Your reasoning suggests that you ought to switch unconditionally.

I'd switch conditionally as explained here to reap the .25 gain. — Michael

It is absolutely true that in iterated play you can learn stuff about the sample space and its distribution, and that you can develop strategies that build on this knowledge, even simple strategies like using an arbitrary cut-off will work.

Cool article here.

I just don't think of the paradox itself this way.

Yeah, the effectiveness of the conditional switching strategy is a red herring that is irrelevant to the puzzle, so I didn't want to touch it. The thing to address now is that this strategy is not even actually what's recommended by Michael's position. If he is right, you ought to switch no matter what.

Yeah, the effectiveness of the conditional switching strategy is a red herring that is irrelevant to the puzzle, so I didn't want to touch it. The thing to address now is that this strategy is not even actually what's recommended by Michael's position. If he is right, you ought to switch no matter what. — Snakes Alive

If I'm right about what? That if there's £10 in my envelope then either X = 10 and the other envelope contains £20 or X = 5 and the other envelope contains £5, and that each is equally probable? Or that from this we can deduce an expected return of £2.50 from switching, given that Expected Return = SUM (Return_i * Probability_i) = (10 * 0.5) + (-5 * 0.5)?

Honestly this take on the problem is far more interesting than the "paradox" and of real-world use. I'd rather be thinking about that.

But it's also clear to me there's a conceptual muddle in the argument for switching in the non-iterative case, so I can't get past that.

Expect for one to be possible the other one has to be impossible and therefore can not be included in the same sample space as possible outcomes. R and S are mutually exclusive events.

What are you talking about? Heads and tails are mutually exclusive outcomes for a coin toss, yet the sample space for the outcome of a coin toss is [Heads, Tails].

You claim the sample space is [5,20]

The sample space is defined as all possible outcomes, but for 5 to be a possible outcome then 20 must be an impossible outcome. For 20 to be a possible outcomes 5 must be a impossible outcome. Therefore by the definition of a sample space your sample space cannot be [5,20].

But it's also clear to me there's a conceptual muddle in the argument for switching in the non-iterative case, so I can't get past that. — Srap Tasmaner

There is no point in even approaching probability and expected returns without properly defining the sample space. As if that is not right, the rest of it won't be right.

We define A and B as: If A=Y=X then B=2X or if A=Y=2X then B=X, where Y is the amount you see opening envelope A and X is the unknown amount originally selected by the facilitator.

My claim is then that the only possible outcomes for B is X or 2X.

Proof:

For all of Y, Y is a positive real number, such that Y=X or Y=2X, where X is some positive real number.

You are handed A and you see Y inside.

There are two cases here:

Case One

A=Y=X

By definition of A and B if A=Y=X then B=2X therefore B=2X

Case Two

A=Y=2X

By definition of A and B if A=Y=2X then B=X therefore B=X.

Those are the only two possible cases for B therefore by the definition of a sample space the sample space for B is [X,2X]

------

Notice how that is for all of Y. A very important concept. Also notice how it says "where X is some positive real number." It absolutely does not matter how X was selected.

Stop repeating this, we get what the argument is, you've said it 5,000 times. — Snakes Alive

And it's not clear to me which part you disagree with, so I was hoping you could clarify the problematic step:

1. There's £10 in my envelope
2. My envelope is either X or 2X
3. X is either 10 or 5
4. If X is 10 then the other envelope contains £20
5. If X is 5 then the other envelope contains £5
6. It is equally probable that X is 10 and that X is 5
7. Expected Return = SUM (Return_i * Probability_i)
8. Expected Return = (10 * 0.5) + (-5 * 0.5) = 2.5

Jeremiah
1k
We define A and B as: If A=Y=X then B=2X or if A=Y=2X then B=X, where Y is the amount you see opening envelope A and X is the unknown amount originally selected by the facilitator.

My claim is then that the only possible outcomes for B is X or 2X.

Proof:

For all of Y, Y is a positive real number, such that Y=X or Y=2X, where X is some positive real number.

You are handed A and you see Y inside.

There are two cases here:

Case One

A=Y=X

By definition of A and B if A=Y=X then B=2X therefore B=2X

Case Two

A=Y=2X

By definition of A and B if A=Y=2X then B=X therefore B=X.

Those are the only two possible cases for B therefore by the definition of a sample space the sample space for B is [X,2X]

------

Notice how that is for all of Y. A very important concept. Also notice how it says "where X is some positive real number." It absolutely does not matter how X was selected. — Jeremiah

You are handed A and you see Y inside.

There are two cases here:

Case One

A=Y=X

By definition of A and B if A=Y=X then B=2X therefore B=2X

Case Two

A=Y=2X

By definition of A and B if A=Y=2X then B=X therefore B=X.

Those are the only two possible cases for B therefore by the definition of a sample space the sample space for B is [X,2X] — Jeremiah

You are handed A and you see £10 inside.

There are two cases here:

Case One

A=10=X

By definition of A and B if A=10=X then B=2X therefore B=20

Case Two

A=10=2X

By definition of A and B if A=10=2X then B=X therefore B=5.

Those are the only two possible cases for B therefore by the definition of a sample space the sample space for B is [5,20]

Like every time previously you're conflating different values of X. You can see it from your own antecedents: in case one, X = 10; in case 2, X = 5.

So just tell me in actual numbers the possible amounts in the other envelope if my envelope contains £10. If I switch, how much money could I walk away with? I say either £5 or £20. What do you say?

We define A and B as: If A=Y=X then B=2X or if A=Y=2X then B=X, where Y is the amount you see opening envelope A and X is the unknown amount originally selected by the facilitator.

My claim is then that the only possible outcomes for B is X or 2X.

Proof:

For all of Y, Y is a positive real number, such that Y=X or Y=2X, where X is some positive real number.

You are handed A and you see Y inside.

There are two cases here:

Case One

A=Y=X

By definition of A and B if A=Y=X then B=2X therefore B=2X

Case Two

A=Y=2X

By definition of A and B if A=Y=2X then B=X therefore B=X.

Those are the only two possible cases for B therefore by the definition of a sample space the sample space for B is [X,2X] — Jeremiah

Replace Y with 10 and you get B = 20 in case 1 and B = 5 in case 2.

We define A and B as: If A=Y=X then B=2X or if A=Y=2X then B=X, where Y is the amount you see opening envelope A and X is the unknown amount originally selected by the facilitator.

My claim is then that the only possible outcomes for B is X or 2X.

Proof:

For all of Y, Y is a positive real number, such that Y=X or Y=2X, where X is some positive real number.

You are handed A and you see Y inside.

There are two cases here:

Case One

A=Y=X

By definition of A and B if A=Y=X then B=2X therefore B=2X

Case Two

A=Y=2X

By definition of A and B if A=Y=2X then B=X therefore B=X.

Those are the only two possible cases for B therefore by the definition of a sample space the sample space for B is [X,2X]

Repeating yourself doesn't make the conclusion go away. If Y is 10 then in case one B = £20 and in case two B = £5.

We define A and B as: If A=Y=X then B=2X or if A=Y=2X then B=X, where Y is the amount you see opening envelope A and X is the unknown amount originally selected by the facilitator.

My claim is then that the only possible outcomes for B is X or 2X.

Proof:

For all of Y, Y is a positive real number, such that Y=X or Y=2X, where X is some positive real number.

You are handed A and you see Y inside.

There are two cases here:

Case One

A=Y=X

By definition of A and B if A=Y=X then B=2X therefore B=2X

Case Two

A=Y=2X

By definition of A and B if A=Y=2X then B=X therefore B=X.

Those are the only two possible cases for B therefore by the definition of a sample space the sample space for B is [X,2X]

@Michael, @andrewk

Nothing here that @Jeremiah and @Snakes Alive haven't already said, I think, just arranged a little differently. Check my math.

Given a sample space $\small [X, 2X]$ of possible values for the selected envelope $\small Y$ and the unselected and unopened envelope $\small U$, we can calculate an expected value for $\small U$:

$\small \begin{align}E(U \mid Y=X) &=P(U=2X \mid Y = X)(2X) + P(U=X \mid Y=X)(X) \\ &= 2X \\ E(U \mid Y=2X) &=P(U=2X \mid Y = 2X)(2X) + P(U=X \mid Y=2X)(X) \\ &= X \\ E(U)&=E(U \mid Y=X)P(Y=X) + E(U \mid Y=2X)P(Y=2X) \\ &= \frac32 X\end{align}$

The question is whether it is legitimate to do something like this instead:

$\small \begin{align}E(U \mid Y=X) &=P(U=2X \mid Y = X)(2Y) + P(U=X \mid Y=X)(Y) \\ &= 2Y \\ E(U \mid Y=2X) &=P(U=2X \mid Y = 2X)(Y) + P(U=X \mid Y=2X)(\frac{Y}{2}) \\ &= \frac{Y}{2}\\ E(U)&=E(U \mid Y=X)P(Y=X) + E(U \mid Y=2X)P(Y=2X) \\ &= \frac54 Y\end{align}$

I think the answer is just no. You can't do it this way. $\small Y$, $\small 2Y$, and $\small \frac{Y}{2}$ are not in the sample space of possible values for $\small Y$ and $\small U$. That space consists entirely of $\small X$ and $\small 2X$. What the switching analysis proposes is "substituting" for $\small X$ when conditioning on $\small Y=X$ or $\small Y=2X$. But that's just not how it works. In this expression

$\small P(U=2X \mid Y=X)(2X)$

$\small (2X)$ is a dollar value, while $\small U=2X$ and $\small Y=X$ are events, with probabilities of occurring, not equations you might use to set the value of $\small X$. There's no line in here anywhere that says $\small Y = X$ as a simple equation.

There's also no place in our expectation formula for something like $\small X \mid Y=X$. The possible values from our sample space are not "conditional" in some way. And it's obvious when you get to the third equation, that you cannot possibly add these "conditional values" together: what is the $\small Y$ in $\small \frac54 Y$ supposed to be? Is that $\small Y \mid Y=X$ or $\small Y \mid Y=2X$?

The sample space is [[10,20],[5,10]]. Notice how there are two 10s. Now show me the math which allows you to eliminate both of them. — Jeremiah

I haven't read past this page and only skimmed the the next two, so if I'm repeating what someone else said, or if that's irrelevant by now, please ignore. But this is complicated and I don't have much time, and I'm sure I'll forget if I don't reply now.

The square brackets represent envelopes - I'm sure of that. In a sample space of [[10,20],[5,10]] you're not defining the envelope we picked; you're defining the envelopes according to "contains 2X [10,20]", "contains X [5,10]". If you defined the envelope as "the one we picked" you'd get "the envelope we picked [10 (=2x), 10 (=x)]", and "the envelope we didn't pick [5 (= X), 20 (= 2X)]". That's because the two envelopes are interdependent, and that's why the events are order-sensitive, i.e. [[10 (=2X), (10 =x)] and [5 (= 2 X), 20 (= X)]] wouldn't work.

Tha sample space in your post is: [[10 (=2 X), 20 (= 2 X)], [5 = X, 10 = X]]. That is the sample space is only correct, if we're not picking an envelope at all, but defining the envelopes according to whether or not they contain X or 2 X. But then the values are arbitrary.

In any case, there is only one "10", and that's the one in the envelope we opened, whether that's X or 2X. The interdependence between the two envelopes determines that the other envelope has either a 5 or 20, and which of those is in there in turn determines whether 10 = X or 2 X (such is the nature of interdependence).

***

Also, the bet is inherently not avaragable, since repetition either immediately makes each subsequent repetition a win (by revealing X, if we check the result), or (if we stack the results without checking the wins) reveals X as soon as we get a dollar bill other than the one we get first (50 % at the start of the first repetition, 75 % at the start of the second repetition, ... = the likelihood of figuring out X).

you're defining the envelopes according to "contains 2X [10,20]", "contains X [5,10]". — Dawnstorm

Sorry no, that was not my intent. In the event of R you have A=10 and B=20. In the event of S you have A=10 and B=5. These are mutually exclusive events, which means in the case of R the amount 5 does not exist at all, and in the event of S the amount 20 does not exist at all. So one of those sample spaces is feeding you false information. The only way to avoid this is to treat X as the unknown variable it is.

Here was the original argument, which I only did because Michael requested that I do it. However, an algerbic solution, which I have posted several times already, models this problem much more accurately.

So we have two cases here and have no clue which one we are in. Earlier I defined these cases as amount X case R, and and amount 2X case S. We have a lot of variables flying around so let's try to be consistent here.

Now that we have listed all possible outcomes we can define our sample space as [R,S], well call this sample space 2.

Now remember by our definitions an event is a subset of our sample space.

In event R X=10 and since in event R B must be 2X then B = 20.

So the sample space of event R is [10,20].

In event S 10=2X and since in event S B must be X then B= 5.

So the sample space of event S is [5,10]. (order does not matter)

So our sample space, which we named as sample space 2, is [R,S] where R is the set [10,20] and S is the set [5,10] or we can express it as [[10,20],[5,10]] — Jeremiah

The problem with the [5,20] argument is that one of those numbers is not possible, therefore calculations based on those numbers will be misleading. To do so is to incorporate false information into your math.

I suppose I could also have thrown in this:

$\small \begin{align} E(U \mid Y=10) &= P(U=2X \mid Y=10)(2X) + P(U=X \mid Y=10)(X) \\ &=\frac12 2X + \frac12 X \\ &= \frac32 X \end{align}$

Sorry no, that was not my intent. In the event of R you have A=10 and B=20. In the event of S you have A=10 and B=5. These are mutually exclusive events, which means in the case of R the amount 5 does not exist at all, and in the event of S the amount 20 does not exist at all. So one of those sample spaces is feeding you false information. The only way to avoid this is to treat X as the unknown variable it is. — Jeremiah

Hm, thinking about it a bit more, I think we're making a basic mistake, here. X/X2 is the relationship of the variables, not the sample space. I'll go at it step by step so we can see if I've made a mistake somewhere:

1. We have two envelopes with two different amounts of money:

Envelope1 = X $
Envelope2 = Y $

At that point E1 and E2 do not refer to specific envelopes, nor do X or Y refer to specific amounts. It's simply two variables with two values, and we have no more information. If we have 10 $ in one envelope and 20 $ in another, it doesn't matter whether we set 10 X and 20 Y, or the other way round. It's completely arbitrary. Both constellations describe the same event.

Both X and Y have the same sample space: any number that makes sense of $. Both sample spaces might be, for example, the natural numbers (weight, space, etc. are complications we don't need).

2. If we learn that one envelope contains exactly double the amount of the other, that tells us more. We now can get rid of one variable. But the sample space isn't X and 2X. It's the natural numbers for one (let's call it X), and depending on one of two assumptions we make about X, the sample space of the second is a transformation of X.

Two assumptions?

2. a) We assume X is greater of the two numbers.

Envelope1 = X $ (natural numbers)
Envelope2 = Y $ (X/2)

2. b) We assume X is the lesser of the two numbers

Envelope1 = X $ (natural numbers)
Envelope2 = Y $ (2*X)

These two assumptions both validly describe situation. We still don't define a real envelope; we merely define wether X is the greater number and Y is the lesser number, or the other way round.

In both these assumptions we don't know the actual value of X. So if someone tells us that one of the envelopes contains 10 $, then we don't know whether X is the greater or lesser number. With regards to the above, we don't know where to put it. But we do know it has to be one of the two: 10 $ is either the greater or the lesser number. This gives us two possibilities:

2. a)

Envelope1 = 10 $
Envelope2 = (10/2) = 5 $

or 2. b)

Envelope1 = 10 $
Envelope2 = (10*2) = 20 $

But what we've done here is twofold: we've set X = 10, and we've set envelope1 as the envelope that contains X. We do not know whether X is the greater or the lesser number. The question we care for is what's in envelope2, and the answer to that is:

If X is greater number, envelope2 contains 5 $.
If X is the lesser number, envelope2 contains 20 $.

The sample space for envelope1 was all the natural numbers, and the event is now 10. Since the sample space of envelope 2 is dependent on the sample space of envelope 1, there are only two possibilities: X/2 or 2X. We simply don't know whether X is the greater number or the lesser number.

It doesn't matter which envelope we open first, we never know which is the greater or the lesser. Because of this, we can set any of the envelopes as 1 or 2, and we always have the same situation:

E1 = 10 and E2 = 5 (X > Y, Y = X/2)
E1 = 10 and E2 = 20 (X < Y, Y = 2 X)

If we only open one envelope, we might open the other envelope first. We wouldn't know about 10, in that case, but either about 5 or 20, depending which is true.

For 20 we'd get:

E1 = 20 and E2 = 10 (X > Y, Y = X/2)
E1 = 20 and E2 = 40 (X < Y, Y = 2X)

The ratio remains a constant, no matter which number you draw, and that's why you alway stand to win twice as much as you would lose. This is a function of what you know about the ratio. However the natural numbers that make up the individual sample spaces differ.

If the envelopes contain 10 and 20 dollars, and you set E1 as the envelope you pick first you get:

For E1 = 10:

the sample space for E2 is [5, 20].

For E1 = 20

the sample space for E2 is [10, 40]

That's because the sample space for E1 is not X. The sample space for E1 is the natural numbers. X is the event.. However, once we know the event for E1, we know that the sample space for E2 is [X/2, 2x], and that's because of the ratio. We can't reduce the sample space to 1 item because we cannot know whether X is the greater or the lesser number until we look at E2. But E1 is chosen at random.

So we get the following:

Envelope1 = [X | € N]
Envelope2 = [Y | € [X/2, 2X]]

I'm not an experienced mathematician, so I might have gotten the notation wrong. But does the reasoning make sense?

I suppose I could also have thrown in this:

$\small \begin{align} E(U \mid Y=10) &= P(U=2X \mid Y=10)(2X) + P(U=X \mid Y=10)(X) \\ &=\frac12 2X + \frac12 X \\ &= \frac32 X \end{align}$ — Srap Tasmaner

You have two different values of X and so this is misleading.

Where you have P(U=2X∣Y=10)(2X), the actual value of X is 10, given that you've defined the smaller envelope as having £10.

Where you have P(U=X∣Y=10)(X), the actual value of X is 5, given that you've defined the larger envelope as having £10.

So it works out as:

$\small \begin{align} E(U \mid Y=10) &= P(U=2X \mid Y=10)(20) + P(U=X \mid Y=10)(5) \\ &=\frac12 20 + \frac12 5 \\ &= 12.5 \end{align}$

To make this clearer, say I flip a coin to determine the value of X to put it an envelope Y. Using your formulation, we have:

$\small \begin{align} E(Y) &= P(H)(X) + P(T)(X) \\ &=\frac12 X + \frac12 X \\ &= X \end{align}$

But let's say I'll put in £10 if it's heads and £100 if it's tails. The expected value, using my formulation, is:

$\small \begin{align} E(Y) &= P(H)(10) + P(T)(100) \\ &=\frac12 10 + \frac12 100 \\ &= 55 \end{align}$

But does the reasoning make sense? — Dawnstorm

It makes sense, but it is wrong, and that is why it is misleading.

Forget the math and just think about this rationally.

Consider:

I have two envelopes and I secretly put 10$ in one and 20$ in the other, then hand you one at random.

You open it and see there is 10$. You have no idea what is in the other envelope so you imagine it could be 20$ or it could be 5$. However, it can't be 5$ ever, as it is 20$. A 5$ outcome would be impossible, so your imagination is feeding you false information. You imagined a number that is not even a possible outcome.

The distributed amounts into the envelopes is X and 2X and it will always be X and 2X no matter what you imagine it might be or how much you know about the envelopes.

Also, all this has been covered in extensive depth.

You have two different values of X and so this is misleading.

Where you have P(U=2X∣Y=10)(2X), the actual value of X is 10, given that you've defined the smaller envelope as having £10.

Where you have P(U=X∣Y=10)(X), the actual value of X is 5, given that you've defined the larger envelope as having £10.

So it works out as:

$\small \begin{align} E(U \mid Y=10) &= P(U=2X \mid Y=10)(20) + P(U=X \mid Y=10)(5) \\ &=\frac12 20 + \frac12 5 \\ &= 12.5 \end{align}$ — Michael

1. Y = 10
2. Y = X → U = 20
3. Y = 2X → U = 5

These statements are all true.

*4. Y = X
*5. ∴ U = 20 (1,2,4)
*6. Y = 2X
*7. ∴ U = 5 (1,3,6)

One of (4) and (6) is true and the other false, and thus one of (5) and (7) is true and the other false.

If (4) is true, then the sample space for Y and U is [10, 20]; 5 is not in the sample space and therefore you cannot use 5 in calculating the expected value of either Y or U.

If (6) is true, then the sample space for Y and U is [5, 10]; 20 is not in the sample space and therefore you cannot use 20 in calculating the expected value of either Y or U.

But let's say I'll put in £10 if it's heads and £100 if it's tails. — Michael

£10 and £100 are both in the sample space for $\small X$: that's why your expected value calculation is correct.

Suppose this was our problem: we know one envelope has £10 and one has £20, and we're just to pick one and that's it.

$\small \begin{align} E(Y) &= P(Y=10)(10) + P(Y=20)(20) \\ &= 15 \end{align}$

Easy peasy.

In our problem, we do not know whether the sample space is [5,10] or [10,20]. It is one or the other, but that doesn't mean our sample space is [5,10,20], not even subjectively. Our space is known to have exactly two values in it. You cannot plug three values into the expected value formula and get the right answer.

Y is 10, therefore the sample space for X is [5, 10]. To work out the expected value of U we multiply the probability that X is 5 (0.5) to the value of U if X is 5 (5), we multiply the probability that X is 10 (0.5) to the value of U if X is 10 (20), and add them together.

It's exactly what we do with my example of the coin toss. The sample space for the toss is [H, T]. To work out the expected value of Y we multiply the probability that the toss is H (0.5) to the value of X if the toss is H (10), we multiply the probability that the toss is T (0.5) to the value of X if the toss is T (100), and add them together.

Another way to phrase your formulation is:

$\small \begin{align} E(U \mid Y=10) &= P(X=10)(2X) + P(X=5)(X) \end{align}$

The left and right side of the addition explicitly define different values of X. It's:

$\small \begin{align} E(U \mid Y=10) &= P(X=10)(2 * 10) + P(X=5)(5) \end{align}$