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Given a natural number $n>1$,

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

and

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.

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    $\begingroup$ What have you tried? $\endgroup$
    – Om3ga
    Dec 18, 2021 at 16:19
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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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

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$n = 20$

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 $\delta=6.5$

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    $\begingroup$ 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

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