If we have a box that contains 1000000 balls numbered from 1 to 1000000.

if we take "K" balls without replacement. let X be the maximum number and y be the minimum number in the K balls.

What is the expected value E(x-y) as function of "K".

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- Apr 30th 2012, 01:44 PMbillobilloExpected value of balls in box (a challenging one)
If we have a box that contains 1000000 balls numbered from 1 to 1000000.

if we take "K" balls without replacement. let X be the maximum number and y be the minimum number in the K balls.

What is the expected value E(x-y) as function of "K". - Apr 30th 2012, 02:27 PMPlatoRe: Expected value of balls in box (a challenging one)
@billobillo

You will never understand this question unless you yourself sit and work with the numbers.

May I suggest that you cut the question down to 100 balls an let K=8.

List the smallest eight numbers. What is $\displaystyle x=?~,~y=?~,~x-y=?$

List the larest eight numbers. What is $\displaystyle x=?~,~y=?~,~x-y=?$

List any eight numbers. What is $\displaystyle x=?~,~y=?~,~x-y=?$

Now what is the maximum for $\displaystyle x$?

Now what is the minimum for $\displaystyle y$?

Now what is $\displaystyle ?\le x-y\le ?$?

The think about number partitions?

When you have done that. Tell us what you find. - Apr 30th 2012, 02:48 PMMooRe: Expected value of balls in box (a challenging one)
- Apr 30th 2012, 02:59 PMhttrRe: Expected value of balls in box (a challenging one)
The title is a bit misleading...

Try it Moo, but maybe not start with K=8...

if K=2, you draw 2 balls out of 100, what is the expected value of the bigger numbered ball?

Now you draw 3 balls...

Then you should get it by now.

prove by using principle of symmetry - Apr 30th 2012, 03:29 PMPlatoRe: Expected value of balls in box (a challenging one)
The set $\displaystyle \{1,2,3,4,5,6,7,8\}$ has $\displaystyle x=8,~y=1,~\&~x-y=7$

The set $\displaystyle \{93,94,95,96,97,98,99,100\}$ has $\displaystyle x=100,~y=93,~\&~x-y=7$.

The set $\displaystyle \{1,24,37,45,50,61,87,100\}$ has $\displaystyle x=100,~y=1,~\&~x-y=99$

Clearly $\displaystyle 7\le~x-y~\le 99$

Now $\displaystyle \mathcal{P}(x-y=99)=\dfrac{\binom{98}{6}}{\binom{100}{8}}$. - Apr 30th 2012, 11:39 PMbillobilloRe: Expected value of balls in box (a challenging one)
Thanks for your replies, before posting this topic, I worked on a sample of 7 balls with K=3 and I was able to find E(x) and E(y) as function of K but not E(x-y),

I'm not a math specialist but I guess that x and y are not independent events, any suggestions? - May 1st 2012, 12:12 AMhttrRe: Expected value of balls in box (a challenging one)
E(x-y) is always E(x)-E(y), you are confusing with E(XY) which required independence to factorize