The grass is greener on the other side, is it?

Suppose that before you lies 2 envelopes; with the following piece of paper and corresponding instructions: In the 2 envelopes, there are money. You are only allowed to take one, and here's some information to help you choose. One envelope contains twice as much money than the other. To further help you decide, once you choose a particular envelope, you can open it to see its contents. At this point, you are given the chance to change to the other envelope, or just take whatever you find in the first envelope. So, which one do you choose?

You reached out for one of the envelopes, and you find $x inside. Now, you stand there, thinking whether you should change. In your head, a small voice appears; The probability that the other envelope has either $2x or $0.5x is 50-50. You feel a little sinking feeling. Perhaps you should be be contented with what you have. At which the small voice came back again; Nonsense, you learned in school about calculating payoffs, did you not? Look again. Your expected payoff from switching to the other envelope is half of 2x + half of 0.5x. Which gives you 1.25x. So you should change. At this, you reach out for the second envelope.... And another voice came in: Wait, you chose the first envelope randomly. If in the first place you had chosen the other envelope, your reasoning just now, would have led you to this envelope that you are holding. So there is really no reason to change. In fact, the reasoning applies irregardless whether whatever sum of money is in your envelope. You do not even have to open the envelopes. You just choose one envelope, and your reasoning will lead you to the other, and once you change to the other, your reasoning will point you back to the first one, and an infinite regress will follow! So.. you are stuck here forever... As you stand there, gripped with indecision, what should you do?

That was the two envelopes paradox, and certainly it poses a good model of grass being greener on the other side. Both arguments seem equally convincing, whether to switch or not to switch. There are more arguments online, arguing the differences between probabilities, plumbing through all the fine nuances within the wording of the problem. It seems that having a look within the envelope that you first chose makes no real difference to your chances, since your chances appear to be resolutely 50-50. In fact, you weren't granted a peek inside the first envelope that you chose, it would not matter which one you chose, or whether you chose to switch. As it turns out, having a look does change matters, for you are granted extra information, in your favour actually.

A search on google found an interesting note by Dov Samet, Iddo Samet, and David Schmeidler on this problem. They recast the problem as follows: Imagine playing a 2 person game as follows, A chooses 2 numbers randomly, B chooses one of the number and then pronounces whether its the larger value or smaller value of the 2. B wins this game if he pronounces correctly. A wins if B loses. Imagine if you are A, how would you prevent B from winning in the long run if you kept playing this game? If you say that the probabilities remain resolutely fixed at 50-50 then your job is pretty simple isn't it? Just anyhow choose 2 random numbers. Since B cant win in the long run, you are going to win. But B do have a strategy for getting more than 50-50. Before B chooses any number, he randomly produces a number t. If the number he chooses is greater than t, then he stays put. If the number he chooses is less than t, he switches. To see how this strategy lets B have chances greater than 50% in the long run, let the 2 values of A be x and y. Assume B chooses x. If x < y < t, B switches correctly. If y < x < t, B switches wrongly. If t < y < x, B stays put correctly. If t < x < y, B stays put wrongly. Therefore, if t is outside range of (x,y), then B has 50-50 chance of getting it correct. If t is within range of (x,y), B is always correct. The game for B has become one of calculating the probability of t landing within range of x,y. If x,y are integers, then A can only make the range of x,y span as little as possible.