Repeatedly selecting a random number from $\mathbb{N}_{≤n}$ always reaches $0$? Let $S$ be the set of natural numbers from zero to $n$, namely $S = \{ s : s∈\mathbb{N} ∧ s ≤ n \}$.
With each turn we pick a random number from the set. If we get one specific number (let's say $0$), the experiment is complete. If we get any other number, we go on with the next turn (hoping to get $0$ and going on with yet another turn otherwise).
Is it possible to prove that such an experiment always ends with a finite number of turns?
 A: The set $\{0,1,2,\ldots,n\}$ has $n+1$ members.
Getting a number other than $0$ on the first trial has probability $n/(n+1).$
Getting a number other than $0$ on all of the first $N$ (not $n$) trials has probability $\left( \dfrac n{n+1} \right)^N.$
Getting a number other than $0$ on all trials in an infinite sequence of trials therefore has probability${}\le\left( \dfrac n{n+1}\right)^N.$
The only nonnegative number that is${}\le\left(\dfrac n {n+1}\right)^N$ for all $N\in\{1,2,3,\ldots\}$ is $0.$
So the probability of getting an infinite sequence of nonzero numbers is $0.$
However, the set of all infinite sequences of numbers in the set $\{1,2,3,\ldots,n\}$ is not empty. This is a nonempty subset of the probability space, but its probability is $0.$
It is analogous to throwing a dart at a square and asking for the probability that it lands exactly on the diagonal. The area of the diagonal is $0$ but the diagonal is not empty.
For many purposes in probability and statistics, sets of probability $0$ can be neglected, just as if they are not there.
A: The probability that the game is still going after $k$ turns is $\left(\frac{n-1}{n}\right)^k$, because you picked a number other than your target number (0 in your example) $k$ times in a row. Let the random variable $X$ denote the number of turns needed to finish the game. You can see that:
$P(X=k) = \left(\frac{n-1}{n}\right)^{k-1} \frac{1}{n}$ for $k>0$.
Now we can compute $P(X < \infty) = \frac{1}{n}\sum_{k=0}^{\infty} \left(1 - \frac{1}{n}\right)^k = \frac{1}{n}\times \frac{1}{1 - \left(1 - \frac{1}{n}\right)} = 1$.
Thus, the number of turns required for the game to finish is almost surely finite.
