or even picking one).

If you are offered the trade after picking, but without opening, you can conclude readily that the value of the other envelope is 5/4 the value of yours, whatever that is. But if you trade, and still don't open, you can conclude that the value of the first envelope is 5/4 the value of the one you traded for.

And you can reason this way even before picking, so that the other envelope is always better. How can you even pick one?

There are two ways to look at this:

(1) Knowing the rules of the game, when you get the $10 envelope, you use your amazing Math Powers to deduce that the other envelope contains $5 or $20. (This is like kids educational TV.)

(2) Upon finding $10 in the first envelope, you start making up fairy tales that convince you to trade your bird in the hand for two in the bush.

I more and more see the soundness of doing the math as you do, as a matter of fact, which bothers me a bit because the answer it produces is patently wrong. It may not be a question of whether the math is being done right, but whether this particular tool is appropriate for the job at hand.

The coins and colored balls thing is different because there are four possible outcomes, you're just choosing in two steps, maybe because you don't have a four-sided coin. Probability of each is 1/4, just takes two steps to get there.

I am still thinking about the math.

This sequence converges:

$\frac{1}{2^n}y-\frac{1}{2^{n+1}}y$

This one doesn't:

$2^{n+1}y-2^ny$

Even using terms like "expectation" or "expected value" is too fancy here. We're just talking about a mean of two values.

What's the mean of A and B? (A + B)/2.
What's the mean of X and 2X, for some X? 3X/2.
What's the mean of U and 10? (U + 10)/2 = U/2 + 5.

You cannot calculate an expectation for X if you do not know what the sample space for X is.

It really should be called the "Grass is Always Greener" problem, because the whole point of that saying is that the greater greenness of the grass on the other side of the fence is an illusion you generate yourself by applying your favorite cognitive fallacy.

Take a step back from all the math and the modeling. Try to see the forest here.

(1) There are two envelopes.
(2) You end up with one of them.

Maybe you just pick one, maybe there's some long drawn out complicated process. Whatever happens between (1) and (2), you end up with one of the two. Your expected gain is just the average of their values.

You don't know the total value of the envelopes, how that value is selected, what it's range might be if it even has one. Right here in this post, I haven't even told you the envelopes necessarily have different values, or that their values have a fixed ratio, or a fixed difference. If I let you look in one, how much more do you know? Almost nothing. You might as well not bother.

There are two objects of unknown value and you end up with one of them. So far as a "decision" goes here, you might as well treat them as equal, or treat the choice as a matter of indifference, just flip a coin. There is not nearly enough information available to base a decision on.

If it's heads you put a red ball in one box and a blue ball in a second box. If it's tails you put a red ball in one box and a green ball in a second box. — Michael

See how red blue and green are all mentioned by name as possibilities.

It is suggestive that the only probabilities ever in play are 50% and the only expected value calculation I have any faith in tells us absolutely nothing.

On the other hand, once you start iterating, there's plenty of cool stuff to do.

The strangest thing about the puzzle to me is that you need only designate an envelope to get into trouble.

Wikipedia says Smullyan thought it was a logic puzzle and had nothing to do with probability.

20 is definitely not possible value of X — Dawnstorm

Just a typo

The more I look at it, the more I miss seeing U=X and U=2X. Maybe it's best to leave them alone. Seeing P(U=5...) when 5 might not even be a possible value for U just feels wrong.

I'm with you. I just hadn't thought of doing it this way before. It gives you a way to acknowledge what you've learned by learning the value of one of the envelopes, while acknowledging that you still don't know which of 5 and 20 is even a possible value of X. It feels close to the way we should think about the problem.

Right, good point. In which case, defining the sample space that way is the mistake -- turns our problem into the Ali Baba problem. Didn't mean to do that.

How would you feel about something like this?

E(U | Y=10) = P(U=5 | Y=10)(x) + P(U=20 | Y=10)(2X)
= 3X/2

Okay to describe the events that way?

For bogarting your L.

Nooooooo — andrewk

Sorry, man.

the player's expected gain from blind switching is zero, but her expected gain from switching based on comparing the value of Y to L/2 is strongly positive. — andrewk

I think a more complete, realistic answer is around here, yes. Still, for a single trial, your guess would have to be awfully lucky to be any help.

Here's another slightly different version with a little twist.

As above, we'll have an envelope loading event, this time [L = 2R, R = 2L].

I'm just going to do the case for Y = L = 10. In addition to dropping all the terms that include U=L, I'm going to leave U=R (=1) out of the terms I keep.

E(U|Y=L=10) = P(L=2R | L=10)(R) + P(R=2L | L=10)(R)

Now the question is, can I do this:

= P(R=5 | L=10)(R) + P(R=20 | L=10)(R)
= R

That is, maybe there's no harm in doing a little algebra in how you describe the events whose probabilities you're looking at, but all that stays inside the parentheses-- you still don't get to touch (R) because there's no (R | L=10) or something.

I feel a little weird about it, but maybe there's room for a partially subjective description here, so long as you're careful not to average values that are not known to be in the sample space.

What do you think @Jeremiah? Is this okay?

You saying that there's a 50% chance that U is 2X given that Y is £10 is you saying that there's a 50% chance that U is £20.

You saying that there's a 50% chance that U is X given that Y is £10 is you saying that there's a 50% chance that U is £5. — Michael

I'm going to meet you half-way, Michael.

In essence what I'm saying is that P(U = 2X | Y = 10) = P(U = 2X), which entails that (U = 2X ∩ Y = 10) = ∅. The trouble here, which we should have fixed long ago, is just abuse of notation. I can look at that and understand it as saying that the values loaded into the envelopes are independent of my choice of envelope. But it also suggests that the value of U and the value of Y are unrelated, which is plainly false. It's much more cumbersome to do it the long way, but I guess there's nothing for it.

We'll label the envelopes L and R. There are two independent events:
(1) [L = x, R = x], the loading of the envelopes; and
(2) [Y = L, Y = R]. the choosing of an envelope by the player.
(There are several equivalent ways of defining these.) We're also going to assign probabilities Y = L = 1/2 = Y = R, and L = x = 1/2 = R = x. (The defense for this is that since the player cannot deliberately choose L even if she wants to, she is indifferent, and since she is indifferent then there is no point in the loader of the envelopes not being indifferent. There is no gamesmanship attached to which indistinguishable envelope is which.)

The blind expectation looks like this:

$\small \begin{align} E(U) &= P(U=x)(x) + P(U=2x)(2x) \\ &= P(U=L)P(L=x)(x) + P(L=R)P(R=x)(x) + P(U=L)P(L=2x)(2x) + P(U=R)P(R=2x)(2x) \\ &= \frac12\cdot\frac12(x) + \frac12\cdot\frac12(x) + \frac12\cdot\frac12(2x) + \frac12\cdot\frac12(2x) \\ &= \frac32x\end{align}$

What happens when we have selected and opened an envelope with value 10?

There are two possibilities: Y = L & L = 10 or Y = R & R = 10. It seems safe to treat these as equally likely. But now we have complications. Y = L and L = 10 are independent; the sort of events we'll be conditioning on this pair are U = R and R = x, which are also independent. But Y = L and U = R are not independent and L = 10 and R = x are not independent.

Case 1: Y = L & L = 10
$\small \begin{align} E(U\mid C_1) &= P(U=x\mid C_1)(x) + P(U=2x\mid C_1)(2x) \\ &= P(U=R\mid Y=L)P(R=x\mid L=10)(x) + P(U=R\mid Y=L)P(R=2x\mid L=10)(2x) \\ &= 1\cdot\frac12(x) + 1\cdot\frac12(2x) \\ &= \frac32x \end{align}$

Obviously Case 2 will be the same and E(U) = 3X/2. This looks a whole lot like what I did in this post.

But is it right? I know what you're going to say: (x) should be 5 and (2x) should be 20. After all, if L = 10 and R = x, then R = 5.

But consider this: I dropped some terms from the expansion, namely all the terms in which U = L, because we're doing the case in which Y = L. All those terms just go to 0, so I dropped them. What if it turns out x = 10? Then P(R = x | L = 10) would also go to 0, and we'd end up with U = 20, which is fine.

Your way would have the term containing (5) disappear when x = 10. Does that make sense?

*** ADDED: Yeah, actually it does. That's what makes this so misleading. There's no avoiding the sample space issue. ***

We could even, if we wanted, go ahead and do more cases in which we set x = 10, now that we know that's a possibility. The only problem with that is that we cannot do a complementary case that's x =/= 10. What would you plug in for that?

So now we're back to the same issue: there is one unknown value in our sample space for L and R. It's either 5 or 20, but two values don't go in one slot. And you cannot calculate an expected value using values not known to be in your sample space. So we're stuck with x.

((Actually we don't need x. I'll be posting a version in which the loading sample space it's just [L = 2R, R = 2L], and our expected value is just (L + R)/2.))

No problemo.

so please point out exactly which step it is you disagree with — Michael

You are taking a weighted average of one value which is not only possible but actual, and one value that is not possible. You cannot calculate an expected value for a variable using values not in the sample space for that variable.

The sample space for U has two values in it. One of them is 10, and that's taken by Y. That leaves exactly one slot open in the sample space for either 5 or 20, and it certainly will be one of those, but we don't know which. If you know for a fact that one of those two values cannot be a value of U, because it cannot be in the sample space from which the values of U are drawn, then you cannot use both of those values when calculating the expected value of U. One of them has to go.

Since we don't know which one has to go, the only thing to do is continue, reluctantly perhaps and with some disappointment, using X and 2X. And what you find is that your expectation for U, in terms of X, does not change when you learn that Y = 10. Knowing only one of the two values is just not enough information.

That it is not enough information should be obvious. Look again at this example:

If I present envelopes containing either £10 or £20 to a hundred switchers and let them open one, half of them will think there might be envelopes with £5 out there and half of them will think there might be envelopes with £40 out there. Half of them will trade their £10 for the £20, and half of them, sad to say, will trade their £20 for £10. Those with £10 did not risk losing only £5. Those with £20 had no chance of getting £40.

There is a perfect symmetry here: half were lucky in the draft but unlucky in the trade; the other half were unlucky in the draft but lucky in the trade. Not one of them actually had any idea which they were, whether they saw £10 in their envelope or £20. And they all traded for no reason at all.

There's no such thing as a common language when using ambiguous terms. The commonality of language is based entirely on mutual agreement. Where there's no mutual agreement, there's no commonality. — Pseudonym

But this can't be it. You'd accept something short of a Vulcan mind-meld as communication, yes?

Here's a couple thoughts.

Suppose you're a lifelong faithful Christian, a deacon in your church, you volunteer your time in your church's charities and so on, and some 19-year-old comes up to you on the sidewalk and offers to tell you about Jesus Christ. The right reaction here is "Fuck off!" It has nothing to do with whether you might agree with the kid's likely limited understanding of the gospel. The problem here is that he's broadcasting rather than communicating. You're just a pair of ears for him to talk into.

What's admirable about verificationism was never that it might stamp these words "meaningful" and those words "meaningless", or these propositions "good" and those "bad". It's that verificationism engages. It recognizes that talking is only part of the story and that it can be part of the story of how we learn things and share what we learn. It assumes there might be some practical point to the things we say to each other, and that those connections to our wider cognitive lives might actually inform how we talk.

There are facts that represent other facts. They do this, roughly, by their elements being arranged the same way the elements of the facts represented are. This is pretty intuitive. But it also leads directly to the point that a representing fact, a picture, cannot have as an element any of these: something that indicates it is a picture, something that indicates what fact it is a picture of, something that explains how it represents what it represents, any indication that it is a picture at all. None of those are present in the fact represented, so there is no element in what is represented for any such element of the picture to correspond to. (If there are such things, they will have to be immanent in the picture, but not an element.)

Thus we get 2.172: "The picture, however, cannot represent its form of representation; it shows it forth." (More below.)

We're still working through whether and how pictures are veridical. We begin with pictures tied to facts and reality. Thus at 2.15: "That the elements of the picture are combined with one another in a definite way, represents that the things are so combined with one another." That's are. No question that the elements of a picture are combined in a definite way. Of course they are. So what does a picture represent? Obviously a veridical picture shows things combined as they are. What about a non-veridical picture? Doesn't it show things combined in a way that they aren't? (More on this in a moment.)

We get the rest of the terms we need in the remainder of 2.15: "The connection of the elements of the picture is called its structure, and the possibility of this structure is called the form of representation." Structure makes perfect sense, but form of representation is the possibility of this structure? And then in 2.151 he says: "The form of representation is the possibility that the things are combined with one another as are the elements of the picture."

So this is what 2.172 says the picture cannot represent. It represents a possible situation in logical space, atomic facts existing or not, and it represents (vorstellen this time instead of darstellen) that things are combined the way its elements are, but it does not represent the possibility of itself having the structure it shows the things it represents having.

The form of representation is the possibility of this fact, the picture, having a certain structure; what's notable about this structure is that its elements can be coordinated with the things it represents. This is the representing relation. The representing relation is what makes a fact a picture, and this relation is immanent in the picture. (Evidently pictures are not signs at all, are not arbitrary — that will come later. Pictures are models.)

All this leads up to 2.17: "What the picture must have in common with reality in order to be able to represent it after its manner — rightly or falsely — is its form of representation." So its form of representation — the possibility of the picture having the structure that it does, its elements being combined as they are — this possibility is what the picture has in common with reality.

So now we can come back to the true and false problem. We can look at this backwards: what the picture does not represent, it cannot get wrong; what no picture represents is its own form of representation. This it cannot help but get "right". But what is it?

((Timeout. Possible that "form of representation" is not the best translation here. P&M use "pictorial form" which is scarcely better. The key word here is Abbildung which seems to cover lots of stuff related to projection — reproductions (as of pictures), mappings and such in mathematics. As a matter of fact, it seems likely that this idea of projection — from what is represented to the picture — is exactly what's missing here. LW has all these descriptions that run from the picture to reality — the feelers and all that — but almost nothing in the other direction, which is quite strange.))

It seems like he was on the verge of saying that a picture has the same structure as what it represents, but he doesn't — he says it has in common with what it represents this form of representation, or projection or mapping. So we can conclude that this possibility of the structure the picture has is also a possibility found in what is pictured.

Remember my marbles? Physically sorting the red and blue marbles into separate boxes is a way of physically representing the logical partition of the marbles by color, using the marbles themselves. The physical arrangement of them into separate boxes creates a correspondence, a systematic correspondence like a mapping or a projection, between how the marbles are combined or separated and the logical partition. This is a physical model of a logical possibility. That's suggestive anyway.

I'll stop here and wait for your input, @Posty McPostface. Need clarity on the form of representation, and then we'll make some sense of the true and false business. Maybe that won't be clear until we push into 3 and propositions.

You can't condition on Y=X and then multiply by 2X where 2X≠2Y. That's just a contradiction. — Michael

It's really not.

There are two different sorts of things here. One is the conditional probability of an event occurring, one is the value in our sample space for U that U will have if that event occurs; by abuse of notation we're actually calling that event "U = 2X". The way the game is set up, P(U = 2X | Y = X) = 1 and P(U = X | Y = 2X) = 1. No assertion is made here about the event "Y = X" actually occurring, or about its probability.

Maybe you're imagining we're talking about Y here the same way we're talking about R in this coin toss:

$\small \begin{align} E(R) &= P(R = \mathrm H)(10) + P(R = \mathrm T)(100) \\ &= \frac12(10) + \frac12(100) \\ &= 55\end{align}$

But we're not. Here we're talking about the probability of the events "R = H" and "R = T" occurring, and weighting the values from our sample space, 10 and 100, by those probabilities.

In my expectation calculations, we're talking about U = ... occurring, not Y = ..., so you need to read them as

$\small \begin{align} E(U \dots) &= P(U = 2X \dots)(2X) + P(U = X \dots)(X) \end{align}$

We're just matching up events of U taking a value from the sample space with that value.

Okay, so pictures and reality.

A picture represents a way things might stand in logical space. I've been calling that sort of thing a possible partition of logical space into existing, or obtaining, atomic facts and non-existing atomic facts. There are many, many such partitions possible.

Before going on, is this right? Should we think of a way things might stand in logical space as a way logical space might be partitioned into obtaining and not-obtaining atomic facts? I really can't think of an alternative. That makes some of LW's grammar seem a little weird to me, but I'm hoping that's mainly a linguistic issue. The phrase he uses in 2.06 and again in 2.11, "the existence and non-existence of atomic facts" — how are we take that if not as a partition?

Maybe we're inclined to say, no, it's not like a list of atomic facts with a check mark next to the ones that exist. But is that a partition? Or is that a representation of a partition? What is the partition "itself"? Suppose I have a collection of red marbles and blue marbles. I can make a rule to sort the marbles by color. Is that rule the partition? Or is it the rule we follow in constructing the partition? What if I get two boxes and put all the reds in one and all the blues in the other? Is that physical separation the partition? I think, again, and oddly, that's a representation of the partition, a representation that happens to use the very things it represents. If I explained to someone that I could sort the marbles by color and they didn't know what I meant, I could do the boxes and say, "I mean like that." But was physically separating the marbles into boxes necessary? I don't think so.

So a partition is purely logical, formal. In most usages it is simplest to identify the partition as the rule you use to create it, but that's slightly abusive. We need something like the rule and the application to a given space, or the rule applied to a given a space. It is a function, defined on a specific domain, and assigning to each member of that domain some single value from some finite list.

Now, Wittgenstein says that what a picture represents is its sense, a possible partition of logical space, and that its sense, this partition, can agree or disagree with reality. If the sense of the picture corresponds with reality, then the picture is true. It is the picture itself that is truth-apt, not its sense. He does not say that the sense of the picture is true or false but the picture itself. Agreement with reality would appear to consist in representing the partition "actually in effect" — I don't have handy another way to say this.

Next installment: structure, form of representation and all that.

And yet my use of X and 2X leads to the correct conclusion, while your substituting Y and Y/2 leads to the wrong conclusion. Since that is the only difference, you ought to conclude that you cannot substitute this way.

((This is fun, but I have to go work on the Tractatus. I'll check back here later today.))

But your own argument made this same assumption. — Michael

That's not an argument, it's just a calculation, and it didn't. It weights the possible values known to be in the sample space for $\small U$, namely $\small X$ and $\small 2X$, by the probability of the event of $\small U$ containing the larger of the two values, given that the value of one envelope is known to be 10, and the probability of the event of $\small U$ containing the smaller of the two values, given that one envelope is known to contain 10.

Knowing that one envelope contains 10 does not tell you whether $\small U$ is the larger or the smaller envelope, and does not tell you whether 5 or 20 are in the sample space. Not knowing whether 5 or 20 are in the sample space for $\small U$ we cannot use them. We could use 10, but 10 is the value of the envelope we've already got.

Y is 10, therefore the sample space for X is [5, 10] — Michael

I know it seems that way to you. That is your subjective view of the situation. But we are told no such thing. We are not told X is selected by some random process or even by a choice. There is no such thing, so far as we know, as a "sample space for X".

Here's an analogy. Suppose a friend tells me he's going to flip either his red/blue coin or his yellow/green coin. What's my expectation of red? Is it 1/4? How would I know that? He didn't tell me how he was going to choose which coin he was going to flip. We do have — unlike in the envelope problem — a space known to have four elements: [[red, blue], [yellow, green]]. Maybe what he meant was that he was
always going to flip the red/blue coin. That counts as flipping either. Maybe he meant he would randomly choose one, maybe some other method. I have no way of knowing. I should probably treat the two possible spaces indifferently, because what else can I do?

In our situation, I know it's tempting to say we should average the chance that our sample space is [10, 20] and the chance that it's [5, 10], but we can't. Even if we choose to treat the two spaces indifferently, we do not know there are two such spaces. This is the difference.

@andrewk I think argued that we have to treat X as a random variable, and that we have to treat any value we cannot exclude as a possible value for X. I can see the appeal of that, I really can. But it clearly leads to the wrong answer. If I present envelopes containing either £10 or £20 to a hundred switchers and let them open one, half of them will think there might be envelopes with £5 out there and half of them will think there might be envelopes with £40 out there. Half of them will trade their £10 for the £20, and half of them, sad to say, will trade their £20 for £10. Those with £10 did not risk losing only £5. Those with £20 had no chance of getting £40.

Here we go. Thinking I'll post bits as I get them done, then you can read and comment while I'm doing the next bit. (Here and there you'll notice me futzing with the translation a bit.)

The picture theory is presented in 2.1-2.225. It's broken into chunks as follows:

2.1-2.141 deals with facts.
2.15-2.1515 is structure, form of representation, and the representing relation.
2.17-2.182 picks up the strand of the link to reality from 2.1511-2.1512
2.19 throws in the world
2.2-2.225 is a bit of a kitchen sink summation, with some emphasis on truth and falsehood.

We have already talked a bit about 2.11: a picture presents "how things stand", "what the situation is" in logical space. There's trouble right off the bat. Why doesn't he say "how things might stand"? Or "what the situation might be"? To drive home this is not an oversight, remember 2.11 is the comment on 2.1: "We make to ourselves pictures of facts ((Tatsachen))." That's facts, not possible facts, possible atomic facts, whatever.

Sticking with 2.11: what the picture presents is the obtaining and not-obtaining of atomic facts; this is the gloss on "how things stand in logical space". And we have seen exactly this formula before, in 2.06: "The obtaining and not-obtaining of atomic facts is reality." Alright, then, a picture presents reality. (And here we remember that even though Tatsache was originally introduced simply as the obtaining of an atomic fact, we have since refined that as a positive fact, and there are also negative facts.)

Wait, so does that mean pictures are by definition "accurate" or "true" or something?

If a picture, a picture of facts, presents reality, then what are we to make of 2.01-2.02 2.201-2.202, which seems to say exactly what I pointed out he didn't say at the start: "The picture depicts ((bilden)) reality by representing ((dartellen)) a possibility of the obtaining and not-obtaining of atomic facts. The picture represents a way things might stand, a possible situation, in logical space."

Now there's a little variation in the verbs here — hard to say yet whether this is a technical usage or not. In 2.11 it's plain stellen: a picture presents, puts before you, offers maybe. In 2.202 it's darstellen, which is more like "depicts" or "represents" — it's "present or put there", so I guess it becomes "depict" or "represent" because some sense of displacement is built in.

Leaving the usage of verbs aside as uncertain, we'd have to conclude that reality is a possible situation in logical space. That doesn't seem so bad. The way things are is a way things might be. But there's more: 2.21 says "The picture agrees with reality or not; it is right or wrong, true or false."

So a picture, of facts, presents reality, by representing a possible situation in logical space. What it represents is its sense and the picture is true if its sense agrees with reality. (2.221-2.222)

You have two different values of X and so this is misleading.

Where you have P(U=2X∣Y=10)(2X), the actual value of X is 10, given that you've defined the smaller envelope as having £10.

Where you have P(U=X∣Y=10)(X), the actual value of X is 5, given that you've defined the larger envelope as having £10.

So it works out as:

$\small \begin{align} E(U \mid Y=10) &= P(U=2X \mid Y=10)(20) + P(U=X \mid Y=10)(5) \\ &=\frac12 20 + \frac12 5 \\ &= 12.5 \end{align}$ — Michael

1. Y = 10
2. Y = X → U = 20
3. Y = 2X → U = 5

These statements are all true.

*4. Y = X
*5. ∴ U = 20 (1,2,4)
*6. Y = 2X
*7. ∴ U = 5 (1,3,6)

One of (4) and (6) is true and the other false, and thus one of (5) and (7) is true and the other false.

If (4) is true, then the sample space for Y and U is [10, 20]; 5 is not in the sample space and therefore you cannot use 5 in calculating the expected value of either Y or U.

If (6) is true, then the sample space for Y and U is [5, 10]; 20 is not in the sample space and therefore you cannot use 20 in calculating the expected value of either Y or U.

But let's say I'll put in £10 if it's heads and £100 if it's tails. — Michael

£10 and £100 are both in the sample space for $\small X$: that's why your expected value calculation is correct.

Suppose this was our problem: we know one envelope has £10 and one has £20, and we're just to pick one and that's it.

$\small \begin{align} E(Y) &= P(Y=10)(10) + P(Y=20)(20) \\ &= 15 \end{align}$

Easy peasy.

In our problem, we do not know whether the sample space is [5,10] or [10,20]. It is one or the other, but that doesn't mean our sample space is [5,10,20], not even subjectively. Our space is known to have exactly two values in it. You cannot plug three values into the expected value formula and get the right answer.

I suppose I could also have thrown in this:

$\small \begin{align} E(U \mid Y=10) &= P(U=2X \mid Y=10)(2X) + P(U=X \mid Y=10)(X) \\ &=\frac12 2X + \frac12 X \\ &= \frac32 X \end{align}$

Two Universal Eliminations and I get to p(x,y); don't know what to do next? — Rayan

Usually a good idea to use different variable names when you add or remove quantifiers, so you remember they're not the same variables.

((And use the reply button so people helping you know you answered.))

@Michael, @andrewk

Nothing here that @Jeremiah and @Snakes Alive haven't already said, I think, just arranged a little differently. Check my math.

Given a sample space $\small [X, 2X]$ of possible values for the selected envelope $\small Y$ and the unselected and unopened envelope $\small U$, we can calculate an expected value for $\small U$:

$\small \begin{align}E(U \mid Y=X) &=P(U=2X \mid Y = X)(2X) + P(U=X \mid Y=X)(X) \\ &= 2X \\ E(U \mid Y=2X) &=P(U=2X \mid Y = 2X)(2X) + P(U=X \mid Y=2X)(X) \\ &= X \\ E(U)&=E(U \mid Y=X)P(Y=X) + E(U \mid Y=2X)P(Y=2X) \\ &= \frac32 X\end{align}$

The question is whether it is legitimate to do something like this instead:

$\small \begin{align}E(U \mid Y=X) &=P(U=2X \mid Y = X)(2Y) + P(U=X \mid Y=X)(Y) \\ &= 2Y \\ E(U \mid Y=2X) &=P(U=2X \mid Y = 2X)(Y) + P(U=X \mid Y=2X)(\frac{Y}{2}) \\ &= \frac{Y}{2}\\ E(U)&=E(U \mid Y=X)P(Y=X) + E(U \mid Y=2X)P(Y=2X) \\ &= \frac54 Y\end{align}$

I think the answer is just no. You can't do it this way. $\small Y$, $\small 2Y$, and $\small \frac{Y}{2}$ are not in the sample space of possible values for $\small Y$ and $\small U$. That space consists entirely of $\small X$ and $\small 2X$. What the switching analysis proposes is "substituting" for $\small X$ when conditioning on $\small Y=X$ or $\small Y=2X$. But that's just not how it works. In this expression

$\small P(U=2X \mid Y=X)(2X)$

$\small (2X)$ is a dollar value, while $\small U=2X$ and $\small Y=X$ are events, with probabilities of occurring, not equations you might use to set the value of $\small X$. There's no line in here anywhere that says $\small Y = X$ as a simple equation.

There's also no place in our expectation formula for something like $\small X \mid Y=X$. The possible values from our sample space are not "conditional" in some way. And it's obvious when you get to the third equation, that you cannot possibly add these "conditional values" together: what is the $\small Y$ in $\small \frac54 Y$ supposed to be? Is that $\small Y \mid Y=X$ or $\small Y \mid Y=2X$?

Honestly this take on the problem is far more interesting than the "paradox" and of real-world use. I'd rather be thinking about that.

But it's also clear to me there's a conceptual muddle in the argument for switching in the non-iterative case, so I can't get past that.

I'd switch conditionally as explained here to reap the .25 gain. — Michael

It is absolutely true that in iterated play you can learn stuff about the sample space and its distribution, and that you can develop strategies that build on this knowledge, even simple strategies like using an arbitrary cut-off will work.

Cool article here.

I just don't think of the paradox itself this way.

once the player has seen the amount Y in the first envelope, they have narrowed the possibilities for X down to two possible values: Y or Y/2. So if that interpretation is correct I would say the code does not reflect the player's expectations. — andrewk

And in every single case, whatever the value of Y, the player will choose to switch. I model the results of switching, which are quite clearly the same as not switching, contradicting the player's stated reason for switching, namely the expectation of gain. No such expectation is fulfilled. Half the time switching is a mistake for the most obvious possible reasons.

I don't believe that description correctly represents the analysis. — andrewk

Your reasons?

I'll rephrase.

Did you imagine that players might have some other goal besides maximizing their monetary gain?

Suppose I am a player and I accept your analysis. Suppose also the envelopes are helpfully labeled L and R. Time for me to choose.

I consider L. But whatever value L has, R has 5/4 of that value. So I consider R. But whatever value R has, L has 5/4 of that value. So I consider L again.

Is there a third option besides:
(1) the two envelopes have the same expected value;
(2) one has a higher expected value than the other.

reduces the problem to the player selecting the option from which she expects the highest gain, based on the available information. — andrewk

From the OP:

You are playing a game for money.

Did you think something else was relevant? Really? Like what?

Btw, I added some functionality to the simulation andrew doesn't like. Counts some more stuff on each run just for extra confirmation of what's going on:

Number of trials: 500,000

Switch take: 187,394,381,897
Switch payout ratio: 1.4996906679464
Switch fails: 250,148
Total gain by switching: 62,439,025,501
Percentage of total: 0.33319582406328
Total loss by switching: 62,516,330,895

No Switch take: 187,471,687,291
No switch payout ratio: 1.5003093320536

Half the time you get X and win by switching to 2X; half the time you got 2X and lose by switching to X. Over a decent number of trials, your net gain is zilch.