# 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?

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.

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.