The probability that $x$ is divisible by $p$ where $x\in\mathbb{Z}$ and $p\in\mathbb{P}$

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The probability that $x$ is divisible by $p$ is $\frac{1}{p}$ for an integer $x$ and a prime $p$. Firstly, the probability should be some decreasing function, like $2^{-p}$ or $\frac{p}{p-1}$, but I don't understand why it's only just $\frac{1}{p}$.

Secondly, I don't understand why the probability doesn't depend on $x$. It means that the probability that $1$ is divisible by $2$ is the same as the probability that $1024$ is divisible by $2$, which I find nonsensical.

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It's safe to say that this is a heuristic and that your criticism of the heuristic is reasonable. However, another reasonable interpretation of "the probability that a random natural number is divisible by $p$" is $$ \lim_{N \to \infty} \frac{\text{numbers less than }N \text{ divisible by p}}{\text{all numbers less than }N} = \frac{1}{p} $$ which is of course true. Since the heuristic can be useful, it's worth entertaining it, keeping in mind that its meaningfulness depends on some pretty deep questions in philosophy of math.