1. ## Ball and Urn Model: Please spot the flaws!

I have 4 urns and 4 balls.
With each turn, I randomly place 1 ball in 1 urn.
I take a total of 4 turns.

What is the probability of exactly 0 empty urns?
P(0 empty urns) = P(all 4 balls are placed in 1 urn each) =
P(1st ball placed anywhere).P(2nd ball placed anywhere of the 3 urns left) ....P(and so on)=
${4 \over 4}.{3 \over 4}. {2 \over 4}.{1 \over 4} = {24 \over 256}$

What is the probability of exactly 4 empty urns?
Impossible. Probability is ZERO.

What is the probability of exactly 3 empty urns?
P(3 empty urns) = P(all 4 balls are placed in the same urn) =
${4 \over 4}.{1 \over 4}. {1 \over 4}.{1 \over 4} = {4 \over 256}$

What is the probability of exactly 1 empty urn?
P(1 empty urn) = P(1st ball anywhere).P(2nd ball in any of the 3 urns left).P(3rd ball in any of the 2 urns left).P(4th ball in any of the 3 urns originally chosen)=
${4 \over 4}.{3 \over 4}. {2 \over 4}.{3 \over 4} = {72 \over 256}$

What is the probability of 2 empty urns?
P(2 empty urns) = P(1st ball anywhere).P(2nd ball in any of the 3 urns left).P(3rd ball in any of the 2 urns originally chosen).P(4th ball in any of the 2 urns originally chosen)=
${4 \over 4}.{3 \over 4}. {2 \over 4}.{2 \over 4} = {48 \over 256}$

Ok. Now comes the problem!
I know that they should all add up to the grand probability of 1.
But they don't. Where did I go wrong???

2. The mistakes come in P(1 empty urn) and P(2 empty urns).

Originally Posted by chopet
What is the probability of exactly 1 empty urn?
P(1 empty urn) = P(1st ball anywhere).P(2nd ball in any of the 3 urns left).P(3rd ball in any of the 2 urns left).P(4th ball in any of the 3 urns originally chosen)=
${4 \over 4}.{3 \over 4}. {2 \over 4}.{3 \over 4} = {72 \over 256}$
The mistake here is this: you are assuming that it is not until the fourth ball that one of the urns gets a second ball. But there are other possibilities. For example, the first two balls could go into the same urn, and then the third and fourth balls into empty urns.

One way to calculate this probability is as follows. There are 4 ways of choosing which urn is to remain empty. For each of these, there are 3 ways of choosing which urn gets two balls, and ${4\choose2}=6$ ways of choosing which two of the four balls go into that urn. Finally there are 2 ways of allocating the remaining two balls to the remaining two urns. Total: 4×3×6×2=144 out of 256.

Originally Posted by chopet
What is the probability of 2 empty urns?
P(2 empty urns) = P(1st ball anywhere).P(2nd ball in any of the 3 urns left).P(3rd ball in any of the 2 urns originally chosen).P(4th ball in any of the 2 urns originally chosen)=
${4 \over 4}.{3 \over 4}. {2 \over 4}.{2 \over 4} = {48 \over 256}$
Same sort of problem here: you could put the second ball into the same urn as the first one, then the other two balls into another urn. The total here should be 84 out of 256. Then the probablilities add up to 1.

3. Opalg, I am very grateful to you for pointing out my blind spots.

4. Hi again,

do you agree the number of ways of getting m empty urns among n urns, after putting r balls is :

no of ways of choosing the m empty urns *
no of ways to put r balls into n-m urns *
probability of 0 empty urn in n-m urns

I got this from a textbook, and am confused by their mixing of "No of ways" with "Probability".

5. The whole to topic of occupancy problems has several subcategories.
If we are putting N objects into K cells there are at least seven possible models.
1) objects-distinguishable, cells-distinguishable and $n_i$ objects in the i cell.
2) objects-distinguishable, cells-distinguishable and cells can be empty.
3) objects-distinguishable, cells-distinguishable and cells may not be empty.
4) objects-indistinguishable, cells-distinguishable and cells can be empty.
5) objects-indistinguishable, cells-distinguishable and cells may not be empty.
6) objects-indistinguishable, cells-indistinguishable and cells can be empty.
7) objects-indistinguishable, cells-indistinguishable and cells may not be empty.
It appears to me that you are considering variations of #2 & #3. However, you did so without explicitly saying if the balls and urns are distinguishable or indistinguishable; or maybe some of each. This area is well documented. The counting method used depends upon on which of the models from the list above you are using.