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.

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

I thought putting our ignorance front and center could be a feature rather than a bug.

Also if we do attempt to estimate the shape of the problem as a whole, it will be in terms of X.

For instance we could ask a simplish question like, what is P(3X - a > a)?

We'll end up doing exactly things and not doing exactly the same things.

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. - andrewk
Maybe I was mixing Andrews up. I apologize. — JeffJo

I don't advocate it either. I advocate E = ((v/2) * Pr(v/2,v) + (v*2) * Pr(v,2v)) / (Pr(v/2,v) + Pr(v,2v)).

So what am I supposed to do if I don't know how the values are selected? 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. There's more to gain than there is to lose, and a loss of £5 is an acceptable risk. — Michael

You can make a similar argument for keeping. Suppose you choose an envelope but, in this game, instead of opening your envelope the host opens the other envelope and shows you the amount which is £10. Now your choice is to switch to the £10 envelope or keep your chosen envelope. It seems that there is more to gain by keeping, so you keep.

The arguments are symmetrical. That could be an argument to always select the unknown quantity. Or it could be an argument to be indifferent to keeping or switching. The latter would be an application of the indifference principle based on a symmetry.

That conclusion is also supported by running the game many times which shows that the actual gain from always switching (or always keeping) is more or less zero.

I agree (that we should stick when shown the other envelope).

That conclusion is also supported by running the game many times which shows that the actual gain from always switching (or always keeping) is more or less zero. — Andrew M

And that's because of what I explained here.

Before looking in your envelope, do you have an expectation of gain from swapping?

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

Yes your account of objective probabilities explains it. But that is just probability simpliciter where the player's expected gain calculations are based on her knowledge of the distribution.

The problem is when knowledge of the distribution is absent. There isn't a symmetry between {a/2,a} and {a,2a} (as there is with, say, the two sides of a fair coin) because one of those envelope pairs may not be part of the distribution. And even if they were both present, there might be other relevant differences that make one envelope pair more likely to be selected than the other.

It's like treating a potentially biased coin as fair. Actual results won't reflect calculated expectations if assumptions about the coin weighting are wrong.

I ran across this line while reading one of my books and thought it pertained well to this thread:

The final criterion for deciding whether a model is "good" is whether it yields good and useful information. The motivation for using mathematical models lies primarily in their utility.

Mathematical Statistics with Applications, Wackerly, Mendenhall, Scheaffer.

I will let people interpret that in the context of the thread how they like.

Personally, I think in terms of modeling the actual OP the inclusion of distributions or switching strategies is misguided. If our goal is to assess the situation before us then they move outside that scope. However, these models do provide a useful platform to investigate properties that may be applied to other probabilistic aspects. I know undoubtedly some will take that to mean something I don't mean, 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.

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.

My simulations were there to display the inherit ambiguity in defining a prior distribution. X is an unknown, treat it like an unknown. I have never once in the entire thread expressed that there is no limit or distribution, in fact I expressed the opposite. That doesn't change the fact that X is an unknown.

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

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.

Nothing new here, just checking my understanding. (Or, rather, whether I have shed all my misunderstandings, even recent ones.) Check my math.

If I understand it correctly, our situation is something like this:


The host will choose a value for X before the player chooses an envelope. For any value $x_i$ the host chooses, the player's chance of choosing the smaller envelope is

$\small P(Y=X\mid X=x_i)=\cfrac12$

and you can also sum across all those to get the unconditional probability

$\small P(Y = X) = \sum_{x_i}P(Y=X\mid X=x_i)P(X=x_i)=\cfrac12$

And similarly the chance of picking the 2X envelope is 1/2 for any value of X or for all of them together.

Also, for any given value a we can say that the probability of picking a is

$\small\begin{align} P(Y=a)&=P(Y=X\mid X=a)P(X=a)+P(Y=2X\mid X=\cfrac{a}{2})P(X=\cfrac{a}{2})\\&=\cfrac{P(X=a)+P(X=\cfrac{a}{2})}{2}\end{align}$

And naturally P(X = a | Y = a) is just

$\small \begin{align} P(X=a\mid Y=a)&=P(Y=a\mid X=a)\cfrac{P(X=a)}{P(Y=a)}\\&=\cfrac{\cfrac12 P(X=a)}{\cfrac12\left( P(X=a)+P(X=\cfrac{a}{2})\right)}\\&=\cfrac{P(X=a)}{P(X=a)+P(X=\cfrac{a}{2})} \end{align}$

In considering the chance of choosing the smaller envelope, the choice of X drops out, but here we have the opposite: the equiprobability of choices made the the chance of picking some value a a simple mean of the probabilities of X being a and X being a/2; now those chances of choosing cancel out, and we're only comparing probabilities of X values.

Our expectation for the unpicked envelope:

$\small \begin{align} E(U\mid Y=a)&=\cfrac{a}{2}\cdot P(Y=2X\mid X=\cfrac{a}{2}, Y=a)+2a\cdot P(Y=X\mid X=a, Y=a)\\&=\cfrac{\cfrac{a}{2}\cdot P(X=\cfrac{a}{2})+2a\cdot P(X=a)}{P(X=\cfrac{a}{2})+P(X=a)} \end{align}$

And again, choice has dropped out completely.

What do we know about P(X = a) and P(X = a/2)? We know that

$\small P(X=a\mid Y=a)+P(X=\cfrac{a}{2}\mid Y=a) = 1$

nearly by definition, although really there are choices canceling out here.

Do we know that both P(X = a) and P(X = a/2) are non-zero? We know that at least one of P(X = a | Y = a) and P(X = a/2 | Y = a) is non-zero, but we do not know that both are. Without Y = a, we wouldn't even know that at least one of P(X = a) and P(X = a/2) are non-zero. Without knowing that both are non-zero, we can't even safely talk about the odds P(X = a):P(X = a/2).

I was never the one advocating for the use of a prior. I am not sure where you got that notion.

you can't define/calculate the prior distribution, and that it was a misguided effort to even try (as you did) — JeffJo

FWIW, my memory is that @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.

Your post suggests you read those old posts as showing Jeremiah is invested in some of these statistical models of the problem and he never has been.

Personally, I think in terms of modeling the actual OP the inclusion of distributions or switching strategies is misguided. If our goal is to assess the situation before us then they move outside that scope. However, these models do provide a useful platform to investigate properties that may be applied to other probabilistic aspects. — Jeremiah

I agree. Regarding the OP, the math provides no reason to be anything other than indifferent to sticking or switching.

(Of course, people may wish to stick or switch for individual reasons related to the specific context, such as utility, but that is a separate issue.)

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) — JeffJo

Isn't 2X just a transformation of X that doubles the possible values in X? So Pr(2X=a) would be equivalent to Pr(X=a/2). Here's an academic example using P(X^2 <= y).

Do we know that both P(X = a) and P(X = a/2) are non-zero? We know that at least one of P(X = a | Y = a) and P(X = a/2 | Y = a) is non-zero, but we do not know that both are. Without Y = a, we wouldn't even know that at least one of P(X = a) and P(X = a/2) are non-zero. Without knowing that both are non-zero, we can't even safely talk about the odds P(X = a):P(X = a/2). — Srap Tasmaner

Yes. You learn something about the distribution when you open an envelope (namely, that it had an envelope with that seen amount). But not enough to calculate anything useful. It's like getting a bicycle with one wheel. You might wonder whether you could get somewhere with it, but you probably can't.

I'm still working on it.

We can also say that

P(X = a) + P(X = a/2) <= 1

but other than that, their values can range freely.*** (It is in some sense a coincidence that their sum can also be described as the event Y = a.)

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

*** Urk. Forgetting that at least one of them has to be non-zero.

Yes. You learn something about the distribution when you open an envelope (namely, that it had an envelope with that seen amount). But not enough to calculate anything useful. It's like getting a bicycle with one wheel. You might wonder whether you could get somewhere with it, but you probably can't. — Andrew M

I disagree. Suppose you were to engage in one million iterations of that game and find that the seen envelope contents converge on a specific and roughly uniform distribution of amounts all belonging in the discrete range ($1,$2,$4,$8,$16), with deviations from 1/5 frequencies that aren't very statistically significant. (I am assuming that the game only allows for amounts in whole dollars, for simplicity). Wouldn't that information be useful? I would argue that the useful knowledge that you thus gain is being accrued progressively, one little bit at a time, and could be represented by the successive updating from an allegedly very tentative long tailed initial prior distribution that represents your initial expectation that the dealer likely doesn't have access to an amount of money in excess of one million dollars, say, with the peak of the distribution somewhere around $50, say. This initial prior could be very wrong, but through (Bayesian) updating it after each iteration of the game, will lead to a prior that converges towards the 'real' distribution. Hence, each envelope that you look at provides some useful information since it is more likely than not to lead you, at each step, to updating your initial prior in the direction of a more reliable one.

(I am assuming that the game only allows for amounts in whole dollars, for simplicity — Pierre-Normand

Interestingly, if the game only allows whole dollars - or even whole cents, and the player knows that, they can use it as the basis for another strategy: if the number is odd, switch, otherwise don't. That's because if the number is odd it cannot be the doubled value, so the other one must be.

To remove that option, I recast the problem with the envelopes containing IOUs rather than cash, for an amount that is a real number of cents, with an arbitrary but large number of decimal places shown. The amount is only rounded to the nearest cent (or dollar) when the IOU is cashed in.

(...) To remove that option, I recast the problem with the envelopes containing IOUs rather than cash, for an amount that is a real number of cents, with an arbitrary but large number of decimal places shown. The amount is only rounded to the nearest cent (or dollar) when the IOU is cashed in. — andrewk

That's clever.

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

*** Urk. Forgetting that at least one of them has to be non-zero. — Srap Tasmaner

Yes, I think it's reasonable to assume discreteness (with a finite but unspecified precision and range).

Wouldn't that information be useful? — Pierre-Normand

Yes. But I think the OP is asking for a general solution for one run with no special assumptions about the context (such as whole dollar amounts or million dollar limits).

Considering multiple runs as blind runs with no knowledge acquired from previous runs should be OK. Perhaps using amnesia per the Sleeping Beauty problem...

To remove that option, I recast the problem with the envelopes containing IOUs rather than cash, for an amount that is a real number of cents, with an arbitrary but large number of decimal places shown. The amount is only rounded to the nearest cent (or dollar) when the IOU is cashed in. — andrewk

:up:

Yes. But I think the OP is asking for a general solution for one run with no special assumptions about the context (such as whole dollar amounts or million dollar limits). — Andrew M

For sure, I agree, since any non-uniform prior would violate the equiprobability condition that alone grounds the derivation of the unconditional 1.25X expectation from switching.

the equiprobability condition that alone grounds the derivation of the unconditional 1.25X expectation from switching. — Pierre-Normand

I'm still confused. This makes it sound like the switching argument isn't fallacious -- it just makes an unwarranted assumption. So if every value of X were equally probable, then it would be true that you can expect a .25 gain from switching? I see how the math works, but if that's true, isn't it true whether you know the value of your envelope or not? And if that's true, isn't it also true for the other envelope as well?

If that's how the calculation goes, then something's wrong because in any pair of envelopes, trading is a gain for one side and a loss for the other.

I'm still confused. This makes it sound like the switching argument isn't fallacious -- it just makes an unwarranted assumption. So if every value of X were equally probable, then it would be true that you can expect a .25 gain from switching? I see how the math works, but if that's true, isn't it true whether you know the value of your envelope or not? And if that's true, isn't it also true for the other envelope as well? — Srap Tasmaner

Yes, yes and yes, but the assumption isn't merely unwarranted, it is impossible that it be true in any real world instantiation of the problem where the available amounts that can figure in the distribution all are smaller than some finite upper bound. It follows from the equiprobability assumption not only that every possible amount in the distribution are equiprobable but also that the distribution is unbounded(*). In that case, your prior expectation, before you open the envelope, is infinite. The probability that you would be dealt an amount X that is lower than some finite upper bound M is vanishingly small however big M might be. But if you do wind up with some finite amount X in your envelope, then your conditional expectation upon switching is 1.25X. That doesn't make the unconditional expectation from the always-switching strategy any larger than the unconditional expectation of the 'always-sticking' strategy since 1.25 times aleph-zero is aleph-zero. The situation is rather akin to Cantor's Hotel where all the (countably infinitely many) rooms are filled up and there nevertheless still is 'room' for accommodating twice as many guests.

(*) Some distributions, such as the Gauss or Poisson distributions, are unbounded and well behaved (i.e. yielding finite expectations) but they are not uniform and so don't satisfy the equiprobability assumption.

Thanks. You've told me this before -- and I appreciate your patience. I'll mull it over some more.

I think I'm just reluctant to see the simple situation of choosing between two envelopes of different values in terms of the strange behavior of infinity.

I keep thinking of switching as just being a positive or negative change, but the switching argument accepts that!

Every time I think I've got a handle on this, it slips away.

I think I'm just reluctant to see the simple situation of choosing between two envelopes of different values in terms of the strange behavior of infinity. — Srap Tasmaner

Yes, it is indeed the strange behavior of infinity that generates the paradox.

(Edit: Opps, there are some mistakes below that I'll correct later today)
((Edit again: I'll correct them in a separate reply to this post))

Maybe I can get some more mileage from my analogy with Cantor's Hotel(*). Suppose you are hosted in this Hotel in some random even numbered room X. (The rooms are labeled 1,2,3,...). You are then offered to toss a coin and move to room X/2 if you flip tails, and move to room 2X if you flip heads. You can also choose not to toss the coin and stay in your room. Whatever you do, when you leave the hotel you are being awarded your room's number worth of dollars. Suppose everyone who occupies an even numbered room gets offered the same deal and everyone chooses to flip the coin. After the move, some rooms will be empty and some rooms will have one or (at most) two occupants. But that's immaterial. The point is that guests can expect to increase their rewards by 1.25 on average. It is also true that if they had chosen to flip the coin without looking at their room number, and only know the number Y of the room where they relocated as dictated by the result of the coin toss, they would still expect to increase their reward expectation to 1.25Y if offered to move back to whatever initial room they came from (assuming they had forgotten what it is that they flipped and hence couldn't deduce where it is that they came from before choosing whether to move back).

In this case, it's the fact that the Hotel has countably infinitely many rooms that enables the assumption of equiprobability to hold. Before the coin flip, each occupant is equally likely to double her reward than she is to halve it and hence is warranted to flip. But conversely, after she moved, she is equally likely to have moved there from a rooms that was twice as valuable than from a room that was half as valuable and hence is warranted to move back if given the opportunity, assuming only that she has forgotten where she came from!

(*) On edit: What I referred to as Cantor's Hotel is better known as Hilbert's Grand Hotel, or Hilbert's Infinite Hotel.