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Imagine you have a machine, that spits out a perfect random natual number $c$.

What is the probability that this number $c$ is divisible by a number $n \in \mathbb{N}$?

I have 2 ideas and for me, both seem to be logical.

1. The probability is $\frac 1n$

because evere $n$-th number is divisible by $n$...

2. The probability is $\frac 12$

If we divide the set $\mathbb{N}$ into two subsets, the first with all numbers that are divisible by $n$ and the second set with all numbers that are not divisible by $n$. Because the set $\mathbb{N}$ is a countable infinite set, both subsets are countable infinite sets too and hence they are the same size (I'm not quite sure that's right?)...

Second Question: Does the probability change if the machine spits out integers?

CodeCrafter 1
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    "perfectly random" doesn't have a clear meaning here. If you mean something like "choose uniformly between ${0,1,\cdots, N}$ for some large $N$" then the first result is a good approximation. Or did you mean something else? – lulu Dec 03 '19 at 18:04
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    Before continuing... we have to stop you and inform you that there is no such thing as a uniform distribution over a countably infinite set. So, we need to first correct and clarify what sort of distribution you are actually interested in. One option, you could talk about this in terms of limits, select a number from $1$ to $N$ uniformly at random, talk about the probabilities, and see how this changes as we look at the limit as $N\to\infty$. Another option, you could pick an actually valid distribution over the natural numbers such as $P(X=n) = \frac{1}{2^n}$ – JMoravitz Dec 03 '19 at 18:06
  • What you seem to be wanting to talk about is the relative density of the sets in question. While yes, when talking about subsets of natural numbers the set of multiples of $3$ and the set of non-multiples of $3$ are both countably infinite and are of the same cardinality, the relative density of the multiples of $3$ is $\frac{1}{3}$. – JMoravitz Dec 03 '19 at 18:08
  • Just to indicate a different sort of problem with your second idea; Fix the divisor, let's say $4$ and look at the possible remainder classes. That defines four different sets of natural numbers each of which is countably infinite. Any "randomly chosen natural number" (however you define it) would be in exactly one of those sets. Would you argue that they each had probability $\frac 12$? – lulu Dec 03 '19 at 18:09
  • Thus with respect to your link and if we describe this kind of infinite set with limits, to get a uniform distribution the answer to the problem should be the first one? – CodeCrafter 1 Dec 03 '19 at 18:17
  • Yes. The set of natural numbers which are multiples of $n$ have relative density $\frac{1}{n}$ with respect to the natural numbers (as well as with respect to the integers overall) – JMoravitz Dec 03 '19 at 18:22
  • Ok. Thank you very much. – CodeCrafter 1 Dec 03 '19 at 18:23

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