If you could prove that always switching is the best strategy over the long term, doesn't that amount to proving that you are more likely to have chosen the smaller envelope? Why doesn't that bother you? — Srap Tasmaner

There is no long term in this setup. There is a single offering. If we want to consider a long series of such activities, it needs to be set up precisely, and that has not been done. A number of crucial conditions will need to be specified, such as whether X is always the same and if not, how it varies from one play to the next. The best strategy will vary according to how those conditions are set up.

In any case, a setup with repeated plays is a completely different probability space, and does not entail anything for this one, or vice versa.

I'm going to reprise my proof, because three pages have been added and so far nobody seems to have attempted to point out an invalid step in it:

we use conditional probabilities to calculate the gain G from switching as follows, where the envelope we looked in contained £10:

E[G] = 0.5 * E[G | X= 5] + 0.5 * E[G | X= 10] ] . . . . . . . . . . . (0)

= 0.5 * E[-X | X= 5] + 0.5 * E[+X | X= 10] . . . . . . . . . . . (1)
= 0.5 * E[-5 | X= 5] + 0.5 * E[+10 | X= 10] . . . . . . . . . . . (2)
= 0.5 * (-5) + 0.5 * (+10) . . . . . . . . . . . (3)
= £2.50 . . . . . . . . . . . . (4) — andrewk

If we want to consider a long series of such activities, it needs to be set up precisely, and that has not been done. A number of crucial conditions will need to be specified, such as whether X is always the same and if not, how it varies from one play to the next. The best strategy will vary according to how those conditions are set up. — andrewk

Such as this?

This isn't relevant to the question anyway. The switching strategy you're discussing has nothing to do with the OP.

If we want to consider a long series of such activities, it needs to be set up precisely, and that has not been done. A number of crucial conditions will need to be specified, such as whether X is always the same and if not, how it varies from one play to the next. The best strategy will vary according to how those conditions are set up. — andrewk

This isn't relevant. It simply doesn't understand the crux of the puzzle.

On a related note there's this. Apparently one can use something called "Cover’s switching strategy" even if there is no upper bound for the amount that can be placed in the envelope. It's too technical for me, so I don't know how it compares to my strategy.

Sorry, late for the party, here is my 2 cents. I agree with Michael.

Decision Theory is ∑[(amount of gain) x (probability of gain)] + ∑[(amount of loss) x (probability of loss)]
There is no loss because the game is free.
Choices are: (1) Pick Envelop 1 containing $Y, or (2) Pick Envelop 2.

Decision (1) = Y x 100% = Y
Decision (2) = 2Y x 50% + Y/2 x 50% = 1.25Y or 5/4Y
Decision (2) is better than Decision (1).

That is far from a proof and the only thing it displayed is your lack of understanding of what expected gains are.

Expected gains are an average over the long run. Probability is the frequency of occurrences of repeated random events and that includes only possible outcomes.

If you have to limit your number of trials to leverage your prior then you are essentially doing the equivalence of p-hacking.

You are presented with two envelopes, one valued at X and one valued at 2X; the average value of an envelope is 3X/2.

You choose an envelope, and do not look at the contents. You are asked if you would like to swap.

Your envelope has a definite value, call it Y.

If you have the larger of the two envelopes, the value of the other envelope is Y/2; the average value of an envelope is then (Y + Y/2)/2 = 3Y/4.

If you have the smaller of the two envelopes, the value of the other envelope is 2Y; the average value of an envelope is then (Y + 2Y)/2 = 3Y/2.

It is absurd that the average value should change depending on which of the two envelopes you have, therefore they must be equal:

3Y/2 = 3Y/4
6Y = 3Y
6 = 3

Hmmm. Let's go back.

If you have the larger of the two envelopes, then

• Y = 2X;
• the other envelope has Y/2 = X;
• the average value of an envelope is 3Y/4 = 3X/2.

If you have the smaller of the two envelopes, then

• Y = X;
• the other envelope has 2Y = 2X;
• the average value of an envelope is 3Y/2 = 3X/2.

Since Y is defined to be whatever is in the envelope you selected, then Y has a different value depending on which envelope you selected. It is not a constant, as X is. And therefore you cannot, after all, set the average values equal to each other: the LHS Y has one value and the RHS Y has another. X ≠ 2X.

Neither can you make this calculation:
If I have the larger valued envelope, I risk losing only Y/2 by switching, while if I have the smaller valued envelope I stand to gain Y. Y has two different values in this sentence. By definition.


Now suppose you are allowed to look, so you learn, say, that Y = 10. Won't that prevent the problem of Y changing values? It will be 10 in every equation.

Suppose Y = 10. Then either 10 = X, or 10 = 2X. Where before we had equations with two unknowns, now we have equations with one.

Suppose you reason as follows:
If I have the larger envelope, then the other has 5 and the average value of an envelope is 7.5.
If I have the smaller, then the other has 20 and the average value of an envelope is 15.

(It is prima facie absurd that the average value of an envelope changes depending on whether you have the larger or the smaller of the two.)

The only way this can be is if our only remaining unknown, X, changes its value depending on whether I have the larger or the smaller envelope. Since we have fixed Y, which was defined to change depending on your choice, we are forced to make X vary with your choice in order to preserve the equations. But X is a constant. Unknown, but a constant.


So what is the right way to reason, once you know that Y = 10? The simple answer is, don't. You haven't learned anything you can act on. It will turn out X = 5 or X = 10, but Y = 10 does not help you figure that out, and there is no way to calculate using Y = 10 that does not force X to vary with your choice of envelope, which is absurd.

A number of crucial conditions will need to be specified, such as whether X is always the same and if not, how it varies from one play to the next. The best strategy will vary according to how those conditions are set up. — andrewk

My version of Michael's sim was intended to be quite simple. I imagined it as the question being offered once to 2 x 10^6 participants, half of whom switch and half of whom don't. Then we aggregate the results to compare the two strategies.

What's wrong with this?

We could get into a discussion of what's wrong with it, but we'd be discussing the wrong thing. Computer programs - with very few exceptions like dedicated theorem-proving programs, which is a very niche area - can only ever be used to do calculations and generate intuitions. They cannot prove anything. To do that we need to use mathematics, which is what my proof does. If we stray away from mathematics, we just end up trading word salads, which is what most of this thread is.

If the computer program has any valid insights, it should be able to be expressed as a mathematical proof, that is presented as a series of numbered formal propositions, each being either a premise or else having a justification as to how it can be deduced from one or more previous propositions in the sequence. If you can produce such a proof, I promise to engage with it.

And here is mine. I haven't written out the justifications for the steps yet because most of them are obvious, so I'll wait until somebody engages it and challenges a line, before taking the trouble to write out the justification for it.

We use conditional probabilities to calculate the gain G from switching as follows, where the envelope we looked in contained £10:

E[G] = 0.5 * E[G | X= 5] + 0.5 * E[G | X= 10] ] . . . . . . . . . . . (0 - Premise)
= 0.5 * E[-X | X= 5] + 0.5 * E[+X | X= 10] . . . . . . . . . . . (1)
= 0.5 * E[-5 | X= 5] + 0.5 * E[+10 | X= 10] . . . . . . . . . . . (2)
= 0.5 * (-5) + 0.5 * (+10) . . . . . . . . . . . (3)
= £2.50 . . . . . . . . . . . . (4)

X does not vary. You are wrong from the start.

You can literally just check this by performing it live. Empirically, the switching strategy doesn't help. The running of the program simply does the performance, and demonstrates this.

Thus we need to start with the recognition that switchers are empirically just wrong. Whether they know the error they've committed is irrelevant.

That is a coherent response. Whether you realised it or not, you have sided with the Frequentist approach, and a Frequentist will indeed reject my calculations, saying that there are no probabilities for values of X.

I am a Bayesian, so the calculation is valid for me.

I would love to explore the Frequentist alternative, but nobody has so far (to my knowledge) offered a formal presentation of it - the Frequentist counterpart to my Bayesian calculation.

My current expectation is that there is no Frequentist alternative, because one cannot use any probability distributions at all. But I would be delighted to be shown to be wrong. I suggest you (implore you to?) have a go at writing a formal argument from the Frequentist perspective, in which none of the items are treated as random variables.

I already provided a refutation for this, actually, in my first long post. Several people have repeated this or something similar, but it's fallacious. Can you read that post and get back to me?

Your error, BTW, is having the two situations are that X = 5 or that X = 10. The correct way to think about it is that in one case, you've drawn X, and in the other, 2X.

I can spell out why what you've posted here is equivalent to what I went over in the long post, but I don't want to if you understand anyway.

Probability distributions are irrelevant.

I doubt you could even explain the differences.

Classical and Bayesian analyses are both methods of statistics. Statistics is a data science, as in the analysis of data. They are not used or fit for simple probability problems like this one. Furthermore, as a science, they are empirically based, both of them.

Your error, BTW, is having the two situations are that X = 5 or that X = 10 — Snakes Alive

Good, so you are challenging line 0 of my proof.

Why do you think it is an error? Do you not agree that, if I open the envelope and see 10, the possibilities are that X=5 or that X=10, and that I have no reason to favour one over the other?

The fact that one can think of it another way is neither here nor there. If you want to challenge this line of analysis, you need to find a flaw in it, not just say 'the correct way to think about it is this other way...'

BTW I tried to read your first post a day or so ago, but the embedded images were corrupted, so it was not comprehensible. If that has been fixed, by all means post a link (I don't know where in this long thread it is now) and I'll have another look.

Here is the post:

https://thephilosophyforum.com/discussion/comment/193041

It explains why averaging expected outcome over situations with respect to the amount you drew as constant doesn't work.

Indeed. You have written a program with repeated plays that shows it is best to switch, and Srap appears to have written one that is only slightly different and shows that it makes no difference. That corroborates my contention that if one wishes to make an argument based on repeated plays, the devil will be in the detail of the set-up of the system of repeated plays, and changing the tiniest detail can change the result.

I'm afraid the table(s) is(are) still not showing. It may be a conflict between my system and the web-site.

Perhaps you could enter the table as text rather than an embedded image.

You have written a program with repeated plays that shows it is best to switch, and Srap appears to have written one that is only slightly different and shows that it makes no difference. — andrewk

They are completely different. Michael is allowing participants access to the results of the trials so far performed. This is completely different, and of no relevance to this thread.

I regard all computer simulations as irrelevant to the question at hand. The question is one of mathematics, and must be addressed by mathematics, not programming.

If I have the larger envelope, then the other has 5 and the average value of an envelope is 7.5.
If I have the smaller, then the other has 20 and the average value of an envelope is 15.

(It is prima facie absurd that the average value of an envelope changes depending on whether you have the larger or the smaller of the two.) — Srap Tasmaner

Not if it is in two different games, which is what is happening here.

In the first case, the game has envelopes with 5 and 10 in them, average 7.5, and I got the larger envelope.
In the second case, the game has envelopes with 10 and 20 in them, average 15, and I got the smaller envelope.

What has changed is not just whether I have the larger envelope, but also what the games-mistress has put in the envelopes. So of course the average is different.

What is the probability that the envelopes presented to you are valued at 5 and 10? (P(X = 5).)

What is the conditional probability that the envelopes presented to you are valued at 5 and 10, given that you chose the 10? (P(X = 5 | Y = 10).)

That's what you need to calculate. How will you do it?

The first question is about the prior distribution of X. My understanding is that most posters are assuming this is unknown to the player, and I have gone along with that. So the probability cannot be estimated.

The second question is about conditional probabilities. Given I know I have 10, the two possibilities that are open to me are 5, 10 and 10, 20, and I have no basis on which to distinguish between the two. So, being a Bayesian, I make the probabilities equal at 0.5. That approach is embedded in line 0 of the proof:

Proof: We use conditional probabilities to calculate the gain G from switching as follows, where the envelope we looked in contained £10:

E[G] = 0.5 * E[G | X= 5] + 0.5 * E[G | X= 10] ] . . . . . . . . . . . (0)
= 0.5 * E[-X | X= 5] + 0.5 * E[+X | X= 10] . . . . . . . . . . . (1)
= 0.5 * E[-5 | X= 5] + 0.5 * E[+10 | X= 10] . . . . . . . . . . . (2)
= 0.5 * (-5) + 0.5 * (+10) . . . . . . . . . . . (3)
= £2.50 . . . . . . . . . . . . (4)

I have no idea what a Frequentist would do.

the two possibilities that are open to me — andrewk

What does that mean? You know one of them must be in the distribution of X, but you don't know which. Are you claiming to know that both are in the distribution?

Deleting my post doesn't make you right.

What does that mean? You know one of them must be in the distribution of X, but you don't know which — Srap Tasmaner

I mean that I know my envelope has 10 and that the other envelope has either 5 or 20.

I don't know what you mean by 'one of them must be in the distribution of X'.