What is the probability of guessing the right number $n$ from all numbers $\mathbb{N}$?

Question

As we all know, $\mathbb{N}$ contains infinitely many numbers. What is the probability of guessing the right number $n \in \mathbb{N}$, i. e. what is $\frac{1}{\infty}$? It is clear that there is a chance of guessing the right number but it is very unlikely.

Answer 1

"It is clear that there is a chance of guessing the right number" It becomes less clear the more you think about it. Can you imagine any kind of physical process which could, even in principle, return any natural number with positive probability (let alone equal probability)?

This is an example of why mathematicians have resorted to reasoning axiomatically when it comes to the infinite: intuition is often not up to the task. In this case it is easy to see that it is impossible to set things up in a coherent, nontrivial way. Namely:

Suppose we want to assign a non-negative "probability" to every subset $S$ of $\mathbb{N}$. What properties do we want? Surely at least the following:

$\bullet$ for all subsets $X$ of $\mathbb{N}$, $0 \leq P(X) \leq 1$.
$\bullet$ For all $x,y \in \mathbb{N}$, $P(\{x\}) = P(\{y\})$.
$\bullet$ For any disjoint subsets $X$ and $Y$, $P(X \cup Y) = P(X) + P(Y)$.

From these we quickly deduce that for every finite subset $X$ of $\mathbb{N}$, $P(X) = 0$. Namely, suppose $P(\{1\}) = \alpha > 0$. There is some natural number $n$ such that $\frac{1}{n} < \alpha$, and then $P( \{1,\ldots,n\}) = n \alpha > 1$: oops!

Decreeing that every finite set has probability zero is thus far consistent but not a very interesting answer (at least not if you only care about finite sets!). Moreover, note how little we asked for: in modern mathematics to speak of "probabilities" at all one usually requires also

$\bullet$ $P(\mathbb{N}) = 1$.
$\bullet$ If $\{X_n\}_{n=1}^{\infty}$ is a sequence of pairwise disjoint subsets of $\mathbb{N}$ then $P(\bigcup_{n=1}^{\infty} X_n) = \sum_{n=1}^{\infty} P(X_n)$.

But now these axioms are strictly contradictory: we get

$1 = P(\mathbb{N}) = P( \bigcup_{n=1}^{\infty} \{n\}) = \sum_{n=1}^{\infty} P(\{n\}) = \sum_{n=1}^{\infty} 0 = 0$.

Note: in view of the above, something's gotta give. From a mathematical perspective the best thing to do is throw away the second axiom that all singleton sets arise with equal probability. Thus you can endow the natural numbers with a perfectly good "probability measure" by choosing an infinite sequence $p_n$ of non-negative real numbers with $\sum_{n=1}^{\infty} p_n = 1$ -- e.g. we could take $p_n = \frac{1}{2^n}$. Then for any $n \in \mathbb{N}$ we put $P(\{n\}) = p_n$, and for any subset $S \subset \mathbb{N}$ we put $P(S) = \sum_{n \in S} P(\{n\})$. Such "discrete probability measures" can be quite useful in mathematics...but they do not formalize the romantic allure of picking a positive integer at random.