You may entertain yourself by switching and call that a reason, but there is no expected gain from switching.

And if I see £10 then I stand to gain £10 and I stand to lose £5. — Michael

If you see £10 then either you stand to gain £10 or you stand to lose £5, but not both.

I have two pairs of envelopes A = {5, 10} and B = {10, 20}. I'm going to offer you a choice from one pair or the other. What is your chance of getting 10?

P(10 | A) = P(10 | B) = 1/2

if we assume you're an average chooser without "10 seeking" talent. Clear enough.

But what is P(10)?

P(10) = P(10 | A)P(A) + P(10 | B)P(B) = P(10 | A)P(A) + P(10 | B)(1 - P(A)) = P(A)/2 + 1/2 - P(A)/2 = 1/2.

So your chance of picking 10 is 1/2, no matter what P(A) is. P(A) drops out.

What's your chance of picking 5?

P(5) = P(5 | A)P(A) = P(A)/2

What's your chance of picking 20?

P(20) = P(20 | B)(1 - P(A)) = (1 - P(A))/2

No idea in either case, because P(A) does not drop out.

If you got a 10, what's your expected value for the other envelope U? You can go two ways here. You could say

E(U | 10) = 5 * P(A)/2 + 20 * (1 - P(A))/2

and that would be true, but it somewhat misses the point. I choose before you choose. "The other envelope" is not well-defined until I have chosen A or B, at which point you can say P(A) = 1 or P(A) = 0. You never get to pick from all four envelopes; you only get to pick from the pair I have chosen. We ignored this when calculating P(10) because my choice didn't matter. Now it does.

E(U | A, 10) = 5 and E(U | B, 10) = 20.

You'll still want to do this

E(U | 10) = E(U | A, 10)P(A) + E(U | B, 10)(1 - P(A))

and then say that since you know nothing about P(A), you can only apply the principle of indifference and assume P(A) = 1/2. You might be wrong; I may be unaccountably inclined toward A to the tune of 1000:1 but you have no way of knowing that and the rational thing to do is go with indifference.

But this only makes sense because I've told you that I was choosing from two sets of envelopes in the first place. What if I didn't tell you that? What if there only ever was one pair? What if there were thousands? Maybe some of those have duplicate amounts, maybe not. Maybe there's only a single pair with 10 in it. (This is @JeffJo's thing, and it's worth thinking about. You can't really even count on using 1 - P(A), much less assume P(A) = 1/2.)


Here's a real life version. Suppose I have some cash and two envelopes, and I'm going to split my cash in such a way that one envelope has twice as much as the other. Suppose I have one $10, one $5 and 6 $1's. What are my options?

$\small \begin{array}{ccc}&Envelope\ 1&Envelope\ 2\\A&1&2\\B&2&4\\C&3&6=5+1\\D&5&10\\E&6=5+1&12=10+2\\F&7=5+2&14=10+4\end{array}$

There are some combinations I can't make because I don't have enough of the right denominations.

We could talk about this table as we talked about the collection {{5, 10}, {10,20}}. If you knew how much money I had and in what denominations, there are several cases in which, upon opening an envelope, you'd already know whether you have the larger or the smaller.

But let's suppose you don't know any of that. You could also figure that if you got an odd number it must be the smaller (because I'm only using bills, no coins) so I'll cleverly not choose one of those; I'll choose only from

$\small \begin{array}{ccc}&Envelope\ 1&Envelope\ 2\\B&2&4\\E&6=5+1&12=10+2\end{array}$

If I choose B, and you draw the 2, you can reason that I would have excluded {1, 2}, so the other must be 4. Similarly, if I choose E, and you draw the 6, then you can reason that I would have excluded {3, 6} and so the other must be 12. Ah well. I'd have to have more money to make the game perfect.

But what if I choose B and you draw the 4? 8 is mathematically possible, but there's no {4, 8} here. Similarly, if I choose E and you draw the 12; 24 is mathematically possible, but there's no {12, 24} here.

So what is your expectation before getting an envelope? Unknown. Something less than half the total cash I have on me, which you don't know, but there are other constraints based on the denominations and some gamesmanship.

Again, there's no game until I choose. Say I choose B. You don't know it, but the average value of B is 3. If you draw the 2, trading gains you 2; if you choose the 4, trading costs you 2. Say I choose E. You don't know it, but the average value of E is 9. If you choose 6, trading gains you 6; if you choose 12, trading costs you 6.

Once I have chosen, what you stand to gain or lose by switching is always a fixed amount, without regard to how you choose. Even taking the god's-eye-view of all the possibilities, as we did above with {{5, 10}, {10, 20}}, there is no case in which you stand both to gain and to lose.

You may still think it's rational to assume there is. That is, on drawing 4, to assume the envelopes might very well be {4, 8} rather than {2, 4}, and even to assume the chances of {2, 4} and {4, 8} are equal.

That's a lot of assuming. (And it will convince you trade your 4, which is a mistake.) You could instead recognize that all of your choices are conditional on my choice: my choice determines how much is to be gained or lost; your choice determines whether you gain or lose. There are some cases where you can guess whether you have the lower value or the higher, but that's just a guess. (If you draw a 6, do you know for a fact that the envelopes aren't {3, 6}? Of course not. I may have chosen {3, 6} just on the chance that you wouldn't expect me to include any odd numbers.)

So what is the rational expectation for the other envelope, given that I have chosen and given that you have chosen? There is no chance left once we've both chosen, though there is knowledge on my side and ignorance on yours. Does the other envelope contain either half or twice the amount in yours? Yes, of course. Are there non-zero chances of both? No. Should you assume there are anyway? No. You should recognize that I have fixed the amount you will gain or lose by switching; you cannot know whether you chose the larger or the smaller, so you cannot know whether you will gain or lose that fixed amount by switching, so there is no reason either to switch or to stick.

(Note also that we get here without assigning P(A) or P(B) or P(any letter) a value. I choose, then you choose. That's it.)

EDIT: Table typos.

It's just that I would go further that I think he does, and reject the notion of a picture as a model that is distinct from reality. — Banno

A picture is a fact, and thus part of reality, part of the world.

It still feels to me like we're circling around the difference between

P(picking larger)

and

P(I picked larger | I picked)

All of us agree the first is just 1/2.** But the second is troublesome. Once you've picked, you definitely have the larger or the smaller, but you don't know which. It might be safe to continue to treat this the same as just the chance of picking larger, so long as you don't use the observed value of what you picked. But if you want to use the observed value, you have to be very careful to avoid saying things that amount to there being a 1/2 chance that 10 > 20.


** Although maybe it needn't be. Suppose you actually had data on individuals picking, and one individual is consistently "lucky". We don't need to know why or how, but we could still say this individual's chance of picking the larger is better than average.

knowledge is (ontologically) mental phenomena based upon experience — numberjohnny5

But a super special kind of mental phenomena. If you want to pick out some of your beliefs and call them "knowledge", you do that by saying something about the connection between those beliefs, the mental phenomena, and the content of those beliefs, what the beliefs are about, and what the beliefs are about is not (necessarily) mental.

The point is that, in the correct version for your calculation E=($V/2)*P1 + ($2V)*P2, the probability P1 is not the probability of picking the larger value. It is the probability of picking the larger value, given that the larger value is $10. In my example, that 90%. In the OP, you do not have information that will allow you to say what it is. — JeffJo

This is absolutely right. I think the confusion comes when you switch from

E(other) = (larger)P(picked smaller) + (smaller)P(picked larger)

where the probabilities of picking smaller and larger are equal, to

E(other | picked = a) = (2a)P(picked smaller | picked = a) + (a/2)P(picked larger | picked = a)

because it's tempting to think these conditional probabilities are equal, just like the unconditional probabilities above, but this we do not know.

(Philosophical aside: I think this is close to my concern that there is a difference between "There's a 1/2 chance of my picking the smaller envelope" and "There's a 1/2 chance that the value of the envelope I picked is the smaller.")

What is true is that

P(picked smaller | smaller = c) = P(picked larger | smaller = c) = 1/2

but that's completely different.

But averaged over all possible values of V, there will be no expected gain. — JeffJo

I would still like to know more about how this works, though it may be over my head.

If the initial set up calls for randomly assigning values for the two envelopes in the finite range ((1,2),(2,4),(4,8)) for instance, then, in that case, assuming the player knows this to be the initial set up (and hence uses it as his prior) then the posterior probability conditionally on observing any value of M that isn't either 1 or 8 (that is, conditionally on values 2 or 4 being observed) p will indeed be 1/2. — Pierre-Normand

In one sense, yes, because we can say E(N | M=a) = (3*E(p) +1)/2, where p = P(S=a | M=a).

But how do we calculate E(p)? I think the player in your example can, but can a player with a lot less information than yours?

The equiprobability that I am talking about is the posterior equiprobability between the two possible contents of the second envelope: either X/2 or 2*X. — Pierre-Normand

Okay, tell me if I'm doing this wrong.

Let S be the smaller of the two values, M be your envelope, and N the other. This is certainly true:

$\small \begin{align} E(N) &= (2S)P(M=S)+(S)P(M=2S)\end{align}$

And so is this:

$\small \begin{align} E(N\mid M=a) &= (2S)P(M=S\mid M=a)+(S)P(M=2S\mid M=a)\end{align}$

Let $\small p=P(S=a\mid M=a)$; then $\small 1-p=P(S=\cfrac{a}{2}\mid M=a)$. Now we can say this:

$\small \begin{align} E(N\mid M=a) &=2ap+{a\over 2}(1-p) \\&= a\cfrac{3p+1}{2}\end{align}$
Possible values of p:

$\small \begin{align} p&=1\to E(N\mid M=a)=2a\\p&=0\to E(N\mid M=a)=\frac{a}{2} \end{align}$

So far so good. But we cannot do this:

$\small p=\cfrac12\to E(N\mid M=a)=\cfrac{5a}{4}$

anymore than we can do this:

$\small p=\cfrac13\to E(N\mid M=a)=a$

There are only two possible values for N, given an observed value for M, and the only two values for p that produce possible values for N are 0 and 1.

How am I to interpret that result?

I agree with all of those.

*** You might have (3) and (4) a little wrong but I can't judge. The McDonnell & Abbott paper makes noises about the player using Cover's strategy having no knowledge of the PDF of X.

One interesting point about the Arbitrary Cutoff strategy is that Never Switch and Always Switch can be seen as the degenerate cases: Never sets the cutoff to 0; Always to, well, "infinity".

(Btw, good find on the article, @Jeremiah. Addressed my bafflement over how assigning variables leads to trouble.)

The formula just explains why the gain from the strategy is 0.25; the expected value of the other envelope is 5Y/4. — Michael

Not all Sometimes Switch strategies produce an expected gain of Y/4.

None of them calculate their expected gain using your formula.

If that expected value calculation is correct, then Always Switch should produce the expected gain, shouldn't it?

What, in that formula, suggests that a Sometimes Switch strategy is correct?

But that argument, that "calculation", is not based on using any particular strategy. It's just this:

E(U)=.5(2Y) + .5(Y/2)

Do you believe that the success of the various switching strategies available shows that this expectation is correct?

There are Sometimes Switch strategies that work, so far as I can tell.

Do you believe that shows that your original argument, which concludes that the value of whatever envelope you don't have is 1/4 more than the one you do have, is valid?

Unfortunately, we don't (and can't) know the probabilities that remain. For some values of v, it may be that you gain by switching; but then for some others, you must lose. The average over all possible values of v is no gain or loss.

What you did, was assume Pr(X=v/2) = Pr(X=v) for every value of v. That can never be true. — JeffJo

My first post in this thread three weeks ago:

You're right that seeing $2 tells you the possibilities are {1,2} and {2,4}. But on what basis would you conclude that about half the time a participant sees $2 they are in {1,2}, and half the time they are in {2,4}? That is the step that needs to be justified. — Srap Tasmaner

But I still need help with this.

Yesterday I posted this and then took it down:

$\small \begin{align} O(X=a:X=\frac{a}{2}\mid Y=a)&=O(X=a:X=\frac{a}{2})\frac{P(Y=a\mid X=a)}{P(Y=a\mid X=\frac{a}{2})} \\&=O(X=a:X=\frac{a}{2}) \end{align}$

Bayes's rule in odds form, which shows that knowing the value of the envelope selected (Y), provides no information at all that could tell you whether you're in a [a/2, a] situation or [a, 2a].

I took it down because the Always Switcher is fine with this, but then proceeds to treat all the possible values as part of one big sample space, and then to apply the principle of indifference. This is the step that you claim is illegitimate, yes? Not the enlarging of the sample space.

Essentially, we include the impossible values that may come up in calculations in the range, and make them impossible in the probability distribution. — JeffJo

Is this the approach that makes it all work?

I kept thinking that Michael's mistake was assuming the sample space includes values it doesn't. (That is, upon seeing Y=a, you know that a or a/2 is in the sample space for X, but you don't know that they both are.) But I could never quite figure out how to justify this -- and that's because it's a mistaken approach?

Unfortunately, we don't (and can't) know the probabilities that remain. For some values of v, it may be that you gain by switching; but then for some others, you must lose. The average over all possible values of v is no gain or loss. — JeffJo

Right and that's what I saw above -- the odds X=a:X=a/2 are still whatever they are, and still unknown.

Your last step, averaging across all values of V, I'm just trusting you on. Stuff I don't know yet. Can you sketch in how you handle the probability distribution for V?

If the game is iterated, so that you can accumulate data about the sample space and its probability distribution, then it's an interesting but completely different problem. We can still talk about strategies in the non-iterative case.

The three choices are:

1. Never Switch,
2. Always Switch, and
3. Sometimes Switch.

There is some evidence that Sometimes Switch increases your expected gain. (I've played around a tiny bit with this, and the results were all over the place but almost always positive. I don't really know how to simulate this.) That's interesting but not quite the "puzzle" here.

If you were trying to find a strategy that is nearly indistinguishable from Never Switch over a large number of trials (not iterated, just repeated), then it would be Always Switch. If, for instance, you fix the value of X, so that there is a single pair of envelopes used over and over again, then we're just talking about coin flips. A Never Switch strategy might result in HTHHTHTT... while Always Switch would be THTTHTHH... and it couldn't be more obvious they'll end up equivalent.

So then the puzzle is what to do about the Always Switch argument, which appears to show that given any value for an envelope you can expect the other envelope to be worth 1/4 more, so over a large number of trials you should realize a gain by always switching. This is patently false, so the puzzle is to figure out what's wrong with the argument.

The closest thing I have to answer is this:
If you define the values of the envelopes as X and 2X for some unknown X, you're fine.
If you want instead to use variables for the value of the envelope you selected, Y, and the value of the one you didn't, U, you can do that. But if you want to eliminate one of them, so you can calculate an expectation only in terms of Y or U, you have to know which one is larger. (Note that there is no ambiguity with X and 2X.) Which one is larger you do not and cannot know.

some knowledge of the bounded probability distribution of the possible contents of the two envelopes — Pierre-Normand

I'm having trouble imagining what the source of this knowledge might be.

(Never mind.)

it seems pretty relevant to me. — Moliere

Absolutely. In fact, since posting it occurs to me that the concept of "lying" belongs to one level -- the person level, where we hold individuals responsible for their words -- while "self deception" belongs to another level, where we try to understand how we and others think.

I think that's probably right, but we've become so sophisticated that now we hold people responsible for fooling themselves. Which is not completely crazy -- as I said above, I think there are related norms in play here. Both lying and self-deception are violations; they're just not exactly the same violation of exactly the same norm.

what would make this singular self picture a better picture than a split self picture? — Moliere

For some purposes we ignore what's going on under the hood. You, the single individual person, are responsible for what you say, and for the consequences of your decisions. Looking under the hood provides a more nuanced description, but it's really changing the subject.

If you want your qualitative experience to be the foundation of your knowledge, then I think you need to be able to say something like this eventually:

(F) Because I have experience of A, I have knowledge of B.

The question is how to fill in A and B, and whether one determines the other, so that (F) -- is true? must be true? If is the latter, what's the nature of that necessity?

Lets say I'm at a singles bar looking for a date, and I know that statistically my chances of being successful are low — VagabondSpectre

This is a point that should have been made earlier. Beliefs are almost always best thought of as partial, as confidences. You believe you're unlikely to be successful and that you have a chance of being successful. Alcohol either suppresses the former completely, or just futzes with the numbers, so that your chances look better with every drink. You're not going from believing P to believing ~P or something, because you believe both, partially, from the start. Typical self-deception is deliberately mis-calibrating your confidences.

What's wrong with saying that they believed nothing was wrong, but after all night passing without change in their condition, they began to believe that something was wrong. That is, they changed their belief, as compared/contrasted to misrepresenting it to themselves. — creativesoul

Because that happens too, and it's a different phenomenon. What you're missing is that self-deception is usually strongly motivated and irrational.

Here's a salient example: relationships. Self-deception often involves manipulation of evidence, but when it comes to figuring out what are other people think and feel, a lot of that evidence is subtle and ephemeral. We're good at picking up on these tiny tells, almost unnoticeable variations in inflection, expression, eye movement and focus, tone of voice -- all of that stuff we process without usually being consciously aware of it. We just know.

My point is this: it's particularly easy to get away with fooling yourself in this context because your "judgment" was arrived at automatically based on "evidence" you probably couldn't articulate. And that makes it all too easy to dismiss. You don't want to believe something's bothering your spouse? No problem: there's not much you could really point to as evidence anyway. (It was just a feeling you had.) But anyone who's ever done this knows they were fooling themselves.

Or, from the other side, want to believe that cute girl in your homeroom, or at work, or making your coffee, is into you? You can probably find something to count as "evidence". For most of us, enough contrary evidence arrives and quickly enough that a restraining order is unnecessary.

Are the quality of an experience and its content related?

having this experience qua this experience cannot be reasonably refuted, even if what my experience is of cannot be known absolutely known (that is, even though I can be certain I am having an experience, it's possible that the content of my experience is false — numberjohnny5

Your experience has a quality and a content: the quality you know ("know"?) infallibly, but the content -- maybe not infallibly? Maybe not at all? Are you sure there's a foundation for knowledge here?

My god, that's brutal. I had forgotten.

Chomsky said somewhere that his life's work was organized around two complementary problems that he called "Plato's Problem" and "Orwell's Problem". Plato's problem is: How do we know so much, given so little evidence? While Orwell's Problem is: How do we know so little, given so much evidence?

It is humanly impossible to knowingly believe a falsehood. — creativesoul

I daresay you haven't had much practice. When I was your age, I always did it for half-an-hour a day. Why, sometimes I've believed as many as six falsehoods before breakfast.

They're nearly two sides of the same coin, the same virtue and the same fault: falling to aim for truth in what you say, failing to aim for truth in what you think.

Aren't you shifting the focus from what counts as belief to what counts as acting rationally? — creativesoul

I'm insisting on another aspect of the connection between beliefs and actions. Banno has talked about our use of "belief" in giving post facto explanations of our behavior.

I'm just pointing out that we also say, if you want to find your keys and you think they're in the kitchen, you should want to look for your keys in the kitchen, and anyone who didn't must either have some good reason not to or they just don't think the way we do.

I don't see anything on offer that can take up the role of belief (or expectation, or something else in this neighborhood) in such judgments.

Yes, I understand that as far as you're concerned the phrase "lying to yourself" is just a contradiction. But it's a phrase we all use, so what are the options?

• It's just an idiom and the word "lying" is not meant literally.
• People do literally deceive themselves, even though you, and maybe none of us, don't quite understand how that's possible.
• Your criterion for lying is too narrow and leaves out this case and perhaps others, like "lying by omission".

Stage magic and storytelling both include techniques that rely on our capacity for self-deception. Sometimes the magician, instead of trying to hide how a trick is done, can get the audience members themselves to dismiss the solution, and this is much more effective. A movie can present a character that's a little "off" but not make a big deal about it, and the viewers will mostly decide not to worry about him, until the third reel when it turns out he's the killer.

You could say these are cases of deception, but really it's just giving us the opportunity to deceive ourselves and most of us are generally quite prepared to do so.

All else being equal. It is, as I keep saying, a question of our norms of rationality. In my first presentation of the lost keys, I put a man with a knife in my kitchen. Then it's rational not to look for my keys just then.

I remember hearing years ago that it's common for emergency rooms to have a spike in admissions just before dawn. The explanation was people lying awake all night telling themselves "It's nothing" and eventually accepting that something was terribly wrong.

Are you really not familiar with any of these phenomena?

I would have thought that things like "Pp" and "PPp" were our symbols — Banno

So how do you read 3.1431?

Let me put it a different way.

If you can show that we can, using the perhaps unknown value of our selected envelope, get the same answer we get by ignoring that value and doing some simple algebra, then that would count as a solution to the two envelopes problem. So you're offering a solution and I'm really not.

We haven't actually been talking about solutions much in this thread because some participants did not accept that the simple algebraic solution is actually right.

If, at that point, you postulate a value in an envelope, you need to postulate a probability distribution that covers all possible ways that value could be in an envelope. Even if it is unknown. — JeffJo

I'm just trying to understand the "need to" in that sentence.

Did you miss the part where I said — JeffJo

Don't be that guy. I'm reading your posts and asking about what I don't understand.

(Btw, MathJax is available here, so it's possible to make your equations more readable.)

There is an easier way, but it applies only if you don't know what is in your envelope. — JeffJo

If, even before selecting an envelope, you see that there is no reason to prefer one envelope over the other, what compels you to discard that analysis upon selecting and opening an envelope? Is it no longer true that if you drew X you'd trade for 2X and if you drew 2X you'd trade for X? How could selecting or even looking in an envelope make that false?

Can you give an example of how you'd analyze a typical knowledge claim? Anything about that analysis that makes you uncomfortable?

But confabulation is a little of each.

You know how it's impossible to walk any great distance** if your stride with one leg differs slightly from your stride with the other? Now tell yourself at each step that it's only a little different, and that can't make much difference. It's like that: you relax your cognitive standard just a bit, and indeed it does not make the inferential step you're taking invalid, but if you keep compounding this little compromise you end up in the wrong place. I'd call this a kind of lying to yourself and it's incredibly pervasive.

** in a straight line

Do you have numbers you could share? I've always assumed post deletion and thread deletion were pretty rare, a tiny fraction of the posts submitted and threads created.

It's curious that Plush offers a member reputation system (which I understand there was a decision not to use) but no ability to upvote and downvote threads, which could allow moderators to police policy instead of quality.

How many deletions are for low quality rather than forbidden content? Do we also get a lot of spam?

I guess I'm just wondering if we're really under siege in our little corner of the internet. I remember Usenet becoming, well, unusable, and the internet's response was tools that allowed communities to self-police a bit. Here we're forced to rely on (so far as we know) human moderators. I very much agree with what @TheWillowOfDarkness posted, but I still think allowing posters a way to freely self-sort could be helpful, and it looks like the only tool we have that might reduce the workload of the moderators, whatever that is.

Oh, and "expressed" can be glossed as "becoming perceptible by the senses," like a propositional sign. That's cool.