Probability Number A greater Number B

Given $$a \in \{1,2,...,250\}$$ and $$b\in \{0,1,...,1000\}$$

What is the probability that $$a > b$$?

I know there already is this question with a very similar question, however it's over an interval of $$\mathbb{R}$$ so I hope there is another way opposed to the Integral used in the formal solution.

How does one go about calculating said probability?

You can use the same idea as the integral, counting lattice points. There are $$250 \cdot 1001$$ choices for $$a,b$$. If $$a$$ is $$1$$ there is $$1$$ successful choice, if $$a$$ is $$2$$ there are $$2$$, and so on, so the number of successful choices is $$\sum_{i=1}^{250}i=\frac 12(250)(251)$$ The probability is then $$\frac{\frac 12(250)(251)}{250 \cdot 1001}=\frac {251}{2002}$$
The possible $$(a,b)$$ pairs cover a rectangular lattice of $$250\times1001$$ points. Among these, the ones that do fulfill the condition cover a right triangle of base and height $$250$$, counting $$\dfrac{250\times251}2$$ points.
$$\frac{250\times251}{2\times250\times1001},$$ roughly one eighth.
\begin{align} \sum_{a = 1}^{250}{1 \over 250}\sum_{b = 0}^{1000}{1 \over 1001}\left[a > b\right] &= {1 \over 250 \times 1001} \sum_{a = 1}^{250}\sum_{b = 0}^{a - 1}1 = {1 \over 250 \times 1001}\sum_{a = 1}^{250}a \\[5mm] & = {1 \over 250 \times 1001}{250\left(250 + 1\right) \over 2} = \bbox[10px,border:1px groove navy]{251 \over 2002} \approx0.1254 \end{align}