Why does this strategy work? Game:
You have two envelopes. Each one contains a real number. 
Choose one envelope and you get to see the number on it. Now you have two choices: select the same envelope, or swap. 
At the end, if you have chosen the greater number, You win.
Problem: 
Find a strategy that gives you a probability of winning that is more than $0.5$

Solution: 
Choose one real number $x$. If envelope one contains $y>x$, keep it, else swap.

My Question: Why does the strategy work? Can we prove that probability-theoretically?

(Previously looked at it from the wrong point of view...)
If the two envelope numbers, and YOUR number, are all picked using the same random number generator, and all you can ask is is A > B,
then, for all that it matters, the random number generator could be handing out 1, 2, and 3 for each set.
So if we call the numbers in the envelopes Y and Z, and your number X,
then one of these is true, with equal probability:
Y<Z<X
Z<Y<X
Y<X<Z
Z<X<Y
X<Y<Z
X<Z<Y

In the first three cases, you elect the second envelope, as Y<X.  In the last three cases you elected the first envelope as X<Y.
And you win with those selections in: 1, 3, 4, and 6.  Thus 2/3 probability of a win.
One might claim that one could get equal numbers.  I claim that this is negligible, as we are talking about real numbers, and presumably a continuous distribution.
