But there is a well-defined mean here. The average of 2X and .5X is 1.25X. Perhaps I’m misunderstanding “well-defined mean.”
The mean profit of switching, for any particular value of X, is well-defined. It is equal to 1.25X, just like you said. This doesn’t conflict with any other calculation.
The mean profit of switching, over all X, is not well-defined. And all the conflicts are regarding this ill-defined quantity.
Just like “The mean of all the integers with absolute value X” is well-defined (and equal to 0) for any particular X, but “The mean of all the integers” is not well-defined.
Perhaps this simpler analogue will help clarify:
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| | | | | | | | | | |-10| ...
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| | | | | | | | | |-8 | 9 |
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| | | | | | | | |-6 | 7 | |
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| | | | | | | |-4 | 5 | | |
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| | | | | | |-2 | 3 | | | |
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| | | | | | 0 | 1 | | | | |
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| | | | | 2 |-1 | | | | | |
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| | | | 4 |-3 | | | | | | |
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| | | 6 |-5 | | | | | | | |
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| | 8 |-7 | | | | | | | | |
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|10 |-9 | | | | | | | | | |
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...
There is a well-defined mean of 1/2 in any particular row. And nothing conflicts with this. No matter how you try to calculate the mean for any particular row, you will get 1/2. This would imply that the overall mean, over all the rows, is 1/2.
There is a well-defined mean of -1/2 in any particular column. And nothing conflicts with this. No matter how you try to calculate the mean for any particular column, you will get -1/2. This would imply that the overall mean, over all the columns, is -1/2.
Now, we have a conflict: Is the mean over the entire grid 1/2 or -1/2? Well, it’s simply not well-defined.
But each row’s mean is well-defined and each column’s mean is well-defined. It’s just that the mean over the entire grid is not well-defined, so the mean of the rows’ means does not have to match the mean of the columns’ means.
Wow this thead is still going. Without going into probability theory, here’s an explanation I found helpful for the open envelope situation. Dunno if others will find it helpful.
First, do we all agree on the closed envelope problems? i.e. if we don’t open an envelope, the expected value is the same, and it’s pointless to switch? Most people have an easier time accepting this one, even though the math is pretty similar.
The open envelope situation is tougher. As Cheesesteak asked, shouldn’t you always switch? If it were an isolated question, the answer seems to be yes. If there was a 50/50 shot of having double or half the money you currently have, you should switch every time. So why doesn’t this result in a paradox?
The problem is, when you read this problem it comes with some implicit assumptions. Let’s imagine for a second we know implicitly that the amount is somewhere between 1-100$. If you get any amount 50 or over in your envelope. Should you switch? No, obviously not, the other envelope could only contain less. If instead you have 48$ it might be 24$ or 96$, and switching makes sense.
In Cheesesteak’s example, when you find 100$ in your envelope, the question isn’t “What is the expected value of the other envelope?” it’s “What is the expected value of the other envelope, given that I know I already have 100$?” If we know the maximum value is 100$, this is easy. Now, since we didn’t state an explicit upper bound, it gets a little confusing but surely the man couldn’t stuff a million in there could he? Ten million? The actual number doesn’t matter, just that there’s some finite limit.
But what if there actually is no limit, if the upper limit is truly infinite? Then that first comparison, between what you found and the expected value of the other envelope, can never happen. Sort of makes sense: it will always make sense to trade a finite amount of money for a possibly infinite amount of money. Infinity results in some weird paradoxes but that’s an idea most people can wrap their heads around.
Question for Indistinguishable: Care to add any insight to the below?
Where exactly does this go wrong?
Ah, yes, the actual question from the OP which we’ve all ignored…
Well, the thing this calculation does correctly is that it shows that, conditioned on any particular value of X, for the mean profit from switching to be zero, it must be the case that the frequency of the other envelope containing twice as much is 1/3.
The conclusion we can draw from this is that it cannot, in fact, be the case that, for every particular value of X, the mean profit from switching is zero (given the natural assumption that you choose envelopes at random from the two available; i.e., there is no bias on your part to psychically select the larger envelope). Indeed, on the simplest setup, the frequency of the other envelope containing twice as much is 1/2, for each particular X, and the mean profit from switching is X/4, as we’ve all noted to our chagrin. There are more complicated frequency distributions we could imagine as well, but none will allow it to be the case that the mean profit from switching, conditioned on every particular value of X, is zero.
Allow me to re-arrange some words in that last part:
The conclusion we can draw* from this is that it cannot, in fact, be the case that, for every particular value of X, the mean profit from switching is zero. Indeed, on the simplest setup, the frequency of the other envelope containing twice as much is 1/2, for each particular X, and the mean profit from switching is X/4, as we’ve all noted to our chagrin. There are more complicated frequency distributions we could imagine as well, but none will allow it to be the case that the mean profit from switching, conditioned on every particular value of X, is zero.
[*: Given the natural assumption that you choose envelopes at uniform random from the two available, independently of anything else; i.e., there is no bias on your part to, for example, psychically select the larger envelope at a 2/3 rate. Without that implicit assumption, all kinds of stories about whether to switch or not make themselves plausible…]
Are you saying that the problem is in the first line: “Clearly the expected value (in reality) of the other envelope is the same as the one in your hand”?
Cos if that’s what you’re saying then I’m giving up on this problem forever!
Overall, both envelopes have an infinite average value.
This allows their average values to be equal without forcing the expected profit from switching to be zero. (Since infinity + anything = infinity, the anything doesn’t have to be zero, so to speak).
As it happens, the average profit from switching is ill-defined, overall.
It is in fact the case, though, that if you restrict attention to those particular cases where the value of your envelope is some particular constant, you will find (on the simplest frequency distribution with which to interpret the problem) that the average value of the other envelope is greater than that constant.
That may be shocking, but it is what it is. The reason it seems shocking is because it seems to conflict with the symmetry of the situation. But, in fact, there is no asymmetry: if you restrict attention to those particular cases where the value of the other envelope is some particular constant, you will symmetrically find that the average value of your envelope is greater than that constant.
That may seem terribly inconsistent, but there is no actual contradiction; these are, after all, two different averages we have calculated, with attention restricted to different subsets of cases.
The only lingering apparent inconsistency is in the fact that the universality of the first result (holding true no matter what constant we pick) would seem to imply that the overall average profit from switching is positive, while the universality of the second result would seem to imply that the overall average profit from switching is negative, and, of course, the symmetry of the situation implies that the overall average profit from switching is its own negation and thus zero. But all this actually just reveals that the overall average profit from switching is ill-defined, which is a thing that can happen sometimes. It doesn’t happen in cases with only finitely many possibilities, but it does happen sometimes when there are infinitely many possibilities, as we’ve seen just here and can see in many other examples as well.
I (mostly) get what you are saying.
Just wanted to thank you for putting in the effort to explain it so well. 
Put another way, whenever you open the first envelope, you’re always going to be disappointed. Before you opened either envelope, the expected value for both was infinite. Then, you opened one, and found some finite value. No matter what finite value it was, it was certainly less than infinite, and so you always do worse than your expectation.
Speak for yourself…
I’ve been thinking about this problem for years, and I find it really fascinating. The thing that’s so tricky about it is that when you present the problem to someone with a reasonable amount of math or probability knowledge, and ask them where the “trick” is, they’ll almost always fail to find it, because the problem sounds like a trick question, when in fact, it really isn’t, if you see what I’m saying. That is, people expect it to be something like the monty hall problem, where something about the wording or about psychology or about whether or not they can “choose” and what “choose” means causes the math that has been presented to them to be wrong, when in fact it’s way more subtle than that.
I think the easiest way to really grok what’s going on is one presented by someone earlier in the thread. If you know that there’s a range of numbers that the original value (before doubling) is in, then in fact there’s no paradox… if the number you see is smaller than the maximum value, you should switch. If it isn’t, you shouldn’t. What’s really tricky is that that same argument stops it from being a paradox even if you don’t know the range of numbers that might be presented to you, because there IS more information once you look at the first number, even if it’s information that isn’t fully available to you.
Giving a really good and clean explanation of it is beyond me, but I think that’s where the key “trick” is.
Okay so there are two issues that need to be dealt with separately.
Bolding mine. Yes, the expected value of the unchosen envelope is greater than X, where X is the value of the chosen envelope. The misstep in the analysis is that the expected value of the chosen envelope is also greater than X, even though X is defined to be the value of that envelope. The decision to swap needs to be based on comparing the expected values, which are equal like we’d expect. This is important because it leads to the flaw with the 1/3 issue below.
This line of reasoning is conflating the expected value with the defined value, which is the problem with the original paradox. If we use the expected value in your equation instead and solve for p:
p(2X) + (1-p)(X/2) = (5/4)X
2P + (1-p)/2 = 5/4 [divided both sides by X]
4P + (1-p) = 5/2 [multiplied both sides by 2]
8P + 2 - 2P = 5 [multiplied both sides by 2 again, yay we have integers]
6P + 2 = 5
6P = 3
P = 6/3 = 1/2, the result you’d expect
Not everyone. I pointed out that OP’s “paradox” exists even in the simplest case, where there is only a single possibility: envelopes of $1 and $2. Discussion of probability distributions, infinities, etc. are completely unnecessary for the question OP asks.
It is Indistinguishable who, just as he did when a simple question about Two Envelopes arose a year ago, insists on bringing up the matter of non-Kolmogorovian probability theory.
Using Google, I see that non-Kolmogorovian probability is a topic in quantum physics. Perhaps Indistinguishable can start a thread explaining that and its relationship, if any, to Two Envelopes.
Huh? You wrote: the following:
This is incorrect. The expected value of the other envelope is not higher than the (unopened) envelope in your hand. The reason no one addressed it is because it many other people very early in the thread had pointed out that the 5/4 calculation is wrong.
This problem, and the associated St. Petersburg paradox, the two wallets problem, etc. have been interesting to mathematicians for decades. There is an established literature about it and many interesting papers have been generated trying to figure it out. It’s not because they couldn’t figure out E1 =5/4 * E2
Huh, yourself. OP presents an argument he knows to be incorrect and asks why it is incorrect.
I merely show that the same incorrect argument appears in the trivial case ($1,$2). Neither OP nor myself (nor anyone else here) believes that argument is correct. The goal is to explain the fallacy most simply.
The fallacy is that conditional expectations are incorrectly made absolute expectations. I’ll admit I didn’t explain this lucidly. My “contribution” was to demonstrate that anyone introducing infinity, non-trivial distributions or (gasp!) non-Kolmogorovian probability was obfuscating rather than addressing OP.
Hope this helps.
I’ll admit I misunderstood your post but you are being very unclear as to your argument. In fact I still don’t understand it. The OP specifically states: “Im sure there have been many threads asking to explain how to resolve this problem (i.e. the 5/4X solution), this is not one of those. Clearly the expected value (in reality) of the other envelope is the same as the one in your hand.”
He then goes on to ask the question rephrased as a probability question:
If people find it helpful, that’s great, but I personally am not sure how your example of 1 and 2$ envelopes helps the OP with this problem. Nor does it address the open envelope problem, which other posters have brought up and is in many ways a far more interesting question. Defining the possible amounts present in the envelopes eliminates the paradox but fundamentally changes the setup.
Even more important, always switching your (open) envelope is not incorrect if the amounts in the envelopes have no upper bound. That’s an interesting insight that’s not captured without resorting to probability theory.
It is not a fallacy to conclude from the conditional expectation “E[Y | X] = 1.25X (for each X)” the absolute expectation “E[Y] = 1.25E”, or such things. Absolute expectations are just the average of conditional expectations, whenever they are well-defined.
For what it’s worth, I only brought up “non-Kolmogorovian probability theory” in response to those who objected to my objecting to bringing up technicalities of probability theory… I never wanted to discuss probability theory in the first place. Hell, I don’t even want to discuss probabilities; I’m the one who wants us to accept the problem at its word that everything is 50-50.
All I ever wanted to say was “Yup. The averages do indeed work out that way. Just look. Just write out a diagram and look. None of the calculations make missteps, but it turns out that there’s no well-defined average switching profit. That’s a thing that can happen, as you’ve discovered.” And, also, yes, “(In case you are interested in understanding this phenomenon better, it can only happen in infinite contexts)”.
It’s true that most of my posts have been about the two-envelope problem in general, rather than the OP’s question in particular, but I think it’s to be expected that a thread on the two-envelope problem and confusions must inevitably become a thread on understanding of the whole problem, not just any particular point which may be questioned. Understanding the whole problem helps one cut through the fog of particular questions.
Note, though, that in my response to the OP’s particular question, I never once mentioned probability, Kolmogorovian or not. I did mention the infinite nature of the setup, because it is integral to the paradox. I am sure the OP is well aware what happens in finite truncations of the setup, which puzzle no one. The paradox isn’t about the finite truncations. The paradox is about a situation in which, no matter what is in one envelope, the other envelope is equally likely to contain half or twice that. To confront the paradox is to confront that situation.
Come now, this is a cheap shot. Using Google, I see that expected values, arithmetic, and cash are all used in quantum physics. Perhaps no one should mention those, either.