My long-promised take on pictures will happen tomorrow. I'm leaning toward pushing on into 3 pretty quickly too. Stuff there that rounds out what we've been working at.

The confusion only gets clarified about this whole world/reality thing in propositions 5.6, and so on, with the limits of my language being the limits of my world and so on. — Posty McPostface

Cool, so there's only about two thirds of the book between where we are now and that.

You can't use Y as a value defined independently of X — Snakes Alive

Something is wrong with the way Y is being used, clearly, but I'm not sure this is it. You can get the (b) method out of the (a) method just by substituting the possible value of Y in terms of X. (And you can fix (b), if you start mistakenly there, by substituting to get (a).)

The thread seem to be going slowly — Banno

This is a Good Thing™.

The point being that atomism is explicitly rejected in the investigations. That has got to have some relevance. — Banno

<shrug> Whitman wrecked his early work by revising it. Later doesn't equal better.

I'm not sure we're far enough along to begin some Grand Appraisal. It's a strange book. Hard to tell what's premise and what's conclusion, or, maybe better, what's a speculation and what a consequence of that speculation.

I intend to continue trying to take it on its own terms. Insofar as I have an agenda, that's it.

It would be a start.

I don't know what you're on about it, but it looks off topic to me.

can the Tractatus be understood apart from the Investigations? — Banno

My working assumption is that Wittgenstein understood it when he wrote it, so yes, it is possible.

I should not wish to be spared the trouble of thinking.

Cool, thanks.

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. — andrewk

I'm still not seeing it. We've done to death the example of finding £10 and you calculate an expected gain of £2.50. Or you can do a generic calculation and show an expected gain of £Y/4, for any value of Y.

In general, it's best to avoid arguments based on 'absurdity — andrewk

Then let's go with "inconsistency": given such a pair of envelopes, you can readily construct a pair of arguments that tell you each has a higher expected value than the other. I would prefer finding that such arguments are invalid.

Thanks, bro.

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.

Post what you've tried so far, pointing out what you're not sure about.

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).

One little bit of off-the-cuff chitchat

What I find exciting about the Tractatus is the intimation that everything there is to say about the world is immanent in stating facts. This is hard to express clearly, but it's as if he anticipates Gödel, Tarski, etc. There is no extra bit for "true" or "provable" or "constructibe" or even "member of a class"; there are only the atomic facts. All such ways of describing how things stand are immanent in saying how things stand with the atomic facts themselves.

Hey Posty. (Been busy and what time I've had here has gone to the damn two envelopes paradox, but I have not forgotten about TLP.)

States of affairs are combinations or an amalgamate of atomic facts in logical space and are observer dependent, that are denoted by an observer creating a reality of their own. — Posty McPostface

Maybe this will turn out to be right, but I just don't think it's in what we've read so far. The possible realities are built in, there from the start. What you're talking about is picking one. As far as props 1-2 are concerned, we're still just establishing what representation is, how it works, how it's possible.

Here's one thing I keep thinking about: can we think "state of affairs" as always short for "state of affairs in logical space"?

A state of affairs is a function defined on logical space that assigns the value obtains or the value doesn't obtain to possible atomic facts, the elements of logical space. But there's an oddity here: must such a function be defined over the entirety of logical space? Why not just some subspace? When we consider pictures, it is inconceivable that a picture would present how things (could) stand in all of logical space; a picture presents how things (could) stand in some subspace of logical space.

Here's an analogy. Given a deck of cards, either the ace of spades is on top or it isn't. If you define a state as [ace of spades on top], that picks out an equivalence class of many possible states of the deck, in each of which the ace of spades is on top, but with the other 51 cards distributed in all the other possible ways. You have the option here of saying [ace of spades on top] is a complete description of part of the deck, or a partial description of the complete deck.

Which gets us back to my question. Which way you go could matter to you, epistemically, but if it matters to LW he hasn't said yet. For instance, objects contain within themselves all possible ways things could stand in logical space; looked at from object-side, there are only complete realities, and in each there are atomic facts this object could be part of that obtain or don't. Or start with atomic facts: each divides logical space into those states of affairs in which it obtains, and those in which it doesn't, and there is somewhere a pair in which all other atomic facts have the same value.

We get "world" for all obtaining atomic facts; "reality" for all obtaining and not obtaining atomic facts; I think it turns out "state of affairs" is kept around for its useful ambiguity: it covers the case where you only have a subspace defined, the case where only the positive facts are defined, and the case where absolutely everything is defined.

I would add this: the extreme realism of the TLP suggests that every partial state of affairs, up to and including the partial state of affairs that is the world, is one and only one complete state of affairs, one reality, whether you know it or not. We, picture makers, only ever deal with complete realities, but we always fail to completely specify them.

I cannot bring a world quite round,
Although I patch it as I can.

The short answer is that logical equivalence is just a matter of truth value, which in turn is just a matter of extension. All the other nuances of language are deliberately left out.

I'm going to keep thinking about your approach, and I hope you will as well. As things stand, the game theory approach has been the only game in town, so it would be nice to have an alternative.

Really glad you posted this.

Oh yes, everyone decides. Didn't mean to suggest otherwise. But each holds a preference for participating only if all the others participate, and rightly so. It is not rational to hunt stag solo.

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.

The person who makes the first move grants the other the ability to decide, putting himself under that persons power. — Bliss

That's not bad, but you make the first move with an expectation, possibly mistaken, that the other will reciprocate. An even better way to put this is that you only make the first move if you believe the other wants you to make the first move.

With the stag hunt, the idea is that each is willing to participate if all the others participate. No one is willing to try and hunt a stag alone. Your choice of what to do depends on what you expect others will do; what they will do depends on what they expect you will do. There are mutual expectations and conditional decisions binding all to each and each to all. —When it works, and no one goes chasing a hare.

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.)

Are you sure that was me? It doesn't sound familiar, but maybe I've just forgotten. Can you link the exchange? — andrewk

What in the rules of the game or in the knowledge you have acquired (i.e., that Y = 10) assures you that the games mistress includes both 5 and 10 in the sample space for X? Maybe a tenner is the smallest bill she has. — Srap Tasmaner

It's not about the games mistress's sample space. She doesn't have one. She has two amounts that she has always intended to put into the envelopes. It's about my sample space, which consists of the events that I cannot rule out based on my knowledge to date.

That knowledge - based on seeing 10 in my envelope - allows me to rule out every other event (an event being a particular sum being in the other envelope) except for 5 and 20. So that's my sample space - just two events. — andrewk

[Y, Y/2]. — Michael

That's $\small \{ \cfrac{y}{2},2y\}$.

This can't be written as B = 1/2 (x + 2x) without conflating different values of x. — Michael

You just write it as $\small B = \cfrac{\cfrac{y}{2} + 2y}{2}$.

$\small y$ is whatever's in the envelope in your hand, whether you've looked at it or not. Looking makes no difference.

The point about the maximum possible amount has been made before, but has been lost in the length of the thread. I mentioned it somewhere early on, and in the interests of simplicity, declined to probe further because (1) I thought digging into it would have required some fancy notation like integral signs, and I couldn't remember how to invoke latex on this site and (2) I was pretty confident it would not make a material difference. — andrewk

And I asked you last night about the minimum possible value of X. Your response then was that what matters is not what might or might not be put in the envelope, but whether you the player have reason to exclude a given value.

Yes, and such debates are meaningless. — Pseudonym

I like now & then to quote Ramsey:

I think we realize too little how often our arguments are of the form:-- A.: "I went to Grantchester this afternoon." B: "No I didn't."

Suppose the envelopes on offer contain $10 and $20. You pick one.

If your envelope contains $10, then you will be offered the chance to buy the remaining envelope for $10. To you this looks like an envelope with an average value of $12.50. You swap and get $20.

If your envelope contains $20, then you will be offered the chance to buy the remaining envelope for $20. To you this looks like an envelope with an average value of $25. You swap and get $10.

This is very strange. @Snakes Alive noted that your initial choice is a matter of indifference because whichever envelope you pick, there's a story you can tell yourself to justify swapping. (Half these stories are damned lies.) @BlueBanana noted that if two players take an envelope each, they'll both want to trade.

I see three possibilities here:

(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.

(2) @Michael and @andrewk are right about the expected value of the other envelope, and this is a genuine paradox. (What they're wrong about is thinking swapping works, since it's really obvious that it does not and cannot.)

(3) @Michael and @andrewk have part of the right analysis, but something more is needed.

(2) has to be avoided at all cost, I'd say.

I must to bed. Cheers.

That's more answer than I'm asking for. What in the rules of the game or in the knowledge you have acquired (i.e., that Y = 10) assures you that the games mistress includes both 5 and 10 in the sample space for X? Maybe a tenner is the smallest bill she has.

I just don't understand how you know that.

Sorry, yes, I mean Y = 10 → . X = 5 ∨ X = 10.

You know that at least one of 5 and 10 are possible values of X. Do you know that both are?

One of them must be a possible value of X. Not necessarily both.

the two possibilities that are open to me — andrewk

What does that mean? You know one of them must be in the distribution of X, but you don't know which. Are you claiming to know that both are in the distribution?

What is the probability that the envelopes presented to you are valued at 5 and 10? (P(X = 5).)

What is the conditional probability that the envelopes presented to you are valued at 5 and 10, given that you chose the 10? (P(X = 5 | Y = 10).)

That's what you need to calculate. How will you do it?

You have written a program with repeated plays that shows it is best to switch, and Srap appears to have written one that is only slightly different and shows that it makes no difference. — andrewk

They are completely different. Michael is allowing participants access to the results of the trials so far performed. This is completely different, and of no relevance to this thread.

X does not vary. You are wrong from the start.

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

My version of Michael's sim was intended to be quite simple. I imagined it as the question being offered once to 2 x 10^6 participants, half of whom switch and half of whom don't. Then we aggregate the results to compare the two strategies.

What's wrong with this?

You are presented with two envelopes, one valued at X and one valued at 2X; the average value of an envelope is 3X/2.

You choose an envelope, and do not look at the contents. You are asked if you would like to swap.

Your envelope has a definite value, call it Y.

If you have the larger of the two envelopes, the value of the other envelope is Y/2; the average value of an envelope is then (Y + Y/2)/2 = 3Y/4.

If you have the smaller of the two envelopes, the value of the other envelope is 2Y; the average value of an envelope is then (Y + 2Y)/2 = 3Y/2.

It is absurd that the average value should change depending on which of the two envelopes you have, therefore they must be equal:

3Y/2 = 3Y/4
6Y = 3Y
6 = 3

Hmmm. Let's go back.

If you have the larger of the two envelopes, then

• Y = 2X;
• the other envelope has Y/2 = X;
• the average value of an envelope is 3Y/4 = 3X/2.

If you have the smaller of the two envelopes, then

• Y = X;
• the other envelope has 2Y = 2X;
• the average value of an envelope is 3Y/2 = 3X/2.

Since Y is defined to be whatever is in the envelope you selected, then Y has a different value depending on which envelope you selected. It is not a constant, as X is. And therefore you cannot, after all, set the average values equal to each other: the LHS Y has one value and the RHS Y has another. X ≠ 2X.

Neither can you make this calculation:
If I have the larger valued envelope, I risk losing only Y/2 by switching, while if I have the smaller valued envelope I stand to gain Y. Y has two different values in this sentence. By definition.


Now suppose you are allowed to look, so you learn, say, that Y = 10. Won't that prevent the problem of Y changing values? It will be 10 in every equation.

Suppose Y = 10. Then either 10 = X, or 10 = 2X. Where before we had equations with two unknowns, now we have equations with one.

Suppose you reason as follows:
If I have the larger envelope, then the other has 5 and the average value of an envelope is 7.5.
If I have the smaller, then the other has 20 and the average value of an envelope is 15.

(It is prima facie absurd that the average value of an envelope changes depending on whether you have the larger or the smaller of the two.)

The only way this can be is if our only remaining unknown, X, changes its value depending on whether I have the larger or the smaller envelope. Since we have fixed Y, which was defined to change depending on your choice, we are forced to make X vary with your choice in order to preserve the equations. But X is a constant. Unknown, but a constant.


So what is the right way to reason, once you know that Y = 10? The simple answer is, don't. You haven't learned anything you can act on. It will turn out X = 5 or X = 10, but Y = 10 does not help you figure that out, and there is no way to calculate using Y = 10 that does not force X to vary with your choice of envelope, which is absurd.

Here's a new version that also calculates the average payouts as ratios to X.