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Mike1170 science forum addict
Joined: 17 Sep 2005
Posts: 74

Posted: Fri Jun 23, 2006 3:05 am Post subject:
Re: Buckets and balls... help please.



"Mike" wrote in message
Quote:  First, put some tighter restrictions on m and n:
1 < m < N, 1 < t < N and m + t >= N
If m or t >= N, then probability is 0, or if m + t < N, then probability
is > 1.

Oops, tighter restrictions should be on m and t
and they should each be greater than or equal to 1. 

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Mike1170 science forum addict
Joined: 17 Sep 2005
Posts: 74

Posted: Fri Jun 23, 2006 2:51 am Post subject:
Re: Buckets and balls... help please.



"BEZMan" wrote in message
Quote:  ...
Let say I have N buckets, m (<=N) red ball and t (<=N) blue balls. I
randomly put all my red balls in different buckets (such that no two
red balls are in the same bucket), and then do the same with my blue
balls. Buckets will then contain [no ball], [one red ball], [one blue
ball] or [one red and one blue ball] at the end.
When I am done, what are odds that I end up with an empty bucket.
Also, if I repeat this operation S times, what are the final odds that
I obtain empty bucket at least once?

First, put some tighter restrictions on m and n:
1 < m < N, 1 < t < N and m + t >= N
(If m or t >= N, then probability is 0, or if m + t < N, then probability is
1.)
After red balls are placed, there are m occupied and Nm unoccupied buckets.
The required probability is the complement of the probability that all of
the
Nm empty buckets get filled when placing the t blue balls, i.e., that Nm
of
the t blue balls are placed in the Nm empty buckets and the rest are placed
in
or among the m occupied buckets.
Probability = P = 1  C(Nm,Nm) x C(m,tN+m) / C(N,t) where
C(A,B) = the number of combinations of A things taken B at a time
C(A,B) = A! / [ B! x (AB)! ], 0! = 1, 1! = 1, 2!=2, 3!=6, 4!=24, 5!=120,
....
P = 1  m! t! / N! x (m+tN)!
(Odds [against] = (1/P  1) to 1. Thus, if P = 0.25, then odds [against]
are 3 to 1.)
The probability of obtaining an empty bucket at least once in S repititions
of the
operation is the complement of not obtaining an empty bucket in S
operations.
P(S) = 1  (1P)^S 

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Mike1170 science forum addict
Joined: 17 Sep 2005
Posts: 74

Posted: Sat Jun 17, 2006 7:19 pm Post subject:
Re: Buckets and balls... help please.



"BEZMan" wrote
Quote:  Hi all,
I am struggling with this problem, and was hoping somebody would help
me.
Let say I have N buckets, m (<=N) red ball and t (<=N) blue balls. I
randomly put all my red balls in different buckets (such that no two
red balls are in the same bucket), and then do the same with my blue
balls. Buckets will then contain [no ball], [one red ball], [one blue
ball] or [one red and one blue ball] at the end.
When I am done, what are odds that I end up with an empty bucket.
Also, if I repeat this operation S times, what are the final odds that
I obtain empty bucket at least once?

Given that red and blue balls are each restricted to buckets not already
occupied by a ball of the same color, try using a "sampling without
replacement" approach instead of the ball and urn model. Instead of N
buckets, think of a population of N items (say the integers from 1 to N) and
instead of randomly putting balls in buckets (in a way that prevents two
balls of the same color in the same bucket) think of sampling the N integers
without replacement. Randomly putting all of the red balls in different
buckets is equivalent to drawing a sample of size m from a population of
size N. Ditto the blue balls. Randomly putting all of the blue balls in
different buckets is equivalent to drawing a sample of size t from the same
population of size N. With this appreach, I think you'll have better luck
figuring the probabilities of unsampled, singlysampled and doublysampled
items (buckets, integers or whatever). 

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Pavel314 science forum addict
Joined: 29 Apr 2005
Posts: 78

Posted: Fri Jun 16, 2006 9:57 pm Post subject:
Re: Buckets and balls... help please.



"BEZMan" <nonobezman@yahoo.com> wrote in message
news:1150441053.038019.203340@y41g2000cwy.googlegroups.com...
Quote:  Hi all,
I am struggling with this problem, and was hoping somebody would help
me.
Let say I have N buckets, m (<=N) red ball and t (<=N) blue balls. I
randomly put all my red balls in different buckets (such that no two
red balls are in the same bucket), and then do the same with my blue
balls. Buckets will then contain [no ball], [one red ball], [one blue
ball] or [one red and one blue ball] at the end.

After you put all the red balls into the buckets, do you have to put the
blue balls into empty buckets first, filling all empty buckets before you
can but a blue ball in with a red one? Or do you put the blue balls randomly
into the buckets, just avoiding two blue balls in the same bucket? Depending
on the rules governing the blue ball placement, you will get different
answers.
Quote:  When I am done, what are odds that I end up with an empty bucket.

If you have to fill the empty buckets with blue balls first, it depends on
the relationship of N, m, and t.
1. m + t < N, you get N  m  t empty buckets so the odds are 100%.
2. m + t >= N, no empty buckets.
Quote: 
Also, if I repeat this operation S times, what are the final odds that
I obtain empty bucket at least once?

This seems to indicate that the blue balls don't have to go into empty
buckets, otherwise the results would always be the same.
Paul


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BEZMan science forum beginner
Joined: 16 Jun 2006
Posts: 3

Posted: Fri Jun 16, 2006 7:06 am Post subject:
Re: Buckets and balls... help please.



BEZMan wrote:
Quote:  BEZMan wrote:
Hi all,
I am struggling with this problem, and was hoping somebody would help
me.
Let say I have N buckets, m (<=N) red ball and t (<=N) blue balls. I
randomly put all my red balls in different buckets (such that no two
red balls are in the same bucket), and then do the same with my blue
balls. Buckets will then contain [no ball], [one red ball], [one blue
ball] or [one red and one blue ball] at the end.
When I am done, what are odds that I end up with an empty bucket.
Also, if I repeat this operation S times, what are the final odds that
I obtain empty bucket at least once?
Thanx.
What I meant was in fact:
"When I am done, what are odds that I end up with AT LEAST one empty
bucket."
Sorry....

Also, what are the odds of having NO bucket containing 2 balls: that is
all bucket have at most one ball?
Again, thank you very much in advance. 

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BEZMan science forum beginner
Joined: 16 Jun 2006
Posts: 3

Posted: Fri Jun 16, 2006 7:00 am Post subject:
Re: Buckets and balls... help please.



BEZMan wrote:
Quote:  Hi all,
I am struggling with this problem, and was hoping somebody would help
me.
Let say I have N buckets, m (<=N) red ball and t (<=N) blue balls. I
randomly put all my red balls in different buckets (such that no two
red balls are in the same bucket), and then do the same with my blue
balls. Buckets will then contain [no ball], [one red ball], [one blue
ball] or [one red and one blue ball] at the end.
When I am done, what are odds that I end up with an empty bucket.
Also, if I repeat this operation S times, what are the final odds that
I obtain empty bucket at least once?
Thanx.

What I meant was in fact:
"When I am done, what are odds that I end up with AT LEAST one empty
bucket."
Sorry.... 

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BEZMan science forum beginner
Joined: 16 Jun 2006
Posts: 3

Posted: Fri Jun 16, 2006 6:57 am Post subject:
Buckets and balls... help please.



Hi all,
I am struggling with this problem, and was hoping somebody would help
me.
Let say I have N buckets, m (<=N) red ball and t (<=N) blue balls. I
randomly put all my red balls in different buckets (such that no two
red balls are in the same bucket), and then do the same with my blue
balls. Buckets will then contain [no ball], [one red ball], [one blue
ball] or [one red and one blue ball] at the end.
When I am done, what are odds that I end up with an empty bucket.
Also, if I repeat this operation S times, what are the final odds that
I obtain empty bucket at least once?
Thanx. 

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