I think learning the value of your envelope is an uninformative posterior and so gives you no information with which to reassess the prior probability.

I tell you that one side of a fair coin is worth more than the other.

What is the probability that it will land on heads? 1/2. What is the probability that it will land on the more valuable side? 1/2.

After flipping it, but before looking at the result, what is the probability that it landed on heads? 1/2. What is the probability that it landed on the more valuable side? 1/2.

You check the coin and see that it landed on heads. What is the probability that it landed on the more valuable side? 1/2. You knowing that it landed on heads makes no difference as you don't know which of heads and tails is the more valuable.

I tell you that heads is worth £30. What is the probability that it landed on the more valuable side? 1/2. You knowing that it landed on heads and that heads is worth £30 makes no difference as you don't know which of heads and tails is the more valuable or how valuable the more valuable side is.

I think you either have to say that after flipping the coin, but before looking, the probability is undecidable (or, rather, "is either 1 or 0"), or you have to accept that the probability after looking and learning the value is 1/2. I think it's inconsistent to say anything else.

I tell you that heads is worth £30. What is the probability that it landed on the more valuable side? 1/2. You knowing that it landed on heads and that heads is worth £30 makes no difference as you don't know which of heads and tails is the more valuable or how valuable the more valuable side is. — Michael

In general I think it's one of these cases where describing exactly what is random, what isn't, and what is conditioned upon is completely necessary. The debate in thread consists in different ways of setting up the randomness in the problem.

I'm of the opinion that knowing the value of your envelope tells you nothing about what you should do, so long as you can assume the probabilities of each value being in each envelope are equal. Here is what I think describes the randomness in the situation.

1 ) There are two pairs of envelopes. One of them contains (5,10), one of them contains (10,20).
2 ) Someone flips a coin and assigns you one of these pairs of envelopes.

Then depending upon the formulation, either you open it or don't.
Case A, you open it:
A3 ) If you open it and see 10, you don't know if your 10 is in the (5,10) pair or the (10,20) pair.
A4 ) Each of those is equally likely.
A5 ) Assume you're in the (5,10) pair, switching has 0 gain there under equal probability and expected loss.
A6 ) Assume you're in the (10,20) pair, switching has 0 gain under the same assumptions.
A7 ) The expected value of switching is 0.

Edit: A5 is a conditioning step (let your envelope pair be 5,10), A6 is a conditioning step (let your envelope pair by 10,20), A7 is a calculation using the law of total expectation

Case B, you don't open it:
B3 ) If you don't open it, you don't know if your pair is the (5,10) pair or the (10,20) pair.
The reasoning is exactly the same.

Open or don't open, with that framing, there's no gain from switching

If you want to assume the unknown envelope contains (5,20) given that you observe 10 in your envelope, the randomness mechanism looks like:

C1) Someone assigns 10 to your envelope and you open it.
C2 ) Someone assigns 5 or 20 to the other envelope with equal probability
C3 ) If you switch, you either lose 5 or gain 10, with equal probability
C4 ) You'd gain 2.5 by switching.

The major contrast is between the randomness coming in C2 with the randomness coming in the steps 1) and A3 or B3. Those are totally distinct sampling mechanisms (A3 and B3 are the same).

Edit: there actually isn't a conditioning step in this presentation, since the assignment of 10 to the envelope you see is nonrandom. You simply stipulate you open an envelope with 10 in it. If you want to frame this as conditioning, then the probability distribution is just assigning probability 1 to the value 10, for the value in your envelope. (And strictly speaking that isn't a probability distribution which comes with its own subdiscussion... And its support isn't clear, which comes with its own subdiscussion)

Open or don't open, with that framing, there's no gain from switching — fdrake

I've mentioned this before, but from 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.

It doesn't matter if you can frame the situation in such a way that there is no rational reason to switch. What matters is whether or not the switching argument in the paradox is sound. Specifically, is this formula correct?

$E(z) = {1\over2}2y + {1\over2}({y\over2}) = {5\over4}y$

I don't think the problem with this formula is with its probability assignments. If I know that one envelope contains more money than the other, and if I pick one at random, then whether I open it or not the probability that I picked the envelope with the more money is $1\over2$. That seems perfectly correct to me.

I don't think the problem with this formula is with its probability assignments. If I know that one envelope contains more money than the other, and if I pick one at random, then whether I open it or not the probability that I picked the envelope with the more money is — Michael

Probability assignments are done with respect to a space of events. The event spaces in case A and C are different.

Probability assignments are done with respect to a space of events. — fdrake

I'm not sure what you mean. Perhaps you could answer the questions I posed earlier?

If I flip a coin and don't look at the results then what is the probability that it landed heads?

If heads is twice as valuable as tails then what is the probability that it landed on the more valuable side?

I say that the answer to both is $1\over2$.

I'm not sure what you mean. Perhaps you could answer the questions I posed earlier? — Michael

It's a "wu" thing, the presumptions in your questions already define away how to dissolve the problem IMO. So I don't think it's wise of me to answer them. I do agree, given what you just said and your framing, that the calculation of gain is correct. What's wrong is the framing, not the calculation.

What I'm saying: case C makes the assignment of (5,20) to envelopes random. Even though (5,20) could never be an assignment of envelopes. Case C has a perfectly cromulent way of calculating expectations, it's just not the random assignment mechanism to the envelopes.

If I gave you an envelope containing 10 pounds, and I told you I had an envelope containing (5 or 20), you should switch. Just... That's not what situation we're in. We're in a situation where we don't know whether the pairs of envelopes are (5,10) or (10,20).

I do agree, given what you just said and your framing, that the calculation of gain is correct. What's wrong is the framing, not the calculation. — fdrake

The framing is the paradox. I pick one of two envelopes at random. One is twice the value of the other. Given that the probability that I picked the more valuable envelope is $1\over2$, it is rational to switch. But also to then switch back.

The framing is the paradox. I pick one of two envelopes at random. One is twice the value of the other. Given that the probability that I picked the more valuable envelope, it is rational to switch. But also to then switch back. — Michael

And the distinction between Case C and Case A is my solution to it. Ambiguous phrasing suggests we're in Case C, whereas we're actually in case A or B.

I don't see how your cases solve the problem.

Case A, you open it:
A3 ) If you open it and see 10, you don't know if your 10 is in the (5,10) pair or the (10,20) pair.
A4 ) Each of those is equally likely.
A5 ) Assume you're in the (5,10) pair, switching has 0 gain there under equal probability and expected loss.
A6 ) Assume you're in the (10,20) pair, switching has 0 gain under the same assumptions.
A7 ) The expected value of switching is 0. — fdrake

If there is a 50% chance that I am in (5, 10) and a 50% chance that I am in (10, 20), and if I have £10, then there is a 50% chance that the other envelope contains £5 and a 50% chance that the other envelope contains £20. According to $E(z)$ it is rational to switch.

B3 ) If you don't open it, you don't know if your pair is the (5,10) pair or the (10,20) pair.
The reasoning is exactly the same. — fdrake

Yes, the reasoning is the same. There is a 50% chance that the other envelope contains twice as much as what's in my envelope and a 50% chance that the other envelope contains half as much as what's in my envelope. According to $E(z)$ it is rational to switch.

If there is a 50% chance that I am in (5, 10) and a 50% chance that I am in (10, 20), and if I have £10, then there is a 50% chance that the other envelope contains £5 and a 50% chance that the other envelope contains £20. According to — Michael

That changes the sample space. The subcase in A5's sample space is (5,10) - IE an envelope contains 5 or 10, the subject just doesn't know that. The subcase in A6's sample space is (10,20), the subject just doesn't know that.

When you assign (5,20) to the other envelope, that's not an event in the sample space in A5 or A7. It's only an event in the sample space in case C.

If I have a red ball, a blue ball, and a white ball, and if I pick two at random and put one in one package and another in another package, and if you open one and find a red ball, it is perfectly correct for you to say that the probability that the ball in the other package is blue is 50% and that the probability that the ball in the other package is white is 50%, even though the sample space, given that one ball is known to be red, is either (red, blue) or (red, white).

↪fdrake If I have a red ball, a blue ball, and a white ball, and if I pick two at random and put one in one package and another in another package, and if you open one and find a red ball, it is perfectly correct for you to say that the probability that the ball in the other package is blue is 50% and that the probability that the ball in the other package is white is 50%, even though the sample space, given that one ball is known to be red, is either (red, blue) or (red, white). — Michael

That's case C! Drawing a new ball is the same as assigning a value to the other envelope. It's just not assigned at that step in the envelope set up. Hence, the distinction between (A and B) and C. It seems to me you're having difficulty seeing the distinction between A and C?

There is no drawing a new ball. I picked two at random, then put them in packages, and then asked you to pick one.

Exactly. You asked me to pick one, then treated that like drawing a ball from the bag.

Exactly. You asked me to pick one, then treated that like drawing a ball from the bag. — fdrake

I don't understand what you're saying or how this is any different to the envelopes.

I put a coloured ball in one envelope and a coloured ball in another envelope. You pick an envelope, open it, and see the ball to be red. What is the probability that the ball in the other envelope is white, given that it must be either white or blue?

If I have a red ball, a blue ball, and a white ball, and if I pick two at random and put one in one package and another in another package, and if you open one and find a red ball, it is perfectly correct for you to say that the probability that the ball in the other package is blue is 50% and that the probability that the ball in the other package is white is 50%, even though the sample space, given that one ball is known to be red, is either (red, blue) or (red, white). — Michael

To break it down, this comes with the following steps:
1) The possible ball types are red, blue, white.
2) Two distinct types of ball are taken.
3) They're put in two distinct envelopes.
4) I choose an envelope.
5 ) I know that the only balls I could have chosen are red, blue, white
6 ) I observe white.
7 ) I know that the only balls remaining are red, blue
8 ) Each occurs with probability half

5 ) There corresponds to telling the agent, in the envelope case, that the possible values in the other envelope are 5 or 20. This makes your framing case C.

Case A, however, does not have the agent aware that the possible values in the other envelope are 5 or 20. Precisely because they are epistemically indifferent to the cases where the pair is (5,10) or (5,20). The randomness there is epistemic. But it's equivalent to randomly assigning red or blue to the other package at that time by a coin flip. They represent exactly the same set up. But no coin flip is done at that stage; IE, nothing actually happens to the agent in the envelope set up in case A. But they would in case C or your ball example.

Case A, however, does not have the agent aware that the possible values in the other envelope are 5 or 20. — fdrake

They are aware. They are told before the experiment starts that one envelope contains twice as much as the other. They open their envelope to find £10. They know that the possible values in the other envelope are £5 or £20.

They are aware. They are told before the experiment starts that one envelope contains twice as much as the other. They open their envelope to find £10. They know that the possible values in the other envelope are £5 or £20. — Michael

Their awareness corresponds to a different fact. I don't think we're making any progress here. Do you understand the difference between case A and case C?

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

Re-wording my argument:

The puzzle states that one envelope contains twice as much as the other. Let one envelope contain x euros and the other 2x euros.
The player selects one envelope without opening it. This envelope contains either x euros or 2x euros
The puzzle states that the amount in the selected envelope is A

Possibility one
If the selected envelope contains x euros, then A = x
The puzzle goes on to state that the other envelope may contain either 2A or A/2, meaning either 2x euros or x/2 euros.
But the puzzle had previously established that the only amounts in the two envelopes are x euros and 2x euros.
Therefore, the statement "The other envelope may contain either 2A or A/2" leads to a contradiction.

Possibility two
If the selected envelope contains 2x euros, then A = 2x
The puzzle goes on to state that the other envelope may contain either 2A or A/2, meaning either 4x euros or x euros.
But the puzzle had previously established that the only amounts in the two envelopes are x euros and 2x euros.
Therefore, the statement "The other envelope may contain either 2A or A/2" leads to a contradiction.

Conclusion
It seems to me that the statement "The other envelope may contain either 2A or A/2" is where the flaw in the line of reasoning lies, as it leads to a contradiction.

Do you understand the difference between case A and case C? — fdrake

Yes, and I think it is perfectly correct to say, in case A, that the probability that the other envelope contains £20 is 50%.

As I mentioned before, knowing that there is £10 in my envelope is an uninformative posterior. I know that one envelope contains twice as much as the other, I know that the probability that I will pick the more valuable envelope is 50%, and I know that after picking one the probability that I did pick the more valuable envelope is 50%, and so I know that the probability that the envelope I didn't pick is the more valuable envelope is 50%. That then leads to the E(z) calculation.

Opening the envelope and finding £10 or £20 or £60 provides me with no information that will lead me to reassess that prior probability.

Yes, and I think it is perfectly correct to say, in case A, that the probability that the other envelope contains £20 is 50%. — Michael

It isn't though. "The other envelope contains 20" in case A subcase 1 has probability 0. It just isn't an event in that case. The only events in that case are "my envelope contains 5 and the other envelope contains 10" and "my envelope contains 10 and the other envelope contains 5". In case A subcase 2, "the other envelope contains 20" has probability half, the events in that one are "my envelope contains 10 and the other envelope contains 20" and "my envelope contains 20 and the other envelope contains 10".

Nowhere in that set up is there ever a way of considering that the other envelope contains 5 or 20! That's only an available move in case C.

There is a difference between these two claims:

1. I have a 1 in 2 chance of picking the larger of two envelopes.

2. The envelope I have chosen has a 1 in 2 chance of being the larger of the two envelopes.

Suppose the pair is (5, 10). I have a 1 in 2 chance of picking 10. But 10 does not have a 1 in 2 chance of being greater than 5; it _is_ greater than 5.

I flip a coin but don't look at the result. The probability that it landed heads is $1\over2$.

You both seem to be mixing up the participant's subjective assessment and some God's eye objective assessment.

You both seem to be mixing up the participant's subjective assessment and some God's eye view objective assessment. — Michael

Not true, a different subjective assessment. Your uncertainty concerns the values (5,20), which is case C. Case A's uncertainty concerns the pairs (5,10) and (10,20). When you switch, you don't know if you're in case A subcase 1 or case A subcase 2. So you average the gain of switching over each of those. Which is 0.

When you switch, you don't know if you're in case A subcase 1 or case A subcase 2. So you average the gain of switching over each of those. Which is 0. — fdrake

I don't know how you get 0.

If you're in subcase 1 then you lose £5 by switching. If you're in subcase 2 then you gain £10 by switching. There is a 50% chance that you're in subcase 1 and a 50% chance that you're in subcase 2. So according to the E(z) calculation, it is rational to switch.

If you're in subcase 1 then you lose £5 by switching. If you're in subcase 2 then you gain £10 by switching. There is a 50% chance that you're in subcase 1 and a 50% chance that you're in subcase 2. So according to the E(z) calculation, it is rational to switch. — Michael

You don't know whether you're in Case A subcase 1 or case A subcase 2. Each of those has probability half. If you're in case A subcase 1, if you switch you gain 5 or lose 5. If you're in case A subcase 2, if you switch you gain 10 or lose 10. Each of those has 0 gain.

The relevant question is; let's say you condition on having 10, does that give you any information on what pair you're in (IE, whether you're in subcase A 1 or subcase A 2), and the answer is no. If you knew what case you're in, knowing that you have 10 makes the choice deterministic. The conflation between case C and case A turns precisely on this point.

If you know you have 10, it becomes very tempting to say that you know the other envelope has 5 or 20. Whereas you don't know that, you just know that either (your envelope is 10 and the other is 20) or (your envelope is 10 and the other envelope is 5), and that tells you nothing about whether you're in case A subcase 1 or case A subcase 2.

When you're stipulating that you "have 10", you're using that to resolve uncertainty within subcases. Whereas the point you're shown it, you could only gain information about which subcase you're in - and it provides no information there.

What's the probability of your envelope being 10? Why can you condition on it?

You don't know whether you're in Case A subcase 1 or case A subcase 2. Each of those has probability half. If you're in case A subcase 1, if you switch you gain 5 or lose 5. If you're in case A subcase 2, if you switch you gain 10 or lose 10. Each of those has 0. — fdrake

If you're in subcase 1 and you have £10 and you switch then you lose £5, if you're in subcase 2 and you have £10 and you switch then you gain £10. Each of these is equally likely, and nothing else is possible, hence E(z) suggesting you should switch.

If you're in subcase 1 and you have £10 and you switch then you lose 5, if you're in subcase 2 and you have £10 then you gain £10. Each of these is equally likely, hence E(z) suggesting you should switch. — Michael

Eh, we're just asserting the same thing over and over again at this point. I stop.

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. [...] — fdrake

Your assertion that 'only two values are possible' for the contents of the envelopes in the two-envelope paradox deserves further exploration. If we consider that the potential amounts are $(5, 10, 20), we might postulate some prior probabilities as follows:

P_1 = P(a) = P(($5, $10)) = 3/4,
P_2 = P(b) = P(($10, $20)) = 1/4,

which translates into priors for the unopened envelope:

P_3 = P(A) = P(($5)) = 3/8,
P_4 = P(B) = P(($10)) = 1/2,
P_5 = P(C) = P(($20)) = 1/8.

This distribution could reflect an informed guess about Joe, the envelope-filler, who is more likely to risk a smaller rather than a larger amount.

Suppose Ann chooses an envelope. If it contains either $5 or $20, she can unambiguously update her priors to 1 and 0, or 0 and 1, respectively. The decision to switch or not becomes trivial. If, however, her envelope contains $10, she must update her beliefs about the contents of the other envelope using Bayes' theorem:

P_updated(A) = P_updated((unseen=$5)) = P((unseen=$5) | (seen=$10)) = (1 * 3/8) / (1/2) = 3/4.

Given this posterior, if Ann sees $10 in her envelope, the expected value (EV) for switching is negative:

(3/4)$5 + (1/4)$20 - $10 = -$1.25.

Therefore, she should retain her $10, as her prior for Joe having included $20 is sufficiently low. Regardless, before she inspects the second envelope, both outcomes ($5 or $20) remain possible.

If we return to the original problem scenario (addressing @Michael's concern), where the first envelope remains sealed, the initial value probabilities become (3/8, 1/2, 1/8) for $5, $10, and $20 respectively. This gives an initial expected value of:

3/8 * $5 + 1/2 * $10 + 1/8 * $20 = $9.375.

The expected value if Ann switches relies on the weighted sum of the expected values for the unopened envelope, conditional on the potential contents of the chosen envelope. As choices of $5 and $20 guarantee $10 in the other envelope, while a choice of $10 leads to an expected value of $8.75 for the other envelope, this calculates to:

3/8 * $10 + 1/2 * $8.75 + 1/8 * $10 = $9.375. (Lo and behold!)

@Michael

Simulation in R to demonstrate the different sampling mechanisms.


Onto something that sounds like Case_A, and something you can to do it to make it superficially resemble Case_C.


You notice that when you condition on "having your envelope be 10" in that set up, you're subsetting to cases where the gain is -5 or 10, and those occur with equal probability since the cases (5,10) and (10,20) occur with equal probability. If you were to condition on your envelope being 10 in the loop (case index=1), you end up with exactly the same gain numerically but it represents a subjectively different belief state for an agent. Why? The subsetting done at the end lets you look at the whole ensemble of cases where the first envelope was 10, where the assignment of 10 was random. Fixing the envelope as 10 within case makes the assignment of 10 nonrandom.


The thing we're butting heads on, in my view, is the Case_A inner loop line 1, which is where the randomness comes in through the allocation of pairs. If at any point, in the loop, the agent *knows* what case they're in, their gain is deterministic. When you grant that knowledge hypothetically, you either enter subcase A 1 (the first gain line) or subcase A 2 (the second gain line). Once you've done those hypothetical calculations, you reintroduce the randomness of allocating envelopes in the next line to choose the received gain.