I would return the envelope and pick the other. Whether I get more or less, at least, I would have eliminated curiosity because there would be no more uncertainty.

I'm learning as I go here. — Srap Tasmaner

These are the lecture notes from one of my professors, which should give you some idea of what Bayesian statistics is suppose to look like.

http://www.math.montana.edu/ahoegh/teaching/stat491/notes/index.html

When you open the envelope and look at what's inside, your numéraire is dollars. When you don't open it your numéraire is 'value of contents of the first envelope', ie Y. So it's a different calculation that can validly have a different result. — andrewk

I'm still not seeing it. We've done to death the example of finding £10 and you calculate an expected gain of £2.50. Or you can do a generic calculation and show an expected gain of £Y/4, for any value of Y.

In general, it's best to avoid arguments based on 'absurdity — andrewk

Then let's go with "inconsistency": given such a pair of envelopes, you can readily construct a pair of arguments that tell you each has a higher expected value than the other. I would prefer finding that such arguments are invalid.

Cool, thanks.

Then let's go with "inconsistency": given such a pair of envelopes, you can readily construct a pair of arguments that tell you each has a higher expected value than the other. I would prefer finding that such arguments are invalid. — Srap Tasmaner

Those results are both correct, and it's because each is done from the point of view of observing that envelope - two opposite points of view. It's analogous to how two spaceships travelling at half the speed of light relative to one another both measure time as going more slowly in the other spaceship. That 'feels' inconsistent but when you dig into it, trying to locate and fix the inconsistency, you find there isn't one.

I'm still not seeing it. We've done to death the example of finding £10 and you calculate an expected gain of £2.50. Or you can do a generic calculation and show an expected gain of £Y/4, for any value of Y. — Srap Tasmaner

That's right, in this case using either dollars or the value of Y as numéraire gives the same result - an expected gain equal to Y/4. That's why I said 'can' rather than 'does'. But using X as numéraire gives a different result - an expected gain of zero. It really is worth spending the time to come to grips with the numéraire concept. It has many more important applications than just in probability.

Also, the Bayesian calculation now has the graphs in. I think they look rather nice.

You can't use Y as a value defined independently of X and average across possibilities using that value.

You can't use Y as a value defined independently of X — Snakes Alive

Something is wrong with the way Y is being used, clearly, but I'm not sure this is it. You can get the (b) method out of the (a) method just by substituting the possible value of Y in terms of X. (And you can fix (b), if you start mistakenly there, by substituting to get (a).)

Yeah, I realized this after writing my original post, but I don't have a good way to exposit the issue. I suspect that epistemic and metaphysical possibilities are being confused in some inappropriate way.

That is, given some fixed value of X (and we know there to be some such), it is not metaphysically possible for the unopened envelope to be 'either 5 or 20' (read as an exhaustive disjunction), since neither of these is twice the other (and so this is inconsistent with what we know to be metaphysically so, that the unopened envelope is either X or 2X, for some X, while 5 and 20 are not X and 2X for any X), and so our knowledge of the situation should bar this space of possibilities, since our knowledge that this is metaphysically impossible should rule out our averaging over this epistemic space as regards our outcomes.

it is not metaphysically possible for the unopened envelope to be 'either 5 or 20' (read as an exhaustive disjunction), since neither of these is twice the other ( — Snakes Alive

Can you justify that 'since'? There is no justification provided in the sentence in which it occurs, because the words following it have no logical relation to the words before it. It sounds like you're saying that, having observed 10 in the envelope, you now believe that the other envelope might contain some amount other than 5 or 20. If that's not what you're saying, what are you saying?

Also, what is 'metaphysically possible' and how does that differ from 'possible'?

You can't use Y as a value defined independently of X and average across possibilities using that value. — Snakes Alive

We agree on that. Y and X are interdependent. That's why I define Y as

$\quad\quad\quad\quad\quad Y = X\cdot (1+B)$

where B is a Bernoulli random variable with probability 0.5, that is independent of X. B is independent of X, but Y is not.

It sounds like you're saying that, having observed 10 in the envelope, you now believe that the other envelope might contain some amount other than 5 or 20. — andrewk

No. The point is that it is not a possibility that the exhaustive disjunctive possibilities of what's contained in the other envelope are 5 and 20. This is because it's known already that for some X, the amount in the other envelope is X or 2X. But 5 and 20 are not X and 2X for any value of X. Thus it cannot be that the exhaustive disjunctive possibilities are these, and cannot be that in determining the average expected value of switching, we average across those possibilities, since our knowledge state rules out this way of setting things up.

If you define Y in this way, then there are two possible states: you have drawn X (where B = 0), or you have drawn 2X (where B = 1). Now eliminate the Y altogether, do the calculation, and you'll see that your expected value is 1.5X regardless of whether you switch or not.

The point is that it is not a possibility that the exhaustive disjunctive possibilities of what's contained in the other envelope are 5 and 20. — Snakes Alive

What do you mean by 'exhaustive disjunctive possibilities'?

it's known already that for some X, the amount in the other envelope is X or 2X. But 5 and 20 are not X and 2X for any value of X — Snakes Alive

That's all correct, but is not inconsistent with my note. Nowhere does it say that 5 and 20 are X and 2X, given we have seen that Y=10. Rather, we know that X is either 5 or 10, so either

• we have X in the envelope we opened, so that X = Y = 10; OR
• X is in the other envelope, so we have Y = 2X = 10 and X=5 is in the other envelope.

What do you mean by 'exhaustive disjunctive possibilities'? — andrewk

Exhaustive disjunctive possibilities are those that are mutually exclusive and jointly exhaustive, viz. those that form a partition over the space of possible outcomes.

That's all correct, but is not inconsistent with my note. Nowhere does it say that 5 and 20 are X and 2X, given we have seen that Y=10. Rather, we know that X is either 5 or 10, so either

we have X in the envelope we opened, so that X = Y = 10; OR
X is in the other envelope, so we have Y = 2X = 10 and X=5 is in the other envelope. — andrewk

No, this doesn't work. If you define Y in terms of X, then everything you write with Y can (must) be rewritten in terms of X. Do this and the illusion that Y is a value you can average over multiple situations disappears. You are acting like Y is some other value that you can now double or half in terms of what you draw. But it is not: it is, for some fixed X, either X or 2X. It follows that whatever the hidden amount is, the two possibilities over which we average must be such that one is double the other. 5, 20 violates this and so cannot be what we average over – it is not consistent with the way the problem is set up, because we know there is some value of X fixed at the time of the choosing of the two envelopes, and the only possibilities are that the hidden amount is either X or 2X, regardless of what amount you look at in one envelope. The fact that you even see one envelope at all is a complete red herring, only thrown into the problem to seduce you into making this very fallacy.

Now eliminate the Y altogether, do the calculation, and you'll see that your expected value is 1.5X regardless of whether you switch or not. — Snakes Alive

If you define Y in terms of X, then everything you write with Y can (must) be rewritten in terms of X. — Snakes Alive

Eliminating Y is making X the numéraire. That's why you need to address the numéraire issue, as explained in this post. When we use X as numéraire, the expected gain from switching is zero units of X, but when the numéraire is dollars, it is a gain.

You can't use dollars in that way.

You've agreed that Y must be defined in terms of X. X is fixed. Therefore, Y = 2X, or Y = X. It simply doesn't matter what the dollar value of Y is. What you want to do is on the one hand agree that Y is defined in terms of X, and agree that there is some fixed value of X, such that X and 2X are the values of the envelopes, and calculate expected utility solely in reference to Y (a dollar amount) not expressed in terms of X. You cannot do this. If you do, you will be averaging over situations with respect to which X is different in each. But you cannot average over outcomes in which X is different, for we know that X has a fixed value, and so whatever it is, it must remain constant over the outcomes that we use to average.

You are being misled by naming a variable. Just replace Y with its definition in terms of X, which you must do anyway since that's how you've defined it, and the illusion disappears.

You've agreed that Y must be defined in terms of X. X is fixed. — Snakes Alive

That doesn't prevent us from modelling our uncertainty about it by representing it as a random variable. In Bayesian analysis we model a fixed, unknown population parameter like X as a random variable from an assume distribution we call the 'prior'. We then use new information to update that distribution to a more accurate 'posterior distribution'.

I did say when presenting the Bayesian analysis that people who don't accept Bayesian methods won't agree.

This has nothing to do with whether an analysis is Bayesian or not. Your error is a basic conceptual one.

This is reflected in the fact that you are empirically wrong about the results of actually playing the game.

I suspect you are basing that claim on some computer simulation. Firstly, a computer simulation cannot prove anything in probability theory. Secondly, if it even suggests what you believe it does, I am convinced it contains an inappropriate assumption. Present your code and we can discuss it.

If you play the game, switching on average will not afford you any gains. You predict it will. You are wrong.

What evidence do you have for that claim? By all means present it and we can discuss it.

The prior simulations demonstrated this.

If you don't accept that, literally just go out and play the game. Do trials where one person always switches, and another never does, giving them randomized values for X. The switcher will not receive 1.25 times the money on average that the non-switcher does.

Btw, I added some functionality to the simulation andrew doesn't like. Counts some more stuff on each run just for extra confirmation of what's going on:

Number of trials: 500,000

Switch take: 187,394,381,897
Switch payout ratio: 1.4996906679464
Switch fails: 250,148
Total gain by switching: 62,439,025,501
Percentage of total: 0.33319582406328
Total loss by switching: 62,516,330,895

No Switch take: 187,471,687,291
No switch payout ratio: 1.5003093320536

Half the time you get X and win by switching to 2X; half the time you got 2X and lose by switching to X. Over a decent number of trials, your net gain is zilch.

That is correct, but has no bearing on the problem. The problem asks about the player's expectation, not the game host's expectation, which is what the procedure you suggest would reflect.

No, the problem asks what you should do.

I covered that on the first page, but that was before you joined the discussion. What the player 'should do' depends on her utility function, and none of the analysis to date has addressed the question of utility functions. Implicitly, they have assumed the utility function is the identity function, rather than the usual concave-down assumption, which reduces the problem to the player selecting the option from which she expects the highest gain, based on the available information.

reduces the problem to the player selecting the option from which she expects the highest gain, based on the available information. — andrewk

From the OP:

You are playing a game for money.

Did you think something else was relevant? Really? Like what?

I don't understand how your question relates to the quote. Can you elaborate?

Suppose I am a player and I accept your analysis. Suppose also the envelopes are helpfully labeled L and R. Time for me to choose.

I consider L. But whatever value L has, R has 5/4 of that value. So I consider R. But whatever value R has, L has 5/4 of that value. So I consider L again.

Is there a third option besides:
(1) the two envelopes have the same expected value;
(2) one has a higher expected value than the other.

But whatever value L has, R has 5/4 of that value. So I consider R. But whatever value R has, L has 5/4 of that value. So I consider L again. — Srap Tasmaner

I don't believe that description correctly represents the analysis.