There are two reasonable ways to assign variables to the set from which you will select your envelope: {x, 2x} and {x, x/2}. Either works so long as you stick with it, but if you backtrack over your variable assignment, you have to completely switch to the other assignment scheme. — Srap Tasmaner

I agree with that. I think I wrote something similar, but with more words, in my reply here. I think that's the right generating mechanism for what's random in the scenario. The agent doesn't know which of (x,2x) they have. But they know (x,2x) are the amounts in the envelopes. Rather than the agent doesn't know which of (x/2, 2x) is in the other envelope - since (x/2, 2x) isn't a pair of envelopes, it's a representation of the agent's epistemic uncertainty that doesn't reflect the uncertainty of allocating amounts to the envelope pair. The variable they ascribe to the other envelope doesn't describe the allocation of amounts to envelopes at all; so their expectation, based on that variable, can't be expected to reflect the true expectation of switching.

But if in reality, someone told them they had 10 in their envelope, flipped a coin to choose between 5 and 20, shoved the resultant amount into another envelope, and presented the choice to switch to the agent - in that scenario they should switch.

Do we have the same take?

I don't see any disagreement.

I don't see any disagreement. — Srap Tasmaner

Same. Alright, I'm a bit less confused now. Thank you.

What are we supposed to learn here?

I mainly take it as a logic puzzle: there's a bit of fallacious reasoning, you have to spot it, explain why it's fallacious, and why it's attractive, easy to fall into.

On the other hand there is a way of taking it as an interesting probability problem, where you have to model several different sorts of issues, and under some additional conditions there are interesting switching systems available. (I don't have the background for that stuff anyway, as you know.)

So there's kinda an old school and a new school response possible.

The old school take is just another case study in faulty reasoning, but might have some interest in how people elide different sorts of uncertainty and chance. And last time I recall speculating about some use for the new school approach. (Something about improving performance without feedback -- it was pretty half-baked.)

Besides an opportunity to play with our respective toys, do you get anything out of this?

Besides an opportunity to play with our respective toys, do you get anything out of this? — Srap Tasmaner

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!

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.

, there's still no reason to switch, so the paradox has nothing to do with probability at all. — Michael

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

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.

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.

The second set of equations is different.

Are the situations described in the following questions the same? If not, what's the difference?

(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}?

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.

But given the set {10,20}, E(z)=15=(5/4)12, and 12 isn't the value of the chosen envelope. — Michael

That's true. It's not in the set. Neither is 15. But 15 is, for all that, the expected value of both envelopes.

The total value of any pair of envelopes is 3x, where x is the smaller of the two. I choose one, and call it y; if I got the smaller then y is x, if the larger then y is 2x; so x is either y or y/2. The total value of the envelopes is then either 3y or 3y/2, so the average total value is 9y/4. Alternatively, you could just say that since x is y or it's y/2, the average value of x is 3y/4, again making a total of 9y/4, on average. By definition, my envelope is y, so the other envelope must be worth 5y/4, on average.

Where have I gone wrong?

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.

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.

You can be sure that the expected value for the other envelope is 5/4 of that of the one you have.

The math is completely OK.

But the experiment cannot be set up because the distribution of the envelopes is not producible.

Give an example of a set of envelopes and I'll demonstrate.

You can be sure that the expected value for the other envelope is 5/4 of that of the one you have. — SolarWind

I can be sure it isn't as per the post immediately before yours.

I don't understand.

The task is quite simple. Name a few amounts that are in a set of envelopes and it becomes quite clear where the paradox is.

I don't understand. — SolarWind

This and this explain it quite clearly I think.

I think you're conflating two different expectations. I think your post should read :

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

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

2. $E[z | y ] = {5\over4}y$ and $E[y | x] = {3\over2}x$

This is the reasoning that leads to the switching argument:

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

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

But then by the exact same logic (or by performing the appropriate substitions):

$E(z) = {1\over2}2x + {1\over2}x = {3\over2}x$

Therefore:

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

Yes, my previous criticism was directed more towards the confusing use of notation with respect to expectation values. It is clearer to condition expectation values on the variables and information they assume.

But the switching argument isn't invalid; It's contradictory expectation values don't appeal to faulty reasoning given acceptance of the premises. Rather the switching argument is unsound, for among it's premises is an improper prior distribution over x, the smallest amount of money in an envelope. And this premise isn't possible in a finite universe.

Intuitively, it's contradictory conclusions makes sense; if the smallest amount of money in an envelope could be any amount of money, and if the prior distribution over the smallest amount of money is sufficiently uniform, then whatever value is revealed in your envelope, the value of the other envelope is likelier to be higher.

It's contradictory expectation values don't appeal to faulty reasoning given acceptance of the premises. — sime

I believe it does, as I showed above. It covertly redefines $y$ such that when it concludes $E(z) = {5\over4}y$, $y$ is no longer the value of the chosen envelope.

Rather the switching argument is unsound, for among it's premises is an improper prior distribution over x, the smallest amount of money in an envelope. And this premise isn't possible in a finite universe.

Intuitively, it's contradictory conclusions makes sense; if the smallest amount of money in an envelope could be any amount of money, and if the prior distribution over the smallest amount of money is sufficiently uniform, then whatever value is revealed in your envelope, the value of the other envelope is likelier to be higher. — sime

It doesn't require anything like that. The only premises are that one envelope contains twice as much as the other and that I pick one at random. The probability that I will pick the envelope with the smaller amount is $1\over2$.

I believe it does, as I showed above. It covertly redefines y
such that when it concludes E(z)=54y is no longer the value of the chosen envelope. — Michael

No covert redefinitions of y are happening, just covert pathological expectations over infinity. For the problem implicitly assumes

$P(x) = c$

This is an improper prior that has a divergent mean.

$E [x] = \infty$

The distribution of envelope values conditioned on the smallest value is symmetric, consisting of two spikes on the infinite plane at (x,2x) and (2x,x)

$P(y , z| x) = 0.5 \delta(x,2x) + 0.5 \delta (2x,x)$

Summing out, say, z gives

$P(y | x) = 0.5 \delta(x) + 0.5 \delta (2x)$

The two previous formulas imply the conditional

$P(z | y, x) = 0.5 \delta(2x)^{y=x} + 0.5 \delta(x)^{y=2x}$

since $P(x)$ is improper, one cannot "integrate out" x from the last equation and get a well-defined distribution. Instead one gets a function defined up to proportionality.

$P(z | y) \propto 0.5 \delta(2y)+ 0.5\delta((1/2)y)$

Taking the expectation of $y$ over this function as if it were a distribution gives

$E [ z | y ] = \frac 5 4 y$

Hence by symmetry

$E [ y | z ] = \frac 5 4 z$

No redefinitions, covert or otherwise, of any variable were involved here.

But I perhaps should correct myself a bit; the expectation values aren't contradictory in the sense of

$E [ y | z ] \gt E [ z | y ] \gt E [ y | z ]$

rather, the expectations imply contradictory strategies in the context of utility maximisation.

You may have missed my edit.

It doesn't require anything like that. The only premises are that one envelope contains twice as much as the other and that I pick one at random. The probability that I will pick the envelope with the smaller amount is 1/2.

No covert redefinitions of y are happening — sime

There is. I explained it above. I'll do it again.

Assume, for the sake of argument, that one envelope contains £10 and one envelope contains £20, and that I pick an envelope at random.

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

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

Notice that in $E(z)$ the variable $y$ stands for 3 different values. In the first case it stands for the value of the smaller envelope (10), in the second case it stands for the value of the larger envelope (20), and in the third case it stands for a different value entirely (12).

That third value (12) isn't the value of the chosen envelope, given that the chosen envelope contains either £10 or £20.

This is true for every possible pair of values and is true even when we don't assume the values of the two envelopes (it's just harder to see).

And it should be obvious that the probability assignments are correct. If one envelope contains £10 and one envelope contains £20 and I pick one at random then the probability that I will pick the envelope that contains £10 (the smaller amount) is $1\over2$ and the probability that I will pick the envelope that contains £20 (the larger amount) is $1\over2$.

Notice that in E(z) the variable y stands for 3 different values. In one case it stands for the value of the smaller envelope (10), in another case it stands for the value of the larger envelope (20), and in the final case it stands for a different value entirely (12). — Michael

Yes, I see that. So why are you redefining y?

Your two definitions of E[z] aren't equivalent. The first one is implicitly referring to
E[ z | y], the conditional expectation of z given the value of envelope y. Whereas the second definition refers to E[z | x] the conditional expectation of z given the value of the smallest envelope x.

Yes, I see that. So why are you redefining y? — sime

I'm not redefining y, the switching argument is. I'm showing you what it covertly does.

I'm not redefining y, the switching argument is. I'm showing you what it covertly does. — Michael

I haven't redefined y, and still derive the switching argument from it's premises.

Your two definitions of E[z] aren't equivalent. — sime

They are equivalent.

There are two, equally probable, situations that $E(z)$ uses:

1. $z = 2y$ and $y = 10$
2. $z = {y\over2}$ and $y = 20$

So it's doing this:

$E(z) = {1\over2}2y[where\space y=10] + {1\over2}({y\over2})[where\space y=20]$

I'm just making this more explicit:

$E(z) = {1\over2}(2\cdot10) + {1\over2}({20\over2})$

Which leads to:

$E(z) = 15 = {5\over4}12$

Hence the redefinition of $y$. It no longer stands for the value of the chosen envelope given that the value of the chosen envelope isn't 12.

I haven't kept up with this thread, but when I read the original statement in the OP, what comes to mind is Yogi Berra's advice: When you come to a fork in the road, take it.