Once you’ve opened one of the envelopes, then you need more information to make any meaningful judgment. If, for instance, you know that the person offering you the envelopes has a maximum available budget of less than $300, and your envelope has $100, then you should always keep your envelope, because you know that it must be the larger of the two.
What happens when you open the envelope, and see that it contains $100? Is $100 A, 2A or 1.5A?
One way to “resolve” the paradox is to say “Hey, let’s suppose that there are actually only finitely many possible monetary values” or “Let’s suppose that the values aren’t all equiprobable”. But these aren’t really confronting the core issue.
The paradox is essentially just the phenomenon of conditionally convergent series. It’s the same as the paradox that 1 - 1 + 1 - 1 + 1 - 1 + … can be made to look positive, negative, or zero, depending on how you re-arrange and group the terms.
To make this clear, let’s frame the problem in a different way:
Suppose we place an equal weight at every point (X, Y) where X and Y are adjacent powers of 2 [thinking of X as representing the value of your envelope and Y as representing the value of the other envelope]. Like so:
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
+---+---+---+---+---+---+---+ +--
(I’ve drawn the axes on log-scale for convenience)
Where is the center of mass of this system?
One argument says: Well, this system is symmetric under reflection across the line Y = X, so the center of mass must be on that line.
Another argument says: For each particular column, the center of mass is on the line Y = 1.25X, so the overall center of mass must also be on that line.
Another argument says: For each particular row, the center of mass is on the line Y = 0.8X, so the overall center of mass must also be on that line.
But these are three different lines!
Yes, troubling… But no more so than the ambiguity in the question “Where is the center of the integers?”. One might say “0”, because the integers are symmetric around 0… but the integers are just as symmetric around 17, or -1/2.
We can see the connection to conditionally convergent series even more clearly if we look at the profit from switching envelopes:
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, on average, 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 to. 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?
(Note: This is largely a repost from here)
You know that with p(0.5), A = $50, and with p(0.5), A = $100.
In other words, the expected value of A is $75.
If A = $50, then the expected value was $75. If you stick, you are $25 up on the EV, if you switch, you are $25 down.
If A = $100, then the EV was $150. If you stick, you are $50 down on the EV, if you switch, you are $50 up.
Regardless of which case is true, you have an equal chance of winning or losing.
I see now what the “paradox” is - the average EV would seem to be more than $100, suggesting you should always switch.
But that only works once you know what the amount is and there is no hidden variable. It’s like saying, “Here’s $100. Would you like to swap that for an even chance of $50 or $200?” Well of course you would - the upside is bigger than the downside.
You can’t extend that to say “you should always keep switching”, because clearly you can’t. Either you switch and you win, or you switch and you lose.
The paradox comes from the apparent difference in outcomes from the player’s perspective vs. the game maker’s perspective. Those arguing that A (or X) stands for two different things are seeing this exclusively from the game maker’s perspective, i.e., someone who know exactly what the two amounts are, but perhaps not which envelope contains which amount.
The player OTOH has no idea at all about what the two amounts are. Imagine you go into the home of an eccentric infinataire (who has an infinite amount of money). He hands you an envelope with some money in it, and holds up a second envelope. He says,
“I put some amount of money in the envelope you’re holding. Then I flipped a coin. If the coin turned up heads, I put double that amount of money in this envelope. If it turned up tails, I put half in this envelope. Would you like to trade envelopes with me?”
Suppose you look into the envelope before making your decision, and you found $100. Then there is a 50-50 shot that the other envelope contains $50 or $200. Then the expected value of the other envelope is (1/2)*50 + (1/2)*200 = $125, which is more than your $100, so you switch.
Suppose instead you found $450 in your envelope. In this case the other envelope either contains $225 or $900, with expected value $562.50. So again, you would want to switch.
In fact, no matter what your envelope contains, the other envelope has an expected value of 25% more. You’ll switch no matter what your envelope contains, so why even bother opening it?
So the infinitaire leaves and comes back with a third envelope, and tells you the same thing about the coin flip, so this envelope either contains half or double the amount you currently have, and offers to switch with you. By the same analysis, you would switch without even looking in your envelope, right?
Let’s change things a bit. Same scenario, but instead of giving you a long explanation about coin flips, the infinitaire tersely says, “This envelope either has half or double the amount in your envelope, with equal probability.” Surely the same analysis applies, and you would switch without even opening your own envelope? And when he goes back to get a third envelope and gives the same terse explanation, you would switch again without opening the envelope?
Finally, suppose that before the Infinitaire gives you an envelope, he thoroughly mixes the first two up so that even he doesn’t know which is which. So he hands you one and tells you the other contains either half or double that amount, with equal probability. This statement is still true, yes? So you switch without looking in your own envelope, and he leaves, but this time, unbeknown to you, he comes back with the same envelope you just gave him. He says “his envelope either has half or double the amount in your envelope, with equal probability.” Neither you nor he knows which one holds more money, so it’s 50-50 that he has the larger envelope. So, you do the same analysis in your head, conclude that the envelope he holds has an expected value 25% higher than the one you hold, and decide that you should switch envelopes.
Thus the paradox. It seems that the player would be made better off by continually switching envelopes, but we know that it cannot be so.
The thing is once you open one envelope, and discovered the amount of money inside, it probably makes sense to trade for the other one.
For example, if you open the envelope and it holds $100 and someone gives the option of trading it, you should take it.
Think about. There two possibilities, that the other envelope contains $50 or that it contains $200.
It’s like being offered to play a game with a 50% chance to win $200 and a 50% to win $50, with a play price of $100. That’s clearly a good game.
Another way to look at it is even odds to lose $50 (going from $100 to $50), or to gain $100 (going from $100 to $200). Again, this is a good deal.
But you may not want to make the trade in very extreme circumstances, as the relationship between utility and money can become strange at the extremes and the ultimate goal is to maximize utility, not dollars and so you might need to incorporate a risk adjusted point of view. For example, if you opened the envelope and it contained 2 million dollars, you might not want to risk losing 1 million dollars for the chance of winning another $2 million even though the game a has a positive expected value. It would all depend on the curve of utility to dollar ratio.
But the point is that once you have a piece of information, the paradox goes away. Trading for the other envelope makes sense. In the absence of that information, the expected value is equal. It’s also different in that once you open one envelope you have a constrained view of the options and you know that once you switch, you now have a new game. In that when you have an unopened envelope in your hand and the $100 as the other option, you know that if you trade you have a 50% chance to gain $50 and 50% to lose $100 - so the proposed trade is non-paradoxical in that you would have no inclination to trade back.
That’s not really a paradox - that’s playing with incomplete knowledge of the game. If the man is truly an infinitaire, and can keep preparing fresh envelopes, each of them checked against the previous envelope and either doubled or halved with equal probability, then the more times you play the better off you are likely to be. That’s surely a fact, not a paradox. At each step your expected win increases by 25%, but that’s OK because the man has infinite wealth and can keep stuffing new envelopes with more cash.
The paradox only arises if he switches back to your original envelope without your knowledge, in which case all bets are off. And that’s effectively what happens if there are only ever two envelopes to start off with - if you keep switching, you’re just getting the same envelope “switcherooed” on you.
If you know that seeing the amount is going to make you want to switch, then why not just say right at the start “I’m going to open the envelope to look, and then switch”? And if you’re going to do that anyway, then what’s the point of looking in the envelope?
Logically, you can look at the situation as a whole and it’s clear that you should just pick an envelope at random and take its contents. You have no information on which to make a choice that will be more informed than a random guess.
But when you try to prove this answer is correct, it falls apart. No matter what amount of money there is in your envelope, by switching you will either gain an amount equal to what you have or lose an amount equal to half of what you have. So it appears that the possible gain is twice as large as the possible loss and the possibilities of a gain or loss are equal so you logically should take the risk of switching.
The paradox is that you shouldn’t get this result. You “know” that there’s no reason why one envelope is “better” than the other. So when you examine it in detail, you should confirm this intuitive expectation. But somehow you don’t and one envelope somehow appears to be a better choice than the other.
This is why I prefer the geometric rephrasing in terms of center of mass… so we don’t get stuck on the epistemological bits of the presentation, which are irrelevant to the underlying phenomenon of the paradox.
To close the loop. Once you look in one envelope, you know all of the possible outcomes.
Do you get to keep the money if you chose the one with LESS cash?
Yes, if you switch you get the contents of the second envelope regardless of whether it’s twice the amount or half the amount in the first.
If you didn’t get to keep the lesser amount the paradox wouldn’t exist. If that was the case then you have an envelope with a known quantity X and you’d have the option of switching to a second envelope which had an effective quantity of either 2X or 0 (the X/2 amount wouldn’t matter if you don’t get to keep it). Under that situation, the expected value of the second envelope would be equal to X and there’d be no reason to prefer either envelope over the other. The proof would match the intuitive answer.
The paradox would still exist; it just wouldn’t be relevant to your decision-making. You could still ask yourself “Do I expect, on average, the other envelope to have more money than this one?”, instead of “Do I expect, on average, the other envelope to yield more money for me to keep than this one?”, whether or not you expected to do anything with that information.
Like I said, though, I think all the decision-theoretic framing just obfuscates what’s going on.
As others have pointed out, if you open the envelope and it contains 100, you know the other contains either 50 or 200. If you open the envelope and see it contains X, the other contains either 2X or X/2. So why bother opening the envelope? Please don’t offer me any Schroedinger’s cat type explanations either 
It seems to me that the problem is not how ‘X’ is used in the equation, but rather the use of 50-50 probability. The only way for it to be 50-50 is for the guy bankrolling this to have an infinite amount of money.
So what I did was set the expected value of swapping to be equal to X, and then solve for p. I ended up with p(2X) = 1/3.
This seems to be just about as suspect as gaining from swapping indefinitely, when we assume that p = 1/2.
In other words, what I did was start with the fact that the expected values are the same and then work backwards from there.
Why? It’s a one-time deal. The only reason an infinite amount of money would be required would be if one envelope contained infinity/3 and the other contained 2infinity/3. That would fit the criteria but so would any lesser sum that is evenly divisible by six.
Lets say the bankroller had 500 moneys to make the game and he tells you so. (in other words the max he can lose it 500)
If you open the envelope and it contains 400. You know for sure that the other contains 200 because is can’t possibly contain more than the banker can afford to lose.
If you open the envelope and see that it contains 250, then it is possible that the other can contain 500.
Now consider that you don’t know what the bankers limit is, but you know that it is finite. Imagine a function that randomly picks a number from (but not including) 0 to the limit. He puts that in envelope 1. He then flips a coin to determine whether envelope 2 will contain twice or half envelope 1. However, if envelope 1 contains more than half the limit, there is no point in flipping the coin.
The limit itself is now a random variable but because it is finite, the chances of the other envelope containing the lower amount seem to be higher. The only way for it to be 50-50 is if the limit is infinite.
But in both of these examples, you’re adding new information in order to make the problem easier to solve.
Thats true but in the real world no-one has an infinite amount of money so we can assume that information.