This is why I refuse to use betting arguments.

Suppose that SB gets paid $1 if the coin lands tails, otherwise she must pay $1. Furthermore, suppose that before the experiment begins she is given the choice as to whether or not she will have amnesia during the course of the experiment. According to thirder reasoning, she should choose to have amnesia in order to raise the probability of tails to 2/3 — sime

Is/does she paid/pay this $1 on both days, or on Wednesday after the experiment is over? In the latter case, can she choose not to have amnesia, and then choose "Heads" if she recalls no other waling but change that to Tails if she does?

This is why I tried to use a different setup, where this illogical juxtaposition of two different states is not an issue.

+++++

Probability is not a property of the system, it is a property of what is known about the system. Say I draw a card. After I look at it, I tell Andy that it is a black card, Betty that it is a spade, Cindy that its value is less than 10, and David that it is a seven (all separately). I ask each what they think the probability is that it is the Seven of Spades. Andy says 1/26, Betty says 1/13, Cindy says 1/32, and David says 1/4. All are right, but that does not affect my draw. I had a 1/52 chance to draw it.

The thirder argument is that SB knows of only three possible states. To her knowledge, there is no functional difference between them, so each has the same probability. Since they must sum to 1, each is 1/3. The fault they find with the halfer argument, is that it implies that to her, Monday&Tails and Tuesday&Tails represent the same state of possible knowledge. And they don't. They represent the same future, as determined on Sunday, but not the same state on a waking day.

Given these four possible experimental runs following the four possible initial coin flip results, we find that when Sleeping Beauty awakens, she can certainly rule out HH as the current state of the two coins during that specific awakening episode. However, this does not eliminate the possibility of being in either of the last two experimental runs (in addition to, of course, either of the first two). — Pierre-Normand

The points of the variation are:

• There is an "internal" probability experiment that begins when the researchers look at the two coins, and ends when either they see HH or they put SB back to sleep after seeing anything else.
• The effect of the amnesia drug not just that no information is transmitted between "this" SB and a possible version of herself in another internal experiment. It is that there are two such internal experiments in the full experiment, and each is a well-defined, self-contained experiment unto itself.
• The question in the internal probability experiment is about the state of the coins during it, not whether it is the first or second possible internal experiment. So the issue of "indexical 'today'" is completely irrelevant.

But as you point out, halfers are very good at making up reasons for why the answer they intuitively believe must be correct, could be. The "indexical" argument is one. Your example, where they might try to argue that SB must consider the version of herself in the other internal experiment is another - I've seen it used in the classical version.

So, if you need, make one slight change to mine. Always wake SB. If the coins are showing HH, take her on a shopping spree at Saks Fifth Avenue. If not, ask the question. This gets around your suggestion, since there are always two wakings. It is being in a "question" waking that allows an update: Pr(HH|Q)=0 and Pr(HT|Q)=Pr(TH|Q)=Pr(TT|Q)=1/3.

Now ask yourself whether it matters, to the reasons why these updates are possible, if the researchers had simply said "something other than a waking question will happen." The update is valid because SB has observed that the consequence of HH did not happen, regardless of what that consequence could be.

+++++

When I tell you that the card I drew is a Spade, 39 of 52 outcomes are "eliminated" as possibilities. So you can update Pr(Ace of Spades)=1/52 to Pr(Ace of Spades|Spade)=1/13. But "Diamonds" are still a part of the sample space, and their (prior) probabilities are still used for that calculation.

The difficulty with the classic version of the SB Problem arises when we try to physically remove one possibility from the sample space, not just eliminate it from the current instance of the process. That can't be done when the two "internal" parts of the whole problem are different, but SB has to view them as the same. That is the motivation for my variation, to make the internal parts identical in the knowledge basis used o answer the question.

Being left asleep does not remove Tuesday+Heads from whatever sample space you think is appropriate. It is still a possibility, and being awake constitutes an observation that it did not happen.

No, the thirder answer is not based on "who am I?" That appears to be just an excuse to reject the logic. In fact, in that post you linked, that question is neither asked nor answered. But there are equivalent ways to get the right answer, with only one SB.

Before I describe one, I need to point out that what most people think is the Sleeping Beauty Problem is actually a more contrived modification of it, invented by Adam Elga to enable his thirder solution. And the controversy is entirely about the parts he added, not the original.

This is the problem as it first ever appeared publicly, in the paper "Self-locating belief and the Sleeping Beauty problem". The two modifications alter words that clearly indicate Elga was thinking about his solution, since they have no impact on this text:

Some researchers are going to put you to sleep. During the [time] that your sleep will last, they will briefly wake you up either once or twice, depending on the toss of a fair coin (Heads: once; Tails: twice). After each waking, they will put you to back to sleep with a drug that makes you forget that waking. [While you are awake], to what degree ought you believe that the outcome of the coin toss is Heads? — Adam Elga

Elga made the two wakings occur on Monday and Tuesday, and made them different by having Monday be a mandatory waking, and Tuesday's be optional. The first issue you must face, is to decide whether you think Elga's changes alter the correct response.

But there is another way to implement the original problem, as worded, where there is an easy answer. After putting SB to sleep, flip two coins. Call them C1 and C2. If ether is showing Tails, wake SB and ask her "to what degree do you believe that coin C1 is showing Heads?"

After she answers, put her back to sleep with amnesia and turn coin C2 over to show its other side. Then repeat the steps in the previous paragraph, starting with "If either...".

The issue with the SB problem, is whether to consider the two potential wakings as the same experiment, or different ones. This version resolves that. SB knows that when the researchers looked at the coins, there are four possible arrangements with probability 1/4 each: {HH, HT, TH, TT}. She also knows that, since she is awake, HH is eliminated. She can update her beliefs in the other three to 1/3 each.

Look, we've got your point, Jeffjo. — ssu

No, I really don't think you do. Or at least, you have shown no evidence of it.

The point of the evolution you misinterpret is to determine if the set T is internally consistent. — JeffJo

No. It's the inconsistency between two or more axioms in the axiomatic system, which make the system inconsistent.

And how is that not what I said?

But my point, that you have not shown you understand, is that finding such an inconsistency does not mean any of these two-or-more axioms is false.

There is a thing called the philosophy of mathematics and there are various schools of thought in philosophy of math, you know.

And it is that there are no pre-determined truths, only truths that follow from one's axioms which are assumed to be true without proof.

And in just what category would you put your idea presented here btw

It is a statement about philosophy, not a statement in math. "True" statments in math are either axioms, or theorems that follow from axioms. Unlike what you want here:

Axiom of Infinity is anything but established and self-evidently true. The discussion here ought to show it. — ssu

Even if math follows it's own logic (no pun intended), it's still something that people do and it does evolve. — ssu

Nobody has said otherwise. (Well, other than "what people do" is completely ambiguous.)

What was said, is that Math accepts no absolute truths.The entire point is that there should be no need to discuss what is, or is not, self-evidently true.

Math says "If the statements in the set of axioms A are accepted as true, then the statements in the set of theorems T follow logically from them. The point of the evolution you misinterpret is to determine if the set T is internally consistent. If it is not, then the set A is invalid as a set of Axioms. But no one axiom in A is invalid, and any one of them can be in another set A' that is consistent.

And notice the word "could". Could doesn't have the same meaning as is. I've only said it could be a possibility that in the future it is shown to be inconsistent. — ssu

Did you notice the word "could" also? Anything "could" happen.The sun could explode tomorrow, ending life as we know it. Do you discuss that possibility? I personally am not even considering emptying my bank account, to use it before I lose it. mathematics is not about what "could" be true, it is about what is shown to be true within a given framework of axioms.

And did you understand that finding that a ****SET**** of axioms is inconsistent does not mean that any one axiom in the set is "false"? Or that claiming that an axiom could be false is an oxymoron?

I don't blame the axiom, in my view Infinity (and hence an axiom for it) is an integral part of mathematics. All I've said that we haven't understood infinity well.

And all I've said is that this is a nonsensical statement. You, on the other hand, said:

Axiom of Infinity is anything but established and self-evidently true. The discussion here ought to show it. — ssu

That would be a pointless discussion. An axiom is not, and cannot be, inherently "self-evidently true." We cannot "prove" it, and no amount of discussion will shed any light on it. It is because it cannot be shown to be self-evidently true, or false, that we assume it is self-evidently true. So we can lay the groundwork for a specific field of mathematics.

we have gotten new insights on mathematics in history and our understanding of math has greatly changed from what it was during Ancient times and what it is now.

The insights you refer to all apply within a particular set of axiom, The only insights on mathematics in general that we have gained, are how to demonstrate that a field is internally consistent, and that there is no such thing as truth *in* a field from supposed truths *outside of* that field.

No, the set of axioms are inconsistent when they aren't consistent with each other. You don't compare two different axiomatic systems to each other. — ssu

Which is what I have been saying. When the set of axioms lead to an inconsistency, it is the set is that is inconsistent. No one axiom is inconsistent, or false. Nor is any one axiom inconsistent with another. The set itself is inconsistent.

And I never compared two systems to each other. I listed three different, consistent sets that include three contradictory versions of the parallel postulate. The point wasn't to compare the sets, it was to show that none of these parallel postulates could be called "true" or "false." But I'm beginning to suspect you know this, and are deliberately arguing around the point.

And you still have not demonstrated an inconsistency with Zermelo–Fraenkel set theory, You have supposed it could be inconsistent, and blamed it on the Axiom of Infinity possibly being false. Which is preposterous.

At least I'm trying to understand your point. — ssu

There is no evidence of it.

What I think, is that you only try to see what contradicts your predetermined idea of universal truth. And then you say the first thing that comes to your mind, that seems to imply what I said as wrong. Example:

That geometry is different in two dimensions and more dimensions is evident yes. Yet we do speak of Geometry, even when there is Euclidean and non-Euclidean geometry.

This is a classic example of a strawman argument.

I described three different examples of two-dimensional geometry. They can be expanded into more dimensions. Yet I mentioned no such higher dimensions, and their existence is completely irrelevant. There are ***THREE*** ***DIFFERENT***, mutually contradictory versions of ***PLANE*** geometry. Each is a consistent field of mathematics. And the point is that there is no such thing as this universal, capital-G Geometry as you imply. Just consistent fields of geometry based on different sets of axioms. One axiom in particular is different in each field, making the other two false in that field. Yet none of them is true, or false, outside of a set of axioms that describes it.

If you truly are trying to understand my point, understand that one. And then try to see that your claim is preposterous, because it says the opposite. The claim that the Axiom of Infinity can be considered to be true, or false, outside of a field of mathematics that has either accepts it as true, or as false, without justification or proof.

So how much do you do with "inconsistent" axiomatic systems, or as you wrote, "a *****SET***** of Axioms" that is inconsistent?

?????
I don't "do" any quantity of whatever it is you are implying with them. I also don't suppose that they could be inconsistent because they contradict a "universal truth" that I want to others to accept as blindly as you do, and dismiss an individual Axiom in the set solely on the basis of that unsupported supposition. Which is exactly what you are doing.

So changing the axioms isn't changing the way think about math? — ssu

That's the first thing you've gotten right. And the fact that you will disagree is why you won't ever understand what I am saying.

Axioms don't define "the way we think about math." They define areas ("fields") within the framework of "how we think about math." There are three different fields that use contradictory Parallel Axioms (hyperbolic, elliptic, and Euclidean geometry), yet the way we think about them in math is the same. Each field is valid, and each Parallel Axiom is true within its field, and false in the others.. Each becomes a false statement (not an Axiom) in the others.

So are against something the idea that if something is inconsistent (in math/logic),it is false,...

I can repeat this as often as you ignore it, but I'm running out of ways to make it sound different from what you've ignored before. You keep using the indexical word "something" without indicating what it refers to, the statement or the set. Some part of what you say is clearly wrong each time you use the word, but how it is wrong depends on what you mean. And if you understand the difference.

Even if you prove that a ****SET**** of Axioms is inconsistent, that does not mean that any one Axiom in it is "false." And a ****SET**** can't be false, just inconsistent.

Qualities a *****SET***** of Axioms can have include "consistent" and "inconsistent," but not "true" or "false."

Qualities a ****STATEMENT**** can have include "true in the context of a ****SET**** of Axioms," and "false in the context of a ****SET**** of Axioms," but not "true outside the context of a ****SET**** of Axioms," of "false outside the context of a ****SET**** of Axioms."

An Axiom is a special kind of statement that assumed to be true to define the ****SET***. The only quality an Axiom can have is ""true because we assume it in this ****SET**** of Axioms." Outside the context of any ****SET**** - that is, in the absolute or universal sense you deny you are using - it is neither true nor false. It is merely a set of words.

Quite circular reasoning you have there, Jeffjo. — ssu

Do think you understand the point of Axioms? Maybe you need to explain what you think it is. Because it is your arguments that are circular.

The axiom of infinity could be wrong in the way that it is inconsistent with the other axioms of ZF, for example.

And Santa Claus could visit my house tomorrow night. But I don't draw conclusions from suppositions like that.

It is you that is making the case of some eternal truth ...

You are the one suggesting that statements could be called true, or false, outside of an axiomatic system. All I'm saying th that the AoI can be part of a consistent system, and you can't conclude anything about "Infinity" outside of one.

I'm really not making the case for some universal truth here either.

Yes, you are.

My point is that from the historical perspective we have thought about math one way and because of new theorems or observations we have changed our way of thinking about math.

No, we have not. We may have changed the Axioms.

All I understand is that if something is inconsistent, we can say it's false.

Define what "something" represents here. Because an Axiom, by itself, cannot be this "something" here yet youy keep treating it as though it can.

A ****SET***** of axioms can be inconsistent, which only means that at least one of them disagrees with one or more of the others. Not that any of them is "false." And claiming otherwise is claiming that a universal truth exists.

This is a straw-man argument. — ssu

It is how Mathematics works. Anything that "exists" has to be based on Axioms.

we don't understand Infinity yet clearly.

Now that's a strawman argument. You need the AoI before you can even try to understand this thing you want to call "infinity."

No, the axioms are inconsistent to each other in the defined axiomatic system.

And you have proven this? Or are you just supposing it could be so?

As I said: "I'm not looking for some ultimate truth.

Yes, you did say that. You have also said that the AoI could be "wrong" and that we need to discuss whether it is.These statements contradict each other. This makes your axiomatic system inconsistent, and "false" by your definition.

The question is if a set of axioms, an axiomatic system, is simply consistent. I just happen to be such a logicist that I think that something that is inconsistent in math is in other words false.

Not ultimately false, or absolutely false, but some other kind of "false"? What kind?

Perhaps you didn't understand my point. — ssu

It's clear you don't understand mine. Nor have you tried.

The question is if a set of axioms, an axiomatic system, is simply consistent.

Yes, it is. That is exactly what you have not addressed.

If they aren't consistent, I would in my mind declare then an axiom or axioms to be false

And you would be wrong to do so. All it shows is that the set is inconsistent. Any of the axioms could individually be part of a different, consistent, set. Yet you are calling an axiom, or axioms, "false" in a sense that can only be called "ultimate" or "absolute."

Get this point straight: The Axiom of Infinity cannot be proven to be true, or false, outside of some set of Axioms. I believe your words were that that his discussion should establish whether the AoI is self-evidently true. Nothing is further from the point if this discussion.

Given a consistent set theory that accepts the existence of an infinite set, we require different sizes of such sets. And not believing in infinity cannot change that.

Besides, one shouldn't assume that one school of Mathematical philosophy is correct and another is not. — ssu

And you still haven't grasped the very simple fact that no field of mathematics claims to be "correct", or that another is not. Only that no statement is can be shown to be true without first assuming a set of unsupported Axiom, and proving theorems within that framework.

And it is quite clear that you have no interest in any formalism but your own.

An axiom is a statement - statements are true or false. End of story. — Devans99

Not the end of the story. Definitions are not commutative. An axiom is indeed a "proposition regarded as self-evidently true without proof." That does not mean that every "proposition regarded as self-evidently true without proof" is an axiom.

Looking further at Wolfram, an[ i]Axiomatic System[/i] is a "logical system which possesses an explicitly stated set of axioms from which theorems can be derived." So once again, the statement you claimed was an axiom was never stated as part of such a set, from which theorems could be derived. It was not an axiom, it was a near-religious belief.

Yet what you are stating is a philosophical view of mathematics.

No. What I am saying is that theorems in a field of mathematics need to be based on some set of accepted truths that are called the axioms of that field. Such a set can be demonstrated to be invalid as a set by deriving a contradiction from then, but not by comparing them to other so-called "truths" that you choose to call "self-evident."

What I am explicitly saying is that any arguments not based in the axiomatic system of a field of mathematics cannot not say anything about that field.

I already gave an example of what was thought to be an axiom that wasn't. — ssu

You gave an example of a near-religious belief. It was never an axiom in a consistent mathematics.This is similar to the belief that we can't treat aleph0 as a valid mathematical concept.

And I gave you an example where three different, consistent mathematics have been created based on three different axioms that contradict each other. The point is that neither mathematics, nor the axioms any specific field of mathematics is based on, are intended to represent absolute truth. You cannot win this debate with a nonsense claim like "the axiom is wrong."

My point was that axioms can be possibly false. — ssu

My point is that they can't. That's why they are axioms.

There are no absolute truths in Mathematics, only the concepts we choose to accept as true. While we can find that a set of axioms is not self-consistent, that does not make any of them untrue.

Best example is from Euclid, who first proposed an axiomatic mathematics. He thought that he should be able to prove a self-evident fact from his axioms in plane geometry. That given a line and a point not on that line, then there must exist exactly one line in the plane they define that passes through the point and is parallel to the line. It turns out that this needs to be another axiom. And that you can define consistent geometries if that axiom says there is one, many, or none.

Axioms don't need to be self-evident; in fact, that's what requires them to be axioms.

Yeah well, it can happen that something that we have taken as an axiom isn't actually true. — ssu

With all due respect, if you want "actual truth", then you do not understand the purpose of an axiom in mathematics. The point is that mathematics contains no concept of actual truth. We define different sets of "truths" that we accept without justification. A set is invalid only if it leads to internal inconsistency, not if you think it violates an "actual truth" that is not in that set.

If we accept as true that there exists a set that contains the number 1 and, if it contains the number n then it also contains the number n+1? Then we can prove the existence of the cardinalities we name aleph0, alpeh1, aleph2, etc. Any inconsistencies you may think you have found are due to things you assume are actual truth but are not included in this mathematics.

Just look at how much debate here in this forum there is about infinity. Axiom of Infinity is anything but established and self-evidently true. The discussion here ought to show it. We just don't know yet! Bizarre to think that there are these gaping holes in our understanding of math, but they are there.

What is bizarre is that some don't understand that everything proven in mathematics is based on unproven, and unproveable, axioms.

I wouldn't agree on this. Axioms don't give proofs. Perhaps we are just thinking of this a bit differently. — ssu

I'm sorry, I worded that poorly.

We don't establish the existence of these sets by proof. We do it by the axioms we choose to accept. And since all proofs in mathematics are based on such axioms, there is no difference in the validity of either method.

I think most people understand the reductio ad absurdum proof. What the big problem is what then? — ssu

The problem is that CDA isn't a reductio ad absurdum proof; at least not as people think. The common presentation of it as reductio fails logically. As I described.

It also wasn't intended to be the focus of Cantor;s effort. It was a demonstration of the principle with an explicit set. What is known as Cantor's Theorem uses an abstract set, and proves that its power set must have a greater cardinality. (And, btw, it is a correct use of reductio.)

And we don't prove existence. Axioms do. The Axiom of Infinity establishes that the set of all natural numbers exists. The Axiom of Power Set says the power set of any existing set also exists. With those axioms, Cantor's Theorem proves that an infinite number of Alephs exist.

Just to be clear I will reiterate the proof in slightly more detail

We will map the number 5.7 ... — Umonsarmon

Note that this is rational number; more specifically, a rational number whose proper-form denominator is equal to (2^n)*(5*m) for integers n and m. The important characteristic is that there is a finite number of digits on each side of the decimal point.

... For arguments sake we will say the line terminates here at point E

Now I measure the distance from A to E ... This distance will be some multiple of a 1/2 x some a/b

But what if the process does not terminate? Then this won't always be true.

If I understand your poorly-described algorithm, what you really get is a terminating sum that looks like this, for a set of increasing integers {n1,n2,n3,...,nk}:

2^(-n1)+2^(-n2)+2^(-n3)+...2^(-nk)

If that isn't exactly right, it's something similar. Yes, this is a rational number.

But if the set of integers does not end, it does not have to be rational. And if the number you trace is irrational, your algorithm will not terminate. So no, you did not disprove Cantor.

+++++

A lot of people feel there is something wrong with Cantor's Diagonal Argument. That's because what they were taught was an invalid version of CDA. Some of the issues are just semantics (ex: it didn't use real numbers), and some are correctable (if you use numbers, you need to account for real numbers that have two decimal representations).

But one invalidates what they were taught as a proof. The correct proof is right, though. Here's a rough outline:

• Let t represent any infinite-length binary string. Examples are "1111....", "0000...", and "101010...".
• Let T be the set of all such strings t.
• Let S be any subset of T that can be put into the list s1, s2, s3, ... .
• Use diagonalization to construct a string s0 that is in T but is not in S.

This proves that every listable portion of T is not all of T.

Most of the times CDA is taught, it is assumed that all of T is put into a list. Then, by proving that there is an element that is not in the list, it contradicts that you have a complete list. Thus proving by contradiction that all of T cannot be listed.

This is an invalid proof by contradiction. You have to use all aspects of the assumption to derive the contradiction. The derivation does not use the assumption of a complete list, so it cannot disprove completeness by contradiction. But it did disprove it directly.

What seems to have started the confusion was the second part of Cantor's version. If S=T, then s0 both is, and is not, an element of T. This is a proper proof by contradiction.

Bingo.

Also just a note but probability theory IS a stats course. — Jeremiah

Statistics is a branch of applied mathematics that uses probability theory to analyze, and draw inferences from, data. That's why probability theory is a course required in a statistics curriculum.

This doesn't make it a "stats course", any more than arithmetic is a "calculus course." Even if you do add, subtract, multiply, and divide in calculus.

Nothing in this thread requires concepts that were not taught in your probability theory course, and nothing taught outside of it is appropriate here. So please, stop being petty.

There can be a statistical analysis even without data. There can also be a well defined Bayesian prior. In fact I could exactly define a sigma, mu and range for Y that could be used as a well justified prior. — Jeremiah

Then please, show us one that applies to the problem. And explain how it is a "well justified" anything, and not just a hypothetical.

I'm not saying that for every value in the chosen envelope it is equally likely that the other envelope contains twice as much as half as much. I'm saying that, given the £10 in my envelope, I will assume that the other envelope is equally likely to contain £20 as £5. I accept that there is an upper (and lower) bound. I'm just assuming that my £10 isn't it. — Michael

You seem to think that it is only the highest-possible v where you have an expected loss. Maybe you are confused by the fact that it was the easiest example that shows it.

It isn't so. Each value you see in your envelope can have a different value of Q. (Recall that the expectation is v*(2-3Q/2).) So "assuming my £10 isn't [the highest] " does not justify "assum[ing] that the other envelope is equally likely to contain £20 as 5."

I just want to note that we seem to be in agreement on everything. The only reason why we seemingly disagreed in our recent exchange is because you objected to my stated requirement that the game be conceived as a "real world" problem, — Pierre-Normand

No, I agree it has to be constrained to operate in the real world. That's why there has to be a real-world maximum value, you can't have an arbitrarily-small real-world value, and you can't choose a uniformly-distributed integer in the range (-inf,inf) (which is not the same thing as choosing a uniformly-distributed real number in [0,1) and back-calculating a gaussian random value from it).

What I'm saying is that there is no real-world component present in the OP. You can use real-world examples to illustrate some properties, but that is all you can do. Illustrate. The OP itself is purely theoretical.

All you can do conclusively, is determine what set of properties must apply to any possible real-world distribution that describes the problem, apply the laws of probability to it in a manner that is consistent with the OP, and then draw general conclusions.

Ignoring continuous distributions (which can be made discrete by considering the range A<=v<2A as the same outcome), any instance of the game can be described by the random variable X representing the smaller value in the envelopes (its outcome needs another, for which gets picked). The values x come from the set {x1,x2,x3,...xn}, where x1<x2<x4<...<xn. There is a probability function Pr(*), where Pr(xi)>=0 for all 1<=i<=n, and sum(Pr(xi), i=1 to n)=1.

From this we can prove:

• When you don't know your value v, or consider it to be fixed unknown, the expected gain from switching is 0.
• When you do know your value v, or consider it to be fixed unknown, the expected gain from switching is v*(2-3Q/2), where Q is a function of Pr(v/2) and Pr(v).
  
  • We may not know what the Pr's are, and so we also do not know Q, but we can say that 0<=Q<=1. So the expected gain, given v, is between -v/2 and +v.
  • If v=t1, then Q=0. We may not know what t1 is, but we know there is one.
  • If v=tn, then Q=1. We may not know what tn is, but we know there is one.
  • Exp(v*(2-3Q/2))=0

For Michael: It is true that in any game, the potential gain is bigger than the potential loss. But the possibility of, say, +$12 is always counterbalanced by the possibility of -$12 with the exact same possibility.

THIS IS ALL WE CAN CONCLUDE ABOUT THE OP. There can be no statistical analysis, which includes Bayesian Inference, because they require a population of games played in the real world.

I know it's not necessarily equally likely. But we're assuming it, hence why Srap said we should just flip a coin. — Michael

It cannot be equally likely without postulating a benefactor with (A) an infinite supply of money, (B) the capability to give you an arbitrarily-small amount of money, and (C) a way to select a random number uniformly from the set of all integers from -inf to inf.

All three of which are impossible.

But the reason you should reject the solution you use is because it is not a correctly-formed expectation. You are using the probability of picking the smaller value, where you should use the probability that the pair of values is (v,2v) *AND* you picked the smaller, given that you picked v.

What you are saying is correct in any case (most cases?) where the prior probability distribution of the envelope values isn't unbounded and uniform. In the case where it is, then there is no inconsistency. — Pierre-Normand

But it can't be unbounded and uniform. So it is inconsistent in all possible cases.

[Exp(other) = (v/2)*Pr(picked higher) + (2v)*Pr(picked lower)] is correct in the special case where the prior distribution is uniform and unbounded, — Pierre-Normand

What you are saying, is that if you postulate a distribution where (yes, I did reverse it) Pr(picked higher)=Pr(picked higher|V=v) and Pr(picked lower)=Pr(picked lower|V=v), then the results of the two conceptually-different formulas are the same. What I am saying is that one is conceptually incorrect, and one is conceptually correct. And I keep repeating this, because it is the error people make in the TEP when they perceive a paradox.

We're given a choice between envelopes valued unequally at a and b. We won't know which one we picked. The expected value of switching is

(1/2)(a−b)+(1/2)(b−a)=0

...

Isn't...this true whichever of a and b is larger — Srap Tasmaner

Certainly. It is a complicated way of saying that, before you choose, the expected values of the two envelopes are the same. There is even a simpler way to get there: the total amount of money is (a+b), so the expected value of either envelope is (a+b)/2.

But this applies only if we don't look inside one.

If we do look, and see v, then we are considering two possible pairs of envelopes, not the one pair you described. The pair is either (a,v) or (v,b), where a<v<b. Now there are two random choices - one for which pair was picked, and one for which envelope in the pair was picked.

Let Q be the probability that the pair was (a,v), so the probability that it was (v,b) is (1-Q). Before we look:

1. There is a Q/2 probability that our envelope contains a.
2. There is a Q/2 + (1-Q)/2 = 1/2 probability that our envelope contains v. But this breaks down into
   
   1. A Q/2 probability that we have the high envelope, and it is v.
   2. A (1-Q)/2 probability that we have the low envelope, and it is v.
3. There is a (1-Q)/2 probability that our envelope contains value b.

If our envelope has a, we are in case 1. If our envelope has b, we are in case 3. But if it has v, we know that we are in case 2 BUT IT COULD BE case 2.1. or case 2.2. This still works out as a 50:50 chance that we picked high or low before we look; that is, cases 1 and 2.2 add up to 1/2, as do cases 2.1and 3.

But what if we look?

• If we see a, there is a 100% chance we picked low.
• If we see b, there is a 100% chance we picked high.
• If we see v, the chance is Q that we picked high, and (1-Q) that we picked low. We get this by dividing the probabilities in 2.1 and 2.2, by the one in 2.

The expectation for the other envelope, given that one contains v, is a*Q + b*(1-Q). If a=v/2 and b=2v, this reduces to 2v-3vQ/2. And finally, it is only if Q=1/2 that it reduced further to 5v/4.

That's fine with me. In that case, one must be open to embracing both horns of the dilemma, and realize that there being an unconditional expectation 1.25v for switching, whatever value v one might find in the first envelope, isn't logically inconsistent with ... — Pierre-Normand

But it isn't logically consistent. With anything. That's what I keep trying to say over and over.

1.25v is based on the demonstrably-false assumption that Pr(X=v/2)=Pr(X=v) regardless of what v is. It's like saying that the hypotenuse of every right triangle is 5 because, if the legs were 3 and 4, the hypotenuse would be 5.

• Exp(other) = (v/2)*Pr(picked higher) + (2v)*Pr(picked lower) is a mathematically incorrect formula, because it uses the probabilities of the wrong events.
• Exp(other) = (v/2)*Pr(V=v|picked higher) + (2v)*Pr(V=v|picked lower) is the mathematically correct formula, because it uses the probabilities of the correct events.

The only thing we need to understand to "embrace the dilemma," is why this is so. It isn't simply that one is conditional and one is unconditional, it is that the events used in the first do not represent a value.

It has to do with the sorts of inferences that are warranted on the ground of the assumption that the player still "doesn't know" whether her envelope is or isn't the largest one whatever value she finds in it. — Pierre-Normand

And since the OP does not include information relating to this, it does not reside in this "real world."

Also even without a sample distribution a theoretical model can still be set up. — Jeremiah

The sample distribution of a statistic is the distribution of that statistic, considered as a random variable, when derived from a random sample of size n. Since we have no such sampling, let alone a statistic, there is no sample distribution, or use for one. Period. This just isn't a statistics problem.

A theoretical model of the probability problem can be set up, just not as a statistics problem. But doing so does not address the OP, where we have no information that would allow us to set one up.

The only things we can say about the OP are:

• If you don't look in the envelope, the only valid solutions mentioned in this thread consider one of three functionally equivalent random variables: the smaller value X, the difference D which is equal to X, or the total of the two envelopes T which is equal to 3X.
  
  • A pedant would insist you need to include one probability from the probability distribution of whichever you choose. But it divides out so it isn't necessary in practice.
  • The answer is that the expected value of your envelope is (x)/2 + (2x)/2 = 3x/2, and the other is (2x)/2 + (x)/2 = 3x/2. So switching changes nothing.
• If you look and see value v, you need two probabilities from that distribution: Pr(X=x/2) and Pr(X=x).
  
  • These values are not only completely unknown, they ...
    
    • ... are beyond the scope of Bayesian Inference.
    • ... are beyond the scope of sampling.
    • ... are beyond the scope of anything anybody can contribute here.
    • ... have no "sigma."
  • The only point in mentioning them, is that the expectation calculation requires both.
  • The expectation for the other envelope is v*[Pr(X=x/2)/2 + 2*Pr(X=2x)]/[Pr(X=x/2) + Pr(X=2x)].
  • This is 5v/4 if, and only if, we know that Pr(X=x/2) = Pr(X=2x).
  • This must be greater than v for at least one value of x.
  • This must be less than v for at least one value of x.
  • The expectation of this formula, over the range of V, is the same as the expectation of v over that range. See "If you don't look ...".

A normal distribution does not have to have a mid of 0, nor do they need negative values. — Jeremiah

A normal distribution refers to a random variable whose range is (-inf,inf), and is continuous. The first cannot apply to the TEP, and the second is impractical.

This is the part I'm still struggling with a bit. — Srap Tasmaner

I think this illustrates the issue you are struggling with:

• Say you have a perfectly-balanced cube with the numbers "1" thru "6" painted ion the sides. If you roll it, what are chances that an odd, or even, number ends up on top? Answer: 50% each.
• Say you have an unquantifiable blob of plastic, with an indeterminate set of numbers painted in apparently random places. If you roll it, what are same chances? Answer: "I don't know."

In the second case, if you had to bet on "odd" or "even," you might flip a coin and so have a 50:50 chance of either. That's as good an option as it gets. But that doesn't mean you expect the wager be fair.

There are still exactly two possibilities, but that doesn't mean the chances are the same for each. The Principle of Indifference requires that you make some assessment about the equivalence of the outcomes.

If you look in your envelope and see $10, the question in the OP requires you to make an assessment about expectation. You don't have any information that allows you to do so. And it makes no sense to talk about a "prior" - which actually refers to something else - in this case. The only point in doing so, is if you have the means to update it.

"A random variable is defined by a real world function"

That's a bit like saying that a geometrical circle is defined by a real world cheese wheel. — Pierre-Normand

I'm not sure what "real world" has to do with anything. But...

Probability theory does not tell us how to define outcomes. The outcomes of a coin toss could be called {"Heads", "Tails", "Edge"} or {"Win", "Lose", "Draw"}. But you can't use those in an expectation calculation, can you?

So measure-theoretic Probability Theory requires a way to express outcomes with numbers. So its strict definition of a "random variable" is a function whose argument is an outcome, in whatever form you choose to use, but whose result is a number. So G() might be the function for your gain for a bet on Heads, so G("Heads")=1, G("Tails")=-1, and G("Edge")=0.

Personally, I prefer to call the qualities I use to describe outcomes "random variables." You can still think of them as functions whose arguments are abstract outcomes,whether or not the results are numbers. Especially in problems like the OP, where there is no need to be so formal, and no "real world" significance whatsoever.

I am not so sure about that. — Pierre-Normand

And I'm sure I intended to have a "not" in there somewhere. I'll fix it.

I believe there is not a paradox here but a fallacy. — Srap Tasmaner

Exactly.

Maybe I need to explain Simpson's "Paradox." It is a very similar, not-paradoxical fallacy. It just seems to be a paradox if you use probability naively.

Say there is a fatal disease that 1 in 1,000 have, without showing symptoms. But there is a test for it that is 99.9% accurate. That means it will give a false positive with probability 0.1%, and a false negative with probability 0.1%. Sounds very accurate, right?

Say you are one of 1,000,000 people who take the test. You test positive, and think that there is a 99.9% chance that you have the disease. But...

• 1,000 of these people have the disease.
  
  • 999 of them will test positive.
  • 1 of them will test negative.
• 999,000 of these people do not have the disease.
  
  • 998,001 of them will test negative.
  • 999 of them will test positive.

So fully half of the people who test positive - 999 out of 1,998 - do not have the disease. Yes, you should be worried, but not to the degree suggested by the 99.9% accuracy. The good news is that if you test negative, there is only a 0.0001% chance you have it.

The fallacy is confusing the probability of a correct result in a specific circumstance, with the probability that a specific result is correct. It sounds like the two should be the same thing, but they are not. The latter is influenced by the population who take the test. The 99.9% false negative figure applies only to the population that has the disease, not to the population who test positive. The 99.9% false positive figure applies to the population that does not have the disease, not to the population who test negative.

The exact same thing happens in the TEP. The fallacy is confusing the 50% chance to pick the lower value, with a 50% chance that a specific value is the lower.

Suppose, again, that a die throwing game will be played only once (i.e., there will be only one throw) with a die chosen at random between two oppositely biased dice as earlier described. ... — Pierre-Normand

I know we are reaching an equivalent conclusion. My point is that the framework that it fits into may be different. These concepts can seem ambiguous to many, which is the fuel Bayesians, Frequentists, Subjectivists, Objectivists, Statisticians, and Probablists use to denigrate each other through misrepresentation.

My point was that the difference between "only once" and "many times" has no significance in this discussion. It can only have meaning to a statistician who is (correctly, don't think I am putting them down) trying to create a population through repetition of the game, from which he can use inference to refine his estimates of the properties of your dice.

Probability models what is unknown about a system. At the start of your dice game, we don't know which die will be rolled, or how it will land. After step 1, the gamemaster knows which, but the player does not. Since their knowledge of the system is different, they have different probability spaces. The gamemaster says, for example, that 1 is more likely than 6. The game player says they are equally likely. Both are right, within their knowledge.

Now, what if the 1-biased die is red, and the 6-biased one is green. If the player doesn't know this, only that the two biases exist, his knowledge that he has the red die does not put him in the gamemaster's position.

In the OP, and before we look in the envelope, we are in the role of the game player. The probability that the low value of the envelopes is x is the determined by the distribution of the random variable we have called X. (I'm trying to get away from calling this "initial," since that has other connotations in Probability's derivative fields.)

That distribution is an unknown function F1(x). After picking high/low with 50:50 probability, the value in our envelope is a new random variable V. Its distribution is another unknown function F2(v), but we do know something about it. Probability theory tells us that F2(v) = [F1(v)+F1(2v)]/2. But it also tells us that the distribution of the "other" envelope, random variable Y, is F3(y) = [F1(y)+F1(2y)]/2. Y is, of course, not independent of V. The point is that it isn't F3(v/2)=F3(v)=1/2, either.

Looking in the envelope [correction:] not does change our role from that of the game player, to the gamemaster. Just like seeing the color of your die does not. Simply "knowing" v (and I use quotes because "treat it as an unknown" really means "treat it as if you know the value is v, where v can be any *single* value in the range of V") does not change increase our knowledge in any way.

Yes, it is tempting to say that if the game is only being played once then the shape of the initial distribution(*) isn't relevant to the definition of the player's known problem space. That's a fair point, and I may not have been sufficiently sensitive to it. — Pierre-Normand

No, it isn't fair to say that. No more than saying that the probability of heads is different for a single flip of a random coin, than for the flips of 100 random coins

I'll get to a more qualitative solution to the problem at the end of this post. I hope it helps.

This makes no sense to me. Initial distribution of what? If these are pairs of envelopes from which will be chosen the pair that the player confronts, then not only is this sample space unknown to the player, she never interacts with it. She will face the pair chosen and no other. — Srap Tasmaner

Define "interact."

Say I reach into both of my pockets, pull out a coin in each hand, but show you just one. It's a quarter, worth $0.25. (According to the US mint, their current mint set includes coins worth $0.01, $0.05, $0.10, $0.25, $0.50, and $1.00.) I'll give you either the coin you see, or the one you don't. What do you choose?

Notice I didn't say that I ever carry half-dollar or dollar coins. They are in the current mint set, but most Americans are unfamiliar with them so they won't carry them unless they were just given out as change.

In what way do you not "interact with" with the distribution of coins I carry? If I keep nothing but pennies in one pocket, does that fact not "interact with" what you will get if you take the hidden coin? Even if you do not know the distribution?

How about if I received $9.50 in dollar and half-dollar coins as change for a $10 bill from a vending machine (about the only way I get them), and put them in a different pocket so I can avoid carrying them tomorrow? Does that fact not "interact with" what you will get if you take the hidden coin? Even if you do not know the distribution?

In both the TEP, and this example, the distribution is unknown. You can't "interact" with any specific realization of a distribution, but it most definitely "interacts" with what will happen if you choose the hidden envelope or coin. That's why the answer to OP is:

• If you don't open your envelope, your information about the two is equivalent: you know nothing about either. So your expectation for either has to be the same. You have no idea what "the same" might mean, but you can conclude that there is no difference.
• If you open yours, you do have information about both. But it takes different form for each. You know the value of your envelope, and the two possibilities for the other.
• You need to know how these two different kinds of information interact with each other in order to use them. Specifically, the 50:50 split applies only before you looked. After you look, the split depends on the prior chances of two specific combinations.

Except that you cannot, and you know that you cannot. — Srap Tasmaner

What you "cannot" do is assign probabilities to the cases. You can still - and my point is "must" - treat them as random variables, which have unknown probability distributions.