Expected value of balls in box (a challenging one)

**billobillo** · April 30th 2012, 01:44 PM

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".

**Plato** · April 30th 2012, 02:27 PM

Originally Posted by billobillo

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".

@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 $x=?~,~y=?~,~x-y=?$

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

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

Now what is the maximum for $x$ ?
Now what is the minimum for $y$ ?
Now what is $?\le x-y\le ?$ ?

The think about number partitions?

When you have done that. Tell us what you find.

**Moo** · April 30th 2012, 02:48 PM

Originally Posted by Plato

@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 $x=?~,~y=?~,~x-y=?$

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

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

Now what is the maximum for $x$ ?
Now what is the minimum for $y$ ?
Now what is $?\le x-y\le ?$ ?

The think about number partitions?

When you have done that. Tell us what you find.

Hi Plato,

Sorry I know this is not my problem, but I've been thinking on it lately. I don't quite understand where you're planning to lead our OP, so may I ask for a precision ?

What do you mean by "what is the maximum for x" ? Why would it help ?

Thanks

**httr** · April 30th 2012, 02:59 PM

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

**Plato** · April 30th 2012, 03:29 PM

Originally Posted by Moo

Sorry I know this is not my problem, but I've been thinking on it lately. I don't quite understand where you're planning to lead our OP, so may I ask for a precision ?
What do you mean by "what is the maximum for x" ? Why would it help ?

The set $\{1,2,3,4,5,6,7,8\}$ has $x=8,~y=1,~\&~x-y=7$

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

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

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

Now $\mathcal{P}(x-y=99)=\dfrac{\binom{98}{6}}{\binom{100}{8}}$ .

**billobillo** · April 30th 2012, 11:39 PM

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?

**httr** · May 1st 2012, 12:12 AM

E(x-y) is always E(x)-E(y), you are confusing with E(XY) which required independence to factorize

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