, so this is pointless. — JeffJo

Funny how I already said something very much like this.

You might find this interesting:

https://www.researchgate.net/publication/266706577_The_fallacy_of_the_two_envelopes_problem

That article shows that if you get to open your envelope you can use the knowledge of its value to apply a switching strategy to better your earnings.

So it’s similar to the strategies that andrewk and I have described.

Edit: it’s actually the same as andrewk’s.

If my £10 envelope is Envelope X then switching to Envelope 2X gains me £10 and if my £10 envelope is Envelope 2X then switching to Envelope X loses me £5. — Michael

(This was Michael's response to a post by Baden on p.3 of this thread)

I just wanted to note that, as Michael may have realized by now, the higher expected value of the choice of switching, as compared with the choice of not switching, only is larger than zero if the act of opening the first envelope is assumed not to yield any additional (probabilistic) knowledge regarding the content of the second envelope. But if we assume any knowledge of the prior joint probability distribution of the contents of the two envelopes whatsoever, then, in that case, applying Bayes' Theorem in order to calculate the posterior probability distribution of the content of the second envelope yields an expected value of zero for the act of switching (as compared with not-switching).

On edit: The assumption that no additional knowledge regarding the content of the second envelope would be gained by opening the first only would valid in the case where the two distributions (prior probabilities) were independent, which they aren't.

It seems to me that most of the disagreements aren't actually about how to calculate probabilities given a scenario, but rather on which scenario we're in. In the hope of disambiguating it, I'll consider a generalised envelope problem. Instead of the amounts being X and 2X, we'll generalise to U and V which are assumed to be positive real valued.

This makes the problem, per Jeramiah's OP:

You are playing a game for money. There are two envelopes on a table.
You know that one contains $U and the other $V, [but you do not
know which envelope is which or what the numbers U or V are]. Initially you
are allowed to pick one of the envelopes, to open it, and see that it
contains $Y . You then have a choice: walk away with the $Y or return
the envelope to the table and walk away with whatever is in the other
envelope. What should you do?

The first interpretation consists in:
(1) Being given an envelope.
(2) Opening it doesn't do anything to the sample space, conditioning is irrelevant.



Now for when the card is revealed and the sample space is... to be determined, as it's more complex. This is dealing with the original problem as stated by Jeramiah:

You are playing a game for money. There are two envelopes on a table.
You know that one contains $X and the other $2X, [but you do not
know which envelope is which or what the number X is]. Initially you
are allowed to pick one of the envelopes, to open it, and see that it
contains $Y . You then have a choice: walk away with the $Y or return
the envelope to the table and walk away with whatever is in the other
envelope. What should you do?

The confusions go away when you actually realise there are two possible sample spaces given that you receive X, and there is no information on which you're in.

The second interpretation consists in:
(1) Opening the envelope to receive X.
(2) There is no information over whether you're in the case (X,X/2) or (X,2X).



You end up with expectation 0 from switching and not switching so long as you're attending to the sample space consistently, independent of the interpretation.

Why is this different from the Monty Hall problem? When you're asked 'whether you would like to switch', the probability distribution changes to a distribution yielding 0.5 probability to each door, giving 2/3 probability of winning by Bayes' theorem. In the second interpretation, being told you receive X gives you no information about the resultant gain. Monty Hall does.

's enduring confusion throughout the thread consists in sneaking in some assumption that, in the second interpretation, the subject knows which case they're in. If instead the subject does not, the sample space is not (X/2,2X), it's (X/2,X,2X) if you receive a random envelope (which you need three of for this to make sense). This doesn't resemble anything like the original problem in which there are 2 envelopes.

TLDR: you're never actually getting to the sample space (X/2,2X) from what you're given. If you actually do end up in that case, then the result's an obvious (even deterministic) choice given that you know X. If you interpret everything carefully, there's no preference for switching because there's never any information on which case you're in.

Edit2: Also, this is pretty much the analysis in Jeramiah's linked paper, though it's analysed in terms of conditional expectations rather than probabilities.

You actually have 2 more complicated cases. The first case is when you have X and the amount in the other envelope is X/2, the second case is when you have X and the amount in the other envelope is 2X.

Consider the first case. The envelopes now contain X and X/2. If you have X and switch you gain -X/2, if you have X/2 and switch you gain X/2, which are equally probable, so they cancel in the expectation giving 0.

The second case. The envelopes now contain X and 2X. If you have X and switch you gain X, if you have 2X and switch you lose X, these are equally probable, so they cancel out in the expectation giving 0. — fdrake

What do you mean by X here? Are you using it to refer to the value of the smallest envelope or to refer to the value of the opened envelope? It looks like you're doing both, and so the above doesn't make much sense (especially with "if you have X/2" which is impossible under either definition).

By this do you just mean that if we know that the value of X is to be chosen from a distribution of 1 - 100 then if we open our envelope to find 150 then we know not to switch?

I am already convinced my approach is correct, Michael. I have no doubt about it, and I no longer care about arguing or proving that point. You can interpret this article however you wish. If you learn something from it great; if you decide to think of it as personal validation that is your choice. I have been reading a number of papers on this problem with a number of different approaches and outcomes and I found this guy's approach interesting. And I think it is wise if I keep my thoughts about Andrewk’s approach to myself.

There are 2 envelopes.
You open an envelope and see X.
The envelopes could have been filled with either X and 2X, or X and X/2. These are jointly exhaustive possibilities.
This gives two cases to analyse. Which are labelled (X,2X) and (X,X/2). These correspond to the case when one envelope has X in it and the other envelope has 2X in it. And the case where one envelope has X in it and the other envelope has 2X in it. The probability distribution here is on the sample space:

A=(one envelope contains X and the other contains 2X)
B=(one envelope contains X and the other contains X/2)

Having an envelope means being in case A or being in case B. Opening the envelope does nothing to inform you of whether you are in case A or case B. Let's assume case A and case B are equally likely. That is, it's equally like that A (your envelope contains X and the other contains 2X) or B (your envelope contains X and the other contains X/2).

If you are in case A, having an envelope means that the envelope you hold is either X or 2X.
If you are in case B, having an envelope means that the envelope you hold is either X or X/2.

Assume you're in case A. If you have the lower valued envelope X and switch, you gain X. If you have the higher valued envelope 2X and switch you lose X. It is equally likely that you have X or 2X, so the expected gain conditional on being in case A is:

0.5(X-X)=0

Assume you're in case B. If you have the lower valued envelope X/2 and switch, you gain X/2. If you have the higher valued envelope X and switch, you lose X/2. So the expected gain conditional on being in case A is:

0.5( 0.5X - 0.5X) = 0

Now, it is random whether you are in case A or case B. They are equally likely. This means that the overall expectation of switching is the average of the expectations of switching in each case.

So the average of 0 from case A and 0 from case B is 0.5(0+0)=0.

With your approach, you have the sample space (2X,X/2) from switching. Please answer the following question:

What is the probability of obtaining the envelope containing X? Not the conditional probability of the envelope you have containing X given that you just opened it. This is a trick, your approach does not answer this question - since it cannot be answered using only 2 envelopes unless the probability of obtaining an envelope containing X is 0...

You can interpret this article however you wish. — Jeremiah

I interpreted it as it is literally written:

Suppose we know only that the two amounts in the envelope are different (and no longer that the larger amount is double the smaller one). No information is provided of how the different amounts were chosen. We are allowed to open one randomly selected envelope. Does there exist a strategy that will lead us to accept the larger amount with a probability greater than $\frac{1}{2}$?

The answer is yes. This may come as a surprise to an unprepared reader, but the strategy is very simple.

Let $S$ and $T$ denote the different amounts in the first and second envelopes, respectively, and let $x$ be a fixed positive number. We consider the following strategy:

Choose $S$, if and only if $S \gt x$; choose $T$ otherwise.

...

If the smaller amount equals $M$, say, and the larger $2M$, then a randomized choice yields an expected value of $\frac{M + 2M}{2} = 1.5M$. Strategy (6), however, yields an expected amount of $1.5M(p_1 + p_3) + 2Mp_2 = 1.5M + 0.5Mp_2$.

Having an envelope means being in case A or being in case B. Opening the envelope does nothing to inform you of whether you are in case A or case B. Let's assume case A and case B are equally likely. That is, it's equally like that A (your envelope contains X and the other contains 2X) or B (your envelope contains X and the other contains X/2).

If you are in case A, having an envelope means that the envelope you hold is either X or 2X.
If you are in case B, having an envelope means that the envelope you hold is either X or X/2.

Assume you're in case A. If you have the lower valued envelope X and switch, you gain X. If you have the higher valued envelope 2X and switch you lose X. It is equally likely that you have X or 2X, so the expected gain conditional on being in case A is:

0.5(X-X)=0

Assume you're in case B. If you have the lower valued envelope X/2 and switch, you gain X/2. If you have the higher valued envelope X and switch, you lose X/2. So the expected gain conditional on being in case A is:

0.5( 0.5X - 0.5X) = 0 — fdrake

You're doing it again. Is X the value of my envelope or the value of the smallest envelope? You're switching between both and it doesn't make any sense, especially where you say "if you have the lower value envelope X/2".

Where did I switch between stuff?

Where did I switch between stuff? — fdrake

First you say "You open an envelope and see X" and then you say "If you have the higher valued envelope 2X and switch you lose X".

How can I see X in my envelope but have 2X?

And you also say "If you have the lower valued envelope X/2". But what does X mean here? It can't be the value of my envelope, because you're supposing that my envelope is X/2, and it can't be the value of the smallest envelope, because you're supposing that my envelope is X/2, which is smaller than X.

I read that part. In fact I read it several times.

Ok. I'll say it without the abuse of notation.

There are 2 envelopes.
You open an envelope and see the value U.
So your random variable is U.
The envelopes could have been filled with either X and 2X, or X and X/2. These are jointly exhaustive possibilities. These described the possible states, of either having an envelope twice the value of the other or half the value of another.
This gives two cases to analyse. Which are labelled (X,2X) and (X,X/2). These correspond to the case when one envelope has X in it and the other envelope has 2X in it. And the case where one envelope has X in it and the other envelope has X/2 in it. The probability distribution here is on the sample space:

A=(one envelope contains X and the other contains 2X)
B=(one envelope contains X and the other contains X/2)

Which means that we will analyse the cases of the expectation of switching given A, the expectation of switching given B, then use those to form the overall expectation of switching.

Having an envelope means being in case A or being in case B. Opening the envelope does nothing to inform you of whether you are in case A or case B. Let's assume case A and case B are equally likely. That is, it's equally like that A (your envelope contains X and the other contains 2X) or B (your envelope contains X and the other contains X/2).

If you are in case A, having an envelope means that the envelope you hold is either X or 2X. That is, U=X or U = 2X.
If you are in case B, having an envelope means that the envelope you hold is either X or X/2. That is, U = X or U = X/2.

Case A and Case B are jointly exhaustive. Note that if one envelope is filled with X/2, the other cannot be filled with 2X, it must be filled with X. Similarly, if one envelope is filled with 2X, the other must be filled with X. This is because one envelope must be twice the value of another (or equivalently, one is half the value of the other).

Assume you're in case A. If you have the lower valued envelope (U=X) and switch, you gain X (you now have U=2X). If you have the higher valued envelope U=2X and switch you lose X (since you had 2X and switched to X). It is equally likely that you have X or 2X, so the expected gain conditional on being in case A is:

0.5(X-X)=0

Assume you're in case B. If you have the lower valued envelope X/2 and switch, you gain X/2. If you have the higher valued envelope X and switch, you lose X/2. So the expected gain conditional on being in case A is:

0.5( 0.5X - 0.5X) = 0


Now, it is random whether you are in case A or case B. They are equally likely. This means that the overall expectation of switching is the average of the expectations of switching in each case.

So the average of 0 from case A and 0 from case B is 0.5(0+0)=0.

With your approach, you have the sample space (2X,X/2) from switching. Please answer the following question:

What is the probability of obtaining the envelope containing X? Not the conditional probability of the envelope you have containing X given that you just opened it. This is a trick, your approach does not answer this question - since it cannot be answered using only 2 envelopes unless the probability of obtaining an envelope containing X is 0...

Better?

You open an envelope and see the value U.
So your random variable is U.
The envelopes could have been filled with either X and 2X, or X and X/2. — fdrake

What does X mean here?

X is a fixed dollar amount. Set it equal to 1 if you want, the reasoning proceeds the same way.

X is a fixed dollar amount. Set it equal to 1 if you want, the reasoning proceeds the same way. — fdrake

OK, so how much is the U that I see in my envelope?

You always observe X in your envelope. What was the probability of obtaining U=X? Not the conditional probability of getting X given that you opened the envelope containing it.

You always observe X in your envelope. What was the probability of obtaining U=X? Not the conditional probability of getting X given that you opened the envelope containing it. — fdrake

So now you're back to defining X as the amount in my envelope and then supposing that I could have 2X and lose X. You're not making any sense.

Ok. This is from how expectations work.

Imagine that we have two envelopes, one filled with X and one with 2X.

If I have an envelope containing X, and switch to an envelope containing 2X, I gain X from the switch.
If I have an envelope containing 2X, and switch to an envelope containing X, I lose X from the switch.

Does that make sense to you?

Imagine that we have two envelopes, one filled with X and one with 2X.

If I have an envelope containing X, and switch to an envelope containing 2X, I gain X from the switch.
If I have an envelope containing 2X, and switch to an envelope containing X, I lose X from the switch.

Does that make sense to you? — fdrake

Yes. So let's say that there's £10 in my envelope:

If I have an envelope containing X (10), and switch to an envelope containing 2X (20), I gain X (10) from the switch.
If I have an envelope containing 2X (10), and switch to an envelope containing X (5), I lose X (5) from the switch.

Would you agree that {5,10,20} are the possible values for things in the envelopes?

Would you agree that {5,10,20} are the possible values for things in the envelopes? — fdrake

I agree that {10} is the certain value for the thing in my envelope and that {5, 20} are the possible values for the thing in the other envelope.

I don't know if it's correct to merge them as you have done. Rather it's either {10, 5} or it's {10, 20}.

If 5 and 20 are permissible values for the envelopes, does it stand to reason that I could receive an envelope which is 4 times another envelope?

If 5 and 20 are permissible values for the envelopes, does it stand to reason that I could receive an envelope which is 4 times another envelope? — fdrake

They're permissible values for the other envelope. My envelope must be £10. So either it's {10, 5} for both or it's {10, 20} for both.

Then what's the probability of receiving an envelope containing 10? Not the probability of having 10 given that you just looked,

Then what's the probability of receiving an envelope containing 10? — fdrake

Your question is ambiguous. Are you asking for the probability that I selected the £10 envelope (given that there's £10 in one and either £5 or £20 in the other) or the probability that the host selects £10 as a value to put in one of the envelopes? The first is 50%, the second is unknown.

In your set up. There are three possible values for envelopes, {5,10,20}. Imagine that all envelopes are still closed. And you are given an envelope. What's the probability of having 10?

In your set up. There are three possible values for envelopes, {5,10,20}. Imagine that all envelopes are still closed. And you are given an envelope. What's the probability of having 10? — fdrake

That's not my set up. My set up is that there's £10 in my envelope and that one envelope contains twice as much as the other. I then deduce from this that the other envelope contains either £5 or £20.