Re: Placing Balls in Urns and Expected values

In article <1166352415.987805.284770@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
"kunzmilan" <kunzmilan@xxxxxxxx> wrote:
There are n^n possibilities how to place n indexed balls into n indexed
urns, if balls are indexed, too.
From this number, there are n possibilities how to place all balls into
only one urn.

This agrees with [1] from my previous post; since k = n-1, only the
term with j = n-1 is non-zero:

--- j-k n
> (-1) C(j,n-1) C(n,j) (n-j)
---
j

0 n
= (-1) C(n-1,n-1) C(n,n-1) 1

= n

We have n!/(n-2)!1!1! how to chose two urn to place odd number of balls
in them, and eventually n!/(n-2)!2! , if the number of balls in both
urns is the same. These counts do not count with indexing of balls and
leave (n-2) urns empty.
If only urns are indexed, the sum of all such coefficients for all
partitions reduces to
(2m -1)!/m!(m-1).
The sum of both coefficients divided by this coefficient gives the
probability of (n-2) urns to be empty.

To compute probability, we must index the balls. Consider the case
where n = 2; 2 balls and 2 urns. If we do not index the balls, there
are 3 arrangements:

|o| | | | | | | | | |o|
|o| | | |o| |o| | | |o|
- - - - - -

but they are not equally likely. Coloring the balls (b for blue, r
for red) does not change the probabilities, however, it does show
the proper probabilities; each ball can be in each urn with equal
probability:

|b| | | | | | | | | | | | | |b|
|r| | | |b| |r| |r| |b| | | |r|
- - - - - - - -

That is, each ball will be in the left urn 50% and the right urn
50%, as shown above.

So labelling the balls determines the proper probabilities.

Rob Johnson <rob@xxxxxxxxxxxxxx>
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