I agree with all of those.

*** You might have (3) and (4) a little wrong but I can't judge. The McDonnell & Abbott paper makes noises about the player using Cover's strategy having no knowledge of the PDF of X.

1. If the player does not know the amount in the chosen envelope then the expected gain from switching is zero. — Andrew M

You might not find that everyone agrees on this first claim since, under conditions of equiprobability, the paradox arises whereby (1) acquiring knowledge of the content X of the first envelope yields and expected value of 1.25*X for switching and (2) merely acquiring knowledge of the content of the first envelope ought not to change anything to the already determined expectation of switching. This is what makes the assumption of equiprobability so problematical (since (2) can be inferred from it).

On edit: I looked at earlier posts of yours in this thread, such as this one, and I see that we are on the same page.

Random selection, which means equal probability, mitigates observational bias by treating each n in a population the same. This helps maximize the influence of an unknown population distribution on your sample.

This is also the reason it is used as an uninformed prior in Bayesian statistics, by setting the unknowns equal the prior will have less pull on the posterior, allowing your sample to have more influence on the outcome.

If you don't know the distribution you try to give the population distribution as much room as you can to revel itself. You do this by trying to limit the influence of your personal bias both conscious and unconscious.

Not really something I should have to explain, but it seems it needed clarification.

Random selection, which means equal probability, mitigates observational bias by treating each n in a population the same. — Jeremiah

The equiprobability that I am talking about is the posterior equiprobability between the two possible contents of the second envelope: either X/2 or 2*X. This posterior equiprobability only is guaranteed by an unbounded prior distribution. If one rather assumes a prior distribution that assigns the same probability to a finite population of possible envelope contents, then such a prior isn't uniform since it assigns a zero probability (or zero probability density) to all the conceivable envelope contents that fall outside of this finite discrete (or continuous albeit bounded) range. (One might also consider the case of unbounded albeit convergent, and hence normalisable, probability density functions. And those aren't uniform either).

The equiprobability that I am talking about is the posterior equiprobability between the two possible contents of the second envelope: either X/2 or 2*X. — Pierre-Normand

Okay, tell me if I'm doing this wrong.

Let S be the smaller of the two values, M be your envelope, and N the other. This is certainly true:

$\small \begin{align} E(N) &= (2S)P(M=S)+(S)P(M=2S)\end{align}$

And so is this:

$\small \begin{align} E(N\mid M=a) &= (2S)P(M=S\mid M=a)+(S)P(M=2S\mid M=a)\end{align}$

Let $\small p=P(S=a\mid M=a)$; then $\small 1-p=P(S=\cfrac{a}{2}\mid M=a)$. Now we can say this:

$\small \begin{align} E(N\mid M=a) &=2ap+{a\over 2}(1-p) \\&= a\cfrac{3p+1}{2}\end{align}$
Possible values of p:

$\small \begin{align} p&=1\to E(N\mid M=a)=2a\\p&=0\to E(N\mid M=a)=\frac{a}{2} \end{align}$

So far so good. But we cannot do this:

$\small p=\cfrac12\to E(N\mid M=a)=\cfrac{5a}{4}$

anymore than we can do this:

$\small p=\cfrac13\to E(N\mid M=a)=a$

There are only two possible values for N, given an observed value for M, and the only two values for p that produce possible values for N are 0 and 1.

How am I to interpret that result?

So far so good. But we cannot do this:

$\small p=\cfrac12\to E(N\mid M=a)=\cfrac{5a}{4}$ — Srap Tasmaner

(How should I proceed in order to quote your formulas correctly?)

Why couldn't you do that? If the initial set up calls for randomly assigning values for the two envelopes in the finite range ((1,2),(2,4),(4,8)) for instance, then, in that case, assuming the player knows this to be the initial set up (and hence uses it as his prior) then the posterior probability conditionally on observing any value of M that isn't either 1 or 8 (that is, conditionally on values 2 or 4 being observed) p will indeed be 1/2.

I was arguing that it can't be 1/2 regardless of the value being observed in the first envelope unless the prior being assumed is an infinite and uniform (and hence non-normalisable) distribution.

If you right-click on a TeX formula and select 'Show Math As...' then 'TeX commands', then you can copy and paste the code for that in between [ math ] and [ /math ] delimiters (without the white space inside the brackets) and, all going well, it will appear as nicely displayed TeX in your post.

Unfortunately I don't think there is a single-click facility to quote a post including its TeX properly. It has to be done equation by equation, which is a bother.

If you right-click on a TeX formula and select 'Show Math As...' then 'TeX commands', then you can copy and paste the code for that in between ... — andrewk

Thanks so much. I'll do this from now on.

$\small \begin{align} E(N\mid M=a) &= (2S)P(M=S\mid M=a)+(S)P(M=2S\mid M=a)\end{align}$ — Srap Tasmaner

Isn't this the same as:

$\small \begin{align} E(N\mid M=a) &= (2a)P(S=a\mid M=a)+(\frac{a}{2})P(2S=a\mid M=a)\end{align}$

@andrewk @Pierre-Normand

... Isn't this the same as: ... — Michael

Why are they not the same? In the case where the unseen envelope M is the smaller one, its content is indeed S = a/2.

Sorry, I'm a bit confused by your response. Did you read me as saying "this isn't the same as" rather than "isn't this the same as"?

Sorry, I'm a bit confused by your response. Did you read me as saying "this isn't the same as"? — Michael

Indeed! I misread you precisely in this way. We agree then.

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.

without saying which it is that you think is correct — JeffJo

I have been very clear on my position, and this is why I stonewalled you. It was clear to me that you were not reading my posts. You kept "disagreeing" with me on points I agreed with you on.

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.

The simple truth of this is, if I walked up to you on the street and handed you one envelope and said one of these has twice as much as the other, you'd have no clue as to range, distributions or anything of that sort, even after seeing Y. You could make speculations, but that is all they would be and they would carry a high degree of uncertainty you could never account for in your calculations, as they would just be wild guesses.

You could spin all types of models, but you're still just shooting in the dark. The truth of it is, in that moment, switching or don't, you stand to gain x and you stand to lose x. That is the one thing we know is constant.

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.

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]." — JeffJo

But suppose you don't do this. Suppose you just select some X at random from {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, put it in one envelope, and then put 2X in the other envelope. I select one at random and open it to see £10. Am I right in saying that {£5, £10} and {£10, £20} are equally likely?

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. — JeffJo

If we don't have this information then we should apply the principle of indifference, which will have us say that {£5, £10} and {£10, £20} are equally likely.

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

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.

Still not interested. I don't even read your post, I just kind of skim over them. You know, like you were doing to me.

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.

The simple truth of this is, if I walked up to you on the street and handed you one envelope and said one of these has twice as much as the other, you'd have no clue as to range, distributions or anything of that sort, even after seeing Y. You could make speculations, but that is all they would be and they would carry a high degree of uncertainty you could never account for in your calculations, as they would just be wild guesses.

You could spin all types of models, but you're still just shooting in the dark. The truth of it is, in that moment, switching or don't, you stand to gain x and you stand to lose x. That is the one thing we know is constant. — Jeremiah

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

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.

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.

I already know where you stand, Michael.

↪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.

You don't "guess" a prior. Even if you use an uninformative prior you have to justify that approach.

I already made several comments on the justification behind random samples. Read much?

People who might be reading this thread, who have no exposure to random sampling or Bayesian inference, should not walk away with the idea that using random sampling is a random decision or that Bayesian inference is about guessing.