I already did. — Jeremiah

So you did.

But if we try to model this from the participant's point of view, we don't use X at all; we use Y. Then the expected gain loss sample spaces are [-Y/2, 0] and [2Y, 0], aren't they? That's @Michael's problem.

I'm tempted to think @andrewk's point is relevant here: how do we know Y/2 and 2Y are even in the distribution? We know that Y is, and that Y is either X or 2X, but we have no way of knowing which, and we don't know anything about the distribution of X.

I agree that the algebra works and is correct. Is the objection to Michael's approach not the assignment of the prior but simply that he is using sample spaces he cannot justify using?

The other way to put this is that after learning the envelope has Y in it, we know that either Y/2 or 2Y are in the distribution, but we cannot know which, and this is no different from learning that we have either X or 2X.

I think the objection is that his approach is an error of reason. It's specified that the contents of the envelopes are X and 2X. So whatever value you see has to be either X and the smaller amount (in which case switching only gives you a larger amount of X+X)) or 2X and the larger amount (in which case switching only gives you the smaller amount of 2X-X). @Micheal tries to use maths to break those logical connections creating erroneous possibilities that lead to imaginary gains.

erroneous possibilities that lead to imaginary gains — Baden

Ah, philosophy, your generosity is but an illusion!

So the way to do cases here goes:
If Y = X, then ...
If Y = 2X, then ...
and it's okay to treat each of these as having a 1/2 chance. Then we get @Jeremiah's simple algebra.

But if we try to model this from the participant's point of view — Srap Tasmaner

Up until I introduced the real world example, we both were looking at as if we were participates. Michael and I were looking at the same problem in the same context based on the same information.

we use Y. Then the expected gain loss sample spaces are [-Y/2, 0] and [2Y, 0], aren't they? — Srap Tasmaner

He was looking at it as 2X or X/2. Using Y does not change anything at all as there are still two possible values for Y and that is Y=X or Y=2X. You don't change anything by using Y, the algebra would still come out the same way. By the definition of a function it has to and if it doesn't then you did something wrong.

I'm tempted to think andrewk's point is relevant here: — Srap Tasmaner

It is binary, either it is X or 2X. So if one really needs to use a distribution as a sampling model then use a Bernoulli distribution, with alpha equal to one and beta equal to one as your priors. That is if you want to use a statistical nuke as a fly swatter. It is over the top and completely uncalled for.

Is the objection to Michael's approach not the assignment of the prior but simply that he is using sample spaces he cannot justify using? — Srap Tasmaner

Michael's error is not treating X as an unknown variable in accordance to the definition of a function. You have to trust the algebra and established proofs/definitions more than your intuition.

When he saw the 10 bucks he forgot to consider the uncertainty of his starting position and rewrote the values of Y to Y=X and Y=X/2. Then he creates two possible cases on known values of X. However the 10 bucks does not give us the value of X, as we still don't know if our starting point is 2X or X.

I'm sure if I knew more about the details of the maths, I'd be able to see enough subtleties to allow me to be confused about this. But I'll leave it to the experts, of which I am definitely not one, and just remain confused about the confusion. :D

Thanks for spelling it all out. I think I've got it now.

It is an interesting problem nonetheless.

Let me see if I can sum it up more simply.

You have two envelopes, A and B.

There are two possible amounts for the contents of the two envelopes either X or 2X. You don't know which is which.

We'll call amount X case R, and and amount 2X case S.

You are handed envelope A.

Before you open it, you know that A could be case R or it could be case S. You have no clue which it is, so as an uninformative prior you give each case a 50% likelihood.

Now for the tricky part, you open envelope A and see it has 10 bucks in it.

Intuitively this seems like new information which would call for you to update your prior; however, it still does not tell you if are you in case R or case S. You don't know if that is 10=X or 10=2X. So in truth you can't really update your prior, as your prior was based on the uncertainty of being in case R or case S.

So your prior stands.

You consider swapping. You look at envelope B and realize it could be in case R or case S. You don't know which, so just as before you give it an uninformative prior and you give each case a 50% likelihood.

Now you could swap, but really probabilistically it changes nothing as your initial uncertainty would still stand. You would just trade the uncertainty of A for the uncertainty of B which is the same uncertainty. And the math? Well the math is indifferent to all options. So the decision to swap really comes down to if you feel lucky or not.

Not true. — Jeremiah

You said this in reply to my pointing out that arguments based on expected values prior to opening an envelope are wrong because we don't even know if there is an expected value. To rebut that, you need to show that there is an expected value, eg that we can be confident the lower amount X is not drawn from a Cauchy distribution.

Yet in your answer you have not addressed this at all.

What exactly is it then, that you think is Not True?

I picked this version to generate more discussion. I think this version of the problem forces more conflict of ideas, which in turn generates more discussion. — Jeremiah

It was a good choice. In the version in wikipedia it is easy, and uncontroversial, to conclude that there is no reason to switch. Not so in this case.

There are two possible amounts for the contents of the two envelopes either X or 2X. You don't know which is which — Jeremiah

This is not well-defined. It needs re-stating to make it unambiguous. Here are some options.

(1) A number X is drawn from a distribution f. A second number N is drawn from the Bernoulli distribution with parameter 0.5 (think of a coin flip). If N=1 (2) the amount X is placed in envelope 1 (2) and 2X is placed in envelope 2 (1). None of f, N or X are known to the player.

Note that this includes as a special case the setup where X is a fixed number known to the game operators, but not to the player. In that case the distribution f is just the uniform distribution on interval [X,X].

(2) Same as (1) but the player knows f.

In case 1, the player needs to assume a prior distribution as her guess of f. From then on, the calculation is the same, and it will use the known (case 2) or assumed (case 1) distribution f. The answer of whether expected winnings are increased by switching will depend on the slope and curvature of the PDF of f at the value that is observed when the envelope is opened.

From what I can see by attempts above, the assumption has been made implicitly that f is a uniform continuous distribution on the interval from 0 to some very high number such as the total of all banknotes in the world. That means the slope and curvature of the PDF at the value observed is zero.

Note the distribution must be continuous, which means that rather than banknotes, there must be an IOU specifying a real number that will be paid. If it was discrete, which is necessary if the envelope contains banknotes, then if the value was an odd multiple of the smallest currency unit, we'd know we have the smaller-valued envelope.

I suspect that even with that sparse, uniform distribution assumption, one might be able to make some inference. For instance, if the amount seen is very small, I suspect the Bayesian probability that it is the lower of the two amounts may be more than 0.5. But I've not thought about that much and it could be wrong.

No one said a probability density curve was used to select X. And it won't matter anyway, as X is a real positive value, 2X will always be greater than X, the midpoint will always be 1/2X+X, and your possible outcomes will always be either 2X or X. This has nothing at all to do with distributions. Also the expected value being calculated is not of the density curve, it is the expected value of your take from the contents of the envelopes.

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

This has nothing at all to do with distributions. — Jeremiah

That is not consistent with your use of the term 'expected value' in the following:

In both cases our expected value then is:

1/2(X) + 1/2(2X) = 1/2X+X.

Which is the mid point between X and 2X. — Jeremiah

'Expected value' only has meaning in the context of a probability distribution.

Well technically this is a probability distribution: 1/2(X) + 1/2(2X). 50% is distributed to X and 50% is distributed to 2X.

But what you are talking about is a probability density curve, which is also a probability distribution, but one we would use in estimating parameters of a population from sample data. Typically we shorten the reference by just calling them distributions, but yes you are right, they are all technically distributions. However, your uses of probability density curves is still misplaced.

Well technically this is a probability distribution: 1/2(X) + 1/2(2X). 50% is distributed to X and 50% is distributed to 2X.

That distribution is (1) unknown, as X is unknown and (2) not appropriate for estimating expected gains from switching, since it does not use all the available information after the envelope has been opened. The expected gains from switching, in the absence of any knowledge of how X was selected, is as @Michael calculated it earlier. But the question in the OP does not explicitly ask for expected gains from switching. It just asks 'should you switch?'. What is supposed to be the metric used to determine whether to switch?

That distribution is (1) unknown, — andrewk

It is not unknown at all, as I just laid it out for you, and as I already pointed out 1/2X+X reflected the expected gains from the real world example.

I have come to the conclusion that you just spit out jargon but you really have no clue what you are talking about. I actually came to that conclusion several threads ago. I don't think you understand the concept of a probability distribution and, in relation to statistics, I don't think you know what those density curves are for.

[Jeremiah's] reasoning is:

I have £10. If X is 5 then I lose £X by switching. If X is 10 then I gain £X by switching.

So it's either -£X or +£X. This is symmetrical. — Michael

The conclusion of symmetry is incorrect, because X is £5 in the first case and £10 in the second case. So we can't say that there's no difference because we lose X in one case and gain X in the other.
That would be like saying that if I have two £1 coins in my pocket and a big hole in the pocket it makes no difference whether I take one of the coins out and put it in another, non-holy pocket, because in both cases I lose everything in the holy pocket.

Doing this approach correctly, we use conditional probabilities to calculate the gain G from switching as follows:

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)

So this approach, when done correctly, reaches the same conclusion that you did - that the expected winnings are increased by switching.

I don't think you understand the concept of a probability distribution and, in relation to statistics, I don't think you know what those density curves are for — Jeremiah

You've found me out. Please don't tell my employers that, or I'll lose my job.

Not buying it, as you have displayed a large lack of comprehension here.

This whole approach is wrong though, isn't it? You know that you had a 1/2 chance of picking the envelope with the larger amount. That's just not the same as the amount in the envelope, which you will know once you open it, having a 1/2 chance of being the larger amount.

This is a fun puzzle – I took a crack at it.

Let X be the amount as in the statement of the puzzle. Let Y be the amount revealed in the first envelope.

There are two possible states in which one can be after the revelation: either Y = X (you have pulled the lesser envelope, and the other has twice as much), or Y = 2X (you have pulled the greater envelope, and the other has half as much).

There are also two actions: stay, and switch.

We can then construct the average expected winnings from each move, based on each of the two states, in a 2 x 2 table:



Each of the four cells represents an outcome of the move made based on the corresponding state (states on the horizontal, moves on the vertical). In each case, the value is expressed in terms of both X and Y.

Assuming that there is a 50% chance that one is in either of the two states, we can get the average expected winnings for each move by halving the value given by that move in each state, then adding these two together.

What's interesting is that the results look different depending on whether you express the average result in terms of X or Y.

For Y:

Switch: .5(2Y) + .5(.5Y) = Y + .25Y
= 1.25Y

Stay: .5 Y + .5Y
= Y

Thus, expressed in terms of Y, the greater average output is had by switching, since 1.25Y > Y, regardless of the value of Y.

For X:

Switch: .5(2X) + .5X = X + .5X
= 1.5X

Stay: .5X + .5(2X) = .5X + X
= 1.5X

Thus, expressed in terms of X, both moves have the same average output.

––––

So what's going on? How can the output be equal, but not equal, for the two actions?

It turns out there is an illusion in the way the problem was set up with respect to Y: there is no single value of a variable Y that can be defined independently of X, and then placed next to it in proportion (as either equal to it, or double it). Where Y is the number you drew in the first envelope, such a variable is already defined in terms of X: it is only possible to draw X, or 2X. The illusion results from thinking of 'the value of what I drew' as some distinct variable Y, when this is not the case. This in turn leads to the illusion that one can calculate the average expected return based on this new value, 'what I drew,' and conclude that switching allows a better return on 'what one drew.'

But what you drew is simply either X or 2X in the first place. There are thus two possible situations: either you drew X, or you drew 2X. And there are two possible moves: switch, or stay. Now let's draw the real table, defined this time in terms of the only variable there is, X:



This table simply eliminates the illusory Y (not a well-defined value independent of X to begin with), and we see that as before, switching has no effect on average expected output.

––––

The take-away message is that there is no such thing as a value of 1.25Y, averaged across the two possible situations, because there is no such single value Y that is the same across said situations, since you don't know whether you're drawn the greater or lesser envelope. X, by contrast, is a value we can refer to coherently across both situations, and cast in these terms, it becomes obvious that switching is ineffectual. The illusion comes in thinking that the de dicto description 'what I drew' is equivalent de re to some single numerical value across possible situations, and it is not.

Are you saying that you have reservations about my corrected version set out in that post? It is a fairly straightforward calculation based on conditional expectations, for the two possibilities. Is there a particular part of it that concerns you?

The conclusion of symmetry is incorrect, because X is £5 in the first case and £10 in the second case. — andrewk

Were you just using my post as a launching pad or were you actually addressing me? Because the rest of my post said exactly this.

Interesting. The problem is different. Unlike in the OP, in the Wiki case, the envelope has not been opened before the option to swap. So the player has no new information in the wiki case. I think their analysis in the Simple Case is wrong, because it assumes the existence of an expected value that may not exist, but I think the conclusion that there is no reason to switch may in spite of that error be correct in that case, but not in this one. — andrewk

Before choosing, there is no reason to prefer one envelope to the other. On choosing an envelope and learning the amount in it, there is still no reason to prefer one envelope to the other. So there is no reason to switch.

However if on opening the first envelope, the second envelope were then emptied and randomly filled with either half or twice the amount of the first envelope then the expected value of the second envelope would be (X/2 + X*2)/2 = 1.25X. So the player should switch.

Operationally, they are two different scenarios (requiring different strategies).

Why not? I've opened the envelope and seen that I have $10. That's in the rules as specified in the OP. And knowing the rules of the game I know there's a 50% chance that the other envelope contains $5 and a 50% chance that the other envelope contains $20. — Michael

The problem is that you use the $10 starting amount to generate the random half ($5) or double ($20) amount for the second envelope. That is equivalent to emptying the second envelope and refilling it with half or double the amount of the first chosen envelope. Which doesn't reflect the rules in the OP.

Is there some rule of statistics that says that one or the other is the proper way to assess the best strategy for a single instance of the game (where you know that there's $10 in your envelope)? — Michael

It's that the switching strategy gains nothing if the starting envelope was randomly chosen and the amount of the second envelope remains unchanged. Whereas switching would be correct if the second envelope amount was randomly decided on the basis of the first chosen envelope amount.

I would have thought that if we want to know the best strategy given the information we have then the repeated games we consider require us to have that same information, and the variations are in the possible unknowns – which is what my example does. — Michael

The problem is making the randomizing element dependent on the first chosen envelope amount instead of simply as the choice of the envelope.

The problem is making the randomizing element dependent on the first chosen envelope amount instead of simply as the choice of the envelope. — Andrew M

I'm just doing this:

1. We pick an envelope at random
2. There's a 50% chance that my envelope is the X envelope and a 50% chance that my envelope is the 2X envelope.
3. I open my envelope and see £10
4. From 2 and 3, there's a 50% chance that my £10 envelope is the X envelope and a 50% chance that my £10 envelope is the 2X envelope.
5. From 4, there's a 50% chance that the other envelope contains £20 and a 50% chance that the other envelope contains £5.

We seem to agree on 1-3, but you disagree with 4 and/or 5?

On choosing an envelope and learning the amount in it, there is still no reason to prefer one envelope to the other. So there is no reason to switch. — Andrew M

That claim is not consistent with the following formal calculations from my earlier post:

we use conditional probabilities to calculate the expected value of the gain G from switching [after seeing 10 pounds in the first envelope] as follows:

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

which say that, after opening the first envelope and finding 10 pounds in it, the expected gain from switching is 2 pounds fifty.

I have numbered the steps from 0 to 4. Which step(s) do you not believe?

Do you realize that expected gain is an average between gains and losses? It is stuff like this that makes me doubt your vague claim about your employment.

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?

4. From 2 and 3, there's a 50% chance that my £10 envelope is the X envelope and a 50% chance that my £10 envelope is the 2X envelope. — Michael

This is false. Whether the amount in your envelope is the smaller or the larger of the pair offered you is not a matter of chance. Whether you choose that envelope is.

This is false. Whether the amount in your envelope is the smaller or the larger of the pair offered you is not a matter of chance. Whether you choose that envelope is. — Srap Tasmaner

So you're saying that before I look I can say that there's a 50% chance that my envelope is envelope X but after looking I can't?