Sorry no, that was not my intent. In the event of R you have A=10 and B=20. In the event of S you have A=10 and B=5. These are mutually exclusive events, which means in the case of R the amount 5 does not exist at all, and in the event of S the amount 20 does not exist at all. So one of those sample spaces is feeding you false information. The only way to avoid this is to treat X as the unknown variable it is. — Jeremiah
Hm, thinking about it a bit more, I think we're making a basic mistake, here. X/X2 is the relationship of the variables, not the sample space. I'll go at it step by step so we can see if I've made a mistake somewhere:
1. We have two envelopes with two different amounts of money:
Envelope1 = X $
Envelope2 = Y $
At that point E1 and E2 do not refer to specific envelopes, nor do X or Y refer to specific amounts. It's simply two variables with two values, and we have no more information. If we have 10 $ in one envelope and 20 $ in another, it doesn't matter whether we set 10 X and 20 Y, or the other way round. It's completely arbitrary. Both constellations describe the same event.
Both X and Y have the same sample space: any number that makes sense of $. Both sample spaces might be, for example, the natural numbers (weight, space, etc. are complications we don't need).
2. If we learn that one envelope contains exactly double the amount of the other, that tells us more. We now can get rid of one variable. But the sample space isn't X and 2X. It's the natural numbers for one (let's call it X), and depending on one of two assumptions we make about X, the sample space of the second is a transformation of X.
Two assumptions?
2. a) We assume X is greater of the two numbers.
Envelope1 = X $ (natural numbers)
Envelope2 = Y $ (X/2)
2. b) We assume X is the lesser of the two numbers
Envelope1 = X $ (natural numbers)
Envelope2 = Y $ (2*X)
These two assumptions both validly describe situation. We still don't define a real envelope; we merely define wether X is the greater number and Y is the lesser number, or the other way round.
In both these assumptions we don't know the actual value of X. So if someone tells us that one of the envelopes contains 10 $, then we don't know whether X is the greater or lesser number. With regards to the above, we don't know where to put it. But we do know it has to be one of the two: 10 $ is either the greater or the lesser number. This gives us two possibilities:
2. a)
Envelope1 = 10 $
Envelope2 = (10/2) = 5 $
or 2. b)
Envelope1 = 10 $
Envelope2 = (10*2) = 20 $
But what we've done here is twofold: we've set X = 10, and we've set envelope1 as the envelope that contains X. We do not know whether X is the greater or the lesser number. The question we care for is what's in envelope2, and the answer to that is:
If X is greater number, envelope2 contains 5 $.
If X is the lesser number, envelope2 contains 20 $.
The sample space for envelope1 was all the natural numbers, and the event is now 10. Since the sample space of envelope 2 is dependent on the sample space of envelope 1, there are only two possibilities: X/2 or 2X. We simply don't know whether X is the greater number or the lesser number.
It doesn't matter which envelope we open first, we never know which is the greater or the lesser. Because of this, we can set any of the envelopes as 1 or 2, and we always have the same situation:
E1 = 10 and E2 = 5 (X > Y, Y = X/2)
E1 = 10 and E2 = 20 (X < Y, Y = 2 X)
If we only open one envelope, we might open the other envelope first. We wouldn't know about 10, in that case, but either about 5 or 20, depending which is true.
For 20 we'd get:
E1 = 20 and E2 = 10 (X > Y, Y = X/2)
E1 = 20 and E2 = 40 (X < Y, Y = 2X)
The ratio remains a constant, no matter which number you draw, and that's why you alway stand to win twice as much as you would lose. This is a function of what you know about the ratio. However the natural numbers that make up the individual sample spaces differ.
If the envelopes contain 10 and 20 dollars, and you set E1 as the envelope you pick first you get:
For E1 = 10:
the sample space for E2 is [5, 20].
For E1 = 20
the sample space for E2 is [10, 40]
That's because the sample space for E1 is not X. The sample space for E1 is the natural numbers.
X is the event.. However, once we know the event for E1, we know that the sample space for E2 is [X/2, 2x], and that's because of the ratio. We can't reduce the sample space to 1 item because we cannot know whether X is the greater or the lesser number until we look at E2. But E1 is chosen at random.
So we get the following:
Envelope1 = [X | € N]
Envelope2 = [Y | € [X/2, 2X]]
I'm not an experienced mathematician, so I might have gotten the notation wrong. But does the reasoning make sense?