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Sorry, but no.

No probability distribution has been defined for the amount inside an individual envelope, and you can't just pull one out of thin air. That's not how mathematics works. If you had a probability distribution for that, you could make all sorts of arguments from it. But we're not given one, and we're given no information from which we can imply the existence of one. Therefore there is no such distribution. Period. (Interestingly, assuming such a distribution gives useful information even when it is the wrong distribution, but that is a complicated variant on the problem.)

Now you can argue until the cows come home about what probability "really" means, or what is "really" true in a made up problem. Plenty of philosophers are willing to argue that with you. But the way that mathematicians look at this is quite simple. You can only work with the information you are given, and you can't make up information that you weren't given. Arguing about the fine points of reality within a piece of fiction is, not to put too fine a point on it, just plain silly. After all it is a made up example. So don't do that.

We are given certain information. We look in the one envelope if we wish, and we get a dollar value. Based on that we can draw inferences about what might be in the other envelope. We can create an expected value for said envelope. All that is fine.

But our intuition about what the expected value has to say about what we should do is wrong. Nothing is wrong in the math - it is our intuition that is wrong here. Our intuition is based on what happens if you encounter large numbers of similar, but independent, events. However in this case we have a singular, and extremely unlikely to be repeated, event. Therefore the basis for our intuition is lacking here. And we are mislead.



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:

  P(x=A) = 1/2, P(x=2A) = 1/2, P(x) = 0 for all other x.
and that this is clear from the phrasing of the question. Extracting this distribution from the text is no different than solving any other problem not stated in rigorous mathematical terms. With this distribution, with A properly defined (see the next point) upfront, the whole paradox disappears.

  --
Nothing is wrong in the math [..]

I think the statement in the article that

  4. If A is the smaller amount the other envelope contains 2A.

  5. If A is the larger amount the other envelope contains A/2.
is wrong, because it tries to define A twice with different concrete values and goes on to treat them as the same value. If your point is that saying 'A is the quantity in the envelop you just chose' makes A ill defined, then I guess you are right, but that doesn't mean that A cannot be defined properly and it doesn't mean the intended problem can't be solved.


> > 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.)




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