To remove that option, I recast the problem with the envelopes containing IOUs rather than cash — andrewk

I don't think we really need to agonize over the amounts supposedly being money. We could use real numbers and play competitively. The winner is just whoever ends up with the highest number.

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.

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.

Fair enough.

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.

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.

The result that Pr(X=$10) must be zero is not an issue then, because F(X=$10) might not be. — JeffJo

Right, right. (I am actually studying in my spare time, I swear.)

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. — JeffJo

If I may take advantage of your patience a bit more ...

Suppose I naively approach the problem this way: I think I'm solving for an unknown; I observe the value of y, and I write down my equations:

(1) y + u = 3x
(2) y = 10

I can then get as far as

u = 3x - 10

but lacking, say,

*(3) 3x = 30

I'm stuck with with an equation that still has two unknowns, so I am forced to treat u and x as variables rather than simply unknown values. As you note, instead of using x I could also write

(4) y = ru

where r ∊ {½, 2} but that leaves me with two sets of equations:

{y = 10, u = 2y}, {y = 10, u = y/2}

Either of those can be solved, but not knowing the value of r, it still amounts to an equation

u = 10r

with two unknowns. Since I can't solve that, I'm back to variables instead of unknowns.

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. — JeffJo

If I'm told that one envelope contains twice as much as the other, and if I pick one at random, am I right in saying before I open it that there's a 50% chance that my envelope contains the smaller amount?¹ If so, I must also be right in saying after I open it and see the amount that there's a 50% chance that my envelope contains the smaller amount (assuming I don't know how the values are selected).

If I'm right in saying that there's a 50% chance that my envelope contains the smaller amount then I am right in saying that there's a 50% chance that the other envelope contains twice as much as my envelope (and a 50% chance that the other envelope contains half as much as my envelope).

¹ Consider what you said here:

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.

... [Y]ou have been misinterpreting me since you joined the conversation. ...

If I'm told that one envelope contains twice as much as the other, and if I pick one at random, am I right in saying before I open it that there's a 50% chance that my envelope contains the smaller amount? If so, I must also be right in saying after I open it and see the amount that there's a 50% chance that my envelope contains the smaller amount. — Michael

I share your frustration, Michael.

If I offer you a choice between envelopes containing $5 and $10, you have a 50% chance of picking the envelope that has $5 in it. Having chosen an envelope, you no longer have a chance of picking the $5 envelope -- you either did or didn't.

You still have to express your uncertainty about whether the envelope you did pick was the smaller or the larger, since you don't know the contents of both envelopes, and seeing $10 doesn't tell you whether you got the smaller or the larger. But you cannot say that the amount you observe has a 50% chance of being the smaller. Given more complete knowledge, you would find you had been saying that $10 has a 50% chance of being smaller than $5, which is absurd.

The only safe way to express your uncertainty -- that is, the only way to formulate it so that increasing your knowledge wouldn't render your beliefs absurd -- is conditionally. And this makes sense. If the larger amount is $10, then picking $10 is necessarily picking the larger amount; if the larger amount is $20, then picking $10 is necessarily picking the smaller amount.

When you work through expressing the probabilities conditionally, you find that the 50% associated with your choosing an envelope cancels out. And in a sense, it should -- we're now working out the consequences of your choice. The uncertainty that remains does not derive from your choosing at all -- we're past that. The uncertainty that remains is all down to the host's choice of envelope values.

After that, I'm still a bit murky, I'm sorry to say. Part of it I can see as the sometimes counter-intuitive nature of conditional probabilities, but part of it still eludes me.

This is why I think the issue is related to the debate between frequentists and Bayesians. For objective Bayesians, "probability quantifies the reasonable expectation everyone (even a "robot") sharing the same knowledge should share in accordance with the rules of Bayesian statistics, which can be justified by Cox's theorem"¹. The objective Bayesian will say that an already-flipped coin has a 50% probability of being heads, even if it's actually tails, and that my £10 envelope has a 50% probability of being the smaller amount, even if it's actually the larger amount, whereas the frequentist would deny both of these (as far as I'm aware).

It's only the difference between describing your expectation conditionally and unconditionally. By describing your expectation conditionally, you leave room for the future evidence you would rely on to update.

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.

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

Statistics is used on an experimental data set from repeated trials. — JeffJo

Yes, and it is also used on observational data sets to make generalized inferences about a population.

You people really need to drop this Classical v. Bayesian mind set many of you have. The idea that they are somehow different tool boxes is misleading, they come from the same tool box.

1. For a single trial, the player cannot calculate an expected value for the other envelope, and therefore either (a) they cannot make a rational decision to switch or stick, or (b) they must use some criterion other than expected value.

2. For multiple trials, the Always Switch and Always Stick strategies are, in the long run, indistinguishable.

If I am the player, and I know that (2) is the case, that does not entail, in the case before me, either that I should switch or that I should stick. For any particular trial, either switching or sticking is the right thing to do. But once I know that (1) is the case, I can return to (2) and conclude either that there is no way for me to steer the outcome toward gain, or that I should try some other strategy. (Even if I only get the one chance, I should, if I can, use a method that improves my chances of gain, even if it's a small improvement, unless the disutility of using such a method outweighs the expected gain.) Or I could return to (1) and see if there is anything besides expected value out there.

I believe both (1) and (2), but I am not clear on the relation between them.

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

There is not enough information to calculate expected gain.

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

Assuming your prior is correct, that is.

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

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

you became belligerent started saying you would not read anything I wrote until I had read all of the irrelevant posts. — JeffJo

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

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

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

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

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

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

+++++

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

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

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

+++++

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

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

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

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

+++++

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

Assuming your prior is correct, that is. — Jeremiah

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

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

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

+++++

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

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

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

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

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

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

+++++

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

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

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

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

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

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

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

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

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

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

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

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

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

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

When all I said was that any use of statistics - which you did advocate repeatedly - — JeffJo

Statistics is simply not designed for a problem like this; it is better to just use basic mathematics. Statistics is for analyzing data. — Jeremiah

This is me on page six. You are grossly mistaken on this point, never once did I advocate the use of statistics for addressing the OP. I talked about statistics in response to others but I was very vocal and clear that it should not be used for the OP. Had you read the thread you would have known this and been able to place my posts concerning statistics in proper context.

Already did that. It was not that hard.

You are removing it from its context to make it look bad — JeffJo

Just following your lead.

So that "observational data set" is the "experimental data set," isn't it? — JeffJo

I find it interesting that you didn't pick up on what happened there.