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.

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. The point of a Bayesian analysis is to guess at a prior, and refine it from evidence in the real world.

As an aside, I think we're saying the same thing from different angles. — fdrake

I agree. I think it is most convincing to present multiple angles. But it isn't (just) the sample space that is the issue. It is the probability space which includes:

• A sample space S which lists every indivisible outcome.
• An event space F which lists all of the combinations of outcomes from S that are of interest (with some restrictions we needn't go into).
• A probability distribution function Pr(*) that associates a probability with every member of F.

It is Pr(*) and its relationship to F that is the source of confusion.

+++++

I see the situation as this:

Assume there's £10 in my envelope and that one envelope contains twice as much as the other. The other envelope must contain either £5 or £20. Each is equally likely and so the expected value of the other envelope is £12.50. I can then take advantage of this fact by employing this switching strategy to achieve this .25 gain. — Michael

The error is "Each is equally likely," and it is wrong because we need to work with F, not S. My very first post gave an example. Suppose I fill 9 pairs of envelopes with ($5,$10) and one pair with ($10,$20). I choose a pair at random, and present it to you saying "one contains $X and the other $2X, [but you do not know which envelope is which or what the number X is]."

Aside: Since I am way too familiar with various formulations of the problem, I haven't been looking at the original here. It does use "X" where I used "V", and almost everybody except Michael has been using X to mean something different. This is likely one cause of confusion. So please note that in the future, I will only use V for the value in your envelope. To avoid confusion, I will not use X at all anymore, but use D for the absolute difference between the two envelopes; it is happy coincidence that this is also the lower value of the pair.

Do you really believe, if you see $10, that there is a 50% that this is the smaller of the two envelope presented to you?

No. There is a 10% chance that this is the smaller value, and a 90% chance it is the larger value. Because there was a 90% chance I chose the pair ($5,$10), and a 10% chance I chose the pair ($10,$20). The 50% chances you want to use don't even seem to play a role in this determination.

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

Note that this doesn't mean that the expectation is that there is no change. In my example, it is ($5)*(90%)+($20)*(10%) = $6.50, and you shouldn't switch. In the OP, you still do not have information that will allow you to say what it is. But averaged over all possible values of V, there will be no expected gain.

I will not further debate such specifics. — Jeremiah

Of course you won't. You don't like facts that disagree with your beliefs.

You have made various claims in this thread, I've told you which you are right, and which you wrong. All you can say is "I believe my solution is correct," without saying which it is that you think is correct. And I'm done trying to educate you.

You're going it again. Is X the value of my envelope or the value of the smallest envelope? You're switching between both and it doesn't make any sense, especially where you say "if you have the lower value envelope X/2".

I hope I'm not talking down, that isn't my point. But probability really needs great care with terminology, and it has been noticeably lacking in this thread. This question is a result of that lack.

Let's say that the possible values for envelopes are $5, $10, $20, and $40. This is sometimes called the range of a random variable that I will call V, for the value of your envelope.

A conventional notation for a random variable is that we use the upper case V only to mean the entire set {$5,$10,$20,$40}. If we want to represent just one value, we use the lower case v or an expression like V=$10, or V=v if we want to treat it as unknown.

Each random variable has a probability distribution that goes with it. The problem is, we don't know it for V.

The conditions in the OP mean that there are three possibilities for the pair of envelopes: ($5,$10), ($10,$20), and ($20,$40). Various people have used the random variable X to mean the lower value of the pair; I think you are the only one who has used it to mean what I just called V. So X means the set {$5, $10, $20}. Notice how "$40" is not there - it can't be the low value.

The most obvious random variable is what I have called R (this may be backwards from what I did before). So R=2 means that the other envelope has twice what yours does. Its range is {1/2,2}. It is the most obvious, since the probability for each value is clearly 50%.

From this, we can derive the random variable W, which I use to mean the other envelope. (Others may have confusingly called it Y, or used that for the higher value; they were unclear). Its specific value is w=v*r, so its derived range is {$2.50,$5,$10,$20,$40,$80}. But we know that two of those are impossible. So what we do, is use a probability distribution where Pr(V=$5 & R=1/2)=0, and Pr(V=$20 & R=2)=0. Essentially, we include the impossible values that may come up in calculations in the range, and make them impossible in the probability distribution.

Now, if we don't see a value in an envelope, we know that v-w must be either +x, or -x, with a 50% chance for either. So switching can't help. The point to note is that we don't know what value we might end up with; it could be anything in the full range of V.

Your expectation calculation uses every value of W that is in its range, including the impossible ones. It has to, and that is why they are included in the range. Because of that, you can't use the probabilities for R alone - you need to do something to make $2.50 and $80 impossible.

So you use conditional probabilities for R, not the unconditional ones:

Exp(W|V=v) = (v/2)*Pr(R=1/2|V=v) + (2v)*Pr(R=2|V=v)
= [(v/2)*Pr(R=1/2 & V=v) + (2v)*Pr(R=2 & V=v)] / [Pr(R=1/2 & V=v) + (2v)*Pr(R=2 & V=v)]

We can't separate R and V in those probability expressions, because they are not indepepmndent. But we can if we transform V into X:

= [(v/2)*Pr(R=1/2 & X=v/2) + (2v)*Pr(R=2 & X=v)] / [Pr(R=1/2 & X=v/2) + (2v)*Pr(R=2 & X=v)]
= [(v/2)*Pr(R=1/2)*Pr(X=v/2) + (2v)*Pr(R=2)*(X=v)] / [Pr(R=1/2)*Pr(X=v/2) + (2v)*Pr(R=2)*Pr(X=v)]

And since Pr(R=1/2)=Pr(R=2)=1/2, they divide out:

= [(v/2)*Pr(X=v/2) + (2v)*Pr(X=v)] / [Pr(X=v/2) + (2v)*Pr(X=v)]

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

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

I specifically said that my simulation is not about finding expected results as that entire argument is flawed.

If you did, that statement was completely obfuscated by your sloppy techniques. And it isn't the argument that is flawed, it is the assumption that you know the distribution. As shown by the points you have ignored in my posts.

But if that is what you meant, why do you keep pointing out your simulations, which coudl prove something but, as you used them, do not.

The thing to notice here is that in all cases the absolute value of the difference between column one and column two is always equal to the lesser of the two (save rounding errors). The lesser of the two is X.

The thing to notice here is that you can't use your "X"- the lower value - as a condition unless you see both envelopes, which renders analysis pointless.

I pointed out a few times that the sample space of event R and the sample space of event S are equal subsets of each other, which means mathematically we can treat them the same.

• A sample space is a set of all possible outcomes. An event is a subset of the sample space. Events don't have sample spaces.
• It not possible for two sets to be unequal subsets of each other.
• We are less concerned with the sample spaces, than with the corresponding probability distributions.

So the message of this posts seems to be "I really don't know what I am talking about, but will put words together that I hope will make it sound like I do."

Please, correct me if I am wrong. I have tried to make sense of what you said, but much of it defines all attempts.

R and S is the event you have X or 2X when you get A. By definition of A and B, if A=X then B =2X, or if A =2X then B=X. So by definition if A equals X then B cannot also equal X.

This is unintelligble.

Maybe you meant "R and S are events that are conditioned on you having chosen envelope A and it haing the low value or the high value, respectively. If you associate either with a value X from the sample space of possible values, which has nothing to do with R or S, then R means that B must have 2X, and S means that B must have X/2." But you were trying to dissuade Dawnstorm from what appears to have been a valid interpretation of another of your fractured statements; that X was simultaneously possible for both A and B. Even though he realized you could not have meant that, and said as much.

But this still pointless. If you aren't given a value in X, all you need to do is point out that the potential loss is the same as the potential gain. If you are given a value, you do need the distributions. Which you would know, if you read my posts as well I as have (tried to) read yours.

I also pointed out that as soon as you introduce Y they are no longer equal subsets and therefore mathematically cannot be treated the same.

This may be what you were referring to when about "Y." It still refers to ill-defined sets, not distributions.

The point of this demonstration is to show that the possible distribution of A is the same as the possible distribution of B. ... So we see with a D test statistics of 0.0077 and a 0.92 p-value we don't have strong enough evidence to support the alternative hypothesis that the two distributions are reasonably different.

It is provable without simulation that the the two distributions are the same, so this is pointless. We can accept that the distributions are the same. And it is is obvious you didn't read my posts describing why the simulation is pointless. In short, the "data" you apply "data science" to pertains only to how well your simulation addresses the provable fact.

It is also provable that the distributions are not independent. Since you technically need to use a joint probability distribution for any analyses using two random variables, and you can only separate a joint distribution into individual distributions when they independent, this conclusion can have no bearing on the original problem. It is also obvious that you did not read my posts that explain this.

You conclude that I did not read your posts, because I didn't comment on them. By not reading mine, you missed the fact that I don't need to comment. Conclusions drawn from evidence that has already been discredited do not need to be addressed.

I am happy with my correct results.

And again, you won't say what results you mean.

Your solution from page 1, that ...

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.

... is a correct solution to the original problem when you don't look in the envelope. The problem with it, is that it doesn't explain to why his program doesn't model the OP. That is something you never did correctly, and you refuse to accept that I did.

Your conclusion from page 26, that...

the possible distribution of A is the same as the possible distribution of B

... is also correct, although it is easier to prove it directly, But it is still irrelevant unless you determine that the two distributions are independent. AND THEY ARE NOT.

It is this conclusion from page 26:

the distribution in which X was selected from is not significant when assessing the possible outcome of envelope A and B concerning X or 2X.

... that is incorrect, as I just showed in my last post. The probability that A has the smaller value depends on the relative values of two probabilities in that distribution, so it is significant to the question you address here.

Averaged over the entire distribution, there is no expected gain. Which you can deduce from your page-1 conclusion. For specific values, there can be expected gains or losses, and that depends on the distribution.

The point is that there must be a prior distribution for how the envelopes were filled, but the participant in the game has no knowledge of it. I express it as the probability of a pair, like Pr($5,$10) which means there is $15 split between the two. There is also a trivial prior distribution for whether you picked high, or low; it is 50% each.

The common error is only recognizing the latter.

If you try to calculate the expected value of the other envelope, based on an unknown value X in yours, then you need to know two probabilities from the unknown distribution. The probability that the other envelope contains X/2 is not 1/2, it is Pr(X/2,X)/[Pr(X/2,X)+Pr(X,2X)]. The 50% values from the second distribution are used to get this formula, but they divide out.

The problem with the OP, is that we do not know these values, and can't make any sort of reasonable guess for them. But it turns out that the "you should switch" argument can be true:

In Jeremiah's half-mormal simulation:

• The probability that X=$5 is 0.704%, and that X=$10 is 0.484%.
• The probability that A=$10 is (0.704%+0.484%)/2=0.594%.
• Given that A has $10, the probability that B has $5 is (.704%/2)/0.594% = 59.3%
• Given that A has $10, the probability that B has $20 is (.484%/2)/0.594% = 40.7%
• Given that A has $10, the expected value of B is (59.3%)*$5 + (40.7%)*$20 = $11.11.
• So it seems that if you look in your envelope, and see $10, you should switch.
• In fact, you should switch if you know that you have less than $13.60. And there is a 66.7% chance of that. (Suggestion to Jeremiah: run your simulation three times: Always switch from A to B, always keep A, and switch only if A<$13.60. The first two will average - this is a guess - about $12, and the third will average about $13.60.)
• It is the expected gain over all values of X that is $0, not an instance where X is given.
  
  The naive part of Jeremiah's analysis, is that knowing how A and B have the same distribution is not enough to use those distributions in the OP. He implicitly assumes they are independent, which is not true.

So the OP is not solvable by this method. You can, however, solve it by another. You can calculate the expected gain by switching, based on an unknown difference D. Then you only need to use one probability from the unknown distribution, and it divides out.

Conclusions:

1. If you don't look in your envelope, there can be no benefit from switching.
   
   • This is not because the distributions for the two envelopes are the same ,...
   • ... even though it is trivial to prove that they are the same, without simulation.
   • The two distributions are not independent, so their equivalence is irrelevant.
   • It is because the expected gain by switching from the low value to the high, is the same as the expected loss from switching from high to low.
2. If you do look in your envelope, you need to know the prior distribution of the values to determine the benefit from switching.
   
   • With such knowledge, it is quite possible (even likely) that switching will gain something. If Jeremiah could be bothered to use his simulation, he could prove this to himself.
   • But it is also possible that you could lose, and in the cases where you do, the amounts are greater. The average over all cases will always be no change.

If you want to address the original problem, it matters that by your methods, ignoring their other faults, the two envelopes can contain the same amount. I understand that you are more interested in appearing to be right, than in actually addressing that problem. But I prefer to address it.


Exactly what do you think "[your] correct solution" solves? I told you what it addresses, but you decided you were done before hearing it.

I've tried to explain to you why the issue you addressed is not the OP, but you have chosen not to accept that. If you don't want to "debate in circles," then I suggest you accept the possibility that the contributions of others may have more applicability than yours.

I am well aware 0 was a possible outcome, the code just runs better without the loops, and it was not significant enough to care.

Use a "ceiling" function instead of a "round" function. Or just add 0.0005 before you round.

Sorry my simulation proves you wrong.

Any statistical analysis addresses a question. Yours addresses "Is the distribution of A the same as the distribution as B?" And it can only show what the answer most likely is, not prove it.

But, the answer to that question is pretty easy to prove. As I did. So there wasn't much point in the statistical analysis, was there?

What you didn't address, but could, is whether the two random variables are independent. Which they are not. Since having two identical distributions is meaningless if they are not independent, your simulation did not address anything of significance to the OP.

Or what the expectation of the other envelope is relative to yours, and why the naive apparent gain can't be realized.Answer: With any bounded distribution, there is a smallest possible value where you will always gain a small amount by switching, and a largest value where you will always lose a large amount by switching. In between, you typically gain about 25% by switching once, and lose that 20% (remember, the baseline is now the switched-to value, so the dependency is reversed) by switching back. But over the entire distribution, the large potential loss if you had the largest value makes the overall expectation 0.

Those are the issues that are important to the OP, and your simulation doesn't provide any useful information.

Right. That was after my first post. I read it before the rest of the thread, so I didn't think that was what you were referring to. I didn't want to go into this amount detail about it (hoping that my explanations would suffice), but you insist.

Let X be a set of positive dollar values, and x0 be the lowest value in X.

Let F(x) be a probability distribution function defined for every x in X, and for x0/2 as well. But F(x0/2)=0 since x0/2 isn't in X.

Let Y be chosen uniformly from {1,2}.

Define your random variable A = X*Y. So X=A/Y, where the denominator can be 1 or 2. This makes the probability distribution for A equal to F(A/1)/2+F(A/2)/2 = [F(A)+F(A/2)]/2. Call this F1(A).

Define your random variable B = X*(3-Y). So X=B/(3-Y), where the denominator can be (3-1) or (3-2). This makes the probability distribution for B equal to F(B/(3-1))/2+F(B/(3-2)/2. But simple rearrangement shows that this F1(B).

The point is that we can prove that the distributions for A and B are the same. And all your simulation shows, is that if you can correctly derive the result of a process, then a correct simulation of that process statistically approximates the result you derived. It says nothing about the OP; certainly not that the distribution is unimportant.

And btw, the distribution you used in the version you posted is a discretization of the Half-normal distribution. And it is possible that it could put $0 in both envelopes.

+++++

And your simulation does nothing to explain why the expectation formula E = (X/2)/2 + (2X)/2 = 5X/4 is wrong. It is because the correct formulation, if someone picks your envelope A and it contains X, is:

E = [(X/2) * F(X/2) + (2*X) * F(X)] / [F(X/2) + F(X)]

Your simulation does nothing to show this. But you can in various ways. Choose any of your distributions, but fixing that "zero" problem.

One way to test it, is to chose a range of values like between $10 and $11. If you have an amount in that range (ignore values outside it), your average gain should be about $2.50 if you switch. And this will be true whether you start with envelope A, or B. I can't say the exact amounts, because it depends on how F(10) differs from F(20). The exact amounts are the kind of thing you can find with a statistical simulation like yours.

Or, just calculate the average percentage gain if you switch. You will find that it is about 25%, whether you switch from A to B, or from B to A.

I did look. Maybe you didn't read mine: "I mean examples where actual envelopes have been filled with actual money, and presented to an actual person. That's the only way statistics matter."

If you simulate the problem, then all you are testing is your interpretation of the problem, warts and all. And applying statistics to the results may show how statistics work, but nothing about whether your interpretation is right or about the original problem. Micheal also simulated the problem.

Whether or not you want to address it, the distribution of the possible values in the envelopes is required for the problem. Ignoring it is making a tacit assumption that includes an impossible distribution.

Sorry, I don't know MathJax. With having broken my foot on Friday (did you notice a gap in my responses?), it is taking all the effort I can spare to reply here. If you can point out an on-line tutorial, I'll look at it in my copious "free time."

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

In short, the "need to" include the probability distribution is because the formulation of the "simple algebraic solution" is incorrect if you exclude it.

I'm assuming that you understand the difference between conditional and unconditional probabilities. The expected value of the other envelope, given a certain value in your envelope, requires that you use probabilities conditioned on your envelope having that value. So even if X is unknown:

1. The unconditional probability that the other envelope contains $X/2 is not 1/2, it is Pr(L=$X/2)/2.
2. The unconditional probability that the other envelope contains $2X is not 1/2, it is Pr(L=$X)/2.
3. If you have $X, those are the only possible cases.
4. To get conditional probabilities, you need to divide the probability of each possible case by the sum of them all.

So the expected value of the other envelope, given that you have $X, is:

[($X/2) * Pr(L=$X/2) + (2*$X) * Pr(L=$X)] / [Pr(L=$X/2) + Pr(L=$X)]

Note how this can be different, for different values of X, if Pr(*) varies. I'm not trying to "be that guy," but apparently nobody read my first post either. I pointed out examples of how that can happen: https://thephilosophyforum.com/discussion/comment/196299

The "simple algebraic solution" assumes that the probability of any two possible values of L is the same, which makes this reduce to:

[($X/2) + (2*$X)] / 2]

Note that this assumption means that the probability of getting $10 is the same as getting $5, which is the same as getting $2.50, or $1.25, or $0.625, or $0.3125, etc. And it also implies that any two possible high values are the same, which means that getting $10 is just as likely as $10,240, or
$10,485,760, or $10,737,418,240,etc. THERE IS NO UPPER BOUND TO THE VALUES IMPLIED BY THE "SIMPLE ALGEBRAIC SOLUTION."

This is another reason (besides failing the requirement that the cases be equivalent except in name) that the Principle of Indifference can't be applied.

If, at that point, you postulate a value in an envelope, you need to postulate a probability distribution that covers all possible ways that value could be in an envelope. Even if it is unknown. Did you miss the part where I said that, if you don't know the value, you technically have to integrate over all possible sets? But that you can prove both envelopes have the same expectation with any legitimate distribution?

And if, at that point, you postulate a value for any of these functionally equivalent quantities:

• The lower of the two envelopes.
• The higher of the two envelopes.
• The absolute value of the difference.
• The sum.

... then your solution only uses a probability for one, and it divides out. But if you look in it, you need two.

Honestly, my conclusions are much the same as yours. I just explain them correctly. Why are you arguing? Do you even know what you are arguing about?