I am looking for a formula to determine the probability of a A Die with X faces rolling higher than B die with Y faces.
An example would be: what is the probability of 4d10 rolling greater than 3d12
Edit: I thought the last edit summary would pop up but it didnt, so added here. Corrected the question a bit and added example.
Edit #2: I had come up with a similar conclusion for percentages (x-1/2y) in a case of dx vs dy
But I didnt took into account the difference on whether "x" was larger or smaller than "y".
However what about when there is more than one dice of the same type is it still the same formula? considering the sum is a bit more irregular since some numbers repeat more often than others do I replace "x" and "y" with the total sum of possible outcomes of each die combination? (in xda I assume the total is a^x)
If you get a $k$ on the die with $X$ faces (where obviously $1\le k\le X$), there are $k-1$ numbers that you can beat. The probability that one of these numbers comes up on the die with $Y$ faces is
$$\min\left\{\frac{k-1}Y,1\right\}\;,$$
so the desired probability is
$$\sum_{k=1}^X\frac1X\min\left\{\frac{k-1}Y,1\right\}=\frac1X\sum_{k=1}^{X-1}\min\left\{\frac{k}Y,1\right\}\;.$$
If $X\le Y+1$ this is $$\frac1X\sum_{k=1}^{X-1}\frac{k}Y=\frac1{XY}\sum_{k=1}^{X-1}k=\frac{X(X-1)}{2XY}=\frac{X-1}{2Y}\;.$$ If $X>Y+1$ it’s
$$\frac1X\sum_{k=1}^{Y-1}\frac{k}Y+\sum_{k=Y+1}^X\frac1X=\frac{Y-1}{2X}+\frac{X-Y}X=1-\frac{Y+1}{2X}\;.$$