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.

I'm still confused. This makes it sound like the switching argument isn't fallacious -- it just makes an unwarranted assumption. — Srap Tasmaner

Described this way, the "1.25 expectation" is not fallacious, it just makes an unwarranted assumption. It is the consequences of that assumption that are fallacious.

Again: for any finite distribution - that is, one with a maximum possible value (and a minimum is helpful to assume) - there will be some values where there is an expected gain, and some where there is an expected loss. Jeremiah's simulations can demonstrate this, and that is the only useful purpose they have. It is the expectation over all such values that must be zero for finite distributions.

The reason this is important to note, is that the expectation formula (v/2)*Q(v) + (2v)*(1-Q(v)) = 2v-3Q(v)/2, where Q(v) is a probability function, is correct. Assuming Q(v) is identically 1/2 is not.

That doesn't mean you can't construct a probability space where it is identically 1/2, or others where the expectation is always a gain. You can. They require infinite money to be available, and that is the fallacious part.

Do we need to assume that X is not continuous? If it is, all these probabilities are just 0, aren't they? — Srap Tasmaner

When you deal with continuous random variables, you use the probability density function F(x). You then use events that describe ranges of values, like $5<=X<$10, and integrate F(x) over that range to get a probability. The result that Pr(X=$10) must be zero is not an issue then, because F(X=$10) might not be.

But if you try to use a continuous random variable to make the expectation formula always mean a 25% gain, then you will find that Pr(2^n<=X<2^(n+1)) must be the same for all integers n. That's why that expectation formula implies an infinite supply of money.

Jeremiah only got into the sims & distributions business because everyone was talking about these things and it was his intention to put an end to all the speculation and get the discussion back on track. It seemed to me he did this reluctantly with the intention of showing that even if you made some assumptions you shouldn't -- this has always been his view -- it might not help in the way you think it does. — Srap Tasmaner

My points have been that the results of these simulations can be proven by considering the properties of probability distributions in general, and that the same approach can explain everything that needs to be said about the OP. Including some things that he said which are wrong.

I replied about specifics in the simulations because he asked me to do so. I assumed there was data he considered to be significant in them because he said I didn't see that it was in them.

Isn't 2X just a transformation of X that doubles the possible values in X? — Andrew M

I was being terse. A longer version of what I said is "So '2X' is meaningless if you try to use it as a value." This thread has gone on too long, and I didn't want to have to explain the mathematics of probability theory any more than I already have.

The following comments may seem incredibly pedantic, but understanding them is necessary to avoid the confusions found throughout this thread. The convention I use is that an upper-case letter is a random variable, and the corresponding lower case letter is an unknown value in that range of that random variable. So...

• X is a random variable. It does not, and can not, mean a single value.
• You'd do that with an expression like X=$10 or X=x, which define events not values. What these expressions literally mean is "The event where the instantiated value of the random variable is $10" or "... is the unknown x taken from the range of X."
• A more proper version is "X∈{$10}", which makes it clearer that we are talking about an event.
• Or even "X∈{$5,$10}", but that is not useful in a problem where you only consider one value at a time.
• You can consider a new random variable Y=2X. You find its distribution, not a value, by the transformation methods in your link.
• So "2X" can't be used as a value in an expectation calculation.
• But you could use y, as an instantiated value of the Y I just defined. And x as an instantiated value for X, in which case y=2x.
• That expectation calculate will still have to use either Pr(X∈{x}) or Pr(Y∈{y}). Which we've been simplifying to Pr(x) or Pr(y).

It is very important to understand that you only use the random variable itself to define an event. That is, as the argument of a probability function. And any value, known or unknown, has to be carefully associated with its probability.

In fact, this explains Michael's error from the start:

The amount you have is $x. The other envelope contains either $2x or $x/2. If it's $2x then you gain $x by switching. If it's $x/2 then you lose $x/2 by switching. — Michael

In this expression, Michael is using two new random variables, but doesn't recognize them as such. R represents the palyer's choice, and can be {2,1/2} with probability 50% each. Y is the other envelope, defined by Y=R*X (remember, these are random variables). His error is only recognizing one and not treating it as carefully as he should. So he missed that the distribution for Y is found by the transformation methods in your link, and that he needs to know the probabilities of the pair being ($x/2,$x) and ($x,$2x) to do this.

It also explains what Jeremiah doesn't want to accept as his error. You can't simply treat a random variable as an unknown. You can consider a set of unknown values from its range, but only if you couple that with their probabilities.

My simulations were there to display the inherit ambiguity in defining a prior distribution. X is an unknown, treat it like an unknown. — Jeremiah

I have also already shown that trying to calculate expected returns is a misguided effort. — Jeremiah

And my response to these sentiments has always been that you can't define/calculate the prior distribution, and that it was a misguided effort to even try (as you did).

That doesn't mean there isn't a prior distribution, that you can ignore the fact that there a prior distribution, that it is proper to treat a random variable with a prior distribution as an unknown, or that there isn't useful information that can be obtained from the properties that a prior distribution must have.

Some of that information is that you can prove for a reasonable (i.e., non-infinite) prior distribution, that there must be some values of the amount in your envelope where there is an expected gain given that v, others where there is an expected loss, but that the expectation over V must be zero. This is provable without defining what the prior distribution is, without any of the ambiguity you claim exists.

I think in terms of modeling the actual OP the inclusion of distributions or switching strategies is misguided. — Jeremiah

Choosing any explicit distribution for the OP is indeed misguided, which is why your simulations were misguided. That, and the fact that your conclusions could be proven without such modeling.

But that doesn't mean we can ignore the fact that we need a distribution, if we want to calculate an expectation when three possible values are considered. As in the (v/2)/2+(2v)/2 = 5v/4 expectation formula. We may not know what it is, but there is one. Ignoring the fact that there must be a distribution is a mathematical error, and is why that formula is mathematically incorrect.

And there are properties of all valid probability distributions that we can make use of. That expectation considers three values, v/2, v, and 2v and thus requires probabilities for two sets of envelopes. But we can see that if Pr(v/2,v)=Pr(v,2v), then the formula is correct. And for other possible relationships, there can (and will) be both gains and losses.

If you consider only two values, there is a mathematically correct formula: if D is the difference, the expected gain is (+d)/2 + (-d)/2 = 0. We don't need to to know the distribution to D to see that this is correct.

However, these models do provide a useful platform to investigate properties that may be applied to other probabilistic aspects.

Indeed. Your half-normal distribution shows that there is an expected gain if your envelope contains less than $13.60, and an expected loss if it contains more. The others will show similar properties. But again, you don't really need to simulate them to show this. The only utility would be to convince those who don't want to accept that math works. You will undoubtedly ignore this, but there is not much I can do about that.

And that's because of what I explained here. — Michael

We've already established that the expected gain is

E(B∣A=a)=P(X=a∣A=a)2a+P(2X=a∣A=a)a/2 — Michael

Close. But we can’t all seem to agree on what that means.

I’m going to assume that A means your envelope, B means the “other” envelope, and X means the lower value. But A, B, and X mean the abstract concept of the random variable; it is a, b, and x that mean values. So “2X” is meaningless. What I think you mean here is:

E(B∣A=a) = Pr(X=a∣A=a)*(2a) + Pr(X=a/2∣A=a)*(a/2)

The objective probabilities of X=a∣A=a and 2X=a∣A=a depend on how the host selects the value of X.

And that definition is worthless to the OP. If we have a “host” and know how he “chooses”, then we have an explicit probability distribution. We wouldn’t need any of the other names (objective, subjective, frequentist, Bayesian, epistemic, etc.) mentioned here, whether singly or in combination.

The point is, we don’t. We have no distribution, by any name or definition. That doesn’t change the fact that there has to be one for you to calculate an expectation. It just means that any conclusions you draw have to apply regardless of the name or definition.

If he selects it at random from a distribution that includes a2 and a then the objective probability of X=a∣A=a is 0.5 …

Again, “at random” does not mean “with uniform probability.” You are assuming a uniform probability distribution, and that can’t work. That assumption means that the probability that X is $10 is the same as the probability that X is $20, or $40, or $80, or … to infinity. And it is also the same as the probability that X is $5, or $2.50, or $1.25, or … to 1/infinity. The expectation with this distribution is an infinite amount of money, and each has a probability of 0% (which is not a contradiction, since this is a variation of a continuous distribution).

Does your host have an infinite amount of money? Does he have a way to pick one integer at random from all integers in -inf<N<inf?

So there would be an objective expected gain.

You confuse the fact that there can be an expected gain for specific values in most possible distributions, with the fallacy that it there is a gain any value in any distribution. Again, use my example where there is a 50:50 chance that the envelopes are ($5,$10) or ($10,$20).

• If a=$5, a 25% chance, there is a certain gain of $5.
• If a=$20, a 25% chance, there is a certain loss of $10.
• If a=$10, a 50% chance, there is an expected gain of $10/2-$5/2 = $2.50
• But what this means when you don’t know what a is, is that the expectation is:
  ($5)*(25%) – ($10)*(25%) + ($2.50)*(50%) = $1.25 - $2.50 + $1.25 = $0.

For any valid, finite probability distribution, there will be some values of that produce a gain. But there will also be others that produce a loss. And the expectation over all of them will be $0.

What if we just say that, having observed the value of our envelope to be a, then the expected value of the other is 3X - a for some unknown X? That formula, unlike the expected value formula, doesn't require any probabilities to be filled in. It's uninformative, but so what? — Srap Tasmaner

To what purpose? It doesn't help you to answer any of the questions.

Besides, the correct expectation formula is [(a/2)*P1 + (2a)*P2]/(P1 + P2), where P1 and P2 are the probabilities that the pair had (a/2,a) or (a,2a), respectively. We can set this equal to yours, and solve for X in terms of a, P1, and P2. So whether or not you "fill in" P1 and P2, your X still depends on them.

On the other hand, I think the right way to look at it is what I've been saying lately:

there are two choices;
the host's choice determines how much can be gained or lost by switching;
the player's choices determines whether they gain or lose. — Srap Tasmaner

This is only true if we do not look in an envelope. That *is* the OP, but the other has also been discussed.

It is true because we only need to consider one value for the host's choice, and so it divides out. If we look, we need to consider two. And there is no information about the relative probabilities of those two host-choices.

These are the same two conclusions I have been "harping on" all along, and it is still true that they are the only possible conclusions. If you don't look, the two envelopes have the same expected value. If you do, there is not enough information to say how the probabilities split between having the higher, or lower, value.

So if I pick a number at random from {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} you disagree that the objective probability that I will pick 5 is 0.1? — Michael

Certainly. It isn't "objective." I thought I made that pretty clear.

Before I say whether I agree with the value, you should understand that "pick at random" does not mean "pick with uniform randomness." These two are often confused. Had you said "uniform", of course I would agree with that probability. But it would be neither objective nor subjective, it would be explicit.

The best guess for a subjective probability is that you would pick uniformly. But if you want an objective probability, you need to find a test subject and get them to repeat the experiment many times. That's what objective probability means. And it has been shown over and over that people can't do that with uniformity.

If I know that the odds are even then I will play.

And that's the point. You cannot know this in the two envelope problem, when you know what value you are switching from. Unless, of course, you know how the amounts were determined. Do you?

If I know that the odds aren't even then I might be willing to play, depending on the odds. If I don't know the odds then I will play.

I'm assuming, based on the first sentence here, that you left a "not" out of the second?

You can't know the odds when you look in an envelope and see a value. You can choose to play, not knowing the odds, but your calculation of the expectation is wrong.

And if all he had was a $10 bill and a $20 bill then my chances of picking High = $10 is nil.

Yep. Now, do you know how the odds of having only a $5 and a $10 compare to only having a $10 and a $20? No? Then you can't say that the chances of gaining are the same as the chances of losing.

So what am I supposed to do if I don't know how the values are selected?

Say "I don't know how to compare the chances of gaining to the chance of losing."

As I said here, if I don't have any reason to believe that X = 5 is more likely than X = 10 then why wouldn't I switch? I think the principle of indifference is entirely appropriate in this circumstance.

Nope. The PoI applies only if you can establish that there is a causal equivalence between every member of the set to which you apply it. It is specifically inapplicable here, because there cannot be a strategy for filling the envelopes where it is true for an arbitrary value you see.

There's more to gain than there is to lose, and a loss of £5 is an acceptable risk.

That is a different question. The point is that the risk is unknowable, and probably not 50%. Whether you think that a $5 loss is acceptable regardless of the risk is a subjective decision only you can make.

There is never any case, no matter what envelope you picked from whatever pair, in which you have a non-zero chance of gaining and a non-zero chance of losing — Srap Tasmaner

This emphasizes how a probability space changes based on your knowledge. Or rather, what knowledge is missing.

I just flipped a coin on my desk. I can see whether it is Heads or Tails. So "there is never a case where there is a non-zero chance of Heads, and a non-zero chance of Tails."

But If I ask you to assess the two cases, you should say that each has a 50%. The fact that the outcome is determined does not change how probability works for those who do not know the determination.

Srap Tasmaner is saying that, to someone who knows what is in *both* envelopes, the possibility of gaining or losing is determined. Michael is saying that, to someone who doesn't see both, the two cases should be treated with probabilities that are >=0, and that add up to 1.

The error is thinking that both must be 50%. Your chance of High or Low is 50% if you don't know the value in the one you chose, but it can't be determined if you do.

In broad terms I do not disagree with that characterisation. — andrewk

Well, I can't address your disagreement unless you explain why you feel that way. That characterization is correct. There may be different ways people express their uncertainty, but it still boils down to the same concept.

But there is often more than one way to represent uncertainty, and these lead to different probability spaces. I have referred previously to the observation that in finance many different, mutually incompatible probability spaces can be used to assign a value to a portfolio of derivatives.

What kind of differences are you talking about? There is no single way to express a sample space, and in fact what constitutes an "outcome" is undefined. We've experienced that here: some will use a random variable V that means the value in your envelope, while others represent the same property of the process by the difference D (which is the low value as well).

But the significant difference in those portfolio analyses is the distribution function Pr(). Even though ones subjectivity may be based on past experience, there is no guarantee that the underlying process will be the same in the future, or in the cases being compared. The factors determining the actual values are not as consistent as is required for analysis.

So any assessment is, by necessity, subjective to some degree. Different sample spaces (or probability spaces) simply allow the analysts to apply their subjectivity in different ways.

But we really are getting off the point. Which is that they do use a probability space, not whether it is "real," "absolute," or even "correct." That is irrelevant. The point here is that, over the set of all possible probability spaces for the OP, there will either be some values of v where the expected change is non-zero, or the expected value over V is infinite. And in fact, the only way that the expected change is 5v/4, is if the probability distribution says the two possible pairs are equally likely. Which also implies you are assuming a probability space. The only way the expected change is identically zero, is if you don't know v or you know that Pr(v/2,v)=2*Pr(v,2v).

And what you seem to be avoiding with that attitude, is that the expectation formula (v/2)/2 + (2v)/2 is already assuming: — JeffJo

I am not an advocate for that expectation formula, so I don't see why you'd think I am avoiding those objections to it.

Maybe I was mixing Andrews up. I apologize.

If he selects it at random from a distribution that includes a2 and a then the objective probability of X=a∣A=a is 0.5 and the objective probability of 2X=a∣A=a is 0.5. So there would be an objective expected gain. — Michael

The highlighted assertion is incorrect. First off, "objective probability" means the "likelihood of a specific occurrence, based on repeated random experiments and measurements (subject to verification by independent experts), instead of on subjective evaluations." Se have no such repeated measurements, so any assessment of Pr(X=a∣A=a) is subjective.

The Principle of Indifference itself is a subjective assessment. But to apply it, you first must determine some kind of an equivalence between the origin of the cases to which you apply it. You can't do that to the values here, only the chance of picking High or Low.

If there's £10 in my envelope and I know that the other envelope contains either £5 or £20 because I know that one envelope contains twice as much as the other then I have a reason to switch; I want an extra £10 and am willing to risk losing £5 to get it. — Michael

So, you are willing to risk losing $5 for the chance to gain $10? Regardless of the odds behind that risk?

Say I offer you the chance to play a game. It costs you $5 to play, but if you win I will return your $5, and give you $10. The game is to flip a fair coin, and if you get Heads N times in a row, you win.

1. Say I tell you that N=1. Are you willing to play? I hope so, as the chances of winning are 1 in 2 and your expectation is ($-5)/2+($10)/2=$2.50.
2. Say I tell you that N=10. Are you willing to play? (The chances of winning are now 1 in 1024.)
3. Say we determine N by rolling a fair die. Are you willing to play? You could get odds of 1 in 2 again, but you could also get odds of 1 in 64. (The overall chances are just under 1 in 6.)
4. Say I tell you that there is an integer written on every card in a deck of cards, and you determine N by drawing a card at random. Are you willing to play if I don't tell you what the numbers could be?

You seem to resist accepting that the OP is most similar to game #4, and not at all similar to game #1. You may be able to calculate the chances of drawing a particular card, but if you don't know what is written on any particular card then this does not translate into a 1/52 chance of drawing a "1".

In the OP, you have no way of knowing whether your benefactor was willing to part with more than $10. If all he had was a $5 bill and a $10 bill, then he can't. Your chances of picking Low=$5 or High=$10 were indeed 50% each, but your chances of picking Low=$10 were nil.

But if you don't care about chances, only the possibility of gain? And so answered "yes" to all four game? Then go ahead and switch envelopes in the OP. Just don't expect a gain. That can't be determined from the information.

I differ from that perspective is that I reject the notion that there is such a thing as a 'real' probability (aka 'true', 'raw', 'correct', 'absolute' or 'observer independent' probability). — andrewk

Why? I think you confuse the fact that probabilities are intangible with being unreal.

Probability is, essentially, a measure of our uncertainty about a result. If I know enough about a coin's weight distribution, the forces applied to it, the aerodynamics in the room, and the landing surface, then We can calculate whether it will land on Heads or Tails. As you repeat the flips, some of these details change which will end up in a 50:50 frequency distribution. If we don't know these details, it is a 50:50 probability distribution each time. These are just different, and very real, properties of the same process.

Similarly, it is because we don't know anything except ($D, $2D) about the circumstances behind the amounts in the envelopes that we must treat it as a probability distribution. Could your envelope contain $10? There is no reason to believe that it can't, and it seems to be a reasonable value. Is it certain that it contains $10? Absolutely not. These two facts make it describable by a probability. Don't confuse not being able to determine the probability, with inapplicability of probability.

And what you seem to be avoiding with that attitude, is that the expectation formula (v/2)/2 + (2v)/2 is already assuming:

• That there are at least three values of v that are possible, namely v/2, v, and 2v.
• That neither of the possible pairs is certain. So there is probability distribution.
• That distribution says Pr(v/2,v) = Pr(v,2v), and
• By transitivity, every value of the form v*2^N, where N is any integer in -inf<N<inf, is possible.
• And all the possible pairs have the same probability. Which must be 0.

If you do look, THERE IS PROBABLY AN EXPECTED GAIN OR LOSS, but you have no information that would let you calculate it. This is different from knowing it is 0. — JeffJo

So since you don't know which case you are in after seeing Y and they are not equal you can't really calculate the expected value. Now if you never opened A and never saw Y, that is a different story. — Jeremiah

You did not read the thread. — Jeremiah

I did read the thread. You did not read my replies. Like this one, where I said "you have no information that would let you calculate [the expected gain or loss]" and you replied with "you can't really calculate the expected value" as if I hadn't just said the same thing.

In one of those replies you ignored, I explained to you how your own half-normal simulation will show that there is an expected gain if the value you find is less than $13.60, and an expected loss if it is greater. I'm not claiming that you should know such things in general - in fact, I explicitly said you don't - but THERE MUST BE SOME VALUES WHERE THE EXPECTED GAIN IS NOT ZERO.

Now, you can repeat your unfounded assertions as often as you want. I have proven them to be incorrect.

I think the confusion comes when you switch from

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

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

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

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

But we do know that it can't be true. That's the point.

Either can't be true for all values of this "a", or there are impossible values in the set of possible a's and selection is impossible:

• If there is a minimum value for a, then
  P(picked smaller | picked = amin) = 1 and P(picked larger | picked = amin) = 0
• If there is not, then a can be arbitrarily small, which is impossible (well, impractical).
• If there is a maximum value for a, then
  P(picked smaller | picked = amax) = 0 and P(picked larger | picked = amax) = 1
• If there is not, then a can be arbitrarily large. Which is impossible.

I have agreed that what you said near the beginning of this thread was right. Did you read that?

You have been very reticent to point out what it is you think I have not read, or have misinterpreted. Or you point one out, then say later that it wasn't important. Yet every time you did (and it was ambiguous), I have pointed out why you are wrong, or it is irrelevant. Did you read that?

There are really only two conclusions that can be drawn about the the OP:

• If you don't look in the envelope, it is proven that there is no expected gain by switching.
• If you do look, THERE IS PROBABLY AN EXPECTED GAIN OR LOSS, but you have no information that would let you calculate it. This is different from knowing it is 0.

I keep "harping on" this because you keep implying there are other conclusions that may be possible, and there are not. But you refuse to reply to these facts. I have said this many times. Did you read them?

You are misinterpreting what I said, what your link says and your source is Wikipedia. — Jeremiah

You have not interpreted a single thing I have said correctly; in fact, you've replied to very few of them. Most significantly this time, how Bayesian Inference is inapplicable to the OP. And you can find similar information in any source about Bayesian Inference - it isn't wrong just because it is in Wikipedia.

From Wikipedia:

Bayesian inference is a method of statistical inference in which Bayes' theorem is used to update the probability for a hypothesis as more evidence or information becomes available.
...

H stands for any hypothesis whose probability may be affected by data (called evidence below). Often there are competing hypotheses, and the task is to determine which is the most probable.

Pr(H), the prior probability, is the estimate of the probability of the hypothesis H.

Since there is no provision for "data/information/evidence" in the OP, only a one-time thought problem, Bayesian inference does not, and cannot, apply.

But if you try, you need an hypothesis, and an estimate - which I called a "guess" - of what the probability of that hypothesis is. And an informed prior is just an educated guess.

Yes, I have read this thread; and no, it contains no hypothesis from the OP that can be tested this way.

↪Jeremiah


Why is equiprobability simple but other priors aren't? — fdrake

Not only is it not "simple," you can prove that it makes an invalid probability space.

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

But the probabilities are not the same as the probabilities of having picked the larger, or smaller, value. They are the probabilities of picking the smaller value given that that value is $10, and the probability of picking the larger value given that that value is $10, respectively.

You don't "guess" a prior. Priors have to be justified. If you don't know you use an uninformative prior. — Jeremiah

So an "uninformative prior" is not a "prior" ? And an informative prior that is based on only partial information is not still a "guess" about the rest?

My point is that there is no place to claiming a Bayesian solution to the OP.

I agree in the strictness sense of the definition there is an unknown distribution, but as far was we know it was whatever was in his pocket when he filled the envelopes. We can't select a distribution to use, as we have no way to check it. — Jeremiah

And that unknown value in his pocket has a distribution. We don't need to "check it," as long as the symbolic probability space we use satisfies the requirements of being a probability space.

The OP also doesn't include a provision for repeatability, so devising learning strategies is also outside its scope. The answer is that, if you don;t look, there is no expected gain by switching. If you do, there can be a gain or loss, but you can't calculate it.

If you use a distribution you are making assumptions not included in the OP. I pointed this out before. — Jeremiah

Jeremiah oversimplifies.

If you use a specific distribution, then your results apply only to that distribution and not the OP. And AFAIK nobody but Jeremiah has implied otherwise. So my recent example does not say that the expected value of the other envelope in the OP is $6.50 if you see $10 in yours.

What is true, even in the OP, is that if you see V=v, the expected value of the other envelope is

[(v/2)*Pr(v/2,v) + (2V)*Pr(v,2v)] / (v/2)*Pr(v/2,v) + (2V)*Pr(v,2v)]

But this is a symbolic solution only, not a numeric one. It applies to the OP with an unknown probability space {S, F, Pr(*)}. It implies that you need to know the this space to give an answer for a specific value of V.

But if you apply the requirements placed on Pr(*) for this to be a valid probability space, this symbolic expression can be shown to evaluate to the expected value of your envelope.