Therefore, within the same equation, A is referring to two different amounts. Am I correct in thinking that this is why the equation gives a false result. — RussellA

I would say that is a potential cause of the paradox, but isn't the paradox itself. The paradox is the fact that the switching argument consists of a logically inconsistent set of probabilistic assumptions.

The switching argument, which produces a contradictory strategy for solving the two-envelope problem, starts by subjectively assuming, without evidence, the following conditional distribution, with respect to envelopes A and B whose values are a and b respectively :

P (B = (1/2) a | A = a) = P(B = 2a | A = a) = 1/2 For all values a

which yields the conditional expectation value

E [ B | A = a] = (5/4) a For all values a

So far, terrible reasoning with respect to the two-envelope problem, but no inconsistency.

Next, the switching argument further assumes, without evidence, that

P (A = a | B = (1/2) a) = 1 / 2 for all values a

However, using Bayes Theorem gives

P (A = a | B = (1/2) a) = P (B = (1/2) a | A = a) x P(A = a) / P(B = (1/2) a)

Hence the switching argument assumes that ratio on the right hand side is always 1 :

P(A = a) / P(B = (1/2) a) = 1 for all values a

This is the most deranged part of the argument. It amounts to asserting "If we know nothing about the prior distribution of A and nothing about the prior distribution of B, then we can be certain that the ratio of their probabilities is 1 - for our ignorance cancels out! "

We then derive the contradiction

E [ A | B = b ] = (5/4) b For all values b

The switching argument, which produces a contradictory strategy for solving the two-envelope problem, starts by subjectively assuming, without evidence, the following conditional distribution, with respect to envelopes A and B whose values are a and b respectively :

P (B = (1/2) a | A = a) = P(B = 2a | A = a) = 1/2 For all values a — sime

It just assumes that:

P(A = the smaller envelope) = P(B = the smaller envelope) = 1/2

Which is correct. If either A is smaller than B or B is smaller than A, and if you pick one at random, then the probability that you picked the smaller envelope is 1/2. That's true a priori, is it not? A random selection from a set of two members.

Even if you want to say that your choice of A or B was biased in some way, assume that a true random number generator was used to decide which of A and B was to contain the smaller amount.

In fact, assume that you don't choose an envelope. You're just given envelope A, and a true random number generator was used to determine whether A should contain £10 or £20 (with B containing the other).

What is the probability that your envelope, A, contains £10? It's 1/2.

And this is true even if you don't know beforehand the values of the two envelopes. You're just told that there are two amounts, x and 2x, and that a true number generator was used to determine whether A should contain x or 2x (with B containing the other).

What is the probability that your envelope, A, contains £x? It's 1/2.

It just assumes that:

P(A = the smaller envelope) = P(B = the smaller envelope) = 1/2 — Michael

That expression is used to represent the same set of initial assumptions, but is less explicit with regards to its premises, such as the fact that some distribution is responsible for placing a certain amount of money in each envelope.

For debugging probabilistic arguments and resolving paradoxes it is better to use conditionals so that every premise is made explicit, so that problematic premises are more easily identified.

That expression is used to represent the same set of initial assumptions, but is less explicit with regards to its premises, such as the fact that some distribution is responsible for placing a certain amount of money in each envelope. — sime

I tell you that one envelope contains £20 and the other envelope contains £10. A true random number generator was used to determine which of envelopes A and B should contain which amount.

You are given envelope A.

What is the probability that your envelope contains £20? It's 1/2.

I tell you that one envelope contains twice as much money as the other. A true random number generator was used to determine which of envelopes A and B should contain which amount.

You are given envelope A.

What is the probability that your envelope contains twice as much money as the other? It's 1/2.

There's no reason that the probability in the second case should be different to the probability in the first case. You knowing the actual amounts doesn't affect the probability that you were given the larger amount. It's always going to be 1/2.

The switching argument, which produces a contradictory strategy for solving the two-envelope problem, starts by subjectively assuming, without evidence, the following conditional distribution, with respect to envelopes A and B whose values are a and b respectively — sime

Starting with the Wikipedia Two Envelopes Problem
Imagine you are given two identical envelopes, each containing money. One contains twice as much as the other. You may pick one envelope and keep the money it contains. Having chosen an envelope at will, but before inspecting it, you are given the chance to switch envelopes. Should you switch?

There are two identical envelopes A and B

Envelope A contains the value a, and envelope B contains the value b.

Either i) if the value a is 10 euros, then the value b is 20 euros or ii) if the value a is 20 euros, then the value b is 10 euros.

The probability of my picking an envelope with 10 euros is 50% and the probability of my picking an envelope with 20 euros is 50%.

What subjectivity are you referring to ?

What is the probability that your envelope contains twice as much money as the other? It's 1/2. — Michael

Again, that's an additional subjective premise that isn't objectively implied by the 'physics' of the two-envelope premise. Nevertheless we can assume it for sake of argument and continue the discussion as an exercise in subjective Bayesian probability estimation in which we hope for self-consistency.

There's no reason that the probability in the second case should be different to the probability in the first case. — Michael

Yes there is. That's why conditional probabilities should be used. The conditional probability

P ( B = b | A = a)

that refers to the amount of money in the unopened envelope B when conditioned on the amount of money in opened envelope A, is generally unequal to

P (A =a | B = b)

that refers to the amount of money in unopened envelope A when conditioned on the amount of money in opened envelope B.

Knowing the former conditional distribution upon opening envelope A generally says nothing about the latter distribution unless the ratios of the priors P(A=a) and P(B=b) is assumed to be 1, as indicated by Bayes Theorem.

Only if you take the ratio to be 1 do the calculated subjective conditional expectations come into conflict with respect to decision making.

Assuming your subjective premise:

- Recall the fact that the implied subjective expectation regarding the amount of money gained upon opening either envelope is 1.5x, where x is the smallest amount of money (as i showed in my first analysis).

- Also recall that unlike in my first analysis that referred to the envelopes by labels that were assigned to them prior to them being opened, here we are defining A to refer to the opened envelope. In which case the probability that we will open A is 1!

Doesn't the fact that you will definitely open A first, together with the fact that your expected gain is 1.5x imply that your subjective prior for P(B) should be different to your subjective prior for P(A)?

and

First of all, thank you both for your interesting arguments. I am following them as much as I can understand the OP started by Michael.

I think the main substance of this dilemma is discerning on if it is "rational" or "not" switching one of the envelopes when you chose one already.
Michael showed to us arguments when switching is rational.

Envelope A contains 10 € and Envelope B contains 20 €. Then, according to that dilemma, I have the chance to switch my first option and get the twice I suppose to get in my choice. So, I agree with Michael that it's always going to be a 1/2 probability.
Yet, this scenario only happens if I am expecting a value. I decide to switch the envelopes because I want to expect to double the amount that supposedly is in the envelope. Again, switching is rational or "worthy" because I am expecting a value.

But... What about the indifference on the amount of each envelope? For example: Let the amounts in the envelopes be X and 2X. Now by swapping, the player may gain X or lose X. So the potential gain is equal to the potential loss.

I agree in this point. It could exist the scenario where switching or not is meaningless to the participant if the expected value doesn't exists.

I am sorry If I were not clear in my point. What I intend to say is the fact that this paradox and dilemma only works if the participant expects a value or loss, but it could be indifferent whatever the amount and the action of switching is not motivated by an expected value.

the amount of money in the unopened envelope B when conditioned on the amount of money in opened envelope A — sime

There is no opened envelope:

Imagine you are given two identical envelopes, each containing money. One contains twice as much as the other. You may pick one envelope and keep the money it contains. Having chosen an envelope at will, but before inspecting it, you are given the chance to switch envelopes. Should you switch?

All I know is that one envelope contains twice as much as the other and that I picked one at random. I don't know what's in my envelope. The probability that I picked the more valuable envelope is 1/2.

Michael showed to us arguments when switching is rational. — javi2541997

No I didn't. I showed that the argument which purports to show that switching is rational commits a mathematical fallacy, and that there is no rational reason to switch.

No I didn't. I showed that the argument which purports to show that switching is rational commits a mathematical fallacy, and that there is no rational reason to switch. — Michael

Ok! I see this dilemma and your arguments clearer now. :up:

There is no opened envelope: — Michael

The switching argument begs to differ :

" 1. Denote by A the amount in the player's selected envelope."

The argument's computed expectation value of (5/4) A is a conditional expectation with respect to the variable A denoting the amount of money in the player's selected envelope.

Not opening the envelope and taking an expectation over the envelope's value means that A cannot be mentioned in the expectation value, even as a variable with an unspecified value.

The unconditional expectation of the players envelope value is 0.5 x M + 0.5 x 2M = 1.5M , where M is the mean of the unspecified distribution F for the smallest amount of money in an envelope. No paradox arises from this calculation.

I've subsequently come to realise however, that my initial analysis wasn't quite right. If my calculations are correct (that are too tedious to post here), the switching dilemma is caused by assuming an un-normalised "non-informative prior" for F(M), which causes the conditional expectations for each envelope to diverge.

Intuitively, this can be explained as follows:

if there could be any amount of money in the two envelopes, and if all monetary values are equally likely, then whatever value you observe in one envelope, the other envelope is likely to have an even higher value.

To remove this bug, one has to replace the physically implausible non-informative prior for F with a normalised distribution, so that the conditonal expectation for the unopened envelope gets lower as the observed money in one's opened envelope gets higher.

I agree that there is an error with the calculation of the expected value. That's what I explain in the OP.

My argument with you is over the assigned probabilities. So ignore the expected value. I just want to know an answer to this:

1. One envelope contains £10
2. One envelope contains £20
3. I pick an envelope at random
4. I don't open my envelope
5. What is the probability that I picked the envelope containing £10?

My answer is 1/2. What is yours?

The unconditional expectation of the players envelope value is 0.5 x M + 0.5 x 2M = 1.5M , where M is the mean of the unspecified distribution F for the smallest amount of money in an envelope. No paradox arises from this calculation. — sime

Sure, but as is mentioned on the Wikipedia article:

The puzzle is to find the flaw in the line of reasoning in the switching argument.

...

In particular, the puzzle is not solved by finding another way to calculate the probabilities that does not lead to a contradiction.

My argument with you is over the assigned probabilities. So ignore the expected value. I just want to know an answer to this:

1. One envelope contains £10
2. One envelope contains £20
3. I pick an envelope at random
4. I don't open my envelope
5. What is the probability that I picked the envelope containing £10?

My answer is 1/2. What is yours? — Michael

The paradox doesn't apply in that scenario, since the values of both envelopes are given.

To my understanding , the paradox requires,

1) Knowledge of the value of only one of the envelopes.

2) The assumption of a non-informative prior for P(M) , where M is the smallest (or greatest) money in the envelopes.

In particular, the puzzle is not solved by finding another way to calculate the probabilities that does not lead to a contradiction. — Michael

The contradiction vanishes when a normalised prior is used for P(M), since in that instance knowledge of the value of one envelope is indicative of the value of the other.

To my understanding , the paradox requires,

1) Knowledge of the value of only one of the envelopes. — sime

The paradox is premised on not knowing the value of any.

But do you agree that the probability in my example situation is 1/2?

The paradox is premised on not knowing the value of any. — Michael

That is flat out contradicted by the switching argument. Furthermore, without the premise of knowing the value of one of the envelopes, the paradox disappears.

To have an expectation value containing A on the one hand, and insisting that the envelope isn't opened on the other, is a bait and switch. Wikipedia's article isn't written very well, which might be part of the confusion.

But do you agree that the probability in my example situation is 1/2? — Michael

It is a half if you assume it to be 1/2, but not necessarily. Consider for instance someone sending you the smaller of two envelopes through the post, according to a probability that they have decided. You open the letter and are informed that if you return the envelope and it's contents, you will receive another envelope that has half as much or twice as much.

It is a half if you assume it to be 1/2, but not necessarily. Consider for instance someone sending you the smaller of two envelopes through the post, according to a probability that they have decided. You open the letter and are informed that if you return the envelope and it's contents, you will receive another envelope that has half as much or twice as much. — sime

That’s not what happens in this example. I am shown two envelopes, one containing £10 and one containing £20, and I freely choose one at random. I don’t open it. The probability that I picked the one with £10 is 1/2.

That is flat out contradicted by the switching argument. — sime

It’s not. It’s the premise of the switching argument.

That’s not what happens in this example. I am shown two envelopes, one containing £10 and one containing £20, and I freely choose one at random. I don’t open it. The probability that I picked the one with £10 is 1/2. — Michael

I'll agree for sake of argument . I think the problem is how we are fitting our shared understanding of the problem to probability calculus.

In my preferred description, one of the envelopes is opened to reveal a quantity A, but It isn't known as to whether the other envelope is more than or less than A.

In your preferred description, the quantities of both envelopes is known a priori, but neither of the envelopes are opened.

The problem with your description, is that it runs contrary to how conditional probabilities and expectations are normally interpreted. For the information upon which a probability or expectation is conditioned, is normally interpreted as observed information, rather than possessed information that isn't revealed, compensated by assumed knowledge of other quantities.

I'll agree for sake of argument . I think the problem is how we are fitting our shared understanding of the problem to probability calculus.

In my preferred description, one of the envelopes is opened to reveal a quantity A, but It isn't known as to whether the other envelope is more than or less than A.

In your preferred description, the quantities of both envelopes is known a priori, but neither of the envelopes are opened.

The problem with your description, is that it runs contrary to how conditional probabilities and expectations are normally interpreted. For the information upon which a probability or expectation is conditioned, is normally treated as observed information. — sime

I honestly don't understand your interpretation of probability. This seems very straightforward.

Maybe a different example. I have a red ball hidden in one hand and a blue ball hidden in my other hand. You point to one of my hands at random. What is the probability that you pointed to the hand holding the red ball? It's 1/2.

Maybe a different example. I have a red ball hidden in one hand and a blue ball hidden in my other hand. You point to one of my hands at random. What is the probability that you pointed to the hand holding the red ball? It's 1/2. — Michael

It depends on what interpretation of probability you are appealing to. For those of us who reject Laplace's principle of indifference, the answer is to refrain from asserting a subjective probability.

In any case, it isn't relevant to the two envelopes problem, for It can be reproduced by appealing to a causal interpretation of probability in which a person observes the contents of an envelope he is given, without him making any decisions. A Bayesian analysis reveals that the culprit of the paradox is the assignment of a non-informative prior to the distribution that generates the envelopes contents.

Without that assumption, the conditional expectations involved behave sensibly and the paradox dissolves.

In any case, it isn't relevant to the two envelopes problem — sime

I was leading to explaining why it’s relevant, but if you disagree with me on the red ball probability then it’s not going to go anywhere.

A Bayesian analysis reveals that the culprit of the paradox is the assignment of a non-informative prior to the distribution that generates the envelopes contents. — sime

My understanding and resolution of the paradox is somewhat aligned with this perspective. The paradox was first introduced to me about 30 years ago by a friend who was a professor in statistics at UQAM (Université du Québec à Montréal). After further thought (and only after I was introduced to Bayes' theorem) I realized that the situation where it appears beneficial to switch the initially chosen envelope arises when we make an unrealistic assumption: that our belief about the probability distribution over possible envelope contents is that it is both uniform and infinite.

However, given any reasonably well-defined (and bounded) prior, opening one envelope may indeed inform our decision to either switch or stick to the original choice. This decision would be guided by the Expected Value (EV) of switching, which in turn is dictated by the revised probabilities concerning the potential contents of both envelopes. Notably, there's only one unique amount in the initially chosen envelope that would result in a zero EV for switching, rendering the choice indifferent.

The paradox seems to emerge from the assumption that opening the initial envelope provides equal probabilities for the second envelope containing either 10n or n/10 the amount in the first one, irrespective of the value of n. This is where I believe the core misunderstanding lies.

The paradox seems to emerge from the assumption that opening the initial envelope provides equal probabilities for the second envelope containing either 10n or n/10 the amount in the first one, irrespective of the value of n. This is where I believe the core misunderstanding lies. — Pierre-Normand

In the problem stated in the OP (taken from the Wikipedia article), there is no opening of the initial envelope:

Imagine you are given two identical envelopes, each containing money. One contains twice as much as the other. You may pick one envelope and keep the money it contains. Having chosen an envelope at will, but before inspecting it, you are given the chance to switch envelopes. Should you switch?

Indeed, I concur that the paradox can also manifest when the option to switch envelopes is offered prior to opening the initially chosen one. The resolution I (and @sime also, I think) proposed also applies in this scenario. The apparent rationality of switching, in this case, is predicated on the concept that the Expected Value (EV) of the decision is expressed as (10n + n/10)/2 - n, a value that remains positive irrespective of n. This line of thought, however, is based on the assumption that the probabilities for the second envelope containing either 10n or n/10 are independent of the value of n.

If we adjust this assumption to reflect that these probabilities are indeed conditional upon n (in the manner that Bayesian updating with an informative prior would suggest), then it becomes plausible to hypothesize — and likely not too challenging to demonstrate — that the EV of switching remains zero.

This line of thought, however, is based on the assumption that the probabilities for the second envelope containing either 10n or n/10 are independent of the value of n. — Pierre-Normand

Before I flip a fair coin, what is the probability that it will land on heads? I say 1/2.

After flipping the coin, but before looking at the result, what is the probability that it landed on heads? I say 1/2.

Now imagine that rather than heads and tails, there is a number printed on each side. One is a 10 and one is a 20.

Before I flip the coin, what is the probability that it will land on 10? I say 1/2.

After flipping the coin, but before looking at the result, what is the probability that it landed on 10? I say 1/2.

Now imagine that we don’t know the exact numbers printed on the coin, only that one is twice the value of the other.

Before I flip the coin, what is the probability that it will land on the smaller number? I say 1/2.

After flipping the coin, but before looking at the result, what is the probability that it landed on the smaller number? I say 1/2.

If the probability that it landed on the smaller number is 1/2 then the probability that the other side is the larger number is 1/2, and if the probability that it landed on the larger number is 1/2 then the probability that the other side is the smaller number is 1/2.

So the probability that the other side is the smaller number is 1/2 and the probability that the other side is the larger number is 1/2.

And given that the larger number is twice the value of the smaller number, the probability that the other side is half the value is 1/2 and the probability that the other side is twice the value is 1/2.

Which step in this line of reasoning do you disagree with?

And given that the larger number is twice the value of the smaller number, the probability that the other side is half the value is 1/2 and the probability that the other side is twice the value is 1/2.

Which step in this line of reasoning do you disagree with? — Michael

There is nothing there that I disagree with. But I don't think the paradox arises if the values of the two envelopes are stipulated in advance ($10 and $20, say). The paradox arises when we stipulate that the probability of the second envelope containing the largest amount is 1/2 and is not conditionally dependent on the value of the initially chosen envelope. In the example that you give, the probability of the second envelope containing $20 conditionally on the first envelope containing $10 is 1 (and vice versa).

There is nothing there that I disagree with. — Pierre-Normand

Then the paradox arises. The probability that the other side is half the value is 1/2 and the probability that the other side is twice the value is 1/2. Do I take the value of the coin as it landed (which I still don't know), or do I take the value of the other side?

The paradox arises when we stipulate that the probability of the second envelope containing the largest amount is 1/2 and is not conditionally dependent on the value of the initially chosen envelope. — Pierre-Normand

We don't know the value of the initially chosen envelope because we don't look. All we know is that one envelope contains twice as much as the other. The probability that I will pick the smaller amount is 1/2, and so the probability that I did pick the smaller amount is 1/2, and so the probability that the other envelope contains the larger amount is 1/2, exactly as is the case with the coin toss.

In the example that you give, the probability of the second envelope containing $20 conditionally on the first envelope containing $10 is 1 (and vice versa). — Pierre-Normand

Yes, and the probability of the other side of the coin being tails conditionally on my side of the coin being heads is 1 (and vice versa). But this is irrelevant if we don't look at the result. The probability that the coin will land heads is 1/2, and so the probability that it did land heads is 1/2, and so the probability that the other side is tails is 1/2.

Now assume that tails is worth twice as much as heads. The probability that the other side is worth twice as much as my side is 1/2.

Yes, we're in full agreement. By "non-informative" I was referring to the distributional conditions of both uniformity of probability mass and of infinite support . But you're right in pointing out that a "non informative prior" is often used by Bayesians to refer only to uniformity of probability mass, in which the range of the support is considered to be a separate independent hyper-parameter.

Taken from Wikipedia Two envelopes problem
1) Imagine you are given two identical envelopes, each containing money. One contains twice as much as the other. You may pick one envelope and keep the money it contains. Having chosen an envelope at will, but before inspecting it, you are given the chance to switch envelopes. Should you switch?
2) Denote by A the amount in the player's selected envelope.
3) The probability that A is the smaller amount is 1/2, and that it is the larger amount is also 1/2.
4) The other envelope may contain either 2A or A/2.
5) If A is the smaller amount, then the other envelope contains 2A.
6) If A is the larger amount, then the other envelope contains A/2.
7) So the expected value of the money in the other envelope is: 1/2(2A) + 1/2(A/2) = 5/4A


Item 4) "The other envelope may contain either 2A or A/2" is a problem

1) Two envelopes containing £10 and £20
2) Selected envelope contains A, either £10 or £20
3) The probability that A is the smaller amount is 1/2, and that it is the larger amount is also 1/2.

Item 4) is a problem as it contradicts item 1).

If the selected envelope contains £10, then the other envelope must contain £20. It is not true that the other envelope may contain either £20 or £5. It cannot contain £5.

If the selected envelope contains £20, then the other envelope must contain £10. It is not true that the other envelope may contain either £40 or £10. It cannot contain £40.

Items 4), 5) and 6) should be reworded as: "the other envelope may contain either 2A if A is the smaller amount or A/2 if A is the larger amount."

It then follows that there is no value in switching.

There is nothing there that I disagree with. But I don't think the paradox arises if the values of the two envelopes are stipulated in advance ($10 and $20, say). The paradox arises when we stipulate that the probability of the second envelope containing the largest amount is 1/2 and is not conditionally dependent on the value of the initially chosen envelope. In the example that you give, the probability of the second envelope containing $20 conditionally on the first envelope containing $10 is 1 (and vice versa). — Pierre-Normand

Ultimately the disagreement you're having with @Michael there is about the representation of the agent's belief state. If the agent knew that one envelope contained 10, and one envelope contained 20, and they saw 10, the other envelope has 20 in it with probability 1. Notably, this requires the agent knowing the values in both envelopes. In effect, there is a secondary conditioning step in your analysis; you're also conditioning on the set of values {n=10 or n=20} being the values in the envelope, and aggregating a (probably?) uniform prior on probability (the chance of each amount being in each envelope) into the n=10 and n=20 case.

The ability to condition on {n=10 or n=20} isn't something the agent can do while representing their state of evidence, if they don't know that n=10 or n=20 at the start.

Nevertheless, if they observe n=10 in the first envelope, I still think there's a problem with assigning a probability distribution on the values (5, 20) in the other envelope. This is because that stipulates there being three possible values in the envelopes combined; (5, 10, 20); whereas the agent knows only two are possible.

So IMO the issue with "conditioning" on (n=10) when opening the envelope isn't with the conditioning operation itself, it's with the background specification of the sample space's events. Two envelopes (pairs), where one must be half the other. On the possible values (5,10,20), this just (5,10) and (10,20).

A further illustration of why this is weird; if the agent really thinks that the possible values in the envelopes are (5,10,20), and they observe (10) in theirs, then assign equal probability mass to (5,20) based on that observation, that means initially there will have been a nonzero probability assigned to each (5,10,20). A choice between (5,10,20) needs a sampling mechanism associated with it (whence the randomisation), and there isn't one which produces three values anywhere in the problem.

In other words, given that your envelope is 10, making "the other envelope" (5,20) doesn't give you an event which is possible given the (alleged) randomisation that puts the pairs of values in the envelopes in the first place.

Edited: changed 1/3 to nonzero.