Re: Placing Balls in Urns and Expected values
- From: rob@xxxxxxxxxxxxxx (Rob Johnson)
- Date: Mon, 18 Dec 2006 06:57:26 GMT
In article <1166371192.995942.16570@xxxxxxxxxxxxxxxxxxxxxxxxxxx>,
"R.G. Vickson" <C6L1V@xxxxxxx> wrote:
Rob Johnson wrote:
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.
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.
Very nice little demonstration. Of course, that only holds for
"classical" balls; for indistinguishable quantum-mechanical balls, the
first (unlabelled) version holds for bosonic balls, and only the middle
configuration of the unlabelled version is possible at all for
fermionic balls.
Since the original problem was stated as
Suppose that n balls are randomly placed in n urns in such a
way that each ball is equally [likely] to go into each urn.
What are the expected number of empty urns.
The part about "each ball is equally [likely] to go into each urn"
indicates a labelled ball approach. Your description of "bosonic
balls" counts partitions, where only the number of balls in each
container matters. This requires a different model. Do each of
the partitions of the "bosonic balls" have equal probability?
Rob Johnson <rob@xxxxxxxxxxxxxx>
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- Placing Balls in Urns and Expected values
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