In the SB problem it is 1 for heads and 2 for tails. — PhilosophyRunner

No it's not. It's 1 for heads and 1 for tails. A probability of 2 makes no sense.

Let $y$ be the value of the chosen envelope and $z$ be the value of the unchosen envelope.

1. $y = x$ or $y = 2x$

2. $E(z) = {5\over4}y = {3\over2}x$

3. $y = {6\over5}x$ (solving ${5\over4}y = {3\over2}x$ for $y$)

3 contradicts 1.

This is clearer if we assume the value of $x$ for the sake of argument:

1. $y = 10$ or $y = 20$

2. $E(z) = {5\over4}y = 15$

4. $y = 12$ (solving ${5\over4}y = 15$ for $y$)

3 contradicts 1.

So the conclusion of the switching argument, that $E(z) = {5\over4}y$, where $y$ is the value of the chosen envelope, is false.

Despite the initial definition, the $E(z)$ formula covertly redefines $y$.

And for the case where we know that $y = 10$:

1. $x = 5$ or $x = 10$

2. $E(z) = {3\over2}x = {5\over4}10$

3. $x = 8.\bar{3}$ (solving ${3\over2}x = {5\over4}10$ for $x$)

3 contradicts 1.

The paradox arises because the same variable ($x$ and/or $y$) is used to represent more than one value. It's a disguised fallacy, unrelated to any probability assignments.

Where have I gone wrong? — Srap Tasmaner

I explained it in more detail in that earlier post above. The variable $y$ is used to represent three different values, two of which are the possible values of the chosen envelope, and the third (the one in ${5\over4}y$) isn’t the value of the chosen envelope, and so the conclusion that the unchosen envelope has a greater expected value than the chosen envelope doesn’t follow.

In that case it is more likely that given an instance I wake up I will see the coin has been flipped heads 100 times in a row. — PhilosophyRunner

I think the reasoning that leads you to this conclusion is clearly wrong, given that it’s an absurd conclusion.

I flip a coin and if it lands heads I wake you up tomorrow, if it lands tails you never wake you up. If you wake up and are asked the probability the coin landed heads, what would you say? — PhilosophyRunner

1.

Then if it’s heads 100 times in a row I wake you up 2¹⁰¹ times, otherwise I wake you up once.

I don’t think it reasonable to then conclude, upon waking, that it is more likely that it landed heads 100 times in a row. The fact that you would be woken up far more times if it did happen just doesn’t make it more likely to have happened.

It is only reasonable to understand that it landing heads 100 times in a row is so unlikely that it almost certainly didn’t.

The only thing that matters is the coin flip(s). The rest is a distraction.

It only depends on whether or not the single coin flip landed tails.

Imagine a different scenario. If I flip a coin 100 times and it lands heads every time I will wake you up a million times, otherwise I will wake you up once.

After waking up, is it more likely that I got heads 100 times in a row?

She is more likely to wake up and see a coin showing tails, as she will wake up more often if the coin lands on tails. — PhilosophyRunner

That’s a non sequitur.

That it happens more often isn’t that it’s more likely.

I don't think that table is how to calculate the probabilities. Consider a slight variation where there are no days, just number of awakenings. If heads then woken once, if tails then woken twice. And consider perhaps that the coin isn't tossed until after the first awakening.

P(Heads) is just the prior probability that the coin will land heads, which is 0.5, P(Awake) is just the prior probability that Sleeping Beauty will be woken up, which is 1, and P(Awake|Heads) is just the prior probability that she will be woken up if the coin lands heads, which is 1.

That gives us:

$\begin{aligned}P(Heads | Awake) &= {{P(Awake| Heads) * P(Heads)} \over P(Awake)}\\&={{1*{1\over2}}\over1}\\&={1\over2}\end{aligned}$

Given a set {10, 20}, the expected value of a number selected from that set is 15. There's nothing wrong with your first set of equations, and it gives the right answer. You don't have to go through all that; you just need the average. — Srap Tasmaner

My concern is in explaining where the switching argument goes wrong.

The switching argument says that because $E(z) = {5\over4}y$, where $y$ is the value of the chosen envelope and $z$ the value of the unchosen envelope, it is rational to switch.

But given the set $\{10, 20\}$, $E(z) = 15 = {5\over4}12$, and $12$ isn't the value of the chosen envelope.

The same reasoning holds for all $\{x, 2x\}$ sets (where $x \gt 0$).

What this shows is that the $y$ in $E(z) = {5\over4}y$ isn't the value of the chosen envelope, and so it does not suggest to switch.

(1) What are the chances that y = x and the chances that y = 2x if y is chosen randomly from a set {x, 2x}? (You may, if you like, write it backwards as x = y and x = y/2.)

(2) What are the chances that a y chosen randomly from a set {x, 2x} was chosen from a set {y, 2y} and the chances it was chosen from a set {y/2, y}? — Srap Tasmaner

I think I explained it best here:

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".

And if the above is true then so too is $P(y = 2x|y = 10) = {1\over2}$.

Learning the value of the chosen envelope is an uninformative posterior, and so the prior probability of picking/having picked the envelope containing the smaller amount is maintained.

There are four people, each assigned a number (unknown to them) between 1 and 4. Two of them are to be put to sleep at random, determined by a single coin toss: if heads then 1 and 2 are put to sleep; if tails then 3 and 4 are put to sleep.

After the coin toss one of them is put to sleep first. For each of the remaining three, what is the probability that they will be put to sleep? Is it 1/3, because there are three of them and only one is to be put to sleep, or is it 1/2, because that was the probability before the first was put to sleep?

I think this basically is the Sleeping Beauty problem.

Whereas you know it does, as the "sampling day" of SB's report depends upon the coin flip. — fdrake

Only if it’s Tuesday. She gets interviewed on Monday regardless.

So what if the coin toss doesn’t happen until after the Monday interview? Does that affect your answer?

Eh, probability modelling also includes assigning random variables. It has a lot to do with what random variables you put in play. — fdrake

But in this case they're using a variable to represent more than one value at the same time. It's a fallacy to add $y$ to ${1\over4}y$ for two different values of $y$.

This is most obvious when we assume two values for the sake of argument, e.g. £10 and £20, where $y$ is the value of the chosen envelope and $z$ the value of the unchosen envelope:

$P(y = 10) = P(z = 2y) = {1\over2}\\P(y = 20) = P(z = {y\over2}) = {1\over2}$

$E(z) = {1\over2}2y + {1\over2}({y\over2}) = {5\over4}y\\E(z) = {1\over2}(2\cdot10) + {1\over2}({20\over2}) = {5\over4}12$

Given that, assuming one envelope contains £10 and the other £20, the chosen envelope doesn't contain £12, it is false to assert that $E(z) = {5\over4}y$, where $y$ is the value of the chosen envelope. $y$ is in fact a different value entirely.

This is true for all $x$ and $2x$ pairs.

That's all the paradox is.

With respect to the case where we open the chosen envelope to find £10:

$P(y = x|y =10) = P(z = 2y|y =10) = {1\over2}\\P(y = 2x|y =10) = P(z = {y\over2}|y =10) = {1\over2}$

$E(z) = {1\over2}2y + {1\over2}({y\over2})\\E(z) = {1\over2}2x + {1\over2}({2x\over2}) = {3\over2}x\\E(z) = {1\over2}(2\cdot10) + {1\over2}({2\cdot5\over2}) = {3\over2}8.\bar{3}$

Given that the chosen envelope contains £10, the smaller envelope doesn't contain £8.333..., and so once again $E(z)$ commits a fallacy in using the same variable to represent more than one value at the same time.

I could imagine using it for teaching probability modelling. Get students to analyse the problem. Then do it IRL with both sampling mechanisms. Should be a cool demonstration of "physical" differences between what's seen as a merely "epistemic" probability assignment! — fdrake

Even if we accept the premise (as I do) that $P(z = 2y) = P(z = {y\over2}) = {1\over2}$, there's still no reason to switch, so the paradox has nothing to do with probability at all.

Yes

The $x$ in that case wasn't referring to the smaller value but to the value of the other envelope. I'll rephrase it to make it clearer:

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.

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

$E(z) = {1\over2}2y[where\space y={1\over2}z] + {1\over2}({y\over2})[where\space y=2z]\\E(z) = {1\over2}({2z\over2}) + {1\over2}({2z\over2}) = {z\over2} + {z\over2} = z$

How do you get 1? — Dawnstorm

Because Sleeping Beauty is certain to be questioned during the experiment.

Compare with my alternative scenario here where the probability of being questioned is 1/2.

How do you interpret P(Questioned)? — Dawnstorm

"My credence that I will be questioned."

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?