If you run eight billion trials, giving equal values of X for each trial to each player, will the switcher and non-switcher converge roughly on the same payout?

Will someone, or will they not, get more money on average as a result of choosing the switching strategy, as opposed to choosing the not-switching strategy?

How do you respond to the fact that your analysis is empirically wrong?

Holy shit, fine, so we're on the same page now. So respond to the initial points now that this pointless tangent is over.

This just isn't relevant. Assuming that one wants to make the most money on average, the puzzle remains, and that is clearly the point of the OP. You are missing the point.

No, the problem asks what you should do.

The prior simulations demonstrated this.

If you don't accept that, literally just go out and play the game. Do trials where one person always switches, and another never does, giving them randomized values for X. The switcher will not receive 1.25 times the money on average that the non-switcher does.

If you play the game, switching on average will not afford you any gains. You predict it will. You are wrong.

This has nothing to do with whether an analysis is Bayesian or not. Your error is a basic conceptual one.

This is reflected in the fact that you are empirically wrong about the results of actually playing the game.

You can't use dollars in that way.

You've agreed that Y must be defined in terms of X. X is fixed. Therefore, Y = 2X, or Y = X. It simply doesn't matter what the dollar value of Y is. What you want to do is on the one hand agree that Y is defined in terms of X, and agree that there is some fixed value of X, such that X and 2X are the values of the envelopes, and calculate expected utility solely in reference to Y (a dollar amount) not expressed in terms of X. You cannot do this. If you do, you will be averaging over situations with respect to which X is different in each. But you cannot average over outcomes in which X is different, for we know that X has a fixed value, and so whatever it is, it must remain constant over the outcomes that we use to average.

You are being misled by naming a variable. Just replace Y with its definition in terms of X, which you must do anyway since that's how you've defined it, and the illusion disappears.

What do you mean by 'exhaustive disjunctive possibilities'? — andrewk

Exhaustive disjunctive possibilities are those that are mutually exclusive and jointly exhaustive, viz. those that form a partition over the space of possible outcomes.

That's all correct, but is not inconsistent with my note. Nowhere does it say that 5 and 20 are X and 2X, given we have seen that Y=10. Rather, we know that X is either 5 or 10, so either

we have X in the envelope we opened, so that X = Y = 10; OR
X is in the other envelope, so we have Y = 2X = 10 and X=5 is in the other envelope. — andrewk

No, this doesn't work. If you define Y in terms of X, then everything you write with Y can (must) be rewritten in terms of X. Do this and the illusion that Y is a value you can average over multiple situations disappears. You are acting like Y is some other value that you can now double or half in terms of what you draw. But it is not: it is, for some fixed X, either X or 2X. It follows that whatever the hidden amount is, the two possibilities over which we average must be such that one is double the other. 5, 20 violates this and so cannot be what we average over – it is not consistent with the way the problem is set up, because we know there is some value of X fixed at the time of the choosing of the two envelopes, and the only possibilities are that the hidden amount is either X or 2X, regardless of what amount you look at in one envelope. The fact that you even see one envelope at all is a complete red herring, only thrown into the problem to seduce you into making this very fallacy.

As a side note, the Socratic methodology of just asking, free of context, 'what is X?' is bad methodology. It will invariably lead you to silliness.

If you define Y in this way, then there are two possible states: you have drawn X (where B = 0), or you have drawn 2X (where B = 1). Now eliminate the Y altogether, do the calculation, and you'll see that your expected value is 1.5X regardless of whether you switch or not.

It sounds like you're saying that, having observed 10 in the envelope, you now believe that the other envelope might contain some amount other than 5 or 20. — andrewk

No. The point is that it is not a possibility that the exhaustive disjunctive possibilities of what's contained in the other envelope are 5 and 20. This is because it's known already that for some X, the amount in the other envelope is X or 2X. But 5 and 20 are not X and 2X for any value of X. Thus it cannot be that the exhaustive disjunctive possibilities are these, and cannot be that in determining the average expected value of switching, we average across those possibilities, since our knowledge state rules out this way of setting things up.

Yeah, I realized this after writing my original post, but I don't have a good way to exposit the issue. I suspect that epistemic and metaphysical possibilities are being confused in some inappropriate way.

That is, given some fixed value of X (and we know there to be some such), it is not metaphysically possible for the unopened envelope to be 'either 5 or 20' (read as an exhaustive disjunction), since neither of these is twice the other (and so this is inconsistent with what we know to be metaphysically so, that the unopened envelope is either X or 2X, for some X, while 5 and 20 are not X and 2X for any X), and so our knowledge of the situation should bar this space of possibilities, since our knowledge that this is metaphysically impossible should rule out our averaging over this epistemic space as regards our outcomes.

You can't use Y as a value defined independently of X and average across possibilities using that value.

And I responded to the response.

WTF is with all the questions about how basic dialogue works?

No, my response to that question is as it was before: that one is syntactically derived from the other has nothing to do with epistemology. This was in response to your post above.

So I agree, but that was my whole point. We're accustomed to think our experiences are "of" things, but there's no reason to think that's so. I take Descartes just to have noticed this.

Why don't I try just having this, so there can be no misperception as to what I'm replying to:

It's not true that the sun is small and here. It's large and out there. etc — csalisbury

That's how we are accustomed to think of it – but there's no reason to think that way is right.

Do you understand that?

But it's replying directly to what you said.

I feel like the thing you said about posts being bad was weirdly placed. It seemed to say something like: I don't need to respond to you, because this is a lame place! — csalisbury

I think your posts are pretty good, but the OP is bad. I was just puzzled by why you thought my comments were irrelevant, since they have to do with the topic discussed in the OP exactly. Maybe you wanted to talk about something else, not the OP, and were disappointed the conversation didn't steer the way you wanted?

I try, but it seems like every other place is a Nazi/Communist recruitment center or a place to post pornography of cartoon characters or talk about having bipolar disorder...

Do you know any better ones (not joking)?

OK (not much lost – these threads are of pretty poor quality, OPs are too vague / scatterbrained).

I don't know what you're talking about.

It's not true that the sun is small and here. It's large and out there. etc — csalisbury

It's true that that's how we're accustomed to think of it by default. I don't think there's any possible way to answer transcendent questions about whether that way of seeing it is the right way.

The point is that the Cartesian turn allows one to see it the other way – a way that one initially does not even understand that one can see it. In that sense, it's not like learning a new true proposition, but being able to see where once one was blind. You get a new ability. The Cartesian is also right that in some sense this is the way it was 'all along.' You can of course choose to ignore this new ability and have faith that it is just an aberration, and the old way of seeing things is the 'right way.' But it's just that – faith.

We adjust our perceptions — csalisbury

In general, we do not have the power to adjust our perceptions.

When we see distances, we understand that the thing we're seeing is 'there', not 'here but small.' — csalisbury

Right, and there was an analogous kind of self-awareness when the empiricists noticed that you could come to 'see' things as just rearranged as different sizes in the visual field, instead of representing objective distances. We just naturally see these things as distances, but we only do this by means of the visual field being stimulated in this way, and when one turns to epistemology one 'sees' this again. Usually one sees 'through' it.

When we see veridically (building on your analogy) we understand that the thing is real instead of not-real? — csalisbury

We take the experience to be 'of' something.

I think it's the case because it just happens, in the same way that we see distances, and so on. It's in the structure of experience, if you like. The Cartesian move strikes me as a kind of self-awareness.

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.

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.

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.

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

I don't think we 'come by' a sense of veridicality. It's just how we're hardwired to think about things. There can't ever be 'evidence' ultimately that a perception is veridical.

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.

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).

Still not getting it.

How is it incoherent?

we can also model the other envelope using the sample space [X, 2X], despite the fact that in doing so we're positing two different possible values of X (either 5 or 10). — Michael

Ah, but that's exactly what we can't do. This was the original fallacy (back in my first post).

You seem to be ignoring the fact that distinctions between real and imaginary or hallucinated "seemings" are established intersubjectively. So, your witch example is irrelevant to the context of this discussion. Intersubjective notions of 'is' are founded upon collectively corroborated 'seemings". As Kant pointed out this intersubjectively established understanding of "is" when rationally analyzed leads to the "transcendental illusion' of naive realism. — Janus

Doesn't matter, since the intersubjectivity can't establish anything and faces the same problem.