Undefined Probability Well, the question is:

What's the probability of choosing a natural number (not specific - at random), among all the natural numbers?

It is ascertained by my friends that it is $\frac{1}{\infty}$. But that is so absurd, because this expression equals zero, but there is surely a non-zero chance of it being chosen.
My teacher says that it is infinitesimal and hence the representation. But is there any other convincing explanation?
 A: Probably unrigorous but it convinces me. Note: this argument is analogous to the argument that picking a random real number from $(0,1)$ will never result in a rational number. It relies on the assertion that to include all the rationals in a uniformly random selection, one must allow the sample space to be the reals.

Consider numbers written backwards in base $10$. In other words, start with the ones digit, then the tens digit, then the hundreds digit, etc.
For example, the numbers from zero to ninety-nine would be
$$\begin{align}
&\text{ones}&\text{tens}\\
&0& 0\\
&1&0\\
&\vdots&\vdots\\
&8&9\\
&9&9\\
\end{align}$$
$$\text{Let }n = \sum_{i=0}^M d_i\cdot 10^i$$
We assert that selecting a random number with at most $M$ digits is equivalent to generating a sequence of $M$ random members of $\{0, 1,\dots,9\}$. In order to give all the numbers in $[0, 10^M-1]$ an equal chance of being selected, we have to generate all $M$ digits, letting small numbers be represented with trailing strings of zeros.
We can design an experiment to select random numbers as large as we like. The problem is that infinity is always larger. To include all the natural numbers in the sample space, we must select infinitely many digits. If this were possible, all natural numbers could be represented by the eventual random selection of an infinite sequence of trailing zeros. That this would ever happen is unlikely in the extreme.

To paraphrase @Severin's comment, assume you can uniformly randomly select a natural number and that the probability of selecting a particular number isn't zero. Then the sum of probabilities of all events in the sample space is $\infty$. Surely this is a worse contradiction than a probability of zero.

Now lets design an experiment that actually works:
Flip a coin until you flip heads. Let the number of flips, $n$, be our (non-uniformly randomly selected) natural number. All $n \in \mathbb{N}$ have a non-zero probability of selection and the sum of the probabilities of the outcomes in the sample space is one.
What is a good estimate of the average probability of picking a specific number in this experiment? Anything bigger than zero is too big.
Proof: For any choice $\epsilon>0$, we can construct a finite subset $\mathscr{S} \subset \mathbb{N}$ such that the average probability of selecting a number from $\mathscr{S}$ is less than or equal to $\epsilon$ and the probability of selecting a number not in $\mathscr{S}$ is less than $\epsilon$.
Let $\delta = \lfloor\log_2 \frac{1}{\epsilon}\rfloor$ and let all $n \leq 2^\delta$ be members of $\mathscr{S}$. The average probability of selecting a member of $\mathscr{S}$ is
$$\frac{2^{2^\delta} - 1}{(2^\delta)2^{2^\delta}} < \frac{1}{2^\delta} \leq \epsilon
$$
The probability of selecting a member of $\mathbb{N} - \mathscr{S}$ is at most $\frac{1}{2^{2^\delta+1}} < \epsilon$.
Therefore, any $\epsilon > 0$ overestimates the average probability of selecting a particular natural number.
