Yes, there's no possible world in which the fiver is both the larger and the smaller amount at the same time. 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.

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:

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

It seems to amount to us saying the same thing. We must either be, in your example, in a 10/5 world or a 10/20 world. In neither possible world is switching a rational strategy. (I think we've discussed this one before btw.) But maybe your way of expressing the issue is more accurate technically.

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. — Michael

This seems confused though with you conflating non-overlapping scenarios and adding a poor analogy. If you want to use ordinary language, just say what I said. Your maths is on the ball though.

Maybe another way way to put it is "the conditions under which such a decision can be made negate any possible distribution that can be profitably exploited by switching". This avoids using the word "probability".

You've clearly done your math wrong. 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.

The amount of money in the envelope is irrelevant, so I'm not sure why that's part of your equation. You have two envelopes, and one is an intended outcome. If you pick randomly, its a 50% chance that you get the intended outcome. If you switch, its the same odds. Your x's, y's, and z's are irrelevant to this outcome.

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.

Then you should read this and this. — Michael

Thank you, I missed your link in the original OP as well.

I still don't understand how they apply their math to this situation.
The 5/4 is an expected value, which is an average. But the situation doesn't call for an average because its either or. You either get A, or 2A.

Numbers and their outcomes need to match the representative realities they are equating. Applying an average to a situation in which an average is not in consideration is not a rational application of math. Its like saying people have an average of 1.5 kids and then betting that you'll have half of a kid born. It doesn't make any sense.

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

Thanks Michael.

Ok, I think I see a little clearer what this is trying to do, but it still doesn't quite line up.

So it makes sense to use an average here in our decision whether to play. But it doesn't say anything about us switching between heads and tails repeatedly as our guess before we see the coin, which is the situation with the envelope.

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. But we actually have to play. It doesn't tell us anything about our indecision or which choice we should finally land on.

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.

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. — Michael

Yeah, I don't see that as a paradox, just a misunderstanding of what the math is representing. The average represents the outcomes if you select one of the outcomes. It doesn't apply in any way to whether you should switch your decision before you see the reveal.

This can be easily seen by simply swapping money with a card that say 1, and then 2. Over the course of several selections, the average result will be X. But that's irrelevant to if we want a particular outcome. Lets say we start with wanting "2" to be selected. The calculated average of outcomes is the same. Now lets say that another person wants "1" to be selected. The calculated average of outcomes is the same.

The average of outcomes is irrelevant then to what we want to pick. It tells us no information regarding whether we should switch our choice or not. The addition of money does not change this, it only changes that everyone wants one of the 50% chance outcomes. But wanting a particular outcome has nothing to do with whether we should switch our choice before the reveal. Its always 50/50 no matter if there is money or a simply cards with numbers in them.

Finally, you have to calculate the average using the idea that each is 1/2, or a 50% chance of being selected. Meaning its hard set that its only a 50% chance. It is impossible to go from having a 50% chance, to then telling someone they have a greater chance of winning what they want if they switch their initial choice. Its still 50/50.

We talked about this 5 years ago! Probability calculations are based on models and intuitions about situations. In this case, disagreements are not about calculations, it's about what is an appropriate description of the sampling mechanism of the envelopes.

One way of framing it: You see two envelopes on the table. You open one. It contains X. You know that the remaining envelope contains 2X or X/2. Assuming both are equally likely, half the time you switch and get an extra X, and half the time you switch and lose an X/2, so you stand to gain 0.5X - 0.25X=0.25X by switching.

Another way of framing it: Two envelopes are on the table. You know one is twice the value of the other. You open your envelope and see it contains an amount. That fixes the other envelope to contain the other amount. You open your envelope and see you have the amount X. Do you know whether the other envelope has 2X or X/2 in it? Nope. What situations can this arise in?

Either you were given X and the other envelope contains X/2, xor you were given X and the other envelope contains 2X. Those are separate cases. If you were given X and the other envelope contains X/2, you'd lose X/2. If you were given X/2 and the other envelope contains X, you'd gain X/2. In the second case, if you were given X and the other envelope contains 2X, you'd gain X, if you were given 2X and the other envelope contains X, you'd lose X. If you're in the first case, the expected gain of switching is 0, if you're in the second case, the expected gain of switching is 0, both cases are equally likely, so the expected gain of switching is 0.

The relevant question which determines the accuracy of each of these is whether you model the value of the unobserved card as being "twice or half" for a given observed value of X (2X, 0.5X), or whether you model the value of the unobserved card as being drawn from the two scenarios (X, 0.5X) or (2X, X) and switching a move in two distinct ones.

The first framing construes "the sampling mechanism", which is where randomness arises in this scenario, as being the conditioning step - observing the value in the first envelope. The second framing construes "the sampling mechanism" as the allocation of pairs of values to the envelopes. In the first case, observing the value in the first envelope randomly assigns 0.5X or 2X to the unseen envelope, in the second case, observing the value in the first envelope tells you nothing about whether the unseen envelope and your envelope have possible values (X, X/2) or (X, 2X).

After 5 years I remain of the opinion that the second framing is appropriate, since you really do gain no information about what's in the second envelope given what's in the first.

But, nevertheless, if you relax equal probability assumptions of all cases or use a different loss function than expected loss, it can still be more rational to switch (or not switch) depending upon your problem set up. Like if you really needed $200, but didn't care about $100 at all, you'd switch if you got $100.

Either you were given X and the other envelope contains X/2, xor you were given X and the other envelope contains 2X. Those are separate cases. If you were given X and the other envelope contains X/2, you'd lose X/2. If you were given X/2 and the other envelope contains X, you'd gain X/2. In the second case, if you were given X and the other envelope contains 2X, you'd gain X, if you were given 2X and the other envelope contains X, you'd lose X. If you're in the first case, the expected gain of switching is 0, if you're in the second case, the expected gain of switching is 0, both cases are equally likely, so the expected gain of switching is 0. — fdrake

Said pretty much the same thing five years ago. I think Michael has at least partially come round.

Whether you benefit by switching or not is dependent on what envelope you just chose, which must be Envelope X or Envelope 2X where X is a given sum of money. So, switching can only ever take you from Envelope X to Envelope 2X (+X) or from Envelope 2X to X (-X). — Baden

So, you have 10. But only one scenario applies to your choice, either the one where you lose by switching in which case X is five and 2X is 10, or you gain by switching in which X is 10 and 2X is 20. You can't be in both scenarios at the same time and X has a different value in both — Baden

We know the chosen envelope has £10 in your scenario and we also know that one of the envelopes contained an amount X and one of the envelopes contained an amount 2X, and therefore we know we can only move from X to 2X or from 2X to X by switching. Therefore we know we have a 50% chance of gaining X by switching and a 50% chance of losing X by switching and that therefore there is no point in switching.

You're using X to refer to both the X in the scenario where switching gets you less and in the scenario where switching gets you more. But given any given amount you see in front of you after already choosing an envelope, those are not the same X. — Baden

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 $1 = 2$.

I think you're technically right about the probability but so much the worse for the way we are supposed to talk about probability. The rest we're all solidly on the same page now I reckon.

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. — Michael

I think this is true regardless of framing.

P(the other envelope = my envelope/2)

You can get that by counting the ways this happens. In the first case (X,X/2), the probability is 0.5. In the second case (2X, X), the probability is 0.5. The probability that we're in the first case is 0.5, which means the probability that we're in the first case and the other envelope contains half mine is 0.25 total. You get another 0.25 from the same reasoning in the other case. Law of total probability gives you 0.5 total for that probability.

No values are conflated in this case, as there's a hierarchy of random variables.

The first random variable is a coinflip, it tells you whether your envelope values are (X, X/2) or (2X, X). Within each case, you have a random variable that has probability half for each event. The only "conflation" which arises is when forgetting the distinction between the realisation of a random variable (it attaining a particular fixed value) and the event of a random variable taking a particular value (which may or may not have happened).

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? So it's nonrandom, as soon as you open it. So it doesn't have a probability to BE anything else.

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.

An optimal decision doesn't exist on the basis of the information provided, because the premises fail to specify a well-posed problem :

Let P ( r | x , e) denote the probability of obtaining a value r when opening an envelope labelled e, where x represents the smallest amount of money in the two envelopes. Both e and x are assumed to be hidden variables in the sense that they aren't deducible from a drawn value of r.

The premises of the problem allow the following physical characterisation of P:

P ( r | x , e = 0) := Ind (r ; x)
P ( r | x, e = 1) := Ind (r ; 2x)

Here Ind(r; y) is the indicator function that outputs 1 when r equals y and is otherwise 0. (i.e. we get whatever is in the chosen envelope).

To "objectify" the problem, suppose that instead of choosing an envelope, an envelope is automatically drawn from a prior probability distribution p(e) and then opened. Then we could use Bayes theorem to compute a physically interpretable probability that a given value of r is "caused" by a given envelope:

P (e | r , x) =

P( r | e, x) P( e)
---------------------
P (r | x)

where P(r | x) is obtained by summing over e in the numerator, i.e

P(r|x) = P(r | e= 0, x) P(e = 0) + P(r | e = 1 , x) P(e=1)

But the problem doesn't specify the prior probability P(e) , which implies that P (e | r , x) doesn't have a unique solution, which in turn implies that the decision problem isn't well-posed.

Instead, one might follow the OP and a number of authors, and make the arbitrary selection
P (e =1) = P(e =2) = 1/2. In which case Bayes rule gives

P (r | x) = 0.5 Ind (x ; r +2r)

whose expected value is E [ r | x] = 1.5x

However, the problem is still ill-posed due to the fact that P(r | x) has zero mass at it's expected value of 1.5x :

(r = x) => (r < 1.5x )
(r = 2x) => (r > 1.5x )

Unless we make additional assumptions about P(x) , such that we can gain knowledge of x from knowledge of r, we cannot know which of these cases is the most likely, and so cannot deduce anything about the envelope e from the assumption of a uniform prior for P(e).

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.

Our conclusions might agree. I am saying that only the statements 1,3 4 and 5 are valid in the "switching argument". For any prior probability distribution over envelope choices or envelope contents is consistent with the premise

"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. "

And a chosen prior probability distribution mustn't refer to a person's subjective beliefs about the envelopes, but to a logically consistent set of causal hypotheses concerning the generation of the envelopes and their contents, by which a contradiction isn't derivable.

All that the two-envelope premise concretely specifies is a relation from the cartesian product of booleans and naturals to the naturals

r :: B x N ---> N
r ( 0, a) = a
r ( 1 ,a) = 2a

From which we trivially obtain the inverse relation

ir :: N --> N x N
ir a = { (0,a) , (1 , a/2) }

There is literally nothing else that can be said, deductively or inductively, from the two envelope premise alone. One cannot extract a meaningful notion of probabilities, let alone expectations and averages in relation to that premise when it is stated without additional causal assumptions, which are necessarily to give physical meaning to probabilities in that context.

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 $1\over2$.

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 12
1
2
. — Michael

By "meaningful probabilities" i am referring to epistemically meaningful probabilities that quantify how the real world is expected to behave on the basis of past experience, as opposed to purely mathematical probabilities that merely quantify mathematical properties such as combinations and symmetry.

For example, suppose that the surface area of "heads" on an unknown but symmetric coin is roughly half of it's total surface area. If nothing else is known and assumed about the coin , including how and where it is to be thrown, then one cannot deduce solely only on the basis of the coin's mathematical properties that it's likelihood of landing heads is 50/50. Nevertheless, probabilities are often assigned to such unknown objects purely on the basis of their mathematical properties, leading to the conflation of "mathematical probability" with "physical probability" or "epistemic probability".

The two-envelopes question is epistemic because it concerns decision making in the real-world; so more than mathematics must be appealed to when deciding whether or not a strategy exists for decision-making in that context.

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.

So by your logic we can't even talk about the probability of a coin toss landing heads being 12
1
2
? — Michael

In my view , epistemic probabilities are derived from causal knowledge or assumptions on the basis abductive reasoning and repeated trials. One cannot derive causal knowledge from mathematical concepts as is demanded in the two-envelopes problem.

In my strong opinion, Laplace's principle of Indifference and the principle of maximum entropy are grave misuses of probability calculus that only lead to erroneous inferences and the conflation of ignorance with information.

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 12
1
2
, and we assume that my choice of envelope is truly random. — Michael

It is fair to make probabilistic assumptions, but they shouldn't be arbitrary (unless for the sake of philosophical argument) and they must be explicitly stated, and any conclusion cannot go farther than what is explicitly assumed from the outset. The less one assumes, the less one can conclude.

Since the two-envelopes problem isn't well posed, more must be assumed and/or imprecise probabilities are needed.

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"

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 $1\over2$.

Relative to your use of probabilities, how do you distinguish knowing that an outcome has probability 1/2 from not knowing the likelihood of an outcome?

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. — Michael

From Wikipedia Two Envelopes Problem

1) Denote by A the amount in the player's selected envelope.
2) The probability that A is the smaller amount is 1/2, and that it is the larger amount is also 1/2.
3) The other envelope may contain either 2A or A/2.
4) If A is the smaller amount, then the other envelope contains 2A.
5) If A is the larger amount, then the other envelope contains A/2.
6) Thus the other envelope contains 2A with probability 1/2 and A/2 with probability 1/2.
7) So the expected value of the money in the other envelope is:
$\frac{1}{2}(2A)$ + $\frac{1}{2}(\frac{A}{2})$ = $\frac{5}{4}A$
8) This is greater than A so, on average, the person reasons that they stand to gain by swapping.

In the first part of the equation: $\frac{1}{2}(2A)$ - the A is referring to the situation whereby the person has the smaller amount, say A is referring to £10.
In the second part of the equation: $\frac{1}{2}(\frac{A}{2})$ - the A is referring to the situation whereby the person has the larger amount, say A is referring to £20.

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.

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.

Yes, that’s what I show in the OP. — Michael

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