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?


2 Answers 2


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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .