For bogarting your L.

Let me try rewriting this.

Your sample space is [R =L/2, R = 2L].

Which means your distribution here:

= P(R=5 | L=10)(R) + P(R=20 | L=10)(R)
= R

Could be:

= P(R=5 | L=10)(R) + P(R=20 | L=10)(R)
= L/2

Or

= P(R=5 | L=10)(R) + P(R=20 | L=10)(R)
= 2L


Let me think about it some.

-----

Thoughts in progress. . .

= P(R=5 | L=10)(L/2) + P(R=20 | L=10)(2L)

Sorry, but I need more clarification on where you are going with this.

Let's call the two envelopes A and B. Now envelope A could have X or A could have 2X and likewise B could have X or B could have 2X. Those are all the possible outcomes so by the definition of a sample space our sample space is [A,B] where A is the set [X,2X] and B is the set [X,2X], which means our sample space could also be written as [[X,2X],[X,2X]]. — Jeremiah

I think I got it. We've got two variables, a numerical value X and a binary variable that tells us which letter we picked, the one containing X (smaller value) or the one containing 2X (the bigger value).

Let me explain step by step:

We have a value X. It's a numerical variable, and it describes the value of two letters in such a way that one letter has X, and another letter has 2X.

The second, the binary variable is E, for envelope, and it's values are yes/no. The question it answers is: Did we pick the envelope that contains the value X?

The sample space for X = N (natural numbers); the sample space for E is [yes, no].

Now we pick a letter and open it. We find it has the value 10. We now have new information about X. The sample space for X has shrunken from N to [5, 10], because 10 has to be either X or 2X.

We have no information on the "yes, no" question, but we do know that the letter we picked has the value 10. That leaves us with the following:

X [5, 10], E [yes, no]

We can use a if/then relation to connect the variables:

If E = yes then X = 10 (because if E is yes, then the letter we didn't pick is the larger one, 2X)
If E = no then X = 5 (because if E is no, then the letter we didn't pick is the smaller one, X)

Just for completeness sake:

If E = yes then 2X = 20
If E = no then 2X = 10

And in words:

If we picked the smaller letter X = 10, that means this letter has 10 (X), and the other is 20 (2X).
If we didn't pick the smaller letter X = 5. That means this letter has 10 (2X), and the other letter has 5 (X).

That means that if this letter is 10 (X | E = yes, or 2X | E = no) then the other letter has [5 (X | E = no), 20 (2X | E = yes)].

That's exactly the same situation as your [[10,20],[5,10]], viewed from a different perspective. Let me write it out: [[10 (X | E = yes), 20 (2X | E = yes)], [5 (X | E = no), 10 (2X | E = no)].

It's indisputable that if we pick up a letter and look inside and find a 10 we have:

X [5, 10], E [yes, no]

Everything else is just different groupings:

If we uncover a letter and it has the value Y, then the other letter has a value of [Y/2, 2Y].

Y ... [X | E = yes, 2X | E = no]
Y/2 ... [X | E = no]
2Y ... [2X | E = yes]

I believe this covers all our bases.

What this means for the switchers and the conundrum is currently beyond me. There's certainly something strange going on.

Right, good point. In which case, defining the sample space that way is the mistake -- turns our problem into the Ali Baba problem. Didn't mean to do that.

How would you feel about something like this?

E(U | Y=10) = P(U=5 | Y=10)(x) + P(U=20 | Y=10)(2X)
= 3X/2

Okay to describe the events that way?

My problem with this is that I feel like the numbers are just being bent until you get the answer you want.

There are two things we can use probability models for; making predictions and to better understand relations by creating an accurate model. You got the mid range you are looking for, but does your model explain the problem correctly?

If the player adopts this as the sample space then they cannot use a uniform distribution as their Bayesian prior for X, because there cannot be a uniform distribution on the natural numbers. Hence a 'tailing-off' distribution would need to be selected, such as P(X=n)=2^-n for n=1, 2, ....

Given the adoption of a Bayesian prior distribution, the higher the observed value of Y (the value in the opened envelope), the greater the chance that it is 2X rather than X, and hence the less the expected gains from switching. Since a strategy of compulsory blind switching delivers expected gains of zero, a strategy of choosing an amount H and switching only when Y<H must have an expected positive gain.

I'm with you. I just hadn't thought of doing it this way before. It gives you a way to acknowledge what you've learned by learning the value of one of the envelopes, while acknowledging that you still don't know which of 5 and 20 is even a possible value of X. It feels close to the way we should think about the problem.

The more I look at it, the more I miss seeing U=X and U=2X. Maybe it's best to leave them alone. Seeing P(U=5...) when 5 might not even be a possible value for U just feels wrong.

...which of 5 and 20 is even a possible value of X — Srap Tasmaner

I'm not quite done yet thinking, but 20 is definitely not possible value of X. It's like this:

For Y = 10:

5 is a possible expected value for X (alternative to 10=X).
10 is a definite value, either for X or for 2X
20 is a possible expected value for 2X (alternative to 10=2X).

This symmetry is systematic:

Y/2 = possible expected value for X
Y = definite value, either for X or for 2X
2Y = possible expected value for 2X

Y is definitely in the sample (because you're looking at it). If it's X the other card is 2X (and thus 2Y), and if it's 2X, the other card is X (and thus Y/2).

Or differently put: [5, 20] is [X, 2X] but not of each other - of the respective alternative of 10.

20 is definitely not possible value of X — Dawnstorm

Just a typo

Sorry about the correction. My head is swimming.

Personally, I think there is a general misconception on these forums that Bayesian inference is an excuse to model the subjective rather than the objective; however, that approach is too open to observer bias. Priors assumptions need to be fully justified as does the posterior. Not only do we need to scrutinize the prior but also what information is acceptable for our posterior and perhaps for these reasons a Bayesian inference is just not appropriate for this problem.

The strangest thing about the puzzle to me is that you need only designate an envelope to get into trouble.

Wikipedia says Smullyan thought it was a logic puzzle and had nothing to do with probability.

I think it has more to do with set theory than probability.

It is suggestive that the only probabilities ever in play are 50% and the only expected value calculation I have any faith in tells us absolutely nothing.

On the other hand, once you start iterating, there's plenty of cool stuff to do.

Well the whole expected value approach has some interesting aspects to it.

If A1=Y=X then B1=2X or if A2=Y=2X then B2=X, where Y is the amount you see opening envelope A and X is the unknown amount originally selected by the facilitator.

Then in case one

B1 > Y

In case two

Y > B2

So B1 >Y> B2

Which means

P(A1)Y+P(B1)2X > P(A2)Y+P(B2)X or more importantly P(A1)Y+P(B1)2X is not equal to P(A2)Y+P(B2)X.

So since you don't know which case you are in after seeing Y and they are not equal you can't really calculate the expected value. Now if you never opened A and never saw Y, that is a different story. So the only rational thing to do is pretend Y does not exist, then subjectively you can treat case one and case two with the same algebra.

*Edit

Or rather I should say that the inclusion of Y makes calculating the expect returns impossible.

The distribution in the other letter cannot be [Y/2, 2Y] as one of those values simply does not exist. You have still created a sample space with impossible outcomes. The truth is that Y is not usable information. The error is making new assumptions based on Y.

If you don't change your assumptions upon seeing Y this problem becomes direct and accurate to solve.

I think the issue is that you and I mean something different by an outcome being possible.

Let's say you toss a coin. If it's heads you put a red ball in one box and a blue ball in a second box. If it's tails you put a red ball in one box and a green ball in a second box. I pick a box at random, not knowing the outcome of your coin toss, and find a red ball.

I would say, given the information available to me, that it's possible that there's a blue ball in the second box and that it's possible that there's a green ball in the second box, with a 50% probability of each. The sample space for the other box is [blue, green].

Whereas you would say that the sample space for the other box is either [red, blue] or [red, green] and that we just don't know which it is, and that because we know that it doesn't contain a red ball that either blue is certain and green is impossible or green is certain and blue is impossible.

Maybe @Srap Tasmaner and @andrewk could comment on this too. What is the sample space of the other box? Does the answer depend on whether or not one is a Bayesian?

That's gone a bit over my head. With the above post in mind, this perhaps better explains where I'm coming from, and perhaps you could explain how it's addressed by your post:

I toss a coin. If it's heads then I put a red ball in one box and £5 in the other. If it's tails then I put a red ball in one box and £20 in the other. You pick a box at random, not knowing the outcome of the coin toss, and find a red ball. What is the sample space of the other box and its expected value?

I would say that the sample space of the other box (U) is [£5, £20] and that it's expected value is:

$\small \begin{align} E(U) &= P(U=5)(5) + P(U=20)(20) \\ &=\frac12 5 + \frac12 20 \\ &= 12.5 \end{align}$

Now perform the same calculation but instead of a red ball and £5 if heads and a red ball and £20 if tails, it's £10 and £5 if heads and £10 and £20 if tails (which is the same as X = 5 if heads and X = 10 if tails). The fact that it's £10 rather than a red ball in my box doesn't change the sample space or expected value of the other box.

Paralleling an example can be useful, but it also can be misleading. I rather stay focused on the problem in the OP.

If it's heads you put a red ball in one box and a blue ball in a second box. If it's tails you put a red ball in one box and a green ball in a second box. — Michael

See how red blue and green are all mentioned by name as possibilities.

E(U)=P(U=x)(x)+P(U=2x)(2x) — Srap Tasmaner

What I think we need to do is explain what x means here. Is it "the amount that's actually in the smaller envelope"? So:

E(U)=P(U is the smaller envelope)(the amount that's actually in the smaller envelope)+P(U is the larger envelope)(the amount that's actually in the smaller envelope)

Or is it "the amount that would be in the smaller envelope were this event to be the case"? So:

E(U)=P(U is the smaller envelope)(the amount that would be in U if U is the smaller envelope)+P(U is the larger envelope)(the amount that would be in U if U is the larger envelope)

You seem to be working under the first definition, whereas I'm working under the second definition.

See how red blue and green are all mentioned by name as possibilities. — Srap Tasmaner

I don't understand what you're implying by this response.

Take a step back from all the math and the modeling. Try to see the forest here.

(1) There are two envelopes.
(2) You end up with one of them.

Maybe you just pick one, maybe there's some long drawn out complicated process. Whatever happens between (1) and (2), you end up with one of the two. Your expected gain is just the average of their values.

You don't know the total value of the envelopes, how that value is selected, what it's range might be if it even has one. Right here in this post, I haven't even told you the envelopes necessarily have different values, or that their values have a fixed ratio, or a fixed difference. If I let you look in one, how much more do you know? Almost nothing. You might as well not bother.

There are two objects of unknown value and you end up with one of them. So far as a "decision" goes here, you might as well treat them as equal, or treat the choice as a matter of indifference, just flip a coin. There is not nearly enough information available to base a decision on.

There is not nearly enough information available to base a decision on. — Srap Tasmaner

Yes, because you’ve said that we don’t know anything. But in the case we’ve been discussing we know that one envelope contains twice as much as the other and that mine has £10. This is information that allows us to calculate an expected value as explained here.

You cannot calculate an expectation for X if you do not know what the sample space for X is.

It really should be called the "Grass is Always Greener" problem, because the whole point of that saying is that the greater greenness of the grass on the other side of the fence is an illusion you generate yourself by applying your favorite cognitive fallacy.

Even using terms like "expectation" or "expected value" is too fancy here. We're just talking about a mean of two values.

What's the mean of A and B? (A + B)/2.
What's the mean of X and 2X, for some X? 3X/2.
What's the mean of U and 10? (U + 10)/2 = U/2 + 5.

It's the mean of 5 and 20, given that either there's £5 or £20 in the other envelope.

Do you disagree with the calculation here with the red ball and £5 if heads or red ball and £20 if tails?

The distribution in the other letter cannot be [Y/2, 2Y] as one of those values simply does not exist. You have still created a sample space with impossible outcomes. The truth is that Y is not usable information. The error is making new assumptions based on Y. — Jeremiah

No, I have created a sample space with one impossible and one necessary outcome. It's an either/or situation, and that's appropriate because expectations are based on information rather than on what's actually the case. The statement that for every Y the other envelpe has to contain either Y/2 or 2Y is correct, and remains correct even after you check the other envelope and discover one or the other value inisde.

The distribution will be one of the two scenarios:

Y/2 = 100 %, Y = 0 %
Y/2 = 0 %, Y = 100 %

Y is not a random variable and doesn't have a sample space. That's why I said the random variable is a binary. Y = X? Yes/No. The other envelope has to take one of those values, based on two values:

- The value of this envelope (a fixed value)
- Whether you picked the envelope with X, or the one with 2 X (a random variable)

You need to know the latter to calculate the value of the other envelope, but you won't have the information until you check the other envelope, at which point the calculation becomes pointless.