So what we’re left with is the claim that going through the motion of picking one envelope and then — with absolutely nothing else changing— switching and picking up the other envelope is somehow rational. That’s simply wrong. — Mikie

Yes, the conclusion is certainly wrong. The puzzle is in figuring out which step in the reasoning that leads to it is false.

There's not even agreement among analysts about whether this is a probability problem. (I don't think it is.) — Srap Tasmaner

Neither do I. I believe I showed what the problem is in the OP.

You’re not given any more information, so I really don’t follow the rest of the calculations. — Mikie

The argument is:

Let x be the amount in one envelope and 2x be the amount in the other envelope.

Let y be the amount in the chosen envelope and z be the amount in the unchosen envelope.

$P(y = x) = P(z = 2y) = {1\over2}\\P(y = 2x) = P(z = {y\over2}) = {1\over2}$

The expected value of z is:

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

Given that the expected value of the unchosen envelope is greater than the value of the chosen envelope, it is rational to switch.

Involuntary celibate is a self appointed term to describe men that are celibate against their will because they deem themselves not attractive enough to the opposite sex. — Benj96

The term originated from "Alana's Involuntary Celibacy Project", a website created by a woman to discuss her sexual inactivity.

Not really, as we can only consider this from the perspective of the participant, who only knows that one envelope contains twice as much as the other and that he picked one at random. His assessment of $P(y = x)$ and $P(y = x|y =10)$ can only use that information.

Is it correct that, given what he knows, $P(y = x) = {1\over2}$?
Is it correct that, given what he knows, $P(y = x|y =10) = P(y = x)$?

If so then, given what he knows, $P(y = x|y = 10) = {1\over2}$.

Perhaps this is clearer if we understand that $P(y = x|y = 10)$ means "a rational person's credence that his envelope contains the smaller amount given that he knows that his envelope contains £10".

I'm trying to understand your use of conditioning, so can we start at the very beginning?

Let $x$ be the amount in one envelope and $2x$ be the amount in the other envelope.

I pick an envelope at random but don't look at the contents.

Let $y$ be the amount in my envelope.

$P(y = x) = {1\over2}\\P(y = 2x) = {1\over2}$

Do you agree with this?

Next we look at the contents of our envelope and find it to contain £10.

We then want an answer to this:

$P(y = x|y = 10) =\space?\\P(y = 2x|y = 10) =\space?$

My understanding is that given that we don't know the value of the other envelope, or how the values were initially chosen, $y = 10$ is an uninformative posterior; it provides us with no new information with which to reassess the prior probability. As such, $P(y = x|y = 10) = P(y = x)$.

This seems to be where we disagree?

If we assume that all results are equally likely, the EV of switching given that the chosen envelope was seen to contain n is (2n + n/2)/2 - n = 1.5n. Hence whatever value n might be seen in the initially chosen envelope, it is irrational not to switch (assuming only our goal is to maximize EV). This gives rise to the paradox since if, after the initial dealing, the other envelope had been chosen and its content seen, switching would still be +EV. — Pierre-Normand

I think I showed in the OP why this isn't the case. The EV calculation commits a fallacy, using the same variable to represent more than one value. That's the source of the paradox, not anything to do with probability.

I don’t think your simulation is relevant to our disagreement given that I don’t believe that it is rational to switch.

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.

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.

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.

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.

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.

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?

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

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).

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.

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.

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 $1\over2$.

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 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.

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.

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?

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?

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.

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.

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.

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 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.

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?

Why was the scientific american wasting time on this? — sime

Self-locating belief and the Sleeping Beauty problem, Elga 2000

Sleeping Beauty: reply to Elga, Lewis 2001

An interesting variation taken from here:

Four volunteers will be assigned a random number but each will undergo an experiment that is functionally equivalent to the popular version of the problem. The same sleep and amnesia drugs will be used, and each will be awoken at least once, but maybe twice, based on the same fair coin toss. Only their schedules and the question they are asked will differ, but end up being equivalent to the popular problem. On Monday and Tuesday:

#1 Will be awoken unless it is Tuesday, after Heads.

#2 Will be awoken unless it is Tuesday, after Tails.

#3 Will be awoken unless it is Monday, after Heads.

#4 Will be awoken unless it is Monday, after Tails.

Each will be asked for their credence that this is the only time they will be awoken. For #1 and #3, that means credence in Heads. For #2 and #4, it is credence in Tails. For all four, the answer has to be the same as the correct answer to the popular version of the Sleeping Beauty Problem.

On each day, we can bring the three awake volunteers together to discuss their answers. Of these three, exactly one will not be, or was not, awakened on the other day of the experiment. But none of the three can have more, or less, credence that she is that one instead of one of the others.

So with three awake volunteers, one of whom will be awakened only once, the answer is 1/3.

To be clearer with what each volunteer is considering, it is:

P(1 and Heads or 2 and Tails or 3 and Heads or 4 and Tails | Awake)

Prima facie the answer is $1\over2$, however it is a fact that for $1\over3$ of the awake volunteers, "I will only wake once" is true.

The question, then, is whether or not A entails B:

A. "I will only wake once" is true for $1\over3$ of us
B. The probability that "I will only wake once" is true for me is $1\over3$

Thirders say it does, halfers say it doesn't.

I think this might be a Monty Hall problem. Consider a slight variation which I think is functionally equivalent to the above. All 4 are awoken on each day, are put in a room together, and then one of them is put back to sleep according to the rules. Each person left awake is then asked to consider the probability that they will be awake both days. I think it's a mistake to ignore the person who is put back to sleep, who is comparable to the door Monty opens. It was $1\over2$ before seeing someone get put to sleep, so how does seeing someone get put to sleep make it more or less likely that I will be awake both days?

So if this is equivalent to the original problem then the original problem is a Monty Hall problem as well.

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.

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.

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 :

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.

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

Yes, that’s what I show in the OP.

Let's say that there are three beauties; Michael, Jane, and Jill. They are put to sleep and assigned a random number from {1, 2, 3}.

If the coin lands heads then 1 is woken on Monday. If the coin lands tails then 2 is woken on Monday and 3 is woken on Tuesday.

If Michael is woken then what is his credence that the coin landed heads?

…

Michael's credence before the experiment is P(1) = 1/3, so if woken he ought to continue to have a credence of P(1) = 1/3 since he gains no new relevant evidence if he wakes up during the experiment.

And given that if woken the coin landed heads iff he is 1, he ought to have a credence of P(Heads) = 1/3. — Michael

Does it not stand to reason that the probability that the coin landed heads in this example is less than the probability that the coin landed heads in the original example, given that in this example the coin landed heads iff Michael was assigned the number 1?

Do you agree that Michael’s credence that the coin landed heads is 1/3?

$\begin{aligned}P(Heads | Awake) &= {{P(Awake | Heads) * P(Heads)} \over P(Awake)}\\&={{{1\over3}*{1\over2}}\over{1\over2}}\\&={1\over3}\end{aligned}$

If so then it must be that Sleeping Beauty’s credence in the original example is greater than 1/3, i.e 1/2.

The only rational response to the two-envelopes problem as it is traditionally stated without additional assumptions, is to reply

"The probability of getting a greater or lesser prize when opening the other envelope, is between 0 and 1" — sime

I don't agree with this at all.

I have two envelopes. I have put £5 in one envelope and £10 in the other envelope. You use a true random number generator (which uses some quantum mechanical measurement like radioactive decay) to pick one of the envelopes.

What is the probability that you have picked the envelope containing £5? I say $1\over2$.

Or to be more accurate to the specific puzzle, I don't tell you how much is in each envelope; only that one contains twice as much as the other.

What is the probability that you have picked the envelope containing the smaller amount? I say $1\over2$.

So by your logic we can't even talk about the probability of a coin toss landing heads being $1\over2$?

I don't think that's at all reasonable, or even relevant when we consider puzzles like this. For the sake of puzzles like this we assume a coin toss landing heads has a probability of $1\over2$, and we assume that my choice of envelope is truly random.