Randomly choose a number from 1 to 3 and call it $a_1$;

Randomly choose a number from 1 to 3$a_1$ and call it $a_2$;

Randomly choose a number from 1 to 3$a_2$ and call it $a_3$;


Repeat this process ad infinitum to get a sequence $s=a_1,a_2,a_3,...$

Question: does $s$ contain the number 1 with probability 1?

  • 4
    $\begingroup$ Just for clarity, do you mean an integer and are the endpoints included? $\endgroup$
    – hexomino
    Dec 8, 2023 at 17:47
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    $\begingroup$ @WeatherVane Probability 1 doesn’t mean absolute certainty: if I tell you to pick a random real number, the probability I’m wrong by saying « you picked 5 » is 1, but it’s actually possible for you to have picked 5. $\endgroup$
    – Jujustum
    Dec 8, 2023 at 18:29
  • 1
    $\begingroup$ $0.\dot{9}$ is $1$. See math.stackexchange.com/questions/11/… $\endgroup$
    – Jujustum
    Dec 8, 2023 at 18:50
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    $\begingroup$ Probability 1 actually means almost surely $\endgroup$
    – hexomino
    Dec 8, 2023 at 21:58
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    $\begingroup$ @hexomino Yes, we randomly choose integers and end points are included. $\endgroup$
    – Eric
    Dec 9, 2023 at 4:20

2 Answers 2


Supposing that the number following any integer $n$ is another integer in $[1,3n]$, here's a partial answer, showing that

in the long run, $1$ (or any other fixed number) appears in any fixed position with probability $0$.

Let $p_n$ be the long-run probability that $n$ appears in some position. What came before $n$? If it was $k$, where $k = \lceil\frac{n}{3}\rceil$, then there was a $\frac{1}{3k}$ chance of $n$ following it. If it was $k+m$, then $n$ followed with probability $\frac{1}{3(k+m)}$.

We can thus say, in an informal sense, that $p_n = \frac{p_k}{3k}+\frac{p_{k+1}}{3(k+1)}+\frac{p_{k+2}}{3(k+2)}+\dots$. The first few examples of this are:
$\begin{align} p_1=&\frac{p_1}{3}+&\frac{p_2}{6}+&\frac{p_3}{9}+&\dots\\p_2=&\frac{p_1}{3}+&\frac{p_2}{6}+&\frac{p_3}{9}+&\dots\\p_3=&\frac{p_1}{3}+&\frac{p_2}{6}+&\frac{p_3}{9}+&\dots\\p_4=&&\frac{p_2}{6}+&\frac{p_3}{9}+&\dots\\p_5=&&\frac{p_2}{6}+&\frac{p_3}{9}+&\dots\\p_6=&&\frac{p_2}{6}+&\frac{p_3}{9}+&\dots\\p_7=&&&\frac{p_3}{9}+&\dots\end{align}$
We can solve for later series in terms of earlier ones - a bit of manipulation yields the recurrence $p_{3k+1}=p_{3k+2}=p_{3k+3}= p_{3k}-\frac{p_k}{3k}$. However, there's a problem: if we substitute a multiple of $\frac{1}{n}$ for $p_n$, say, the left-hand side is greater than the right! This implies that the real $p_n$ decrease slower than the harmonic series - and since the sum of the $p_n$ can't diverge, none of them can be positive.

Does this help solve the main problem?

Suppose the answer to the original question was "Yes". Find the next 1 in the sequence, and remove everything up to it - what remains is an identically-generated sequence, which contains another 1 with probability 1. Therefore, a positive answer to the main problem implies a positive answer to the stronger question of whether such a sequence contains infinitely many 1's with probability 1.
However, we've shown that a sequence contains a positive proportion of 1's with probability 0, so the distribution of "repeat times" has infinite expectation. Althiugh this doesn't imply that infinite values appear with positive probability (which would constitute a negative answer), it definitely makes it more likely.


I think the short answer is


Why? Well because logically

If we repeat this process infinitely we have to get $1$ eventually since if we don't get one we just do it again infinitely. So logically the answer is yes.

But let's look at it from a mathematical standpoint

We can define 3 possible outcomes, choosing the number $1$,choosing a number from $[1,2\infty]$, or choosing a number from $[1,3\infty]$. Since $\infty$ is an unreachable value we can consider $\infty *$ any number $= \infty$ effectively. Therefore all in all possible outcomes we are choosing a number from $[1,\infty]$. So, assuming that we never get 1 the probability of getting a 1 the next time $\frac{1}{\infty}$. However, since we know $\infty$ can never be reached essentially the probability is $0$. So mathematically the probability of getting a 1 is p(getting a 1)=1-p(not getting a 1) which we can rewrite as $1-0$ so the probability of getting a 1 is always 1 if we repeat this infinetly.

Note: I'm not a math guy so if someone could look at my math proof and clean it up that would be great!

When I say "Assuming that we never get 1 the probability of getting a 1" what I mean is assuming worse case. Not sure if I can do that when making a mathematical proof. But hopefully someone can make that make sense?

Hopefully this helps someone

  • 1
    $\begingroup$ "If we repeat this process infinitely we have to get 1 eventually since if we don't get one we just do it again infinitely." This, if I understand, is wrong (or insufficient). For example, instead choose $a_i$ to be (independently) $1$ with probability $p_i$ with the infinite product of $(1 - p_i)$ equal to some number between $0$ and $1$. For a less abstract (but less elementary) example, take the probability of returning to the origin in a 3d random walk. $\endgroup$
    – tehtmi
    Dec 9, 2023 at 2:27

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