The Two-envelope Problem

  Well, this is one of the recurring problems we've probably all seen, time and time again, with various "obvious!"-style solutions as well as a nice wall of text explaining away the paradoxes by claiming that it was all a case of confusing phrasing -- with an inspiredly sneaky re-expression of the problem, the paradox dissolves! However, there is more or less no true escape from the true Mathematics of this problem. Wikipedia has a nice and technical exposition on the various "solutions" and solutions of the paradox, and equally sneaky reformulations of the problem to expose the holes in the "solutions". You should read it if you actually want a formal explanation, but I believe mine follows the Wikipedia article somewhat faithfully, up to at least somewhere in the middle.

  The first thing to state is the question in a text-wall format, so that we can demolish it later for clearer insight. Well, here I go: Two envelopes are on a table. One contains twice as much money as the other, and there is no way to tell the contents of the envelope until you open it. You choose one, taking extreme caution not to open it (and cleverly avoid revealing the contents of the other as well, you sneaky person). You think to yourself: If I open this envelope and find X dollars, then the other envelope could contain X/2 dollars or 2X dollars. Those two outcomes have equal chance, since I pick either the larger or the smaller (contentwise) envelope randomly. My expectation of money in the other envelope is hence 5X/4 dollars. This does not make sense, since the expectation of money in my envelope is now also 5Y/4 dollars, where Y is the amount contained in the other envelope. We get X=25/16*X. That's depressing, since it solves to 0. :( In my depression, I claim a paradox!

  Well well, not the most accurate formulation of the problem, but at least somewhat intelligible. Now I shall point out an interesting feature of this problem which is usually ignored in discussion for being somewhat impossible to prove -- Consider, in a train of thought very far removed from the original paragraph, the following paradox: "X is 1. X is 2. Therefore 1=2. Contradiction?!". This is clearly a matter for the logicians out there to name. I'd personally file it under the classification of "Lies" and trash it. But, the thing is... our problem seems self-consistent... up till the paradox. So now let's rephrase it without the envelopes:

  I pick a number (call it X), and map IDs #1 and #2 randomly to the two positive numbers X and 2X. Now, you pick either #1 or #2 and we define the number it maps to be Y. Now the other number could either be Y/2 or 2Y with equal probability, and the expectation is 5Y/4. The problem statement is clearly sound, since these are more or less mathematical formulations, and if a contradiction exists, the subject would fall apart and all I learnt in school would be totally useless!!!!! We therefore call it a confusing argument and try to explain it away.

  Clearer (albeit somewhat spoilt by the ending). So now for "solution" 1. We notice that Y can either be X or 2X. As such, Y is inconsistently defined. We therefore redefine it in terms of X and 2X and it all works out to an expectation of 3X/2. The world is saved!

  Clearly this violates the principle of the matter. It's like saying: "Why don't I get dough when I mix flour with kerosene?" and answering it with "Everything is well if you just buy the dough directly from the supermarket, you twit. After all, they've derived the dough from flour as well.". This resolves the paradox by more or less ignoring it, so let's just make the paradox more annoying and harder to sidestep:

  Both the number mapped from #1 and #2 have equal expectations by symmetry, and the smaller of which is further defined to be a random positive real number. You pick #1. It maps to X. Now, the expectation of the value of #2 is (X/2 + 2X)/2=5X/4, since the probabilities of the pairs [X/2,X] and [X,2X] are equally probable. The problem is: the expectation of the other number is always higher than your no matter what it is, so on average, knowing a number causes the other number's expectation to be 5X/4, which is 5/4 times its own expectation. Based on that, before you even know what #1 maps to (ok call it f(1), and similarly define f(2) -- this is getting tedious), you know that f(2) is on average 5/4*f(1). OMG contradiction!

  The resolution for this is a one-liner (Note: see Zagier's one sentence proof for an idea of how a mathematical line might look): X*5/4=X, where X is the cardinality of positive real numbers (commonly known as infinity, though not the only infinity). Well the thing is that since positive reals stretch indefinitely, the chance of choosing a finite number is infinitesimal. And for infinities, the statement f(1)=5/4*f(1) can make perfect sense.

  One last sneaky trick to throw in (and a solid text wall to help obfuscate matters just that little bit more):

  "We shall now be so nice, going out of the way to help you define just a specific case of the problem, and we'll even babysit you by stating precisely how we chose the numbers. More formally, we shall enlighten you as to the probability distribution of the amounts in the envelopes. We have an old friend here, called Mr. Fair Coin, who is a truly erratic fellow at the best of times and completely random at the worst of times. Now, we all know that Charles Dickens once famously said "It was the best of times, it was the worst of times", but you see, truly erratic and completely random are consistent with each other, so you're not getting a paradox for free. We first choose the smaller number by setting it as 1, and then tossing Mr. Fair Coin until he lands on his head (that clumsy fellow!). For each toss where he does not land on his head, we double the value of our numbers (the charitability in me!). Oh and I unconsciously neglected to mention, since it was so banally trivial, that since Mr Fair Coin is so fair, he does not have a preference to landing on heads, nor does he have a preference to landing on tails, and is most assuredly not inclined to land on sides, but lands on one of the three with equal probability. Then we double our number to generate the second number."

--Awkward seconds pass--

  "That's quite close to a random number generator, isn't it?"

  Disregarding the disconnect in common sense, he makes a fair point. Let's list the argument in point form:

P1.1: If the number you choose turns out to be 1, the other is 2, which is truthfully larger than 1.

P1.2: If the number you choose turns out not to be 1, but instead is X, the expectation is 3/5*X/2 + 2/5*2X = 11X/10. The other number is expected to be larger anyway!.

P1.conclusion: CE[f(2)]>f(1) for all possible values of f(1), where CE denotes conditional expectation.

P2: The chance of f(1) being infinite is limit (2/3)^n as n --> infinity = 0. Therefore f(1) is finite, and CE[f(2)] is finite as well. Therefore the previous argument regarding infinities does not hold.

  Final resolution (for this post): (highlight for spoilers below)

  The number of real numbers between 0 and 1 is a humongously huge infinity, infinitely (lol) greater than the number of positive integers. If we were to pick a random number from 0 to 1, therefore, the probability that it is any particular number is in fact 0. But this does not mean that it is impossible, i.e. P(A)=0 does not mean A is impossible, since the number randomly generated from the first try between 0 and 1 will have occurred, despite having 0 probability of occurring.