I'll rephrase.

Did you imagine that players might have some other goal besides maximizing their monetary gain?

This just isn't relevant. Assuming that one wants to make the most money on average, the puzzle remains, and that is clearly the point of the OP. You are missing the point.

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

Your reasons?

Yes, I imagine they would aim to maximise their expected increase in utility, not their expected monetary gain.

So for instance if the envelope contains $1m and there's something I desperately want that costs $750,000, but nothing much I want that costs between $1m and $2m, I would be silly to swap, because my loss of utility on halving my winnings is greater than my gain in utility from doubling them.

But I am happy to go along with the implicit assumption that has been made by all commenters, that the player's utility function is the identity function.

If the distinction between 'maximise expected winnings' and 'what she should do' is irrelevant (which it is if we accept the implicit assumption that utility=identity, which I've said several times now that I'm prepared to do) it's hard to see why you devoted this post to emphasising the distinction.

Further, my detailed analysis adopts that implicit assumption by asking how the player can maximise her expected winnings, not what she 'should do'.

It bears no resemblance to the analysis. For a start the analysis is about expectations, and your statement doesn't mention them. If you believe it fairly represents the analysis, the onus is on you to justify that belief.

Holy shit, fine, so we're on the same page now. So respond to the initial points now that this pointless tangent is over.

Do you have another question? If so, what is it?

How do you respond to the fact that your analysis is empirically wrong?

I don't answer questions based on false premises. Have you stopped beating your wife?

Will someone, or will they not, get more money on average as a result of choosing the switching strategy, as opposed to choosing the not-switching strategy?

It's a meaningless question. 'on average' is not a meaningful statistical concept. We can only meaningfully talk in terms of expected values. The expected values depend on the distributions of the random variables, and those distributions will depend on the information available to the person that is forming the expectation.

If you run eight billion trials, giving equal values of X for each trial to each player, will the switcher and non-switcher converge roughly on the same payout?

I did my best to read your code. Although I don't speak much PHP I think I can see what it's doing. It appears to me that it does not reflect the information available to the player because it treats X as a variable uniformly distributed over the range from 1 to num_trials whereas, once the player has seen the amount Y in the first envelope, they have narrowed the possibilities for X down to two possible values: Y or Y/2. So if that interpretation is correct I would say the code does not reflect the player's expectations.

As described in the post immediately above, that setup does not reflect the player's knowledge and expectations.

Can you answer the question?

The question is ill-defined. To answer it, you'd have to specify your algorithm, which you have not done. If we take srap's PHP program as the algorithm, the expected value of the winnings from the switching and non-switching strategy would be the same but, as I've pointed out twice now, that simulation does not reflect the information available to the player. All it tells you is that, if you know what X is, a strategy of switching has the same expected reward as a strategy of not switching. But the player doesn't know what X is, so it doesn't reflect the problem.

I have to go to bed now. I'll look in again tomorrow if time permits.

It is not ill-defined. The game described is one we can actually go out and play. I am asking about if we actually go out and play the game, what will actually happen? Will switching get me more money on average?

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... — Snakes Alive

I went over this with Jeremiah.

We have £10 in our envelope. Our envelope is either X or 2X. If our envelope is X then the other envelope is 2X and if our envelope is 2X then the other envelope is X. Therefore the other envelope is either X or 2X. But it is wrong to infer from this that of the possible values in the other envelope, one is twice as big as the other, and that's because our two antecedents (and so consequences) use two different values of X, and your interpretation of the conclusion conflates them.

If our £10 envelope is X then X is 10 and the 2X envelope contains £20. If our £10 envelope is 2X then X is 5 and the X envelope contains £5. Therefore if our £10 envelope is either 2X or X then the other envelope contains either £5 or £20.

Given the premise "my envelope contains £10", the subsequent premise "my envelope is either X or 2X" is identical to the premise "X is either 10 or 5". Your argument doesn't seem to recognize this.

once the player has seen the amount Y in the first envelope, they have narrowed the possibilities for X down to two possible values: Y or Y/2. So if that interpretation is correct I would say the code does not reflect the player's expectations. — andrewk

And in every single case, whatever the value of Y, the player will choose to switch. I model the results of switching, which are quite clearly the same as not switching, contradicting the player's stated reason for switching, namely the expectation of gain. No such expectation is fulfilled. Half the time switching is a mistake for the most obvious possible reasons.

And I proved that you are 100% wrong. — Jeremiah

You didn't.

What's the value of X where:

1. Envelope A contains £10
2. Envelope B contains either X or 2X

As described in the post immediately above, that setup does not reflect the player's knowledge and expectations. — andrewk

None of us has yet to actually play this game; as such, we reflect the player's knowledge and expectations.

It's a meaningless question. 'on average' is not a meaningful statistical concept. We can only meaningfully talk in terms of expected values. The expected values depend on the distributions of the random variables, and those distributions will depend on the information available to the person that is forming the expectation. — andrewk

The definition of a statistic is. . . .

A statistic is any quantity that can be calculated from the observed data.

. . . the word mean when referring to an average calculated over an entire population. A mean is therefore a parameter. When referring to the average in a sample--which is both a statistics and estimate of the population mean--. . .

The Statistical Sleuth, A course in Methods of Data Analysis. By Ramsey/Schafer

The expected value of a random variable is just the mean of the random variable. — Statistics How To

.

You open A and it has 10 bucks, but you don't know if that is 10=X or 10=2X so A is still defined as [X,2X]. If A is X then B is 2X and if A is 2X then B has to be X. So B is still defined as [X,2X]. That part remains the same. — Jeremiah

I addressed this.

If A is X then X is 10 and B is 2X = £20.
If A is 2X then X is 5 and B is X = £5.

You're conflating different values of X when you define B as [X, 2X] and say that of the possible amounts, one must be twice as big as the other.

It's B: [X where X = 5, 2X where X = 10] to compare with A: [X where X = 10, 2X where X = 5].

This is a rerun. . . .

R and S are mutually exclusive events.

R and S are mutually exclusive events. — Jeremiah

I know. When I say that envelope B contains either £5 or £20 I'm saying that either R or S is the case.

Whenever we have a sample space (e.g. [H, T] for a coin toss) we're dealing with mutually exclusive events, so I have no idea what you're trying to get at here.

If there's £10 in my envelope then either X = 10 and the other envelope contains £20 or X = 5 and the other envelope contains £5. We assign a probability of 50% to each event. This is how the Bayesian does his analysis, as shown by @andrewk.

What do you think about the empirical outcome of switching? Does it help?

That is, if you were actually to play the game (say over a huge number of trials), would you try switching every time in hopes of getting more money? Do you think it would work?

That is, if you were actually to play the game (say over a huge number of trials), would you try switching every time in hopes of getting more money? — Snakes Alive

I'd switch conditionally as explained here to reap the .25 gain.