I don’t see how this is correct. I know from the terms of the game that one envelope contains exactly twice as much as the other. And I don’t see any reason why I can’t use X as the unknown amount contained in one envelope. And having done this, there is a relationship - I know that the contents of the other envelope contains either half the amount or twice the amount of the envelope I’m holding. So why can’t I refer to the contents of the other envelope as X/2 or 2X? Those are the only two possible values.
You’re calling this a key point but I’m not seeing what you’re saying here. I think you’re saying that X is the amount in the first envelope and that X’ is the amount in the other envelope if it’s larger than X and X" is the amount in the other envelope if it’s smaller than X.
If that’s the case then your other figures are off. The expected value is not going to be 0.5(2 X’) + 0.5(0.5 x’’) - it would be 0.5 X’ + 0.5 X". And X" would not equal 2X’ - it would equal 4X’.
Ok, lets walk through this:
Envelope 1=X
Envelope 2=X/2 or 2X
That is correct, so far, for this one specific random configuration of the values that envelope 1 and envelope 2 can have. Note that the term “OR” is allowing us to include 2 different values, only 1 of which can ever actually appear in the envelope.
If we define M as the smaller amount, then 2M is the other amount that can appear. When you list X, X/2 and 2X you have included 3 different values - only 2 of which are even possible.
If X=M, then 2X is in the other envelope and X/2 is not even a valid possibility in the game at all. It’s some bizarre number that wasn’t part of the rules that got thrown in there, which means you can’t use it to figure out the average of the other envelope.
When you calculate the expected value, you are only allowed to include the values that can actually appear in the envelope.
Now let’s calculate the expected value of envelope 2 which is just the average if you did this over many trials:
Expected value of envelope 2 = the sum of all possible values divided by the number of possible values
These are the possible values:
M
2M
Note: The possible values do not include 3 different value X, X/2 and 2X, only 2 possible values.
So expected value is (M+2M)/2
But this is the same thing that others have done. I “solved” the problem one way and it was obvious I got the wrong answer. What you’ve done is solve the problem a different way and got the right answer.
But I wasn’t looking for the answer. I already knew that. I want to know why my method of solving the problem didn’t work.
You can’t use X to represent the amount contained in one envelope because the ‘value’ of the envelope changes depending on the amount in the other envelope. The ‘value’ of the envelope is not equal to the dollar amount you see.
If your envelope contains $40, and the other one contains $80, the ‘value’ of yours is more than $40 because you might exchange it for $80.
If your envelope contains $40 and the other one contains $20, the ‘value’ of yours is less than $40 because you might exchange it for $20.
What you are doing is like standing in a casino holding two cards, and assigning them a value without having chosen a game first.
I think the fact that people are so used to only thinking about expected value in terms of “probability” has caused a lot of confusion (particularly about what kinds of mathematical concepts concerning expected value are coherent or not). So I would like to recast the exact same question about expected value in non-probabilistic terms:
Identical weights are placed throughout the XY co-ordinate plane, at all those points whose X and Y coordinates are adjacent powers of 2. Thus, one series of weights is placed along the line Y = 2X, and another is symmetrically placed along the line X = 2Y.
Where’s the center of mass of the resulting system? Is it above, on, or below the axis Y = X? How about the centers of mass of each vertical cross section of the system? Is it above, on, or below the axis Y = X? How about the centers of mass of each horizontal cross section of the system? Is it above, on, or below the axis Y = X?
That is the exact same expected value question being debated in this thread.
The vertical cross section at X = 2[sup]k[/sup] has a well-defined center of mass: it consists of a weight at (2[sup]k[/sup], 2[sup]k + 1[/sup]) and another weight at (2[sup]k[/sup], 2[sup]k - 1[/sup]), so its center of mass is at at (2[sup]k[/sup], 1.25 2[sup]k[/sup]). Thus, every vertical cross section has a center of mass along the line Y = 1.25 X; in particular, every vertical cross section has a center of mass lying above the line Y = X.
Symmetrically, the horizontal cross section at any particular Y = 2[sup]k[/sup] has a well-defined center of mass lying along the line X = 1.25 Y; in particular, every horizontal cross section has a center of mass lying below the line Y = X.
How about the center of mass of the system overall? Is it on, above, or below the line Y = X? Well, carrying out the calculation for the overall center of mass, one finds that both the X and Y coordinates of the overall center of mass are infinite. It becomes difficult to specify where it lies, therefore; however, there certainly is a natural sense in which it can be taken as lying infinitely far out along the line Y = X, neither above it (despite all the vertical cross sections having centers of mass above that line) nor below it (despite all the horizontal cross sections having centers of mass below that line), in accordance with the symmetry of the situation.
The situation is similarly perplexing if one rotates one’s head till the Y = X line becomes horizontal (let us call it the U axis) and the Y = -X line becomes vertical (let us call this the V axis), and then tries to calculate the center of mass in terms of this new coordinate system; it is clear that the U-component of the center of mass is infinite; however, the V-component of the center of mass is the sum of arbitrarily large positive and negative terms which can be rearranged to sum to whatever one likes. But, again, the most natural way to deal with the situation is to consider these terms, by their symmetry, to cancel out, causing the center of mass to lie infinitely far out along the line U axis, with no V component.
Anyway, the point is, there’s nothing incoherent or mathematically insensible about speaking of such things as “The center of mass along a horizontal cross section”. You may or may not be interested in them, but they’re there; the paradox isn’t resolved by prohibiting looking at them. The paradox is resolved by recognizing the ambiguities of combining equal and opposite infinite sums, as in the calculation of the V-component of the center of mass; sorry, that’s what it is.
I just told you - you used 3 values in your equation but your problem only has 2 different monetary values
That is the problem with your method - you are using the values X, X/2 and 2X, but your problem explicitly states that there are only 2 monetary values, not 3
You perfectly can. That’s a perfectly cromulent mathematical calculation of something (what in mathematical jargon would be called the expected value of the other envelope conditional on X). And you’ve calculated it exactly correctly; it is indeed true that this comes out to 1.25X. The only question is what do you intend to do with this value? You can carry out lots of perfectly cromulent mathematical calculations. The question is, so what?
A lot of the discussion in this thread has focused on X and Y as having “real” values; i.e., despite the probability distribution, ultimately, X really is this or that. Well, fine, sure, you can say that, but the technical mathematical operator of “expected value” doesn’t have anything to do with what the real values are or not. It’s just some function of the entire probability distribution. That’s why I prefer the illustration I gave previously in terms of the center of mass of a system of weights at different locations; people are less likely to say things like “Yes, but the only weight that matters is the real location”.
So are you saying that the expected value is not applicable in this type of situation?
How about the situation I described in post 70:
Is expected value being used correctly here?
“the expected value of blah blah” is just an alternative mathematical name for “The weighted average of blah blah”. But this doesn’t imply anything about what you should or shouldn’t do. That’s up to you. Maybe you are very risk-averse; maybe you find even the slightest chance of reward appealing.
The relevance of the weighted average to that bet is not automatic; it’s true that if a large number of people were to independently take up their own independently coin-flipped version of that bet, then about half of them would make 40 bucks profit and about half of them would lose 20 bucks, so that the average profit per person would be about 10 dollars. You are correct in claiming that the expected value of the winnings of the bet is 50 dollars and the expected value of the profit of the bet is 10 dollars. Those are the averages.
But so what? Just because a large number of people taking that bet would even out to an amortized profit per person of 10 dollars doesn’t compel you to take or not take the bet; there are more considerations that you may want to consider than just that particular piece of trivia.
If you were able to work as a team with all those other people, and you had to either all choose to take the bet and split your individual winnings among the team, or all choose to not take the bet and go home, then, sure, you would all choose to take the bet and split your winnings and almost certainly make about 10 bucks a pop. But that’s not the offer being made to you.
It can be perfectly rational to turn down a bet whose profit has a positive “expected value”. If you offered me a once in a lifetime opportunity to enter a game where a coin is flipped and I end up either owing $10,000 or earning $30,000, I’d turn the game down. It’s not appealing to me, and that’s fine. I don’t want to take on that risk, and there’s nothing irrational about that. [If you instead offered me the opportunity to assemble a large team of players who would each play their own version of that game individually, then pool and split their earnings, then, of course, I’d change my mind and go for the near-certain $20k profit].
So, expected value is being used correctly in that the expected value of the bet is indeed $50, since all “expected value” means is “average”, and the average of $80 and $20 is uncontroversially $50. But further leaps as to what this calculation should compel us to do go beyond what the mathematics says.
It’s also worth noting that you could have a bet which is clearly a winner when looking at its expected value one way, and clearly a loser when looking at its expected value another way. For example, suppose your goal in life, for whatever reason, is to have as large a square medal as possible. You already have a medal of side-length 7, but you are offered the opportunity to switch it for a 50/50 choice between medals of side-length 1 and 11. Should you take the offer? Well, on the one hand, no, the expected value of the side-length of the new medal would be only (1 + 11)/2 = 6, which is smaller than your current medal. But on the other hand, yes, the expected value of the area of the new medal would be (1 + 121)/2 = 61, which is bigger than the area 49 of your current medal. So should you switch or not? Well, expected value doesn’t say; it just tells you that the average side-length of the two possible new medals is smaller than your current medal’s side-length, while, at the same time, the average area of the two possible new medals is larger than your current medal’s area. What you do with that information is entirely up to you.
Thank you for these insights; I am seriously out of practice with mathematics and my thinking was well off. This problem is quite tricky, and despite my talk about being rigorous, I was just throwing crap together and assuming it would stick up. Those statements were false and unrelated in my version of the problem, but my version was considerably different. I was at first very confused when you stated E(Y|X) as a function of X, because I was specifically letting X=k, a constant and not a random variable. I was basically doing the open-envelope problem without stating it. Thus, I really wasn’t saying anything new.
All in all, it looks as though the problem lies in the assumption that P(Y=2k | X=k) = .5. Suppose the set up was as follows: I pick a number from a distribution then put that value in one envelope (say A) and put double that value in another identical envelope (B). I then shuffle them and give one to the contestant randomly (X), and keep one(Y). The envelopes have the same probability distribution, one that’s somewhat complex, but clearly the expected value of the envelopes is not infinite if my distribution has finite expected value.
By the def of conditional probability:
P(Y=2k|X=k) = P(Y=2k & X=k)/P(X=k)
To get Y=2k AND X=k, you must pick A=k, then assign A=X. Thus P(Y=2k & X=k) = P(A=k) * P(A=X) = .5 * P(A=k).
For X=k, you could pick A=k then assign X=A or pick A=.5k and assign X=B, and you can’t ever pick any other value for A. Thus P(X=k) = P(A=k).5 + P(A=.5k).5 = .5 * ( P(A=k) + P(A=.5k) )
Thus we have
P(Y=2k|X=k) = .5 * P(A=k) / .5 * ( P(A=k) + P(A=.5k) ) = P(A=k)/(P(A=k) + P(A=.5k))
And so P(Y=2k|X=k) = .5 only when P(A=k) = P(A=.5k) for that value of k. Thus it is only true for all values of k and thus true in general when the distribution is such that the probability of every positive value is equally likely, which leads to an undefined/infinite expected value. This case was already discussed in the open-envelope case, and what I was blind to for a while was that it had to be true given the somewhat unstated assumption that P(Y=2k|X=k) = .5. Assuming that from the given scenario is backwards thinking and only works when you have a truly pathological distribution.
Apologies to everyone for spouting off nonsense without really thinking about it multiple times, and kudos to Indistinguishable for (as usual) being accurate in his assessment.
So how’s this for an overall explanation:
Either the chance the other envelope has double is not actually 50%, or the expected value is infinite. Naively those both sound absurd, but if you work carefully through the probability involved, it falls right out.
Would you be happier if I said the envelopes contained utils instead of money?
I’d like to remind everyone to remember I asked for an explanation in simpler terms that I could understand. Unfortunately explanations that contain the following:
do not qualify as simple explanations that make it easy for me to understand the problem. If these really are the simplest explanations then I’m probably not going to get this.
If you don’t understand random variables or conditional probability, it’s going to be tough to give an explanation that you’ll understand. That said, what do you think of Wikipedia’s article on the necktie paradox?
It seems to be a variation of the wallet bet paradox. I think the issue here is the fact that the men involved have foreknowledge of what they’re betting on. Each is betting on the price of a tie that he’s examined (or the contents of his wallet). So it’s really no different than betting on an upcoming football game. They’re not really betting on a matter of probability - they’re betting on their belief that their assessment of the situation is superior to the other guy’s assessment.
Eh, somewhat, which makes it embarrassingly pedantic, which means I shouldn’t have bothered bringing it up, really. But there’s still some extent to which saying “utils” isn’t enough to make me fully happy. But, let’s ignore that. 
Let me try to give the simple, child-graspable account of what the longer story in my summary post was: The fundamental question is, is switching envelopes a good idea or not? Why is there this paradoxical ambiguity there, that it simultaneously looks like it shouldn’t make any difference, and looks like it would be a great idea, and looks like it would be a terrible idea, all at once?
Well, let’s consider all the possible ways the envelopes could be set up. It looks like this:
YOUR ENVELOPE
| 1 | 1 | 1 | | | | |
| - | - | - | 1 | 2 | 4 | 8 |
| 8 | 4 | 2 | | | | |
+---+---+---+---+---+---+---+
+---+---+---+---+---+---+---+ +--
| | | | | | | | |
| | | | | | @ | | | 8 T
| | | | | | | | | H
+---+---+---+---+---+---+---+ +-- E
| | | | | | | | |
| | | | | @ | | @ | | 4 O
| | | | | | | | | T
+---+---+---+---+---+---+---+ +-- H
| | | | | | | | | E
| | | | @ | | @ | | | 2 R
| | | | | | | | |
+---+---+---+---+---+---+---+ +-- E
| | | | | | | | | N
| | | @ | | @ | | | | 1 V
| | | | | | | | | E
+---+---+---+---+---+---+---+ +-- L
| | | | | | | | | 1 O
| | @ | | @ | | | | | - P
| | | | | | | | | 2 E
+---+---+---+---+---+---+---+ +--
| | | | | | | | | 1
| @ | | @ | | | | | | -
| | | | | | | | | 4
+---+---+---+---+---+---+---+ +--
| | | | | | | | | 1
| | @ | | | | | | | -
| | | | | | | | | 8
+---+---+---+---+---+---+---+ +--
Each column in that grid represents a possible amount of money in your envelope, with the amount doubling as you move to the right. Each row in that grid represents a possible amount of money in the other envelope, with the amount doubling as you move up. Each @ represents one way the envelopes could actually be stuffed; the top left line is the cases where your envelope has half the money of the other envelope, and the bottom right line is the cases where your envelope has twice the money of the other envelope. I’ve only drawn part of the grid, but of course, it extends infinitely.
Ok, well, since we’re interested in the profit or loss of switching, let’s now fill in each @ with the actual profit you’ll make by switching in that scenario:
YOUR ENVELOPE
| 1 | 1 | 1 | | | | |
| - | - | - | 1 | 2 | 4 | 8 |
| 8 | 4 | 2 | | | | |
+---+---+---+---+---+---+---+
+---+---+---+---+---+---+---+ +--
| | | | | | | | |
| | | | | | 4 | | | 8 T
| | | | | | | | | H
+---+---+---+---+---+---+---+ +-- E
| | | | | | | | |
| | | | | 2 | |-4 | | 4 O
| | | | | | | | | T
+---+---+---+---+---+---+---+ +-- H
| | | | | | | | | E
| | | | 1 | |-2 | | | 2 R
| | | | | | | | |
+---+---+---+---+---+---+---+ +-- E
| | | 1 | | | | | | N
| | | - | |-1 | | | | 1 V
| | | 2 | | | | | | E
+---+---+---+---+---+---+---+ +-- L
| | 1 | |-1 | | | | | 1 O
| | - | | - | | | | | - P
| | 4 | | 2 | | | | | 2 E
+---+---+---+---+---+---+---+ +--
| 1 | |-1 | | | | | | 1
| - | | - | | | | | | -
| 8 | | 4 | | | | | | 4
+---+---+---+---+---+---+---+ +--
| |-1 | | | | | | | 1
| | - | | | | | | | -
| | 8 | | | | | | | 8
+---+---+---+---+---+---+---+ +--
Naturally, along the top left line, where the other envelope has more money, you get positive profits, while along the bottom right line, where the other envelope has less money, you get corresponding negative profits.
Now, the question is, is it good, neutral, or bad to switch envelopes? This is essentially the question as to whether the sum of the profits over all cases is positive, zero, or negative. And… there is an ambiguity here. It’s not clear what the value, or even the sign, of the sum of all the boxes comes out. For example, consider the following diagram:
+---+---+---+---+---+---+---+
| | | | | | | |
| | | | | | 4 | | ...
| | | | | | | |
+---+---+---+---+---+---+---+
| | | | | | | |
| | | | | 2 | + |-4 | = -2
| | | | | | | |
+---+---+---+---+---+---+---+
| | | | | | | |
| | | | 1 | + |-2 | | = -1
| | | | | | | |
+---+---+---+---+---+---+---+
| | | 1 | | | | |
| | | - | + |-1 | | | = -1/2
| | | 2 | | | | |
+---+---+---+---+---+---+---+
| | 1 | |-1 | | | |
| | - | + | - | | | | = -1/4
| | 4 | | 2 | | | |
+---+---+---+---+---+---+---+
| 1 | |-1 | | | | |
| - | + | - | | | | | = -1/8
| 8 | | 4 | | | | |
+---+---+---+---+---+---+---+
| |-1 | | | | | |
| | - | | | | | | ...
| | 8 | | | | | |
+---+---+---+---+---+---+---+
= = = = =
... 1/8 1/4 1/2 1 2 ...
This shows that if you add the numbers down each column first, you get a bunch of positive values (which add up to an infinitely large total). On the other hand, if you add the numbers across each row first, you get correspondingly negative values (which add up to an infinitely negative total). And, of course, if you add the numbers along each diagonal pair first [in the \ direction], you’ll just get a bunch of zeros (which add up to a total of zero).
So there is this ambiguity as to what the sum of the numbers in that square comes out to, and that is precisely where all the ambiguity as to whether one should switch comes from. When it looks like switching envelopes is a great idea, this is because one is essentially adding down the columns to get a positive total for the square. When it looks like switching envelopes is a terrible idea, this is because one is essentially adding across the rows to get a negative total for the square. And when it looks like switching envelopes shouldn’t make any difference, this is because one is essentially noting the symmetry and adding diagonally to get a total of zero. Three different ways of totalling that square, which give three different kinds of results. And there’s nothing more to it than looking at that square and observing, yup, that can happen.
Does that help illustrate anything?
2X and X/2 are NOT both possible values, only ONE of those values is a legal value in the game you described.
Therefore you may not use BOTH of those values when you add up all the possible values multiplied by the chance of them occurring.
Here is a concrete example:
The real possible values are $10 and $20
If your envelope has $10, then the other values of 2X and X/2 are $20 and $5.
$5 is clearly not in our original list of $10 and $20 - it’s a third value that simply doesn’t belong - so you can’t use it when you try to calculate expected value - it will cause you to get the wrong answer. Which means you can’t use (2X+X/2)/2 - one of those terms doesn’t belong.
It’s true the other envelope contains either 2X or X/2.
It is also true the other envelope contains either 2X or X/2 or 1,786X or 9,000,000X - it does contain one of those values - but that doesn’t mean you can include 9,000,000X in your expected value calculation.
If you swap, you either lose “X/2”, or gain “X”, with 50% chance of each. However calling them “X/2” and “X” is misleading, since you can only lose if you have the higher amount, and gain if you have the lower. If the envelopes contain a and 2a, then if X = 2a, which has 50% chance, you lose a by swapping (which is only equal to X/2 because X = 2a). Otherwise X = a, with 50% chance, with a gain of a by swapping (which is equal to X only because X = a). The expected gain by swapping is 0.5a* - 0.52a*/2 = 0.
If you want it in terms of the expected value of the other envelope, it is 0.5a* + 0.52a* = 1.5a. This is equal to the expected value of X, 0.5**2a + 0.5**a* = 1.5*a