Part of the issue that the mathematicians have with the problem appears to be that the problem is incompletely defined. Personally, I don't believe it is, because my reasoning ends at "why the heck would you use an expectation value in this case?" But, if we switch our perspective a bit, and assume (because the problem assumes) that an expectation value is OK to use, then it is fair to say that the problem doesn't give enough information to evaluate an expectation value.
Why? Because in the real world, there are upper and lower bounds to the amount of money that might realistically be put in the envelope, and potentially a probability distribution of particular amount pairs. E.g., if you found $0.01 in your envelope, you're not going to find 1/2 cent in the other, it will be 2 cents. (Or, there is a real US-minted half cent in there, which is probably worth a great deal if you can find a coin collector to whom to sell it!) In general, if one assumes even a rudimentary guesswork "money spectrum", if one's envelope amount is on the low end, there is a good chance that the other envelope has more, and as one approaches the high end, switching envelopes gradually becomes more likely to "halve" the initial value you see.
Of course, it all depends on the distribution. If it's a flat distribution with a fixed maximum and minimum, e.g., $1 - $20, no fractional dollars, then you have even more constraints that affect whether you should switch: any odd amount (which will always be < $10) will always be the lower of the pair, and any amount > $10 will always be the higher of the pair. If it's an even number of dollars less than or equal to 10, then, quite interestingly, the expectation value of switching really is 5/4 times the amount you see! Why? Because in this case we have 10 equally likely pairs, each with a 10% chance to come up. Looking at the value in the envelope you select then limits the possibilities down to only two pairs, each 50% likely. So, of the 20 cases (10 pairs, 2 possible choices for each) we know that switching doubles our value in 5 cases, halves our value in 5 cases, and has an expectation value of 5/4 our value in 10 cases. Note that this breaks the symmetry that the paradox implies, that switching back would also have a high expectation value, because one HAS to choose an envelope and open it to get the information needed to calculate the expectation value, and that expectation value isn't ALWAYS 5/4 the original value, but sometimes half or twice.
So, the mathematicians end up saying that if you assume a distribution that is any amount of money (unbounded, not necessarily integers of dollars or pennies), then you have an "improper prior
", in which case an expectation value cannot be calculated. Or, stated more intuitively, if you know "how" the envelopes were filled, then looking at an envelope gives you enough information to determine whether you want to choose the other, and breaks the symmetry of using the same argument to switch. Lacking that knowledge of "how", the expectation value is undefined.