While I think the problem has been adequately solved by the responses in this thread, it's fascinating to see how persistent these sorts of 'transcendental illusions' are. Even reasoning in the simplest cases, we often simply cannot see past some heuristic we usually use to solve problems, convinced they apply to some case in which they do not, and there often appears to be no easy way to demonstrate to people that they are in the grip of such an illusion.

A warning to all those who want to use such logical heuristics to do metaphysics about 'big' questions! If you can't get a toy problem about two envelopes right...

You keep thinking about it as what would happen if you play repeatedly. I'm just talking about playing a single game. — Michael

This is so bizarre to me. The amount of games doesn't matter, everyone keeps telling you this, but you're still convinced it is, and I can't figure out why.

@BlueBanana Every amount has a 50% chance of winning for a 2:1 payout. It’s just that, like every after-the-fact bet, the result is prior to the bet.

It doesn’t matter if my £20 bet on tails even when it’s actually heads is the only bet or if there’s also a £10 bet on heads. It’s 50% for a 2:1 payout.

It seems that the real issue is that you seem to disagree with me on how to interpret after-the-fact betting.

Every amount has a 50% chance of winning for a 2:1 payout. — Michael

That's not how it works in the paradox. You need to end up even. Otherwise you're implying the amount of money combined in the envelopes doesn't equal the amount of money combined in the envelopes.

The envelopes are identical. Learning the amount of money in one of them is not relevant information, as can be seen in the table example, and in what happens when there're two people, each checking one envelope.

Suppose the envelopes on offer contain $10 and $20. You pick one.

If your envelope contains $10, then you will be offered the chance to buy the remaining envelope for $10. To you this looks like an envelope with an average value of $12.50. You swap and get $20.

If your envelope contains $20, then you will be offered the chance to buy the remaining envelope for $20. To you this looks like an envelope with an average value of $25. You swap and get $10.

This is very strange. @Snakes Alive noted that your initial choice is a matter of indifference because whichever envelope you pick, there's a story you can tell yourself to justify swapping. (Half these stories are damned lies.) @BlueBanana noted that if two players take an envelope each, they'll both want to trade.

I see three possibilities here:

(1) There is a flaw somewhere in the way @Michael and @andrewk recommend we should estimate the value of the remaining envelope. If there is, it's not exactly obvious.

(2) @Michael and @andrewk are right about the expected value of the other envelope, and this is a genuine paradox. (What they're wrong about is thinking swapping works, since it's really obvious that it does not and cannot.)

(3) @Michael and @andrewk have part of the right analysis, but something more is needed.

(2) has to be avoided at all cost, I'd say.

(2) Michael and @andrewk are right about the expected value of the other envelope, and this is a genuine paradox. (What they're wrong about is thinking swapping works, since it's really obvious that it does not and cannot.) — Srap Tasmaner

The issue is that there’s an expected gain for each game individually but that because each game that actually gains is offset by an equivalent game that actually loses, the actual (and expected) gain for all games collectively is 0.

(1) There is a flaw somewhere in the way Michael and @andrewk recommend we should estimate the value of the remaining envelope. If there is, it's not exactly obvious. — Srap Tasmaner

I have tried to clarify why this is, but apparently it hasn't worked. But I don't have another method of expositing it. I expect that some way of making yet another distinction, perhaps between epistemic and metaphysical possibility, may make it work, so the probability table can be drawn up again, and the fallacy made obvious to everyone.

Nope. If you were right, the expected return for any number of plays, 1 or a million, should always be 1.25 the amount drawn. Number of plays is irrelevant.

I'm just doing this:

1. We pick an envelope at random
2. There's a 50% chance that my envelope is the X envelope and a 50% chance that my envelope is the 2X envelope.
3. I open my envelope and see £10
4. From 2 and 3, there's a 50% chance that my £10 envelope is the X envelope and a 50% chance that my £10 envelope is the 2X envelope.
5. From 4, there's a 50% chance that the other envelope contains £20 and a 50% chance that the other envelope contains £5.

We seem to agree on 1-3, but you disagree with 4 and/or 5? — Michael

I disagree that the 50% chance can be assumed. If the sample space is finite, then there is a non-zero chance that the $10 amount is the maximum amount. If it is, the actual gain from swapping will always be negative. Accounting for both the minimum and maximum amounts exactly cancels out the 1.25X expected gain for other amounts.

For an infinite sample space, assuming 50% chance and conditioning on any specific amount will always conclude a 1.25X expected gain (the grass always appears greener on the other side). Yet unconditionally, there is no reason to prefer either envelope. That again implies that the 50% chance assumption is mistaken. A higher probability weighting (say 2/3 or a graded weighting) for lower amounts (or perhaps for amounts closer to 1) may bring agreement, but I haven't checked the math.

I have numbered the steps from 0 to 4. Which step(s) do you not believe? — andrewk

See above per the 50% chance assumption.

1. There's a 50% chance that I have envelope X and a 50% chance that I have envelope 2X.
2. I have £10.
3. Therefore, there's a 50% chance that the other envelope contains £5 and a 50% chance that the other envelope contains £20.

You've accepted this as sound.

The expected return is worked out using the equation:

$\begin{align} E[R] = \sum_{i=1}^n R_iP_i \end{align}$

So, given the conclusion above:

$\begin{align} E[R] = \frac{1}{2} \cdot 5 + \frac{1}{2} \cdot 20 = 12.5 \end{align}$

Now, the counter to this, as explained here, is:

Expected value in B = 1/2 ( (Expected value in B, given A is larger than B) + (Expected value in B, given A is smaller than B) )

If we then take the sum in one envelope to be x and the sum in the other to be 2x the expected value calculations becomes:

Expected value in B = 1/2 (x + 2x)

However, consider that we know that there's £10 in our envelope (A):

Expected value in B = 1/2 ( (Expected value in B, given 10 is larger than B) + (Expected value in B, given 10 is smaller than B) )

So plug in the expected values, given that A is 10:

Expected value in B = 1/2 (5 + 20)

This can't be written as B = 1/2 (x + 2x) without conflating different values of x.

We've already been over all this, I'm afraid. We would need a new way of looking at things to move forward, and this doesn't provide it.

(3) Michael and @andrewk have part of the right analysis, but something more is needed. — Srap Tasmaner

I disagree that the 50% chance can be assumed. If the sample space is finite, then there is a non-zero chance that the $10 amount is the maximum amount. — Andrew M

I agree with both of these. The point about the maximum possible amount has been made before, but has been lost in the length of the thread. I mentioned it somewhere early on, and in the interests of simplicity, declined to probe further because (1) I thought digging into it would have required some fancy notation like integral signs, and I couldn't remember how to invoke latex on this site (Edit: I see from the above that Michael has mastered it) and (2) I was pretty confident it would not make a material difference.

But given the continuing uncertainty I'll have a go. If nothing else, it should be able to bring out the intrinsically Bayesian nature of the calculation, which I think is camouflaged in the simplified versions discussed so far. Explicitly recognising a maximum possible value of X is part of recognising a prior distribution for X, that is then used in a Bayesian calculation.

I'll revert if I can either get latex to work, or manage to write the full-Bayesian calc without it.

I thought digging into it would have required some fancy notation like integral signs, and I couldn't remember how to invoke latex on this site (Edit: I see from the above that Michael has mastered it) — andrewk

Ironically, I coped your code from here and amended it as required.

The point about the maximum possible amount has been made before, but has been lost in the length of the thread. I mentioned it somewhere early on, and in the interests of simplicity, declined to probe further because (1) I thought digging into it would have required some fancy notation like integral signs, and I couldn't remember how to invoke latex on this site and (2) I was pretty confident it would not make a material difference. — andrewk

And I asked you last night about the minimum possible value of X. Your response then was that what matters is not what might or might not be put in the envelope, but whether you the player have reason to exclude a given value.

This can't be written as B = 1/2 (x + 2x) without conflating different values of x. — Michael

You just write it as $\small B = \cfrac{\cfrac{y}{2} + 2y}{2}$.

$\small y$ is whatever's in the envelope in your hand, whether you've looked at it or not. Looking makes no difference.

Feigning ignorance does not resolve the conflict. As I have said many times over, the sample space of the other envelope is [X,2X]. That is not that hard to grasp, and since I don't believe you are an idiot, I am forced to conclude you feigning ignorance as a way to avoid this conflict.


Let's review some definitions here:

If A and B are sets, then A is called a subset of B if, and only if, every element of A is also an element of B.

A sample space is the set of all possible outcomes of a random process or experiment. An event is a subset of a sample space.

Discrete Mathematics, An Introduction to Mathematical Reasoning, By Susanna S. Epp

Do you understand that these are not my definitions? They are established definitions that may be used in formal proofs.

Let's call the two envelopes A and B. Now envelope A could have X or A could have 2X and likewise B could have X or B could have 2X. Those are all the possible outcomes so by the definition of a sample space our sample space is [A,B] where A is the set [X,2X] and B is the set [X,2X], which means our sample space could also be written as [[X,2X],[X,2X]].

Did you follow that? They are sets in a set. In this case the elements of our sample space are the set [X, 2X]. So that means since your proposed solution does not follow the format of [X,2X] then, by the definition of a subset, it cannot be a subset of either A or B.

You just write it as $\small B = \cfrac{\cfrac{y}{2} + 2y}{2}$.

$\small y$ is whatever's in the envelope in your hand. — Srap Tasmaner

Yeah, which is $\small B = \cfrac{\cfrac{10}{2} + 2*10}{2} = 12.5$, hence the .25 gain.

As I have said many times over, the sample space of the other envelope is [X,2X]. — Jeremiah

And there's £10 in my envelope. So what are the possible values of X where the sample space of the other envelope is [X, 2X]? It can't be 10, because 10 can't be in the sample space of the other envelope, and it can't be 5 because 10 can't be in the sample space of the other envelope.

Given that there's £10 in my envelope, then the sample space of the other envelope must be [5, 20], which means that it isn't [X, 2X]. It must be [Y/2, 2Y].

Your sample space becomes:

[[10=X,10=2X],[10=X,10=2X]].

Edit* This post contains an error see the next page for clarification.

[Y, Y/2]. — Michael

That's $\small \{ \cfrac{y}{2},2y\}$.

Yes, sorry, typo.

Your sample space becomes:

[[10=X,10=2X],[10=X,10=2X]]. — Jeremiah

Remember what you said earlier:

Therefore your sample space could never be [5,20], as it does not follow the form of [X, 2X]. Your sample space could be [5,10] or [10,20] but there is no way you can mathematically justify a sample space of [5,20] as it is not consistent with [X,2X]. — Jeremiah

Now look how you've contradicted yourself. Because a), in using [10=X,10=2X] you have different values of X, and b) in using [10=X,10=2X] you entail that the other envelope must be [5, 20], given that my [10=X,10=2X] envelope is 10.

Ya, jumped the gun there and made a mistake myself (unlike you I am able to admit that). However, if you going to use my argument to justify your position, my argument, that you referenced there, still completely counters your statement of [5,20] and proves you 100% wrong. By using my argument as proof you have just admitted that you are wrong.

The sample space still would be [[X,2X],[X,2X]].

Which would be consistent with what I said here:

Intuitively this seems like new information which would call for you to update your prior; however, it still does not tell you if are you in case R or case S. You don't know if that is 10=X or 10=2X. So in truth you can't really update your prior, as your prior was based on the uncertainty of being in case R or case S. — Jeremiah

Ya, jumped the gun there and made a mistake myself (unlike you I am able to admit that). However, if you going to use my argument to justify your position, my argument, that you referenced there, still completely counters your statement of [5,20] and proves you 100% wrong. By using my argument as proof you have just admitted that you are wrong. — Jeremiah

I'm saying you contradicted yourself, not that your latest (or first) argument is right.

I already admitted to the mistake, then corrected my position. You just get into rapid fire mode and you miss half of what is being said.

This right here does not go away simply because I miss defined a sample space in a different post. This is still true and proves you wrong. It is still something you have yet to actually address.

Let's review some definitions here:

If A and B are sets, then A is called a subset of B if, and only if, every element of A is also an element of B.

A sample space is the set of all possible outcomes of a random process or experiment. An event is a subset of a sample space.

Discrete Mathematics, An Introduction to Mathematical Reasoning, By Susanna S. Epp

Do you understand that these are not my definitions? They are established definitions that may be used in formal proofs.

Let's call the two envelopes A and B. Now envelope A could have X or A could have 2X and likewise B could have X or B could have 2X. Those are all the possible outcomes so by the definition of a sample space our sample space is [A,B] where A is the set [X,2X] and B is the set [X,2X], which means our sample space could also be written as [[X,2X],[X,2X]].

Did you follow that? They are sets in a set. In this case the elements of our sample space are the set [X, 2X]. So that means since your proposed solution does not follow the format of [X,2X] then, by the definition of a subset, it cannot be a subset of either A or B. — Jeremiah

And I asked you last night about the minimum possible value of X. Your response then was that what matters is not what might or might not be put in the envelope, but whether you the player have reason to exclude a given value. — Srap Tasmaner

Are you sure that was me? It doesn't sound familiar, but maybe I've just forgotten. Can you link the exchange?

FWIW I'm imagining X can be any real number above zero, so there is no minimum value. In order to make that possible, the envelope has to contain an IOU to the value of some real number, not notes and coins, otherwise the min poss value is one penny.

My attempted full Bayesian calc is surprising so far. Assuming a prior distribution for X of uniform on (0,L] for some large L, I'm getting an expected gain from switching, given an observation of y in the envelope, of (5/8)y, which is more than the y/4 from previous calculations. Needless to say, I'm reviewing it fairly critically.

Are you sure that was me? It doesn't sound familiar, but maybe I've just forgotten. Can you link the exchange? — andrewk

What in the rules of the game or in the knowledge you have acquired (i.e., that Y = 10) assures you that the games mistress includes both 5 and 10 in the sample space for X? Maybe a tenner is the smallest bill she has. — Srap Tasmaner

It's not about the games mistress's sample space. She doesn't have one. She has two amounts that she has always intended to put into the envelopes. It's about my sample space, which consists of the events that I cannot rule out based on my knowledge to date.

That knowledge - based on seeing 10 in my envelope - allows me to rule out every other event (an event being a particular sum being in the other envelope) except for 5 and 20. So that's my sample space - just two events. — andrewk

Let's call the two envelopes A and B. Now envelope A could have X or A could have 2X and likewise B could have X or B could have 2X. Those are all the possible outcomes so by the definition of a sample space our sample space is [A,B] where A is the set [X,2X] and B is the set [X,2X], which means our sample space could also be written as [[X,2X],[X,2X]]. — Jeremiah

Then we have:

1. My envelope could be envelope X
2. My envelope could be envelope 2X
3. My envelope contains £10.
4. From 1 and 3, the other envelope could be envelope 2X and contain £20
5. From 2 and 3, the other envelope could be envelope X and contain £5

This is valid.

And just as we can model my envelope using the sample space [X, 2X], despite the fact that in doing so we're positing two different possible values of X (either 10 or 5), we can also model the other envelope using the sample space [X, 2X], despite the fact that in doing so we're positing two different possible values of X (either 5 or 10).

we can also model the other envelope using the sample space [X, 2X], despite the fact that in doing so we're positing two different possible values of X (either 5 or 10). — Michael

Ah, but that's exactly what we can't do. This was the original fallacy (back in my first post).