Ah, but that's exactly what we can't do. — Snakes Alive

This is valid and the first three premises are true.

1. My envelope could be envelope X
2. My envelope could be envelope 2X
3. My envelope contains £10
4. From 1 and 3, the other envelope could be envelope 2X and contain £20
5. From 2 and 3, the other envelope could be envelope X and contain £5

Still not getting it.

Modeling the sample space as [X, 2X] where X changes values doesn't work. This is because, although you don't know which value X is, you do know that the metaphysically possible outcomes relative to which you have to aggregate to determine your average expected winnings do not have different values for X. To say that you might have 5 or 20 in the other envelope is the same as to say that you might have drawn X or 2X. Your mistake is reifying the amount you've drawn into a new amount, compared to which you can average your expected winnings across two situations. This absurdly commits you to the possibility that it is possible that X has two values (not that there are two possible values that X could be epistemically, which is already recorded by your ignorance of whether you've drawn X or 2X).

To say that you might have 5 or 20 in the other envelope... — Snakes Alive

Which is all I'm saying.

My envelope has £10, so:

1. If my envelope is envelope X then the other envelope is envelope 2X and has £20.
2. If my envelope is envelope 2X then the other envelope is envelope X and has £5.

Given each antecedent is equally likely (and the only options), each consequent is equally likely (and the only options). Therefore my sample space for the other envelope is [£5, £20].

It can't be, since neither 5 nor 20 is half the other. Therefore this is not a possible sample space (as said above).

Your error is switching the value of X between situations over which you average. This can't be, since there is some value X determined by the envelopes: it is therefore not possible to average possible outcomes over two distinct values for X. It is "epistemically possible" that the other envelope has 5 or 20, but your knowledge that the value of X is fixed prevents you from averaging over the distinct values in this way, because the real outcomes over which you average in making your decision are to switch or not to switch, and your switching or not cannot change the value of X. You're confusing two epistemic possibilities with the possibilities over which your calculation ought to range.

The idea is this:

You begin knowing only that one of the envelopes on offer is worth twice the other.

Upon drawing an envelope, if you designate its value ('Y'), or learn its value (£10), you have learned that the the pair of envelopes on offer included one of value Y. You do not know whether Y = X or Y = 2X.

Thus there are two possible candidate pairs of envelopes: [Y/2, Y], and [Y, 2Y].

Now your uncertainty is not which of the envelopes originally in front of you was larger, but which pair of envelopes you picked an envelope of value Y from.

(You start thinking you'll get either the larger or the smaller of the pair in front of you. Once you've drawn, you imagine the smaller possibility as the larger of a smaller pair of envelopes, and similarly the larger becomes the smaller of a larger pair of envelopes.)

If you designate the unpicked envelope's value as Z, there are two possible systems of linear equations:

$\small 3x - y - z = 0$
$\small \phantom{3x-} \frac{1}{2}y - z = 0$

and

$\small 3x - y - z = 0$
$\small \phantom{3x -}2y - z = 0$

It's all perfectly consistent. In the first you find that Z = X, and Y = 2X. (This is the smaller pair and the envelope you drew is the bigger there.) In the second, Z = 2X, and Y = X. (This is the bigger pair, and your envelope is the smaller.)

Yeah, we've covered all this already. The pro-switcher posts are saying the same thing over and over.

What you are saying is the equivalence of suggesting 2+2 =5. You can't setp outside the definitions simply because they are an inconvenience you, as then you are no longer doing math.

You need to follow the same steps, list all the possible outcomes and stick with the definitions.

Now let's work through this:

Remember the definition of our sample space is [A,B] where A is the set [X,2X] and B is the set [X,2X]. We'll call this sample space 1.

You open A and it has 10 bucks, but you don't know if that is 10=X or 10=2X so A is still defined as [X,2X]. If A is X then B is 2X and if A is 2X then B has to be X. So B is still defined as [X,2X]. That part remains the same.

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]]


This is no different than any other math, we need to break it down and follow it step by step.

It can't be, since neither 5 nor 20 is half the other. Therefore this is not a possible sample space (as said above). — Snakes Alive

The sample space is the set of all possible outcomes. If there's £10 in my envelope then there are only two possible outcomes; either the other envelope contains £5 or the other envelope contains £20. So the sample space of the other envelope, given my envelope of £10, is [£5, £20]. This assertion that the sample space of the other envelope must have one value twice as big as the other is false.

The problem is that you're making this argument, where the value of X is fixed, and concluding that the sample space of the other envelope is [X, 2X] with a fixed value of X:

1. If my envelope is envelope X then the other envelope is envelope 2X
2. If my envelope is envelope 2X then the other envelope is envelope X

But this is incompatible with our third premise (which I brought up before):

3. My envelope contains £10.

If my envelope contains £10 then the X in "my envelope is envelope X" is 10 and the X in "my envelope is envelope 2X" is 5.

Your error is switching the value of X between situations over which you average. This can't be, since there is some value X determined by the envelopes: it is therefore not possible to average possible outcomes over two distinct values for X. It is "epistemically possible" that the other envelope has 5 or 20, but your knowledge that the value of X is fixed prevents you from averaging over the distinct values in this way, because the real outcomes over which you average in making your decision are to switch or not to switch, and your switching or not cannot change the value of X. You're confusing two epistemic possibilities with the possibilities over which your calculation ought to range. — Snakes Alive

I don't know why you keep bringing up averages.

I don't know why you keep bringing up averages. — Michael

The way that you determine whether a move is worth taking is by calculating the average expected gain from making that move.

You open A and it has 10 bucks, but you don't know if that is 10=X or 10=2X so A is still defined as [X,2X]

If A is X then B is 2X and if A is 2X then B has to be X.

So B is still defined as [X,2X]. — Jeremiah

If A is X and A is 10 then B is 20. If A is 2X and A is 10 then B has to be 5.

Can B still be defined as [X, 2X]?

Either way it has to be defined as [5, 20].

Another angle:

Suppose A is a switcher: he always picks one envelope to see, and then chooses the other to claim.

Suppose B is a stayer: he always picks one envelope, and then claims it.

For both A and B, which envelope they take is determined by which one they pick to see. Therefore, both A and B have two moves available: pick envelope 1 (A does this by choosing 2 to see, B by choosing 1 to see), or pick envelope 2 (done mutatis mutandis).

Therefore, the exact same moves are available to A and B, and they can both be commanded to do the exact same move given a random sample of pairs of envelopes.

According to the switching advocate, A's strategy is superior, despite performing the very same move in each case as B. This is absurd.

Can B still be defined as [X, 2X]? — Michael

Your possible outcomes for A is 2X or X. In the outcome that A=X then B=2X, in the outcome that A=2X then B=X. So yes, it seems, B has two possible outcome of either X or 2X.

Your possible outcomes for A is 2X or X. In the outcome that A=X then B=2X, in the outcome that A=2X then B=X. So yes, it seems, B has two possible outcome of either X or 2X — Jeremiah

And did you read the rest?

If A is X and A is 10 then B is 20. If A is 2X and A is 10 then B has to be 5.

Therefore B has to be defined as [5, 20].

Another way of putting this is that the strategy of switching, and of staying, are the exact same strategy. In each case, one simply picks one of the two envelopes to open. Seeing what is in one of the envelopes is simply a red herring: even on the switcher's account (since we ought to switch no matter what the value seen is), it does nothing.

There are in fact no two distinct strategies at all. Therefore, it is absurd to claim that one is superior, since they are identical.

Did you read the rest of my post?

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 math definitively proves you wrong. This is not a matter of opinion, you are just flat out wrong.

I don't know why I waste my education and time.

Did you read the rest of my post? — Jeremiah

Yes. You said:

1. A is 10.
2. If A is X then B is 2X
3. If A is 2X then B is X.
4. The sample space for B is [X, 2X]

But you're conflating different values of X:

1. A is 10
2. If A is X then B is 2X, where X = 10
3. If A is 2X then B is X, where X = 5
4. The sample space for B is [X where X = 5, 2X where X = 10]

One element is four times as big as the other. It's a sample space of [5, 20].

But you're conflating different values of X: — Michael

No, that is what you are doing. You can't just drop the 10, it is still a possible outcome, as you don't know which 10 you are in. The sample space is [[10,20],[5,10]]. How are you eliminating both 10s and combining the two sets? You have to mathematically justify all of this.

I drop 10 because it’s in my envelope. Therefore it cannot be in the other envelope.

Which 10 are you dropping 10=X or 10=2X? Also like I said I want to see the math.

I’m not dropping either of those. I’m dropping 10 as an outcome for the other envelope given that it’s in mine.

I’m saying that if 10 = X then the other (2X) envelope contains £20 and if 10 = 2X then the other (X) envelope contains £5.

If my envelope contains £10 then the other envelope contains either £5 or £20.

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.

I'm still not sure I understand your question. You've asked about the minimum possible value of X. Do you mean the minimum that I think is possible for X, given that I've just observed Y in the envelope? That minimum is Y/2.

Or do you mean the minimum amount I could imagine having been chosen for X, without my having opened the envelope? I'm pretty sure I already answered that (but the answer has been buried somewhere amongst all the posts since then), and said that the amount cannot be zero or less, so there is no minimum, but 0 is the infimum.

If you mean something other than that, can you clarify and I'll try to answer it? Thanks.

In the meantime, I'm finding some interesting things with my Bayesian calcs, but the answers keep on changing - probably cos I'm scribbling it on scrappy bits of paper, in between meetings, and I keep losing them. Hopefully I'll come up with something coherent and stable and can report it here.

You are shown two envelopes and told one is twice the value of the other. You are then offered the following choice: you may

(a) choose one envelope and keep its contents; or
(b) take both envelopes, open neither, and give back one of your choosing; or
(c) take both envelopes, open one of your choosing, and give back either the unopened envelope or the contents of the one you opened.

If X is the smaller of the two envelope values, your expected gain is 3X/2, no matter which of these procedures you follow.

I've finally found time to work out my full Bayesian analysis of the problem. For those that are interested, it is here. Non-Bayesians may not like it, because it takes the standard Bayesian approach of using a probabilistic distribution for unknown parameters, even when the person setting the parameter did not do so probabilitistically. But if we reject that approach we'll have to reject an enormous part of modern statistical techniques.

The essence is that the information gained by looking at the amount in the envelope allows one to refine one's prior distribution for the possible values of X - the lower of the two amounts. When the observed amount is low the expected gain will be positive and when it is high it will be negative. The note shows how to calculate these expectations based on the prior distribution adopted for X.

In the simplest case, where our prior for X is a uniform distribution on (0,L], where we might choose L as say half the entire budget of the game show, the result is that we should not switch if the observed amount y is greater than L, because it must be the doubled amount. But otherwise we should switch, and the expected gain from doing so is y/2. This is greater than the y/4 expected gain that was discussed above. I think that may have something to do with the fact that we remove the losses that would occur from switching when y>L, thereby increasing the expected gains when we do switch.

I will include some graphs I made if I can work out a way to get latex to cooperate on that. It worked last time I tried that, but not today.
Any observations or comments, @Michael? @Srap Tasmaner?

I looked at your work, but I only really followed page 1 — I'm learning as I go here.

I take it you're just trying to apply known techniques to the problem as presented, as if we don't know the right answer. I'll be curious to see how that comes out, but it's not how I understand our situation.

Here is my understanding of the issue (using X, Y, and U as you do). There are two natural ways to calculate the expected value of switching:

(a) E(U - Y) = P(Y = X)(2X - X) + P(Y = 2X)(X - 2X) = X/2 - X/2 = 0

(b) E(U - Y) = P(Y = X)(2Y - Y) + P(Y = 2X)(Y/2 - Y) = Y/2 - Y/4 =Y/4

There are two questions:

(Q1) Why do the two methods (a) and (b) give different results?
(Q2) Why does method (a) give the right answer and method (b) the wrong answer?

I tried fooling around with pre-labeling the envelopes L and R, so that we have two sample spaces, one for the loading of the envelopes [L = 2R, R = 2L] and one for our choice [Y = L, Y = R]. Removes a little ambiguity but that's about it. (For instance, instead of just saying P(Y = 2X) = 1/2, we can P(Y = L)P(L = 2R) + P(Y = R)P(R = 2L) = 1/2.) So far it just gives me new ways of fleshing out method (a). I can work through this and find no way of producing the method (b) approach, but I don't think it solves the puzzle just to say, "Doing it the right way avoids doing it the wrong way." The whole point is to figure out exactly what's wrong with the wrong way. Why doesn't it reduce to doing it the right way?

Edit: typo in (b).

The difference between these two is not mysterious. Calculation (a) is done from the point of view of the game host, who knows the value of X, and so can treat it as a constant. In this problem X is known and Y is unknown.

Calculation (b) is done from the point of view of the player, for whom X is an unknown, and hence must be treated as a random variable, if any calculation is to be done at all. In this problem Y is known and X is unknown. It's in a sense the reverse (a).

The calculations give different answers because they are different probabilistic scenarios - just as your calculation on the outcome of a coin I toss will be different from the calc I do when I've already looked at the toss result.

Each calc is correct for the perspective to which it pertains.

I'll have another go at inserting the graphs today.

No.

I think it's debatable whether anyone needs to "know" the amounts in the envelopes for the [x,2x] approach to work. If a machine sets the values, you'll be saying we have probabilities "from its point of view". And, okay, whatever, but we're just making the terminology fit now; there's no epistemic situation for the machine, or not one we're interested in.

More importantly, the value of the selected envelope need not be known to you the agent. Simply designating whatever as Y and then doing your calculations in terms of Y is enough to get absurdity. (If you condition your choice on expected value, you end up in an endless loop.)

And furthermore everyone knows the two envelopes have the same expected value and switching is a pointless strategy.

Simply designating whatever as Y and then doing your calculations in terms of Y is enough to get absurdity. — Srap Tasmaner

I haven't seen any genuine absurdities in the thread so far. Assertions of absurdity, yes, but not real ones.

In general, it's best to avoid arguments based on 'absurdity'. If we did that, we wouldn't have quantum mechanics, and so no computers. Best stick to 'contradiction' as the grounds for rejecting a premise.

I have a particularly strong aversion to arguments based on absurdity because they are beloved of Christian fundamentalist apologists like William Craig, who argue that 'absurdities' - by which they just mean something a bit surprising - arise from not believing the fundamentalist dogma. Which is no reason for us to adopt their dogma, because the world is a very surprising place.

More importantly, the value of the selected envelope need not be known to you the agent. — Srap Tasmaner

If we don't open the envelope then the calculation is completely different and in this case only coincidentally gives the same result. This is what is called a Numéraire issue. Changing the numéraire in terms of which a profit is calculated can create a profit out of nothing, or reduce a profit to zero. This is an important principle in derivative pricing.

Say I buy a (troy) ounce of gold for $1000, the price then goes up to $2000 per ounce. In dollars I have made $1000 profit, but in ounces of gold I have made no profit, because I had 1 oz before and I still have 1 oz. I can convert that zero ounces profit to dollars by multiplying by the gold price of $2000/oz, and I still get $0. How can my profit be both $1000 and $0? Because the two calculations use different numéraires, so they are fundamentally fdifferent calculations. Conversely, If I put the $1000 in bank account and earn $1 interest while the price of gold goes from $1000 up to $2000. Then with dollars as a numéraire my profit is $1 but with oz of gold as numéraire I have gone from 1 oz equiv to $1001/2000 $/oz = 0.5005 oz, so I've made a loss of 0.4995 oz, which is equivalent to $999.

The moral is - changing numéraire can change the answer, even if you subsequently convert the result back into the units of the original numéraire. When you open the envelope and look at what's inside, your numéraire is dollars. When you don't open it your numéraire is 'value of contents of the first envelope', ie Y. So it's a different calculation that can validly have a different result.

There's another numéraire we can use, which is X, the unknown amount of the lower-valued IOU. Using that as numéraire, the expected gain is zero, whereas when we use dollars as a numéraire, the expected gain is positive.