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.

n the case of the two envelopes paradox, the case is similar. The player never has the opportunity to chose which branch to take at the first node. So, the player must treat this bifurcation as occurring within a black box, as it were, and assign each branch some probability. But, unlike my example with two equally biased dice, those probabilities are unknown. — Pierre-Normand

I like this explanation. And I thought of a possibly better way explain how "unknown" is used in the TEP, by analogy:

• Say you are given a geometry problem where one angle is represented by the unknown "x", and you are asked to solve for another angle "y."
• Your answer will be expressed in functional form, something like y(x) = 60° + x + arctan(sin(x)/2).
• The unknown "x" can be any angle in some range, and is more properly called an independent variable.
• It is true that the value of "y" is not known. But it is not an "unknown," it is a dependent variable that has a very specific relationship to x. That relationship is determined by the details of the OP and the laws of geometry.

Even if you treat the independent random variable V, for the value v in your envelope, (or the smaller value x of the pair) "as an unknown," the value y in the other envelope is still represented by the dependent random variable Y that has a very specific relationship to X, as determined by the OP and the laws of probability.

The point is that "as an unknown" is a description of your knowledge, not the role played by a value in the problem.

My current, and I think "final", position is that this isn't really a probability puzzle at all. Here are my arguments for my view and against yours. — Srap Tasmaner

The puzzling part is about our understanding the mathematics, not how we use it to solve the problem. But that still makes it a probability problem. People who know only a little have difficulty understanding why the simple 5v/4 answer isn't right, and people who know more tend to over-think it, trying to derive more information from it that is there.

1. The only probability anyone has ever managed to assign any value to is the probability of choosing the larger (or the smaller) envelope -- and even that is only the simplest noninformative prior.

That's because the higher/lower question is the only one we can assign a probability to. There is on;ly one kind of probability that you can place on values. That's "valid," meaning there is a set of possibilities, and their probabilities sum to 1. Any other kind - frequentiest, bayesian, subjective, objective, informative, non-informative, or any other adjective you can think of - is outside teh scope of the problem.

2. All other probabilities used in solutions such as yours are introduced only to immediately say that we do not and cannot know what their values are.

Correct.

4. Much less the PDF on that space.[/quote
Careful. "PDF" usually refers to a "Probability Density Function," which means the sample space is continuous. We have a distribution for a discrete sample space.

Th only thing we can say about it (or the sample space ) is, is that it still must be valid. A valid sample space has a maximum value. A valid distribution implies there are values in the sample space where there is an expected loss.

5. By the time the player chooses, a value for X has been determined.

This is a red herring. It only has meaning if we know the distribution, and we don't. So it has no meaning.

6. We might also describe that as the host's choice of a value for X.

I assume you mean the amounts the benefactor puts in the envelopes (this isn't presented as a game show). That's why I usually speak generically about values. That can apply to the minimum value, which is usually what x refers to in this thread, the difference d which turns out to be the same thing as x but can be more intuitive, the value v in your envelope which can be x or 2x, and the total t (so x=t/3).

The point is that you do need to recognize how they act differently. Assuming you are "given" x means that there are to v's possible, and assuming that you are "given" v means there are two x's.

7. That choice is the very first step of the game and yet it appears nowhere in the probabilistic solutions, which in effect treat X as a function of the player's choices and what the player observes.

Then I'm not sure what you mean - it appears some of mine. If you are given v, and so have two x's, you have to consider the relative probabilities of those two x's.

10. The probabilistic model can safely be abandoned once it's determined that there will never be any evidence upon which to base a prior much less update one.

Please, get "updating" out of your mind here.

what is the advantage of saying that the variable X takes a value from an unknown and unknowable sample space, with an unknown and unknowable PDF, rather than saying X is not a variable but simply an unknown?

The point is that I'm saying both. You need to understand the various kinds of "variables."

• In probability (but not necessarily statistics - this one place where terminology can vary) an experiment is a procedure where the result is not predictable. Not an instance where you perform it. (In statistics, it can refer to repeating the procedure multiple times.)
• A random variable is an abstract concept only, for a measure you can apply to, and get a value from, every possible instance of the procedure. I represent it with an upper case letter like X.
• "Random Variable" and "X" do not technically refer to any actual result, although this definition gets blurred in practice. They represent potential only.
• So a random variable never strictly "has" a specific value. For a given random experiment, the possibilities are listed in a set called its range. So the range for X in our game could be something like {$5,$10,$20}.
• An unknown is a placeholder for a specific value of a instance of the procedure. I use lower case letters, like x.
• When we say X=x, what we mean is the event where the measure represented by X has value x.
• Since X never really has a value, we can use this expression only as the argument for a probability function. We sometimes use the shorthand Pr(x) instead of Pr(X=x), since with the upper/lower case convention it is implied that the unknown x is a value taken from the range of X.

In the now canonical example of Michael's £10, he could say either:

(a) the other envelope must contain £20 or £5, but he doesn't know which; or
(b) there's a "50:50 chance" the other envelope contains £20 or £5, and thus the other envelope is worth £12.50.

I say (a) is true and (b) is false.

(A) is true, and (B) cannot be determined as true or false without more information. We can say that there is an unknown probability 0<=q<=1 where the expectation is E=($5)*q + ($20)*(1-q) = $20-$15*q. Or in general, E(v,q(v)) = 2v-3*v*q(v)/2. (Note that q(v) means a function.)

This is not worthless information, because we can make some deductions about how q varies over the range of V. Specifically, we can say that there must be some values where E(v,q(v)) is less v, others where it must be greater that v, and that the sum of E(v,q(v))*Pr(v) is zero.

What compels us to say that it is probabilistic ...

The fact that you use an expectation formula.

There is an interesting distribution proposed at https://en.wikipedia.org/wiki/Two_envelopes_problem#Second_mathematical_variant . Note that, like all distributions discussed so far in this thread, it is a discrete distribution and not a continuous one. Continuous distributions tend to be messy, and not very realistic.

The envelopes are filled with ($1,$2) with probability 1/3, ($2,$4) with probability 2/9, ($4,$8) with probability 4/27, etc. Unless your envelope has $1, in which case your gain is $1, the expected value of the other envelope always 10% more than yours. But before you get too excited:

• You can't apply my first method above to switching back.
• Even though the expected value is at least (remember the $1 envelope?) 110% of yours, method #2 above is still correct. If you don't consider the value v in yours, the expected value of the two envelopes is the same.
• It is left as an exercise for the reader to determine how the expected value of the other is 110% your value, but the two expected values are the same.

I apologize to this forum for allowing myself to be taken off topic by a troll.
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The difficulty with the field of probability, is that there can be different ways to correctly address the same problem. Because there is no single sample space that describes a problem. Example:

• Rolling two six-sided dice can use a set of 11 outcomes for the sum, a set of 21 unordered pairs of values, or a set of 36 ordered pairs.
• Any of those can be used, but the last one allows you to easily apply the Principle of Indifference to get reasonable probabilities. This is because the PoI requires that we know the causes of the outcomes are all equivalent.

That example doesn't mean there can't be vastly different solution methods that both get the same answer. There can. You can use a different method than I do, and get the same correct answer.

The issues comes when two methods get different answers. If Jack says "I use method A and get answer X," while Jill says "I use method B and get answer Y," all we know for sure is that at least one is wrong. Bickering about why A is right does nothing to prove that it is, or that B is wrong.

To resolve the issue, Jill would need to do two things: find a flaw in A, and identify how B does not make the same mistake. The Two Envelope Problem is trivial once you understand, and apply, these points.

What is wrong with the 5/4 expectation: Any term in an expectation calculation has to be multiplied by a probability for the entirety of the event it represents.

• A term that represents your envelope containing a value v, and the other containing v/2, must be multiplied by probability that represents your envelope containing v, and the other containing v/2.
• A term that represents your envelope containing a value v, and the other containing 2v, must be multiplied by probability that represents your envelope containing v, and the other containing 2v.
• If you are considering v to be a fixed value, even if that value is unknown, then those two possibilities are different outcomes, and may have different probabilities
• Even though the chance of picking the lower or the higher is 50%, once you include a fixed value in the the entirety of the event, it can change.
• Specifically, the unconditional probability that v is in your envelope and v/2 is in the other, is Pr(v/2,v)/2, where Pr(x,y) is the probability that the pair of envelopes were filled with x and y.
• Similarly, the unconditional probability that v is in your envelope and 2v/is in the other, is Pr(v,2v)/2.
• To make these conditional probabilities, you divide each by their sum.
• This gives Pr(v/2,v)/[Pr(v/2,v)+Pr(v,2v)] and Pr(v,2v)/[Pr(v/2,v)+Pr(v,2v)], respectively.
• Check: they add up to 1 and, if Pr(v/2,v)=Pr(v,2v), each is 50%.
• The correct expectation by this method is v*[Pr(v/2,v)/2+2*Pr(v,2v)]/[Pr(v/2,v)+Pr(v,2v)].

This is a correct solution, *if* you know the value in your envelope (even a fixed unknown value v), *and* you know the two probabilities involved (which is much more difficult if v is unknown). For most conceivable distributions, there are even values of v where this correct solution produces a gain.

But the sticky issues of what the probabilities are is a big one. We can't use the PoI because the requirements I mentioned above are not met. If the supply of money is finite, then there must be some values of v where there is an expected loss, and the expected gain over the entire distribution of v will turn out to be 0.

The 5v/4 expectation applies this method, but ignores the required terms Pr(v/2,v) and Pr(v,2v). That is its mistake. It would be right if you know, or could assume, these terms are equal. In the OP, we can't assume anything about the distribution, rendering this method useless.

What is right with my calculation: Say the total in the two envelopes is t. Then one contains t/3, and the other contains 2t/3.

• The unconditional probability that your envelope contains t/3 is Pr(t/3,2t/3)/2. Notice that this is the exact same formula as before, with the modified values.
• The unconditional probability that your envelope contains 2t/3 is Pr(t/3,2t/3)/2.
• To make these conditional probabilities, you divide each by their sum.
• Since they are the same, this gives 50% each.
• The expectation is (t/3)/2 + (-t/3)/2 = 0.
• Even though we wound up not needing Pr(t/3,2t/3), including it made this solution more robust.

This method is robustly correct. Even though it uses a similar outline to the previous one, it applies to the general problem when you don't know t. Because the unknown probability divides out. And, it gives the intuitive result that switching shouldn't matter. Its only flaw, if you want to call it that, is that it does not apply if you know what is in your envelope - then you need to consider different t's.

I already figured out that your field was not statistics, — Jeremiah

And it is even more obvious you want to use statistics anywhere you can, no matter how inappropriate. The lexicon of both probability and statistics is the same, since statistics uses probability. It applies it to the experimental data you keep talking about, and of which we have none.

But if you feel I've mis-used terminology in any way, please point it out. I've pointed out plenty of yours, and you haven't refuted them.

Let's just look at our first interaction here:

You can't just enumerate a set of cases, and claim each is equally likely. If you could, there would be a 50% chance of winning, or losing, the lottery. — JeffJo

By this, I was clearly referring to the valid discrete sample space {"Win", "Lose"}. An event space is a sigma-algebra on this set, and a valid example is {{}, {"Win"}, {"Lose"}, {"Win","Lose"}}. A probability space is the 3-tuple of those two sets and a measure space defined on the event space. By Kolmogorov's axioms, it is valid if it has the form {0,Q,1-Q,1}, where 0<=Q<=1.

My statement above said that you can't simply apportion probability to the members of a valid sample space. It isn't that such a probability space is invalid in any way, it is because it is impractical,

Yet you replied:

That is a very bad understanding of what a sample space and an event is. — Jeremiah

Since my sample space was a perfectly valid sample space, and I never mentioned events at all, it demonstrates your "very bad understanding" of those terms. It was a very bad choice of a sample space for this problem, for the reasons I was trying to point out and stated quite clearly. But you apparently didn't read that.

You are not applying your Principle of Indifference there,

Actually, I was applying it. Improperly with the intent to demonstrate why its restriction is important:

The Principle of Indifference places a restriction on the possibilities that it applies to: they have to be indistinguishable except for their names. You can't just enumerate a set of cases, and claim each is equally likely. — JeffJo

You went on:

Furthermore, it makes no sense to use a probability density curve on this problem, — Jeremiah

I didn't say we should (A) use a probability (B) density (C) curve. I stated correctly that there (A) must be a probability (B) distribution for the (C) set of possible values, and that any expectation formula must take this distribution into account. Even if you don't know it.

The reason your solution in post #6 turns out to be correct, is that this probability turns out to have no effect on the calculation: it divides out. That doesn't make it incorrect to use it - in fact, it is technically incorrect to not use it.

The other way to phrase the difference is that my solution uses the same value for the chosen envelope — Michael

But your expectation uses the value in the other envelope, so this is an incomplete phrasing. That's why it is wrong.

It doesn’t make sense to consider those situations where the chosen envelope doesn’t contain 10.

And you are ignoring my comparison of two different ways we can know something about the values.

Again: if it is valid to use a method that considers one value only, while ignoring how that one value restricts the pair of values, then it is valid to use that method on either the value v in your envelope, or the value t of the total in both.That means that they both should get the same answer.

They don't. So it isn't valid. If you want to debate this further, find what is wrong with the solution using "t".

We should treat what we know as a constant and what we don’t know as a variable.

True. But when that variable is a random variable, we must consider the probability that the variable has the value we are using, at each point where we use one.

Statistics uses repeated observations of outcomes from a defined sample space, to make inference about the probability space associated with that sample space. — JeffJo

I just said that. That is exactly what I said. — Jeremiah

What you said was: "... uses repeated random events to make inference about an unknown distribution." Since an event is a set of possible outcomes, and you used the word to mean an experimental result, what you said had no meaning. The point is that it refers to a probability space itself, and not the actions that produce a result described by such a space.

Since this difference is critical to identifying the misinformation you have been spouting, I corrected it to what you thought you meant. Using correct terminology in place of your incorrect terminology. Which would not have been necessary if what you said meant exactly what I wrote.

What you refuse to understand, is that it is this misuse of terminology that I've been criticizing. Not your solution, which I have repeatedly said was correct.

I already posted the definition of an event from one of my books, which I will refer to you over. I will always go with my training over you.

You may well have. If you did, I accepted it as correct and have forgotten it. If you want to debate what it means, and why that isn't what you said above, "refer it to me over" again. Whatever that means.

But first, I suggest you read it, and try to understand why it was the wrong word to use in your sentence above.

One thing I was taught in my first stats class was that the lexicon was not standardized.

Maybe true in some cases. But "event" is not one of them. Look it up again, and compare it to what I said.

The first ones I found in a google search (plus one I trust) were:

• Wikipedia: "an event is a set of outcomes of an experiment (a subset of the sample space) to which a probability is assigned."
• mathisfun.com: When we say "Event" we mean one (or more) outcomes.
• icoashmath.com: "An Event is a one or more outcome of an experiment."
• ucdavis.edu: An event is a specific collection of sample points.
• intmath.com: Event: This is a subset of the sample space of an experiment.
• statisticshowto.com: An event is just a set of outcomes of an experiment
• faculty.math.illinois.edu: any subset E of the sample space S is called an event.
• mathworld.wolfram.com: An event is a certain subset of a probability space. Events are therefore collections of outcomes on which probabilities have been assigned.

None of these refer to an event as an actual instance of the process. It is always a way to describe the potential results, if you were to run it. As a potential result, there is no way to apply "repeat" to it.

Statistics is a data science and uses repeated random events to make inference about an unknown distribution. We don't have repeated random events, we have one event. Seems like a clear divide to me. You can't learn much of anything about an unknown distribution with just one event. — Jeremiah

Statistics uses repeated observations of outcomes from a defined sample space, to make inference about the probability space associated with that sample space.

In the OP, we don't have an observation. Not even "one event," if that term were correct to use in your context. Which it isn't.

With one exception, this is why every time you have used words like "statistics" or "statistical" in this thread (which is what I mean by "advocating" it; I know you never advocated it as a solution method, but you did advocate associating the word with distributions in the discussion) has been incorrect.

The exception is that you can use statistics to verify that your simulation correctly models a proven result. The repeated observations are the results of a single run of the simulation.
Why you think you should do that is beyond me, but apparently you did.

The OP deals with a conceptual probability problem. There is no no observational data possible. "Statistics" does not belong in any discussion about it. Nor does "Bayesian," "Frequentist," "objective," "subjective," "inference," and many others.

I'm not really sure that this addresses my main question. ... There's a 50% chance of picking the lower-value envelope, and so after having picked an envelope it's in an "unknown state" that has a 50% chance of being either the lower- or the higher-value envelope? — Michael

When all you consider is the relative chances of "low" compared to "high," this is true. When you also consider a value v, you need to use the relative chances of "v is the lower value" compared to "v is the higher value." This requires you to know the distribution of possible values in the envelopes. Since the OP doesn't provide this information, you can't use your solution. and no matter how strongly you feel that there must be a way to get around this, you can't.

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Let's step back, and try to establish this in a simpler way. Compare these two solutions:

1. Say your envelope contains v.
   
   • There is a 50% chance that the other envelope contains v/2.
   • There is a 50% chance that the other envelope contains 2v.
   • The expected value of the other envelope is (v/2)/2 + (2v)/2 = 5v/4.
   • So switching has an expected gain of 5v/4 - v = v/4.
2. Say the total amount of money in the two envelopes is t.
   
   • There is a 50% chance that your envelope contains t/3, and the other contains 2t/3.
   • There is a 50% chance that your envelope contains 2t/3, and the other contains t/3.
   • The expected value of your envelope is (t/3)/2 + (2t/3)/2 = t/2.
   • The expected value of the other envelope is (2t/3)/2 + (t/3)/2 = t/2.
   • So switching has an expected gain of t/2 - t/2 = 0.

The only difference in the theory behind these two solutions, is that #1 uses an approach that implies different sets of values in the two envelopes, where #2 uses an approach that implies the same set of values.

The property of probability that you keep trying to define, is that this difference shouldn't matter. But the paradox established in those two solutions proves that it does.

The objective Bayesian will say that an already-flipped coin has a 50% probability of being heads, even if it's actually tails, and that my £10 envelope has a 50% probability of being the smaller amount, even if it's actually the larger amount, whereas the frequentist would deny both of these (as far as I'm aware). — Michael

Why would you think that?

In my opinion, the debate between Bayesian and Frequentist, or "objective" and "subjective," has nothing whatsoever to to with the definition of "probability." Because there isn't one. The following may be oversimplified, but I believe it addresses your concerns:

Nobody really defines probability as the ratio of successes to trials, for a simple reason. To use such a definition, one would need to perform an infinite number of trials. Since that is impossible, that definition is useless. What all of them do, however, is assume that the frequency of X should approach Pr(X) as the number of trials increases. They just disagree on what X is.

A Frequentist will assume that X is fixed property of his system. So the probability distribution is a fixed property, whether he determines it objectively or subjectively. His goal may be to use experimental data to approximate what it is, by comparing experimental results to what he expects from that fixed system.

A Bayesian's X is an idealized mental system, with a knowable probability distribution. She uses the same experimental data to change her mental system to a less idealized one. So it becomes more objective than before.

The point is that none of these distinction have anything to do with the experiments "I will now flip a coin," "I just flipped a coin," or the Two Envelope Problem. You may believe that your past experience with coin flips (objectively) confirms your (both Frequentist and Bayesian) impression that unflipped coin will land Heads with a 50% probability. But you are really just using the subjective application of the Principle of Indifference, and then accepting that it coens't contradict your experience. There are two cases, neither is preferred over the other either subjectively or objectively (the weak PoI), and there is a subjective similarity between the two (the strong PoI).

When the coin has been flipped, but is not yet revealed, all of these terms apply the same way. It is still a potential result and the same probability factors - whatever you want to claim they are - apply. The coin is not in a state of being 50% heads and 50% tails. It is in an unknown state that has a 50% chance to be either, just like before the flip it had a 50% chance to end up in either. (Many forget that the point of Schroedinger's Cat was that this mixed state is paradoxical by a classical definition of "existence").

The same applies to the Two Envelope Problem. Regardless of which adjectives you want to apply to the definition of probability, the chance that the other one has v/2, given that yours has v, is conditional. It depends on how the (insert the same adjectives here) probability that the pair contained (v/2,v) AND you picked high compared to the (insert the same adjectives here) probability that the pair contained (v,2v) AND you picked low.

You can't assume that both are 50%, because you'd need to assume that it applies to all possible values of v. A true Bayesian would see that such a distribution is impossible.

The solution has always been what I posted on the first page of this thread in post number 6, which has also been my stance this entire thread. A statistical solution has never been a viable option, which has also been my stance this entire thread. The truth is this problem has always been really simple to solve, it is untangling all the speculations and assumptions that confounded it. — Jeremiah

You could have X or 2X. If you have X and you switch then you get 2X but lose X so you gain X; so you get a +1 X. However, if you have 2X and switch then you gain X and lose 2X; so you get a -1 X. — Jeremiah

And few have doubted it. Certainly not I - I said the equivalent many times.

But that solution doesn't explain why 5v/4 is wrong, it just provides a contradictory answer. And unless you can show why one is wrong, then all you have established is a paradox. Not a solution.

Gee, do you think maybe that was my point in my first post? And to try to make that point without criticizing the people who think 54/4 is right? And certainly not anybody who had posted your solution?

The limit does not need to be specified, as the envelopes will never step outside the limit. Mathematically you cannot determine if you have the the smaller amounts or larger amounts as you can never rule out which case you are in. You can speculate on such things, but you can't quantify them. It is pointless to consider the conditional probability since both cases are subjectively equal in probability, it would still boil down to a coin flip. You can do it for completeness, but it really makes no difference. — Jeremiah

The purpose is to show why the formula (v/2)/2 + (2v)/2 = 5v/4 is wrong. The approach behind the formulation is indeed correct; it just makes a mistake that doesn't show up in the formula. And can't, if you accept the assertion "it is pointless to consider the conditional probability."

Call the conditional probability you dismiss so easily Q(v). It is the conditional probability that the other envelope contains v/2, given that yours has v. So the conditional probability that the other envelope contains 2v, given that yours has v, is 1-Q(v). The correct expectation is now:

(v/2)*Q + (2v)*(1-Q(v)) = 2v-3v*Q(v)/2

This reduces to the fallacious 5v/4 iff Q(v)=1/2. If (you do understand what it means when one starts a sentence with "if", don't you?) you could dismiss conditional probability, as you do, 5v/4 actually does become mathematically correct. The only way to prove that is fallacious is to show that Q(v) can't be identically 1/2.

So it seems there is a point to considering Q(v), even if you don't know what it is.

How could anyone who has read this thread possibly concluded I was ever advocating for a statistical solution. I have been very clear that a statistical approach is incorrect. — Jeremiah

How could anyone who has read this thread have possibly concluded that I ever made this conclusion? When all I said was that any use of statistics - which you did advocate repeatedly - was inappropriate for a probability problem or a thought problem?

The only valid use of - or mention of - statistics in this thread would be to verify that a simulation can represent the reality I have proven with mathematics. The only valid use of such a simulation would be to convince doubters (of that proof) that it works. If you would read the tread, and not snippets out of context, you will see that this is what I have said about statistics and simulations.

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"Statistics is used on an experimental data set from repeated trials." — JeffJo
Yes, and it is also used on observational data sets to make generalized inferences about a population. — Jeremiah

So that "observational data set" is the "experimental data set," isn't it? With each sample being an instance of the experiment "how does a single member of population X behave in circumstances Y?"

See how easy this is when you are not trying to find fault that isn't there? But even if you don't want to acknowledge this, do we have either in the OP? No? So why keep bringing it up?

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Then go ahead and switch envelopes in the OP. — JeffJo

There is not enough information to calculate expected gain. — Jeremiah

So read the statement in its context, where I said exactly that. You are removing it from its context to make it look bad:

But if you don't care about chances, only the possibility of gain? ... Then go ahead and switch envelopes in the OP. Just don't expect a gain. That can't be determined from the information.

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it gives you a strategy that works on your assumed prior — JeffJo

Assuming your prior is correct, that is. — Jeremiah

So read the statement in its context, where I said exactly that. You are removing it from its context to make it look bad:

Even in that more general case, the Bayesian approach can give a switching strategy with a positive expected net gain. — andrewk

No, it gives you a strategy that works on your assumed prior, not necessarily on reality.

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The point is that there must be a prior distribution for how the envelopes were filled — JeffJo

True, but you will have no knowledge of what that may be. — Jeremiah

Yes, as I have said repeatedly. And if you read the entire thread, you will see that this has been my point all along. Even though you don't know what the distribution is, you still have to treat whatever value you are using as a random variable with a probability distribution, and not simply "as an unknown." Which is what you have advocated.

I'm not even sure you understand what that means. An unknown "x" can be used in a calculation by itself. The purpose may be to treat it as an independent variable, and draw a plot. But if you want to use a random variable X and assume an unknown value x for it, that calculation must couple the use of x with a probability Pr(X=x).

So the expectation calculation, when you have value v in your envelope, is not (v/2)/2+(2v)/2 = 5v/4. It is (v/2)*Q+(2v)/(1-Q) = 2v-3Q/2 for some unknown probability Q that depends on the unknown distribution which exists even though it is unknown. And it varies with v, so we can use an arbitrary function Q(v).

But there are restrictions we can place on Q(v). From the OP, unless your benefactor has an infinite supply of money, there is a v where Q(v)=1 and the expectation is v/2. And unless he can halve any amount, there is another where Q(v)=0 and the expectation is 2v.

+++++

And yet you didn't read the posts, did you? Not then, maybe you read a few more after I pushed you. I may be an ass, but at least I read a thread before criticizing people. — Jeremiah

And what people did I criticize this way? I simply pointed out that this problem is controversial because of an error that is routinely made everywhere the controversy exists.

But do you even understand what "to criticize" means? It means "to consider the merits and demerits of and judge accordingly". You were the one who chose to take my valid criticism of the methodology personally, and immediately became an ass. I tried very hard to avoid replying to that behavior, but you wouldn't let me.

The simple truth is that you have been misinterpreting me since you joined the conversation. I saw it from your first response to me. I looked at your post and realized you were making false assumptions based on viewing post out of context of the thread. I knew if I enegaged you on that level the conversation would consistent of me untangling all of your misconceptions. — Jeremiah

The *actual* truth is that you have been misinterpreting me from my very first post (), and you continue to demonstrate that here.

In that post, I cited your statement of the OP to address it, and only it. I quite intentionally made no reply to anything that anybody - especially you - had written in the thread. If you would like to tell me how you think the posts I didn't refer to, were referred to out of context, I'm all ears.

I even tried to be polite when you rudely misinterpreted my example of faulty logic as something I was claiming to be true:

You can't just enumerate a set of cases, and claim each is equally likely. If you could, there would be a 50% chance of winning, or losing, the lottery. — JeffJo

That is a very bad understanding of what a sample space and an event is. You are not applying your Principle of Indifference there — Jeremiah

Ya, great math there. — Jeremiah

Yes, my point was that the lottery example is a very bad description of a sample space. — JeffJo

You went on to demonstrate just how insufficient your knowledge of probability is, and the fact that you didn't understand anything I had said. Probably because you didn't read it:

It makes no sense to use a probability density curve[1] on this problem, considering X would only be selected ONCE[2], which means X<2X ALWAYS[3] (given that X is positive and not 0). That means no matter what X is the expected value will always be 1/2X+X[4], in every single case.

If you try to fit X to a statistical[5] distribution you are just piling assumptions on top of assumptions[6]. You are making assumptions about the sampling[7] distribution and the variance[8]. Assumptions in which you do not have the data to justify. You are also making assumptions about how X was even selected.[9] — Jeremiah

1. It's a probability distribution, and it's a set. A density curve is for continuous random variables.
2. The expectation formula considers two. X was the the random variable for the lower of the two envelopes. If v is the value in your envelope (and yes, we only need one v) the expectation considers X=v/2 and X=v
3. Which is what "X was the lower of the two" means.
4. Where did you get this? At first rad, I was willing to accept as a typo. But not anymore.
5. Statistics is used on an experimental data set from repeated trials. We don't have that.
6. I made no assumptions. I described the two random variables that exist in the OP, and used an example.
7. What sampling? Besides, that applies to statistics, and this is a probability problem.
8. What variance? That is a statistical measure, and this is a probability problem. No variance is involved.
9. Which was the point. How X was chosen affects a random variable in the OP, and so it affects the correct expectation formula.

But after my polite reply, where I did not point out any of these misrepresentations of yours, you became belligerent started saying you would not read anything I wrote until I had read all of the irrelevant posts.

I am not doing this, not until you actually read all of my posts in this thread. — Jeremiah

Then, despite the fact that I tried to address only those posts that had a smidgen of relevancy, or ones you pointed out as significant (and later claimed were not), you continued to insist you wouldn't read what I wrote. And it's quite clear you didn't; or at least that you didn't understand any of it.