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.

One cannot extract a meaningful notion of probabilities — sime

Why not?

I know that one envelope contains twice as much as the other. I pick one at random. What is the probability that I picked the smaller envelope? It seems perfectly correct to say $1\over2$.

The same supposed paradox occurs even if we know the possible values.

Assume that one envelope contains £10 and the other envelope contains £20.
Let y be the value of my chosen envelope and z be the value of the other envelope.

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

The expected value of z is allegedly:

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

This commits a mathematical fallacy, conflating three different values of $y$, resulting in a third value.

$\begin{aligned}E(z) &= {1\over2}(2\cdot10) + {1\over2}({20\over2}) = {5\over4}12\end{aligned}$

But my envelope doesn't contain £12, given that one envelope contains £10 and the other envelope contains £20.

This fallacy is committed even when we only know the value of one envelope and even when we don't know the value of any envelope.

The paradox has nothing to do with probability at all. It's just an improper use of algebraic variables when calculating the expected value.

It isn't clear what your random variable z means in your calculation, since its sample space isn't defined. z is a particular fixed value in the first envelope, no? — fdrake

It's explained in the OP:

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

So to bring together my various posts:

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

The expected value of z is allegedly:

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

However, if we assume that the amount in the other envelope (z) is £10 then:

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

Notice that in A, $y = {z\over2} = 5$ and in B, $y = 2z = 20$. That makes the addition performed in E(z) above a mathematical fallacy. You cannot add $y$ to $y\over4$, where each y has a different value, to get ${5\over4}y$.

E(z) is properly represented as:

$\begin{aligned}y_a &= {z\over2}\\\\y_b &= 2z\\\\E(z) &= {1\over2}2y_a + {1\over2}({y_b\over2})\\\\&={1\over2}({2z\over2}) + {1\over2}({2z\over2})\\\\&=z=2y_a= {y_b\over2}\end{aligned}$

Which of course tells us nothing we didn't already know; that the other envelope contains either twice as much or half as much as my envelope.

And so there's no reason to switch, as expected.

I stand by the claim that the probability that the other envelope contains twice as much as my envelope is equal to the probability that the other envelope contains half as much as my envelope, that probability being 1/2.

But what I believe I have shown is that the formula used to calculate the expected value conflates two different values of y:

$\begin{aligned}y_1 &= {z\over2}\\\\y_2 &= 2z\\\\E(z) &= {1\over2}2y_1 + {1\over2}({y_2\over2})\\\\&={1\over2}({2z\over2}) + {1\over2}({2z\over2})\\\\&=z=2y_1= {y_2\over2}\end{aligned}$

And as such it is not more rational to switch.

It is a mathematical fallacy that leads to the conclusion that $E(z) = {5\over4}y$, much like the fallacy that covertly divides by zero to prove that $1 = 2$.

The trouble is on the RHS, the probability that I am being interviewed given that my coin was heads. — Srap Tasmaner

Which is 1. I know that I will be interviewed if the coin lands heads.

Consider a simpler version of the experiment. If heads then I will be interviewed once. If tails then I will be interviewed once. Bayes' theorem is unproblematically:

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

We don't say that because half of all questions are when tails then P(Questioned | Heads) = 0.5. That would give us this very clearly wrong calculation:

$\begin{aligned}P(Heads | Questioned) &= {{P(Questioned | Heads) * P(Heads)} \over P(Questioned)}\\&={{{1\over2}*{1\over2}}\over1}\\&={1\over4}\end{aligned}$

What you want is the odds that this interview is a heads-type interview. — Srap Tasmaner

Yes, that's the left hand side of the theorem that we're trying to solve: P(Heads|Questioned).

We use the known values on the right hand side to determine it.

And how do you expect to apply Bayes's rule without any base rate information? SB can reason as I have described to determine what those base rates would be were the experiment repeated a number of times, and set her subjective probabilities accordingly. — Srap Tasmaner

We have the base rate information.

The probability of a coin landing heads is 0.5. The probability that I will be questioned if the coin lands heads is 1. The probability that I will be questioned is 1. So:

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

Compare with Bayes' theorem as applied to my variation:

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

It should be clear that the reasoning is sound in the second case, and so too is it sound in the first case.

I don't think your numbers are accurate there.

But in this case we're not asking about an outsider's analysis of frequencies over many experiments, but Sleeping Beauty's when it is known that just a single experiment is being run.

The absence of that thing is informative, it amounts to "it was tails or this is your first interview," and this is true as well for stock SB. Being asked is itself information you can condition on. — Srap Tasmaner

How do you condition on such a thing? What values do you place into Bayes' theorem?

$\begin{aligned}P(Heads | Questioned) &= {{P(Questioned | Heads) * P(Heads)} \over P(Questioned)}\end{aligned}$

I switched back shortly after that post; it's right there in the thread. The argument that convinced me was this: consider a variation, "Informative SB", in which Beauty is told she will be awakened twice either way, but if it was heads she will be told at the second interview that it was heads and that this is her second interview; at none of the others will she be given such information. — Srap Tasmaner

I don't understand how this is different to the original. In the original she's woken up on Wednesday and told the result of the coin flip (whether heads or tails). So there are two interviews if heads, three interviews if tails.

All the average return on the envelope tells us is that if we keep playing over the long term, its going to average a return greater than one. — Philosophim

That's the supposed paradox. Switching doesn't increase our expected return, but the reasoning given suggests that it does. So we need to make sense of this contradiction.

Possible answer: because the material conditions for US citizens are conducive to him getting a platform. If the USA didn't have so many poor, didn't have so many people one healthcare invoice away from being poor, then nobody would take Trump seriously. — Benkei

If economic issues were the concern then they'd be voting for Democrats.

It's clearly social issues (the "culture war") that elicit support for Trump and the Republicans.

Given that the coin flip just determines if a Tuesday interview will happen, Beauty is being asked for her credence that a Tuesday interview will happen.

That she's interviewed twice as often if a Tuesday interview will happen isn't that a Tuesday interview will happen twice as often.

That she's interviewed twice as often if it's tails isn't that it's tails twice as often.

If it helps, consider that the coin toss happens after the Monday interview.

No, I'm in the double halfer camp now. The post right above explains my current thinking.

((This is, I don't know, maybe the third time I've argued with Michael about something and then concluded he was right all along.)) — Srap Tasmaner

So you've switched back to being a thirder?

This is straightforwardly true, but from the perspective of an observer of the experiment. But to answer the problem you must adopt SB's perspective. That makes all the difference. — hypericin

Why does it make a difference?

But SB is asked on every wakening, and is woken twice as much on tails. This must influence the odds — hypericin

Why would it influence the odds?

Let's say that I wanted to bet on a coin toss. I bet £100 that it will be tails. To increase the odds that it's tails, I ask you to put me to sleep, wake me up, put me back to sleep, wake me up, put me back to sleep, wake me up, and so on. Does that make any sense?

Influences the gambling odds, even though the coin toss is fair in both cases — hypericin

It doesn't influence the odds. You just get to bet twice, hence twice the payout.

Consider a coin toss.

P(Heads) = 0.5
P(Tails) = 0.5

I bet £1 on a coin toss. If I bet correctly I will win £1. What is my expected return if I bet on heads?

There's a 0.5 chance of losing £1 and a 0.5 chance of winning £1, so the expected return is:

(0.5 * -1) + (0.5 * 1) = 0

So there's no point in betting.

But if I can win £1.50 for betting correctly then the expected return is:

(0.5 * -1) + (0.5 * 1.5) = 0.25

So it's rational to bet.

In the case of the two envelopes paradox, the claim is that there's a 0.5 chance of doubling my winnings and a 0.5 chance of halving my winnings, and so the expected return is:

(0.5 * 2y) + (0.5 * y/2) = (5/4)y

I'm not sure how you're getting the equation that you are, not only in what you've done, but by what the equation stands for. — Philosophim

Then you should read this, this, and this.

So there's no possible world in which the other envelope may with equal probability contain either two fifty or a tenner, which is the only way to make a profit by switching. — Baden

The issue isn't with probability. Given that there's £10 in my envelope, the probability that the other envelope contains £5 is equal to the probability that the other envelope contains £20; that probability being 1/2.

As an analogy, if I flip a coin and hide it from you, it is correct for you to say that the probability that it landed heads is equal to the probability that it landed tails; that probability being 1/2.

The error made in the paradox is with it's calculation of the expected value of switching. It claims that the expected value is 0.5 * 2y + 0.5 * y/2, which is technically correct, but what it fails to explain is that it's 0.5 * 2y where y = x/2 + 0.5 * y/2 where y = 2x.

So a more accurate formulation is:

$E(x) = {1\over2}({2x\over2}) + {1\over2}({2x\over2}) = {x\over2} + {x\over2} = x$

Here's a Venn diagram to show how the probabilities interlink (where "Monday" should be read as "she will be woken on Monday" and "Tuesday” as "she will be woken on Tuesday"):



The probability that the coin will land heads and she will be woken on Monday is 1/2.
The probability that the coin will land tails and she will be woken on Monday is 1/2.
The probability that the coin will land tails and she will be woken on Tuesday is 1/2.

As the Venn diagram shows, there are two (overlapping) probability spaces, hence why the sum of each outcome's probability is greater than 1.

A variation of my variation.

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

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

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

If we use Bayes' theorem then:

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

The number assignment and the coin toss do not affect each other, and so they are independent events.

But there is something different about my example. If we use Bayes' theorem then:

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

Given that P(Heads|Awake) ≠ P(Heads), waking does in fact provide new relevant evidence. And the above shows that the correct answer is 1/3.

Applying this same formula to the original problem gives P(Heads|Awake) = P(Heads), and so is consistent with Lewis' reasoning and his conclusion that the correct answer is 1/2.

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

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?

David Lewis' answer, summarised here, states that:

Sleeping Beauty receives no new non-self-locating information throughout the experiment because she is told the details of the experiment. Since her credence before the experiment is P(Heads) = 1/2, she ought to continue to have a credence of P(Heads) = 1/2 since she gains no new relevant evidence when she wakes up during the experiment.

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.

However, given that if woken he is 1 iff the coin landed heads, he ought to have a credence of either P(Heads) = 1/3 or P(1) = 1/2.

Do you think that when "they let you do it", it is assault? — NOS4A2

If they don't consent then yes. Letting someone do something and consenting to them doing something are different things. The former only implies that they don't object and/or resist.

For example, you let the government tax you, but you've made it clear in the past that you don't consent to it.

And even if Trump meant it in the sense of "consent", that he says that they do isn't that they do, only that he thinks that they do. If he just assumes consent and so "just starts kissing them ... [without] even wait[ing]" then it's assault.

In that regard, either we'd have to rejected that the luminiferous aether isn't possibly physically necessary, or the law of logic which leads to the inference. I'm inclined to reject the latter, since I intuit that things like physical laws are "physically necessary" (whatever that means). — fdrake

Given my previous post that shows that under S5 we cannot assume that ◇∃x□Px is true for any logically consistent Px, the third alternative is that “possibly necessary” under S5 means something different to what it means under other systems (or natural English).