Comparing two die rolls of n-sided dice I'm practicing my coding and designed a simple game that compares two results of a die roll modified by different multipliers. I can easily make these calculations in my code and determine the winner, but am struggling to represent the probability for success.
Let x, y = different percentage multipliers (e.g. .32, .75, 1.00, etc.)
Let RN(n) = roll number N with a given die with n number of sides (e.g. 6, 10, 100, etc.)
What is the probability that...
x * R1(n) >= y * R2(n)
As a practical example from the game, I have a "skill" of .65 (65%), while my opponent has a "skill" of .50 (50%). We are both rolling a 20-sided die. If we each multiply our skill by our individual roll, what are the odds that I would tie or win? (While this is one specific example, skill percentages will change in the game! So they are independent, dynamic variables.)
As a separate but related question, I scoured the internet for an hour trying to figure this out. If you could recommend a few links to sites that explain this in detail, it would also be helpful. (I have a little math background, but am less familiar with combinatorics.)
Thank you!
(Edit - made percentages more clear...)
 A: Probably the simplest is just to count them up for small numbers of sides.  Go through the possible rolls for one person and compute the chance the other will do at least that well.  Using your $x=0.65, y=0.5$ example, the first player has to roll at least $\frac {0.5y}{0.65}$ to win.  If $\frac {0.5y}{0.65}$ is an integer in range, you could have a tie.  So if player $2$ rolls $10$, the first player must roll $8$ to win, which is a chance of $\frac{20-8+1}{20}=0.65$.  If player $2$ rolls $13$, player $1$ must roll $10$ to tie and higher to win, with a winning probability of $0.5$.  
For a die with a large number of sides, you can use a continuous approximation.  Let each die roll be in the range $[0,1]$.  Imagine plotting a point in the unit square representing the two rolls, with the first player horizontal and the second player vertical.  You can draw a line through the origin with slope $\frac xy$.  The first player wins if the point is to the right of the line and the second player wins if the point is to the left of the line.  If $x \gt y$ the area the second player wins is a right triangle with sides $1, \frac yx$, so the second player wins with chance $\frac y{2x}$
