I think the paradox stems from the two different ways we calculate the probabilities, and while one is in line with our intuition, the other isn’t.
One way to think about this is by stating the random variable being sampled is the total value of both envelopes combined, say the random variable is T.
Now, one of the envelopes will contain \frac{T}{3} and the other will contain \frac{2T}{3}.
Now we can apply conditional expectation on the gain from the exchange, G.
E[G] = P(X = \frac{T}{3}) \cdot \text{Gain}(X=\frac{T}{3}) + P(X = \frac{2T}{3}) \cdot \text{Gain}(X=\frac{2T}{3})
Which simplifies to:
E[G] = \frac{1}{2} \cdot \left(\frac{T}{3}\right) + \frac{1}{2} \cdot \left(-\frac{T}{3}\right) = 0
This aligns with our intuition as well, as in a symmetric setup like this, exchanging shouldn’t bring any advantage.
The paradox stems from a different calculation, where we can calculate the gain from the exchange as:
E[G] = P(\text{X = Bigger Value})*(\text{Gain(X is bigger)}) + P(\text{X = Smaller Value})*(\text{Gain(X is smaller)})
E[G] = \frac{1}{2} \cdot \left(\frac{-X}{2}\right) + \frac{1}{2} \cdot \left(X\right) = \frac{X}{4}
SInce getting a positive gain regardless of the value of X, information of the amount we have should have no affect on our decision to switch since it will be positive anyways. Even if we don’t ‘peek’ this leads to the paradox of both players having advantage in a symmetric game. Which seems to go against the intuition that a symmetric setting like this would offer an advantage to switching.
From what I could gather (I could be wrong) the problem stems from:
a) Assigning equal probabilities to X being the bigger value and X being the smaller value, the implicit assumption here is that any X is equally likely to be the bigger value, which is problematic as the sample space is the real line and assigning equal probabilities to all numbers isn’t a valid pdf. Which means this problem is ill defined for this method since we don’t know the distribution of X which will change the chances of a particular X being the bigger value.
b) This is a guess, but there do exist other paradoxes where the way we define the sampling gives different probability values, for eg. the problem of finding the probability a randomly chosen chord is bigger than the side of the inscribed equilateral triangle, the answer changes depending on how you ‘select’ these random chords.
Similarly in this example, I think the first calculation with the T being the random variable reflects the section in which you take a certain amount of money and then put it into two envelopes such that one is double of other (which is what is described in the question).
While for the second calculation you first put X amount in an envelope and then flip a coin to decide whether to put double the amount of half the amount in the other envelope. If the player knows he has the initial envelope then switching makes sense, and this doesn’t lead to infinite switching as he knows the expected value of the unknown envelope is \frac{5X}{4} which is greater than X.
This was my attempt, please do correct me if I’m wrong.