If it was 'seems' all the way down, even the 'evil demon' would make no sense as a cause. — csalisbury

So?

Sure, but we need some sense of what it would mean were there such things as witches. In other words, we need some sense of what veridicality means. Where can we come by such an understanding? — csalisbury

Sure, but who doubts this? Not Descartes. And it implies nothing about our epistemologically 'starting with' veridicality, or having had any veridical experiences.

To say that the non-veridical relies on the veridical is not to say that the seeming of any particular thing must rely on a prior veridical perception of some similar thing. — csalisbury

So if I seem to see a witch, I must have seen something similar to a witch?

I get it, it sounds like a good formula, but if you actually try to apply it to the most banal examples, it doesn't seem to work.

In explaining why there's nothing opposed, you immediately made recourse to existence and its negation i.e "is-talk". Could you answer another way without doing this? — csalisbury

How I could answer has nothing to do with epistemology, but again with syntax of language. It's perfectly possible that there are no veridical experiences whatsoever – that veridicality, however we are attuned to it, is a transcendental illusion of which we're doomed to make use.

Yeah, someone who thinks mental states are purely manifested in behavior doesn't understand what a mental state is in the ordinary use. This isn't an elucidatory but a revisionist account. I feel the tension here between wanting to be the 'common sense' one on the block, and holding a highly uncommonsense position (behaviorism, which no one believes, and which probably makes mince meant of our notions of mental life). I suspect it's because certain philosophers historically have also dealt with these contradictory impulses and OP has read them.

Apologies for trying to read so much into OP's scattered comments so far. I try to jump ahead because there are tells and these beliefs tend to cluster together, and I've heard & gotten tired of them all, so I try to predict where people are going so we can get a move on to more interesting stuff.

We've already been over all this, I'm afraid. We would need a new way of looking at things to move forward, and this doesn't provide it.

Not at all. It only means that non-veridical perceptions can only be understood as non-veridical against the backdrop of a web of other, veridical perceptions. If one were to say that all perceptions were non-veridical, but couldn't explain what he meant by 'veridical', then he'd literally be talking nonsense. — csalisbury

Not so. Compare: it can seem like there is a witch, when there isn't. Must we have veridical witch-perceptions against which to 'compare' for this to be so? No, because it can seem like there are witches (perhaps it even has), yet there are none and have not been (let us assume).

If something 'seemed' to exist, as opposed to what? — csalisbury

There is no 'as opposed to.' Something that seems to exist can actually exist, or it can not.

Also, it's worth noting that the idea that Descartes 'started' with appearances is false. Read the first meditation – he is led to their consideration after spending his life only dealing with things, and not doubting them. Only when he realizes that this has in fact led him into error does he begin the investigation!

Nothing about epistemology follows from that fact that 'seem' statements are syntactically more complex than statements not containing 'seem.'

That from this one can conclude that one must 'start' with veridical perceptions in any way, in the sense that one has to have had any, is nonsense – this would imply that any phenomenon that people say seems to exist must have been met with actually existing, which is not the case. Existence proofs would then be very easy – it something seemed to exist, it would!

Nope. If you were right, the expected return for any number of plays, 1 or a million, should always be 1.25 the amount drawn. Number of plays is irrelevant.

(1) There is a flaw somewhere in the way Michael and @andrewk recommend we should estimate the value of the remaining envelope. If there is, it's not exactly obvious. — Srap Tasmaner

I have tried to clarify why this is, but apparently it hasn't worked. But I don't have another method of expositing it. I expect that some way of making yet another distinction, perhaps between epistemic and metaphysical possibility, may make it work, so the probability table can be drawn up again, and the fallacy made obvious to everyone.

So am I understand this as having no epistemological consequences? In what sense, then, is it a response to Descartes?

I'm not sure what 'conceptual priority' is, but whatever it is, it never gets us to any is-claim, which would seem to be the interesting thing to Descartes.

You keep thinking about it as what would happen if you play repeatedly. I'm just talking about playing a single game. — Michael

This is so bizarre to me. The amount of games doesn't matter, everyone keeps telling you this, but you're still convinced it is, and I can't figure out why.

While I think the problem has been adequately solved by the responses in this thread, it's fascinating to see how persistent these sorts of 'transcendental illusions' are. Even reasoning in the simplest cases, we often simply cannot see past some heuristic we usually use to solve problems, convinced they apply to some case in which they do not, and there often appears to be no easy way to demonstrate to people that they are in the grip of such an illusion.

A warning to all those who want to use such logical heuristics to do metaphysics about 'big' questions! If you can't get a toy problem about two envelopes right...

A couple things:

Statements about seeming are again statements that have truth conditions: one can be wrong about the way things seem (even, depending on the situation, how things seem to oneself).

It's true that since seem is a propositional operator, propositions about seeming are in some sense more complex than the propositions to which they apply (this sense is only relative to our linguistic usage: we require a more complex statement, built out of a simpler one, in order to express such things). However, this says nothing about the epistemological order of things. If talk about seeming implied prior contact with something real, this would lead to the absurd conclusion that we have contact with whatever we can make seeming-claims about. This is just not true, and there looks to be no reason to believe it with respect to whole swaths of perception generally.

Here is the post:

https://thephilosophyforum.com/discussion/comment/193041

It explains why averaging expected outcome over situations with respect to the amount you drew as constant doesn't work.

I already provided a refutation for this, actually, in my first long post. Several people have repeated this or something similar, but it's fallacious. Can you read that post and get back to me?

Your error, BTW, is having the two situations are that X = 5 or that X = 10. The correct way to think about it is that in one case, you've drawn X, and in the other, 2X.

I can spell out why what you've posted here is equivalent to what I went over in the long post, but I don't want to if you understand anyway.

Probability distributions are irrelevant.

You can literally just check this by performing it live. Empirically, the switching strategy doesn't help. The running of the program simply does the performance, and demonstrates this.

Thus we need to start with the recognition that switchers are empirically just wrong. Whether they know the error they've committed is irrelevant.

If we want to consider a long series of such activities, it needs to be set up precisely, and that has not been done. A number of crucial conditions will need to be specified, such as whether X is always the same and if not, how it varies from one play to the next. The best strategy will vary according to how those conditions are set up. — andrewk

This isn't relevant. It simply doesn't understand the crux of the puzzle.

This isn't relevant to the question anyway. The switching strategy you're discussing has nothing to do with the OP.

I would prefer not to get into this topic, because it's off-subject and the errors made here are strictly orthogonal to the errors driving the original misconception.

But it doesn't matter if there's a finite number of games, or how many games there are. What matters is you baked a hard limit into the selection of X. Wherever X can be at most, say 400, then one knows a priori never to switch when what one sees in the envelope >400 (since these can only be the 2X).

It doesn't matter that the strategy doesn't require knowing. Wherever there is some limit, albeit unknown, the strategy will funnel you towards figuring out what that limit is.

The strategy relies on something not present in the original example, and so is irrelevant.

It works because you limited the selection of numbers to below 400, and so funneled the choice to switching only when lower numbers relative to that range appear. This is not a feature of the original example, and if there were no a priori range from which the values were chosen, the strategy would have no benefit.

That's because although it's a 2:1 payout with an even chance of winning the bet when you lose is twice as big as the bet when you win. Obviously you're going to break even. It would be like flipping a coin and hiding the result (heads) and having one person bet £10 on heads and another £20 on tails and offering them both a 2:1 payout. It's an advantageous bet for both people (given that they don't know the result or about each other's offer), despite the fact that the average payout is £0. — Michael

There is some misunderstanding here. The number of trials is irrelevant to the effectiveness of the strategies. If multiple trials cause you to break even on average, so will one.

Also, a 2:1 payout would be getting $30 on a win and nothing on a loss, after betting $10.

I'm trying to think of a new way to explain the fallacy being committed, but I think I've hit a roadblock (not in the matter itself – I understand it, but am not sure how to explain it, given that the above hasn't worked).

Here is as blunt as I can put it.

In deciding that switching is a "good bet," you must average the amount gained (lost) from switching across two situations. This requires you to take the value of what you picked first (say 10) as constant across those two situations, so that you can determine how much you would get as a portion of this (double, or half). The problem is that if you do this, then you are changing the value of X across the two situations, and treating things as if there is one epistemic possibility, say that X = 5, and another, say that X = 10. While this is on one sense so (and is reflected already in saying it the way I've put it: that you may have either drawn X or 2X), the problem comes when you try to use this changed value to average the two situations.

You can't do this, since it presumes there are two epistemic possibilities, each of which has a distinct value for X. This is not so: we know a priori that whatever X is, it is fixed. Hence the epistemic possibilities appropriately described are that, given some value X, with we have drawn that, or double that. They are not that given we have drawn Y, the true value of X is either double that or equal to that. This is fallacious, since it sneaks into our epistemic space two distinct values for X, across which we compare the value of Y as if it were an independent variable. Hence the illusion that we are 'getting' 1.25Y on average, as I explained in my first post.

––––

This conceptual explanation seems to be getting us nowhere, so I wonder if another tack might work. First is the empirical: when the game is modeled, this demonstrates that you are wrong. Multiple trials show the two strategies converging. Hence, you should be able to appreciate that there is an error in your reasoning in fact, though you can't pinpoint it.

Another way to go might be to devise other ways of framing the problem that demonstrate how thinking that a switching strategy is beneficial is absurd. For instance, if we always switched, then which envelope we picked first to look at would always determine which we chose (viz. the other one). But then, looking at 1 would result in picking 2, and looking at 2 would result in picking 1. In either case, what we picked would be advantageous given what we looked at. But then, this leads to the absurd conclusion that beginning with either envelope to look at (and hence de facto picking whichever envelope) is equally advantageous. All switching does is change which envelope is picked; the switcher is thus committed both to thinking that switching is advantageous and indifferent.

That's not the game being played. There is no bet of an independent amount that returns half or double at equal rates. The amount seen first is itself already a function of X.

The number of bets made is irrelevant.

In order to reason this way, we must hold X constant for each case. If we played 100 times, where X = 10 in each game, then switching 50 times will net us on average 15 per game, thus 1500. This is the same payout as it we didn't switch at all, switched 75 times, or any other combination of switching or not switching.

The fallacy is that instead of holding constant X, you are holding constant the amount drawn, and acting as if this variable is independent of X.

Yes it does. It shows that there's a 2:1 payout with an even chance of winning. How is that not advantageous? — Michael

The payout is on average 1.5X. This is the same payout that not switching affords.

When betting the odds, all variables are independent. That is not the case here.

In terms of probability, sure, but not earnings. The average of the above is £12.50, but the average of not switching is £10.00. — Michael

I demonstrated that this is fallacious in my longer post above. It falsely assumes that the amount drawn is a variable independent of X. It is not.

The amount in my envelope is also fixed. So as the amount in my envelope is fixed to £10 and if X is fixed to 10 then it would be false to say that there's a 50% chance of switching into X and a 50% chance of switching into 2X. We can only say that there's a 100% chance of switching into 2X (or if X is 5 then a 100% chance of switching into X). — Michael

No.

We're talking about epistemic possibilities. There are two, equipossible: first, that you have drawn X, and second, that you have drawn 2X. While you know how much you have drawn, you don't know whether this is X or 2X.

Hence, there is a 50% chance of your having drawn each, and a 50% chance of switching into the other, if you decide to switch. This is the same opportunity that not switching gives you.

3. There's a 50% chance that the other envelope contains £5 and a 50% chance that the other envelope contains £20. — Michael

This conclusion does not show what you want it to. From this, it does not follow that switching is advantageous. For if the other envelope contains 5, you have drawn 2X, and if it contains 20, you have drawn X. In either case, your average output as a proportion of X is 1.5. This is the same as if you had not switched.

That doesn't model the case we're considering, which is that we know we have £10. — Michael

The value that you get from the first envelope is irrelevant.

No.

The value of X is fixed. The two possibilities you consider must treat X as the same. You don't know what X is; but you do know that whatever it is, it does not change across the two scenarios you are considering. Your fallacious reasoning assumes that it does.

The shoe is on the other foot; you are assuming that the amount drawn, say 10, is constant in terms of X across both considered scenarios. This is not the case.

No.

If there is a 2:1 payout on a coin toss, then the bet is always worth taking, regardless of the amount bet, and one will not break even on average.

But this is getting far away from the main topic anyway.

What remains true is that there is no benefit to switching, either in the one case or across cases.

The value of X is fixed prior to the decision. Your reasoning switches the value of X across the two scenarios.

On average you will break even, but if we just consider a single game then a 2:1 payout with an even chance of winning is a bet worth making. — Michael

This cannot be.

The post explains why this reasoning is fallacious.

The fallacy is that there is some value, X, determined in each case. You are acting as if there is an independent variable Y, viz. what you drew first, and that switching will get you either .5Y or 2Y, and so on average the "bet" is worth taking, since your average payout is then 1.25Y.

This is fallacious, because there is no such variable: Y is defined in terms of X (either it is X, or 2X), and across the two scenarios you're averaging, the value of Y changes. Hence, there is no single value of Y across the two situations, and the value 1.25Y is a chimera.

If you define Y in terms of X, this illusion disappears. Thus, as you said, you either drew X or 2X. Therefore, there's a 50% chance that switching gets you X, and a 50% chance that it gets you 2X. Likewise if you stay. The average payout for either is 1.5X. Switching does not offer favorable odds.

But take one game in isolation? You have a 2:1payout with an even chance of winning. That's a bet worth making.

So in our case, I have £10 and the other envelope contains either £5 or £20. There's an even chance of winning, but the payout is 2:1. — Michael

This reasoning is fallacious. Did you read the post?

What I am saying is that as a strategy, switching does not increase your chances of earning more money, regardless of how many times the game is played. Your average earnings are the same regardless of whether you switch or not, and regardless of how many times the game is played.

If as a player my goal is to get as much money as possible, the two strategies are indifferently effective.

No, your conclusion doesn't establish what you want it to (that switching is more profitable).

There is no step that's wrong – what's wrong is that in order to use this reasoning to provide an average based on a ratio of the amount drawn from the first envelope, you have to average a single variable ("how much I drew" = Y) across two distinct situations, as if this value would be the same across them as an independent variable (so that I average 1.25 Y by switching). But it isn't. How much you drew is definable only in terms of X, which switches in the two scenarios: in one scenario, it's 5, and in the other it's 10. So it's only coherent to calculate the average based on the odds in terms of X, since Y (how much you drew) is a variable dependent on X. If you do this, you'll see that the average payout is 1.5X, whether you switch or not.

Did you read the post? I can do another one working through your example specifically.

I'm just doing this:

1. We pick an envelope at random
2. There's a 50% chance that my envelope is the X envelope and a 50% chance that my envelope is the 2X envelope.
3. I open my envelope and see £10
4. From 2 and 3, there's a 50% chance that my £10 envelope is the X envelope and a 50% chance that my £10 envelope is the 2X envelope.
5. From 4, there's a 50% chance that the other envelope contains £20 and a 50% chance that the other envelope contains £5. — Michael

Check my previous post – I explained why this reasoning is fallacious.

This is a fun puzzle – I took a crack at it.

Let X be the amount as in the statement of the puzzle. Let Y be the amount revealed in the first envelope.

There are two possible states in which one can be after the revelation: either Y = X (you have pulled the lesser envelope, and the other has twice as much), or Y = 2X (you have pulled the greater envelope, and the other has half as much).

There are also two actions: stay, and switch.

We can then construct the average expected winnings from each move, based on each of the two states, in a 2 x 2 table:



Each of the four cells represents an outcome of the move made based on the corresponding state (states on the horizontal, moves on the vertical). In each case, the value is expressed in terms of both X and Y.

Assuming that there is a 50% chance that one is in either of the two states, we can get the average expected winnings for each move by halving the value given by that move in each state, then adding these two together.

What's interesting is that the results look different depending on whether you express the average result in terms of X or Y.

For Y:

Switch: .5(2Y) + .5(.5Y) = Y + .25Y
= 1.25Y

Stay: .5 Y + .5Y
= Y

Thus, expressed in terms of Y, the greater average output is had by switching, since 1.25Y > Y, regardless of the value of Y.

For X:

Switch: .5(2X) + .5X = X + .5X
= 1.5X

Stay: .5X + .5(2X) = .5X + X
= 1.5X

Thus, expressed in terms of X, both moves have the same average output.

––––

So what's going on? How can the output be equal, but not equal, for the two actions?

It turns out there is an illusion in the way the problem was set up with respect to Y: there is no single value of a variable Y that can be defined independently of X, and then placed next to it in proportion (as either equal to it, or double it). Where Y is the number you drew in the first envelope, such a variable is already defined in terms of X: it is only possible to draw X, or 2X. The illusion results from thinking of 'the value of what I drew' as some distinct variable Y, when this is not the case. This in turn leads to the illusion that one can calculate the average expected return based on this new value, 'what I drew,' and conclude that switching allows a better return on 'what one drew.'

But what you drew is simply either X or 2X in the first place. There are thus two possible situations: either you drew X, or you drew 2X. And there are two possible moves: switch, or stay. Now let's draw the real table, defined this time in terms of the only variable there is, X:



This table simply eliminates the illusory Y (not a well-defined value independent of X to begin with), and we see that as before, switching has no effect on average expected output.

––––

The take-away message is that there is no such thing as a value of 1.25Y, averaged across the two possible situations, because there is no such single value Y that is the same across said situations, since you don't know whether you're drawn the greater or lesser envelope. X, by contrast, is a value we can refer to coherently across both situations, and cast in these terms, it becomes obvious that switching is ineffectual. The illusion comes in thinking that the de dicto description 'what I drew' is equivalent de re to some single numerical value across possible situations, and it is not.