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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?

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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 $$\frac{\frac 12(250)(251)}{250 \cdot 1001}=\frac {251}{2002}$$

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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.

The requested probability is

$$\frac{250\times251}{2\times250\times1001},$$ roughly one eighth.

enter image description here

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\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}

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