How to find the probability of one die roll being higher than a second die roll? And the catch is the first die has $x$ sides, while the second die has $y$ sides, where $x\neq y$
I had this figured out several years ago, but apparently have forgotten both the answer and too much math since then to either work it out or figure it out from Googling.
Similar questions have been asked before, but all the answers are simplified to two dice with the same number of sides on the symmetry that $P(A>B) = P(B>A)$, which does not hold true for my case. What I need to know is:
Given two discrete, uniform distributions, what is the probability that a sample from the first is strictly greater than a sample from the second?
Given $X$ distributed uniformly in $\lbrace0,1,2,3,...,x\rbrace$ and
Given $Y$ distributed uniformly in $\lbrace0,1,2,3,...,y\rbrace$
How can I find $P(X>Y)$?
 A: 
Let these dots be the points $\{(x,y): x\in X, y\in Y\}$. What is the ratio of number of points under the line $y = x$ marked black against the total number of points? This picture is for $x\ge y$. How does the picture look when $y\ge x$?

Alternatively by law of total probability
If we assume $x\ge y$, then
$$P(X>Y) = \sum_{j = 1}^y P(X>j)P(Y = j) = \sum_{j = 1}^y\frac{x-j}{x}\cdot \frac{1}{y} =\frac{1}{xy}(xy-\frac{y(y+1)}{2}) $$
I intentionally didn't simplify the last expression, because $(xy-\frac{y(y+1)}{2})$ is the total number of black points, and we divide by total number of points $xy$. Now you can see the connection between probability method and geometric meaning.
Now for the case $y\ge x$, we have 
$$P(X>Y) = \sum_{j = 1}^x P(X>j)P(Y = j)$$
I encourage you to work this out and draw a picture.

[EDIT]
As discussed in the comments, 
$$P(X>Y) = \sum_{j = 1}^{\max(x,y)} P(X>j)P(Y = j) = \sum_{j = 1}^{\min(x,y)} P(X>j)P(Y = j) $$
because when $j$ goes beyond $\min(x,y)$ then either $P(X>j) = 0$ or $P(Y = j) = 0$. Also this can be verified from picture as well.
