This is essentially a dice roll comparison question, but the range of each dice is not 1 to n, but rather x to n+x-1.
I found this equation to determine the odds that a single roll of a 1d(x) is higher than a single roll of a 1d(y) (assuming x > y) as follows...
1 - (y + 1)/2x
I did not proof this myself, so it may be a bad equation.
My question is, how can I determine the odds that a random number from X1 to X2 is greater than a random number from Y1 to Y2? I think this is a dice question, I am just not certain how to factor in the offset.
Thanks
We will assume that you split ties evenly. The basic idea is that if both people roll within the overlap of the two ranges, each has a $\frac 12$ chance to win. If at least one does not roll within the overlap range, we know who will win independent of the other roll. Let us assume $Y2 \gt X2, Y1 \gt X1$. Then $Y$ is guaranteed a win when he rolls more than $X2$ or when $X$ rolls less than $Y1$. The chance $Y$ rolls greater than $X2$ is $\frac {Y2-X2}{Y2-Y1+1}$ because there are $Y2-Y1+1$ possible rolls, of which $X2-Y1+1$ are in the overlap zone. Similarly, the chance $X$ rolls less than $Y1$ is $\frac {Y1-X1}{X2-X1+1}$. We can't just add these to get the chance we have a guaranteed winner because we double count the events where $Y$ rolls high and $X$ rolls low. The chance at least one is outside the overlap zone is then $$\frac {Y2-X2}{Y2-Y1+1}+\frac {Y1-X1}{X2-X1+1}-\frac {Y2-X2}{Y2-Y1+1}\cdot\frac {Y1-X1}{X2-X1+1}$$ The chance $Y$ wins is this plus half the chance they both roll inside the overlap zone, so $$\frac {Y2-X2}{Y2-Y1+1}+\frac {Y1-X1}{X2-X1+1}-\frac {Y2-X2}{Y2-Y1+1}\cdot\frac {Y1-X1}{X2-X1+1}+\frac 12\cdot \frac {X2-Y1+1}{Y2-Y1+1}\cdot \frac {X2-Y1+1}{X2-X1+1}$$