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The distance between two numbers will be set as $\ | i - j | $ . I pick two numbers without replacement from $\ 0,1,2, \dots ,n $ let $\ X $ be the distance between the numbers. What is the expectancy of $\ X $ ?

there are $\ {n+1 \choose 2} $ options. so $\ {n+1 \choose 2} = \frac{n(n+1)}{2} $ and the probability will be $$\ P\{X = i\} = \frac{2(n+1-i)}{n(n+1)}$$

Then the expectancy $$\ E[X] = \sum_{i=1}^n x_i \cdot p(x_i) = \sum_{i=1}^n x_i \frac{2(n+1-i)}{n(n+1)} $$

I don't really know how to proceed from here?

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    $\begingroup$ I don't know why you are switching to $x_i \cdot p(x_i)$. Your probability $P\{X = i\}$ is correct and the expectation is $E[X] = \sum_{i = 1}^n i \cdot P\{X = i\}$ $\endgroup$ – Lee David Chung Lin Dec 2 '18 at 12:12
  • $\begingroup$ I need to somehow get to this answer : $\ \frac{n+2}{3} $ $\endgroup$ – bm1125 Dec 2 '18 at 13:07
  • $\begingroup$ You have to set $x_i=i$ and then apply the usual techniques to evaluate the sum. $\endgroup$ – Ingix Dec 2 '18 at 13:53
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You are almost there.

$x_i = i$ in your formula for the expectation.

$$E[X]=\sum_1^n\frac{2i(n+1-i)}{n(n+1)}$$

Then, use the identities:

$$\sum_1^n i =\frac{n(n+1)}{2}$$

and

$$\sum_1^n i^2 =\frac{n(n+1)(2n+1)}{6}$$

We get

$$E[X]=\frac{2}{n(n+1)}(n+1)\sum i -\frac{2}{n(n+1)}\sum i^2$$

$$E[X]=\frac{2}{n}\frac{n(n+1)}{2} - \frac{2}{n(n+1)}\frac{n(n+1)(2n+1)}{6}$$

Therefore

$$E[X]=n+1-\frac{2n+1}{3}=\frac{n+2}{3}$$

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