I gave it above. The player would use it. They would estimate L, the maximum possible payout (eg the budget of the game show). The simplest approach is to then set c=L/2 and switch iff Y<c. That strategy delivers a positive expected gain under perspective 2.

Yes, assuming that the player does accurately estimate the maximum possible payout (and the procedure for generating the envelope amounts). If she doesn't, then the positive expected gain calculation is invalid.

She doesn't have to estimate it accurately. There are only three possibilities.

A. X<2X<c. In this case the expected gain from switching is zero.
B. c<X<2X. In this case the expected gain from switching is zero.
C. X<c<2X. In this case the expected gain from switching is X.

So for the competitor to decide not to use the strategy, she would have to be absolutely certain that case C is impossible.

From the God's-eye (ie omniscient) point of view, which is perspective 1 from the quoted post, there is no c, because there is no non-trivial probability distribution of X. X is a fixed quantity, known only to God and to the game show host. — andrewk

And algebra.

The difference between the two envelopes will always be X and you don't need to be God to understand that very simple concept.

((urk. still don't have it.))

I have been messing with this in R, and wrote a function which simulates the game. Note that is not a simple repeated sample from a predefined sample space. It simulates the game, by selecting X randomly then randomly places X or 2X into A. Then X or 2X into B based on what is in A. It then returns the value of both A and B.

The function, which is called two.envelopes, is one go at the game, and then the function replicate can be used to run it several times in a row. Which I did then output those to a matrix.

I ran the game simulation under 4 different conditions: Where X is chosen from a normal distribution, where X is chosen from a uniform distribution, where is X is chosen from a Cauchy distribution and where X is sampled from an interval scale. The function could be used as well to do actual statistical analyses either Classical or Bayesian, by generating enough simulated data to support such an approach.

I know these efforts will be lost on some people, but it does provide a visual summary which demonstrates the distribution in which X was selected from is not significant when assessing the possible outcome of envelope A and B concerning X or 2X.

Also, this gives results that can be reviewed which are not dependent on defining a sample space or on calculating an expected value.

Normal Distribution:




Uniform distribution:


Cauchy distribution:


Interval Scale:


The thing to notice here is that in all cases the absolute value of the difference between column one and column two is always equal to the lesser of the two (save rounding errors). The lesser of the two is X.

I changed the code a bit to come up with another visual demonstration. I changed the two.envelopes function to just output A and then copied it to another function called two.envelopes.s which outputs B.

What this simulates is if you never switch then you walking away with A and if you always switch then you walk away with B. I used a normal distribution for this example but honestly you can do the same thing for any distribution since the content of A and B are determined by the same chance event.

The point of this demonstration is to show that the possible distribution of A is the same as the possible distribution of B so I have included some graphs that can be visually compared, and I use a Kolmogorov-Smirnov Tests, also known as the K-S test. This is a non-parametric test, and if you want the details on how it works just Google it, the concept is actually really simple.

The K-S test compares two distributions to see if they match.

The hypotheses works like this: Let F(x) and S(x) designate some unknown distribution functions of the X's and Y's respectively.

Then our following two-sided null hypothesis is: F(x) = S(x) for all of x
Then our alternative hypothesis is: F(x) does not equal S(x) for at least one value of x

If you have never seen a classical statistical hypotheses test, the short and sweet of it, is if we get a low p-value we consider this evidence against the null hypothesis. The lower the p-value the greater the evidence. P-values range from 0 to 1. They are the probability of a ratio as extreme or more extreme than the observed given the null is true. Note they are not evidence for the null, failing to reject does not prove a null. The null is just that annoying guy that always demands you prove everything you say, but it is the alternative hypothesis that we are really testing for.


Here is the code:


So we see with a D test statistics of 0.0077 and a 0.92 p-value we don't have strong enough evidence to support the alternative hypothesis that the two distributions are reasonably different.

Of course this was an expected outcome and would remain true no matter how X was selected, as once X is selected its distribution in the envelopes is now something separate which depends on how the envelopes themselves are selected.

If you don't like the K-S test here are some plots that allow you to view the similarities:

There are scatter plots of each distribution and histograms of each. They will look very similar.

Scatter Plots:

https://ibb.co/ksr0P8

https://ibb.co/bW6gxT

Histograms:

https://ibb.co/c9w3Bo

https://ibb.co/h30248

3. Treat X as known and Y as unknown. Then the switch gain has a distribution of X or -X with even odds, so the expected switch gain is zero. This is the approach defended by srap. The approach is coherent but it begs the question of why it is valid to model lack of knowledge about Y/X by randomness, but not lack of knowledge about X. — andrewk

This is more or less fair. As far as this part of the problem goes, I haven't gotten past my first comment on this thread, that there is a de dicto/de re problem.

(A) There is a 1/2 chance that I will pick the larger of the two envelopes.
(B) The envelope I pick has a 1/2 chance of being the larger of the two envelopes.

These may usually be functionally or instrumentally equivalent, and we might usually use the same tools to model our uncertainty, but they are still different, and this is the occasion when the difference matters. I have a 1/2 chance of picking the envelope valued at 2X, but that envelope does not have a 1/2 chance of having a larger value than the X envelope. 2X has no chance of being less than X when X > 0.

If you can show me how to respect this difference within a subjective framework, I'd be all for it.

If you can show me how to respect this difference within a subjective framework, I'd be all for it. — Srap Tasmaner

This is how I see the problem:

Objectively, you're in one game, where one envelope contains X and the other contains 2X.

As soon as you pick an envelope, though, you have a potential value Y, which is, again, either X or 2X, but that's a bifurcation point: you have now two games. That should be obvious, because saying that Y could either be X or it could be 2X would mean X=2X, and that would be nonsense in an objective framework. What this means is that you now have two subjectively possible games, only one of which you're actually in. This holds for both values, so you have three possible game in the meta-system, one of which - the objective one you're in - is selected twice: once for each envelope.

So, if you were to dimensionalise your variable for the three games, you get: X(1), X(2), and X(3) - where X(2) is the game you're actually in, and it's the only of the three games that's selected no matter what envelope you pick.

So when you're saying that Y could be either X or 2X, you're not talking about the same X. You're either talking about [X(2), 2X(1)] if you pick X(2), or [X(3), 2X(2)]. if you pick 2X(2).

You know that if you pick X(2) you win by switching, and if you pick 2X(2), you lose by switching. Objectively, the amount you're losing or winning can only be X(2). But all you know is the proportion: if you picked the lower amount you win Y, and if you picked the higher amount you lose Y/2. Even subjectively, the amount you win or lose is always X(2). But your frame of reference differs: If you picked the lower of the two values, the amount you could have lost appears to be X(1) [=X(2)/2], and if you lose, the amount you could have won appears to be X(3) [=2X(2)].

Both those values don't exist in the objective game, but you're problem is that - while playing - you don't know whether X(2) is Y or Y/2. You could be in any of two games, one of which game 2, the real one, and the other is either game 1 (the smaller-sum game), and the other is game 2 (the bigger-sum game), but from value Y alone you can't tell.

Because you can't tell, you have two options: take Y into account anyway, or ignore it. These are two perspectives on decision making, and neither really causes unpleasant surprises, because all that changes is the reference system. You either work with an indefinite certain value (in which case it doesn't matter whether you look into an envelope or not), or with two definite but uncertain values (if you look into one envelope and make that the basis of your decision), [or, for completeness sake, with three indefinite and uncertain values (if you don't look into any envelope and set the value Y as the envelope you currently have - this is the switch-back-and-forth constellation)]. In all three cases, the only value you can win is X(2), but depending on your reference system the value may look proportionally smaller or bigger.

I think the core difference between people here lies in the different ideas of what we should with context, or maybe even what should count as context, when it comes to real-life decisions. And that's something buried fairly deeply in our worldviews, so it's not that easy to untangle.

Because you can't tell, you have two options: take Y into account anyway, or ignore it. These are two perspectives on decision making, and neither really causes unpleasant surprises, because all that changes is the reference system. — Dawnstorm

But doesn't it cause trouble?

Suppose the problem is presented to you this way: one of these envelopes is worth twice the other; you get to pick one, maybe look, are offered the trade. You might begin — as I tried once — by describing the sample space as [L=2R, R=2L]. This leads to trouble. Depending on which one is true you get a different value for |L - R|: if L=2R, then |L - R| = R = L/2, and if R=2L, |L - R| = L = R/2. It gets worse if you not only use two variables but make the variables dependent on your choice. That's our Y and U. What we are quite specifically unable to know is which is bigger, and we've chosen a way of describing the sample space that is only coherent if you know which is bigger.

If you came at the problem from here, you'd realize at some point that the clever thing to do is introduce a single variable X that is orthogonal to your choice and orthogonal to which envelope has which value. |X - 2X| = X, no matter the rest. It gives you an invariant description of the sample space so that you can properly measure the consequences of your decisions.

What's disorienting is that the best way to describe the problem was given to us first, and then we are left to discover the wrong ways all on our own.

That is a very bad understanding of what a sample space and an event is. You are not applying your Principle of Indifference there, which states from your link: "The principle of indifference states that if the n possibilities are indistinguishable except for their names, then each possibility should be assigned a probability equal to 1/n." n in this case would be the total possible combinations of the lottery numbers.

Yes, my point was that the lottery example is a very bad description of a sample space. In fact, It is the archtype for just that. But so is ignoring that you are assuming a distribution of amounts as well as whether you picked high or low. Maybe if you looked at my examples, you'd understand this. That was also a point I made.

When you look in an envelope and see $10, it means that one of two possible events has occurred. The envelopes were filled with ($5,$10) AND you picked the higher, or the envelopes were filled with ($10,$20) AND you picked the lower. The PoI applies to whether you picked high or low, since those outcomes are equivalent except in name. It does not apply to whether the envelopes were filled with ($5,$10) or ($10,$20), yet you treat them as equally likely.

The Two Envelope Problem cannot be solved without a distribution for the amounts, which is why you get a paradox when you ignore it.

I am not doing this, not until you actually read all of my posts in this thread.

I know this is a long thread, but tacking remarks on the end while only skimming though it, or skipping pages just leads to repeated content.

If you came at the problem from here, you'd realize at some point that the clever thing to do is introduce a single variable X that is orthogonal to your choice and orthogonal to which envelope has which value. |X - 2X| = X, no matter the rest. It gives you an invariant description of the sample space so that you can properly measure the consequences of your decisions. — Srap Tasmaner

But you have to remember if you go one envelope has X and the other 2X, then you're defining as the envelope that contains X as the one with the smaller value. So if you look into an envelope, you can't know which envelope you've opened, so the other must contain either twice that of the one you've opened, or half that of the one you've opened: it's the neutral value, split over an either/or situation.

X and Y are commensurable. It's the same thing. I don't see a difference.

I would say that defining the space as [X, 2X] just makes it painfully obvious that knowing the value of one envelope is completely useless, and that you should not bother with some X/2 or 2X conundrum. You can calculate the value of swapping before even choosing, and you will be right.

I pointed out a few times that the sample space of event R and the sample space of event S are equal subsets of each other, which means mathematically we can treat them the same. I also pointed out that as soon as you introduce Y they are no longer equal subsets and therefore mathematically cannot be treated the same.

Here is an example. If Z=M and N=M then Z=N.

I pointed out a few times that the sample space of event R and the sample space of event S are equal subsets of each other, which means mathematically we can treat them the same. I also pointed out that as soon as you introduce Y they are no longer equal subsets and therefore mathematically cannot be treated the same.

Here is an example. If Z=M and N=M then Z=N. — Jeremiah

I'm hopelessly confused.

I read your [[10,20],[5,10]] as: "Given that one envelope has the value 10, either [X = 10 and 2X=20] or [X=5 and 2X=10]". And that describes the sample space of both envelopes A and B.

A sample space of A[X, 2X], B[X,2X] gives you the following possibilities:

A=X, B=X
A=X, B=2X
A=2X, B=X
A=2X, B=2X

A=X, B= X (never selected, due to setup)

A=10, B=10
A=5, B=5

A=X, B=2X

A=10, B=20
A=5, B=10

A=2X, B=X

A=10, B=5
A=20, B=10

A=2X, B=2X (never selected due to setup)

A=10, B=10
A=20, B=20

So if look into A and discover 10 inside, all I have to do is to look through all possible constellations, which are both in the sample space you defined, and selected by the set-up:

A=10, B=20
A=10, B=5

This is the result of [[10,20],[5,10]] under the stipulation that A=/=B.

If this is not the case, I have read you wrong, and I can't for the life of me figure out where or how. Have I missed possible constellations? Have misread your sample space? What?

R and S is the event you have X or 2X when you get A. By definition of A and B, if A=X then B =2X, or if A =2X then B=X. So by definition if A equals X then B cannot also equal X.

We define A and B as: If A=Y=X then B=2X or if A=Y=2X then B=X, where Y is the amount you see opening envelope A and X is the unknown amount originally selected by the facilitator.

My claim is then that the only possible outcomes for B is X or 2X.

Proof:

For all of Y, Y is a positive real number, such that Y=X or Y=2X, where X is some positive real number.

You are handed A and you see Y inside.

There are two cases here:

Case One

A=Y=X

By definition of A and B if A=Y=X then B=2X therefore B=2X

Case Two

A=Y=2X

By definition of A and B if A=Y=2X then B=X therefore B=X.

Those are the only two possible cases for B therefore by the definition of a sample space the sample space for B is [X,2X] — Jeremiah

It gets messy I'll agree, as [R,S] is the same as [X,2X], but you have to remember they represent case one or case two. In case one B cannot equal X, by the definition of B and so on.

We also suffer from sloppy notation, we probably should be using subscript.

It should probably all be reviewed and written up in proper notation, but it has been such a long thread I am not sure I want to spend the time to do that. I feel that my post number 6 sums it up concisely even if informally. Then with the simulations I did on page 26 I feel I have provided empirical proof of that statement. The next step is to formalize everything but since this is informal discourse that seems like unneeded effort to me.

It might be time to consider making a new thread. I kind of feel that everything that needs to be said about this conundrum has been said.

I am not doing this, not until you actually read all of my posts in this thread. — Jeremiah

What? You don't want to address a correct analysis, until I weed through pages of debate that appears to be inconclusive? Because I can guarantee you, that applying my correct analysis can resolve them.

How about I read until I can point out three errors? But some of them will be repeats of what I already have said. In fact, I'll start by stating it another way. Here are two proposed solutions:

1. Your envelope has $X.The other has $X/2 or $2X, 50% chance each. So the expected value is ($X/2)/2+($2X)/2=$5X/4 and the expected gain by switching is $X/4.
2. (You seem to have suggested this one once) The magnitude of the difference between the two envelopes is $D; so there is a 50% chance it is +$D, and a 50% chance it is -$D. The expected gain is 0.

This is paradox, so one must be wrong. Yet, there is only one difference in the theoretical approach taken: #1 uses two possibilities for the contents of the two envelopes, while #2 uses only one. The error must be there.

+++++

You said: "You have to understand that X is a variable." This is incomplete. It's a random variable, representing (above) the (known or unknown) value of the envelope you picked. But some of those pages you want me to read confuse it with D. And even if you don't accept that you need to use it to describe your sample space, you still can represent it as a random variable.

That means it has a range, and a prior distribution. The range has to include $X, $X/2, and $2X. But if the probability of $X/2 (or $2X) is non-zero, the range must include $X/4 (or $4X). And then it must include $X/8 (or $8X). If you don't want this to continue indefinitely - which implies arbitrarily small and large amounts are possible - then there has to be a zero probability in X's distribution for some powers of 2. The first solution above implicitly assumes that all powers of 2 have the same probability.

+++++

Michael said: "By switching there's a 50% chance of gaining an extra £10 and a 50% chance of losing £5." This is wrong. The chance of picking the larger envelope is indeed 50% when you don't consider an amount, but you have to include the probability of having the amount you consider.

His program says that there was a 50% chance that the envelopes contained ($5,$10), but ignored the 25% possibility of picking the $5 envelope from that pair. There was another 50% chance that the envelopes contained ($10,$20), and he ignored the possibility of picking the $20 envelope. He considered only the possibility of picking the $10 envelope.

His error is assuming a distribution for the values. The OP does not provide that information, even if you look and see $10. He calculated the expectation if you are given the 50:50 split between the two sets, and that you picked $10. That conditional expectation is indeed the switching gains. It just isn't the OP.

+++++
NoAxioms said: "My solution is to switch only if the amount in the envelope is an odd number." This recognizes that the probability of ($X/2,$X) must be zero if X is odd, but doesn't recognize that the probabilities of ($X/2,$X) and ($X,$2X) do not have to be the same if X is even.

+++++
You said: "What we have establish[ed] is that -X and X are equally likely to occur." You are misinterpreting how Michael is using X. Yours is the value in the lower envelope,which makes it the difference. His is the value in your envelope, which makes the difference -X/2 or +X. He gets an invalid answer, because he needs to consider the relative probabilities of the two sets of envelopes. You consider only one set, so you don't.

I read more, and it was all thrashing. If you have a specific post you want me to read, point it out.

If you are not willing to read the thread, then I am not willing to read your post.

But I have read it now - the majority is either name calling, or the debating of INCORRECT interpretations (on almost all sides) of various concepts in probability. Regardless, what I have said so far, and will continue to say, applies to all of it, whether of not I cite how in every instance. Some of these concepts are:

1. This problem is not about statistics. What you meant when you called it a "data science," is that it tries to apply the concepts of theoretical probability to real-world situations.

2. In theoretical probability, "probability" is an undefined term. Given a sample space S, any corresponding set of numbers that satisfy the Kolmogorov axioms (all are >=0, if A and B are disjoint events then Pr(A)+Pr(B)=Pr(A&B), and Pr(S)=1), is a set of probabilities for that space. This does not interpret the meaning of the values.

3. A random variable is a measurable quantity in the result of a random process that, within your knowledge, can have any value in some set of values. As a random variable, it has a probability distribution. This means that the lower value in the pair of envelopes is a random variable, that has a probability distribution. A "probability density curve" does not have to be "used to select X.," whatever it is you think that means. Before you look in an envelope, does your knowledge allow L it to have one value in some set, whether or not that set is known to you? Then it is a random variable that has a probability distribution (distributions apply to discrete random variables, as with amounts of money. Densities need a continuous random variable).

A corrected version of your solution, if you don't look in an envelope: Let L be the random variable representing the low value in the envelopes. Say the "given event" is that the (unknown) low-value is L1; that is, one value in the set of possibilities is realized. (Note that L1 is an unknown, and not a random variable. Ask about the difference if it is unclear to you why I point this out.) There is a probability Pr(L=L1) that we can't possibly know, but it will turn out that we don't need to know it.

Similarly, let R be a random variable representing the relative value of the envelope you pick. It can be 1/2, or 2. That is, if R=1/2, you picked the smaller envelope. Note that we can say Pr(R=1/2) = Pr(R=2) = 0.5 by the Principle of Indifference.

The prior probability that L=L1 AND R=1/2 is Pr(L=L1)*Pr(R=1/2) = Pr(L=L1)/2.
The prior probability that L=L1 AND R=2 is Pr(L=L1)*Pr(R=2) = Pr(L=L1)/2.

The definition of the conditional probability for event A, given event B, is Pr(A|B)=Pr(A&B)/Pr(B):

Pr(R=1/2|L=L1) = (Pr(L=L1)/2)/Pr(L=L1) = 1/2.
Pr(R=2|L=L1) = (Pr(L=L1)/2)/Pr(L=L1) = 1/2.

These are the probabilities you should use in your expectation calculation. Since Pr(L=L1) divides out, we don't need to know it. You are confusing the fact you can take a shortcut to get this result, with that shortcut being logically correct.

Note how the value of your envelope different in these two cases: L1 in the first, and 2*L1 in the second. Michael's error is treating the "given" event as a fixed value for your envelope, but using the same shortcut. It no longer gets the right result.

Let V be the value of your envelope. If we treat the "given event" as V=V1, then

Pr(R=1/2|V=V1) = Pr(R=1/2&V=V1)/Pr(V=V1) = Pr(R=1/2&L=V1)/Pr(V=V1) = Pr(L=V1)/Pr(V=V1)/2
Pr(R=2|V=V1) = Pr(R=2&V=V1)/Pr(V=V1) = Pr(R=2&L=V1/2)/Pr(V=V1) = Pr(L=V1/2)/Pr(V=V1)/2
Finally,
Pr(V=V1) = Pr(L=V1&R=1/2)+Pr(L=V1/2&R=2) = [Pr(L=V1)+Pr(L=V1/2)]/2

So the numbers to use in the probability calculation are

Pr(R=1/2|V=V1) = Pr(L=V1)/[Pr(L=V1)+Pr(L=V1/2)]
Pr(R=2|V=V1) = Pr(L=V1/2)/[Pr(L=V1)+Pr(L=V1/2)]

Now the distribution of the values matters. Micheal implicitly assumed Pr(L=V1)=Pr(L=V1/2).

Since Pr(L=L1) divides out, we don't need to know it. You are confusing the fact you can take a shortcut to get this result, with that shortcut being logically correct. — JeffJo

Doesn't this amount to saying that the loading of the envelopes and the selection of an envelope are independent events, in which case conditioning is pointless?

And then given that the player has no choice but to treat which envelope they "choose" as a matter of indifference-- they might as well flip a coin-- what possible reason could the facilitator have for not also treating which value is assigned to which envelope as a matter of indifference? They too could just as well flip a coin. Which leaves me puzzled about the point of your V=V1 stuff, if I'm following what you did there.

What you meant when you called it a "data science," is that it tries to apply the concepts of theoretical probability to real-world situations. — JeffJo

Data science, strangely enough, involves data. As a data scientist your job is to analyze the data, then it is the job of the subject matter experts to interpret how that applies to the real world. Statistics is a data science and in statistic we measure uncertainty, which is why some people call it the science of uncertainty.

I have said all I am going to say about this problem, as at this point the arguments are just recycling themselves. I am fully satisfied with my approach and my solution. It is time to move on.

As we say goodbye to the two envelopes, I'd like to call attention to a couple oddities of the alternative analyses:

(A1) No-switchers imagine the first step as a choice or random selection between two items, and then the second step is trading that item for the one item left-- no choice.

(A2) Switchers imagine the first step as getting some determinate value-- there need not be a "choosing" here at all-- and then the second step is a random selection from two alternatives.

(B1) No-switchers see the options as getting either the smaller or the bigger of two items.

(B2) Switchers redefine smaller as "bigger of a smaller pair" and bigger as "smaller of a bigger pair".

How these symmetries arise seemingly just from the way you assign variables still puzzles me a bit.

Mixing random variables and unknowns can be very un-intuitive, especially to those who are out of practice.

Doesn't this amount to saying that the loading of the envelopes and the selection of an envelope are independent events,

That isn't as useful as you might think. You can assert how that independence is obvious, since the random variable R (where R is 1/2 or 2 if you picked low or high) is chosen without knowledge of the random variable L (the low value of the pair). But all this independence means is, for any values of the unknowns L1 and R1, that:

Pr(L=L1&R=R1) = Pr(L=L1)*Pr(R=R1).

What independence doesn't tell you, is how Pr(L=L1/2&R=2) = Pr(L=L1/2)/2 compares to Pr(L=L1/2&R=1/2) = Pr(L=L1)/2. If you need to compare them, independence does nothing for you.

If you consider a value - like you do whenever you calculate an expectation - you have to condition on that value. Here, V is the random variable for your envelope's value:

E = Exp(Other Envelope|V=V1) = (V1/2)*Pr(R=2|V=V1) + (2V1)*Pr(R=1/2|V=V1)
E = (V1/2)*Pr(R=2&V=V1)/Pr(V=V1) + (2V1)*Pr(R=1/2&V=V1)/Pr(V=V1)
E = V1*[Pr(R=2&V=V1)/2 + 2*Pr(R=1/2&V=V1)]/Pr(V=V1)

Note that it is L that is independent of R, not V. So you can't separate Pr(R=1/2&V=V1) into Pr(R=1/2)*Pr(V=V1) = Pr(V=V1)/2. As a trivial example, if V=$1, then R can't be 2 and Pr(R=1/2&V=$1)=Pr(V=$1).

Changing the random variable in this formulation (that is, once you started this way) from V to L, so you can use independence, doesn't help:

E = V1*[Pr(R=2&V=V1)/2 + 2*Pr(R=1/2&V=V1)]/Pr(V=V1)
E = V1*[Pr(R=2&L=V1/2)/2 + 2*Pr(R=1/2&L=V1)]/[Pr(R=2&L=V1/2) + Pr(R=1/2&L=V1)]
E = 2*V1*[Pr(L=V1/2)/4 + Pr(L=V1)]/[Pr(L=V1/2) + Pr(L=V1)]

This approach to an expectation inherently compares the probabilities of two possible combinations fo envelopes. This result applies whether you do know V1, or don't know V1. If you do, you need to know the relative probabilities of those two possible sets of envelopes, which you don't have. If you don't know V1, then you get the expectation by integrating over all possible values of V, which requires even more knowledge you don't have. With a lot of work (too much), you could prove that this will always be equal to the expectation of your envelope.

There is an easier way, but it applies only if you don't know what is in your envelope. Use L instead of V from the beginning and condition on L=L1. Then you can use independence, you only need to use Pr(L=L1), and it divides out. The disadvantage is, that you can't answer if you look and see $10 in your envelope. WHICH IS CORRECT.

↪Jeremiah

We don't have data in the OP; we have a theoretical problem only.

We don't have data in the OP; we have a theoretical problem only. — JeffJo

You keep demonstrating how few of my posts you have actually read.

There is an easier way, but it applies only if you don't know what is in your envelope. — JeffJo

If, even before selecting an envelope, you see that there is no reason to prefer one envelope over the other, what compels you to discard that analysis upon selecting and opening an envelope? Is it no longer true that if you drew X you'd trade for 2X and if you drew 2X you'd trade for X? How could selecting or even looking in an envelope make that false?