> > No probability distribution has been defined for the amount inside an individual envelope

> I don't understand this assertion. It seems to me that both envelopes have the trivial distribution:

What I mean is that there is no a priori distribution in terms of actual numbers. What are the odds that the amount in the envelope before you open it is in the range $5 - $500? We are given no such information. Before you open the envelope you are given, A is a variable, but not a random variable. Its possible values ideally would have a uniform distribution on an infinite set, which is mathematically impossible. There is no such distribution. (There is an alternate approach which is to claim that that the sum of the two envelopes has an unknown distribution, which creates a distribution for the values of each envelope. This leads to a very different mathematical idealization of the problem. But the other distribution, being unknown, is not available to you. I will trace the reasoning for this through as well.)

If you open the envelope in front of you, A then becomes becomes a number, and there is indeed now a probability distribution for the contents of the other envelope. And that distribution is now, as you said

P(x=A) = 1/2, P(x=2A) = 1/2, P(x) = 0 for all other x.

(Note, the alternate approach would now assert that the number in the the envelope provides unknown information on whether it is larger. Therefore the other envelope now has a different unknown distribution that depends on the actual magnitude of A.)

But now we arrive at a different point. Now we have a probability distribution for the other envelope, but our intuition fails for the simple reason that our intuition of the meaning of expected value is based on what happens after repeated opportunities at similar random events. (Insert the strong law of large numbers, etc.)

(The alternate solution finds that the other distribution causes the other envelope, on average over all possible values you could see, to have the same value as your own. However this is of no obvious direct use.)

From the point of view of the standard mathematical idealization, what you guys have been doing is trying to train your intuition to create arbitrary distinctions that avoid conclusions that bother you. But you're drawing a distinction that doesn't make sense. If you've opened the envelope and you have $10 sitting there, then A is $10. Period. It isn't one of two different numbers, it is the number you see in front of you. That has now become a provided fact. At this point you can draw a distribution for the other envelope. But your intuition about expected value fail. Why? Because this scenario, with this set-up, is not something that can become subject to repeated trials.

(The alternate approach comes to a different conclusion. Your intuition, they claim, is thrown by the fact that you do not have the critical information about the actual distribution of the other envelope, and so cannot come up with the correct numbers to decide. Interestingly in a bizarre twist, it turns out that if you make up a distribution and pretend it is the unknown one, as long as that distribution has some probability of answers falling in every possible range, your decision winds up being right better than half the time. However how much more you are right depends on the unknown distribution, and is therefore unknown to you.)