Probability/Geometry Question 
Why do we need to account for separate instances where $x>y$ and $y>x?$ Why can't we consider them the same (i.e. we have a cut at $1/3$ and $2/3$ is the same as a cut $2/3$ and $1/3?$ For example, we don't consider the roll of two dice $3,3$ as two different outcomes. Thanks!
 A: We consider $3,3$ to be a single outcome of the roll of two dice,
but we consider $2,4$ and $4,2$ to be two different outcomes.
The cases $x < y$ and $x > y$ correspond to $2,4$ and $4,2$.
The case corresponding to $3,3$ is $x = y.$
Since you have infinitely many possible values of $X$ or $Y$ between $0$ and $1,$
not just six values, the probability that $X$ and $Y$ will be equal is zero.
But it is, in fact, perfectly reasonable to define random variables,
let's call them $U$ and $V$, such that
$U = \min\{X,Y\}$ and $V = \max\{X,Y\}.$
Then indeed you will have $U \leq V$ only and no case where $V > U.$
Does that make the problem easier to solve?
When we allow $X$ to range anywhere from $0$ to $1$ with uniform probability regardless of $Y$, and $Y$ to range anywhere from $0$ to $1$ with uniform probability regardless of $X$, it is easy to see that the pair $(X,Y)$ will be somewhere in the square shown in the figure and that it random position in that square is uniformly distributed over the square. How shall we describe the distribution of the pair $(U,V)$? It ranges over some figure in the plane -- is it uniformly distributed in that figure?
(The figure is the right triangle that lies above the line $U=V$ within the square, and yes, the distribution is uniform, but do you know how to prove it? Again, did this really make the problem easier?)
By the way, the figure in the solution is incorrect. The correct figure is below:

The case $x < y$ is the gray triangle in the upper left part of the square,
and the case $x > y$ is the gray triangle in the lower right part of the square.
