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###Brute force calculation

Analysis

Solution

Illustration

Comments

Brute force calculation

The numerator can be simplified to $m(2n-m-1)/2$ for computational purposes, but this simpler formula hides the idea that led to the answer. In case $m$ is the larger number, the sum stops with $n-m=0$ giving a simpler value:

The left hand factor counts the number of outcomes where $Y$ exceeds $X$ while the right hand factor of $1/(nm)$ converts that to a probability. (The count can be simplified to $m(2n-m-1)$ for computational purposes, but that simpler formula obscures the idea that led to the answer.)

In case $m$ is the larger number, the sum stops with $n-m=0$ giving a simpler value:

This can be simplified to $m(2n-m-1)/2$ for computational purposes, but this simpler formula hides the idea that led to the answer. In case $m$ is the larger number, the sum stops with $n-m=0$ giving a simpler value:

The numerator can be simplified to $m(2n-m-1)/2$ for computational purposes, but this simpler formula hides the idea that led to the answer. In case $m$ is the larger number, the sum stops with $n-m=0$ giving a simpler value: