Given a natural number $n>1$,

natural numbers $a$ are chosen uniform randomly between $0$ and $n-1$ i.e.: $$0 \le a \le n - 1$$


natural numbers $b$ are chosen uniform randomly between $1$ and $n-1$ i.e.: $$1 \le b \le n - 1$$ The average $\delta$ of the distance $|a-b|$ between numbers $a$ and $b$ equals $6.5$.

What is the given value of $n$?

First, my true apologies for, initially, not having been conform forum guidelines.

As requested (by @Taladris): some additional information about question.

Have a look at answer provided, also upon request (by @lulu), where formula is derived.

Thanks all for getting me on rails.

I personally believe the formula is strikingly simple. Some people would perhaps expect numerical methods to be required for general reasoning, but not so.

Here is the motivation: average distance in this particular case (note $0 \le a$ and $1 \le b$) turns out to have a very simple formula in terms of $n$, allowing to, conversely, easily find $n$ in function of it. As it so happens, linear Gauss formula, and its quadratic generalization, in this particular case, both contain a common factor that perfectly cancels out for probability computation.

I have used (in disguise) the formulas for two puzzles (112210 112821) on puzzling exchange where also, conversely, some amount is asked in function of some average. So now I thought to present the math.

Have a nice day. I hope to contribute in simplicity.

To be honest: I doubted myself about the validity of such answer and therefore presented the question to the kind exchange community.

  • 3
    $\begingroup$ What have you tried? $\endgroup$
    – Om3ga
    Dec 18, 2021 at 16:19
  • 2
    $\begingroup$ I suggest: pick a number like $n=10$ or whatever and compute the expected value of $|a-b|$. That ought to give you a pretty good idea. $\endgroup$
    – lulu
    Dec 18, 2021 at 17:22
  • $\begingroup$ I would ideally want to see equation(s) that express $n$ in function of the average distance, and/or vice versa, so one can find $n$ by solving such equation(s). $\endgroup$ Dec 18, 2021 at 19:07
  • 1
    $\begingroup$ Please edit your post to include your efforts. I would not expect a simple closed formula, numerical methods will almost certainly be required. $\endgroup$
    – lulu
    Dec 18, 2021 at 21:55
  • 1
    $\begingroup$ I think this question and its answer have some interest, but its format is very unusual and confusing. If it was written in a more standard way, I would vote to re-open. $\endgroup$
    – Taladris
    Dec 21, 2021 at 0:26

1 Answer 1


$n = 20$

To calculate average distance between $a$ and $b$ there are a total amount of $n(n-1)$ pairs $(a,b)$ with: $0<=a<=n-1$ and $1<=b<=n-1$ so consider the $n*(n-1)$ matrix:

$\begin{bmatrix}|0-(n-1)| & .. & |(n-1)-(n-1)|\\: & |a-b| & :\\|0-1| & ... & |(n-1)-1|\end{bmatrix}$

The total of all distances between $a$ and $b$ is the total of the distances between all pairs in the square $n*n$ matrix:

$\begin{bmatrix}|0-(n-1)| & .. & |(n-1)-(n-1)|\\: & |a-b| & :\\|0-0| & ... & |(n-1)-0|\end{bmatrix}$

minus the total of the distances between all pairs from the excluded $n*1$ bottom row:

$\begin{bmatrix}|0-0| & .. & |a-b| & .. & |(n-1)-0|\end{bmatrix}$

giving (using standard formulas) : $$(n-1)n(n+1)/3 - (n-1)n/2$$ where the total amount of pairs $n(n-1)$ nicely cancels out in both terms and the average of the distance between $a$ and $b$ therefore is $$(n+1)/3 - 1/2 = (2n-1)/6$$ such gives $n=20$ for $\delta=6.5$

  • 1
    $\begingroup$ Do we need all these spoiler tags? This is a maths Q&A website. People know they would be "spoiled" when reading an answer. $\endgroup$
    – Taladris
    Dec 21, 2021 at 0:27
  • $\begingroup$ @Taladris ... I am very sorry and apologise ... I come from puzzling exchange where spoilers are 'part of the game' etiquette and the challenge is to find an answer rather than getting one. I will remove spoilers if you think that would help draw attention conform to math exchange policy. Again, sorry for clumsyness, I really wanted to show and have verified an IMO cool mathematical fact. $\endgroup$ Dec 21, 2021 at 0:36
  • $\begingroup$ All right :-) spoilers removed. Thanks for the constructive comments. $\delta = (2n-1)/6$ is (hopefully) the correct formula I humbly wanted to present... $\endgroup$ Dec 21, 2021 at 0:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.