Why does order matter here for throwing $n$ balls into $n$ labelled bins?

Question

I was studying the following practice problem:

When throwing $n$ balls into $m$ distinguishable bins, what's the probability that bin $i$ has exactly $1$ ball in it?

This scenario was modelled by sampling without replacement where order matters, due to the balls being thrown one at a time. However, I am not seeing why order matters here as it is not said whether the balls are labelled or not.

(1) Why should we care about order just because there is an order to which we throw the balls?

(2) If the balls were unlabelled, would we still care about order simply due to how the balls are thrown in a certain order?

(3) If we had thrown the balls at the same time, would we not care about order if they were unlabelled?

Answer 1

Whether thrown one at a time or simultaneously is not the point, neither is it a question of order mattering/not mattering.

• The crux of the matter is (although it has not been explicitly stated) that the throws are made uniformly at random, and the outcome of each throw is independent. $\;\;$ Also, with the italicized stipulations above, it is irrelevant whether the balls are thrown one at a time or together. (So throwing one die ten times, and throwing ten dice simultaneously is the same ) This way creates an equiprobable sample space because each ball has a $1/n$ chance of landing in any bin

• We could instead want to count all the patterns of distribution of the balls. We can count them using stars and bars, but each pattern here is not equiprobable (Obviously, having all the balls in one bin is less probable than a more even distribution of balls)

• To conclude, if we want to compute probabilities of balls occupying various bins, we need to use the first model with equiprobable sample space.

Answer 2

Whether Order matters or not , is a consequence of the Experiment.

In Current Case , we can think that there Intermediate Stages ( before throwing $1$ ball , after throwing Exactly $4$ balls , just before throwing the last ball , ETC ) which we can observe , hence there are many ways the End State can occur. Hence , we want to consider the Order.

In Other Case , we can say that $n$ balls are Pre-Distributed in $n$ bins with "uniformly" & "Equally likely" DISTINCT ways. Hence we will not consider the Order in which those DISTINCT ways will occur.

Here is My Concrete Example :

In Current Experiment , let $n=3$
Initially the bins are Empty :
$(0,0,0)$

Throw $1^{st}$ ball :
$(1,0,0)$ , $(0,1,0)$ , $(0,0,1)$

Throw $2^{nd}$ ball :
$(2,0,0)$ , $(1,1,0)$ , $(1,0,1)$ ,
$(1,1,0)$ , $(0,2,0)$ , $(0,1,1)$ ,
$(1,0,1)$ , $(0,1,1)$ , $(0,0,2)$

Throw $3^{rd}$ ball :
$(3,0,0)$ , $(2,1,0)$ , $(2,0,1)$ ,
$(2,1,0)$ , $(1,2,0)$ , $(1,1,1)$ ,
$(2,0,1)$ , $(1,1,1)$ , $(1,0,2)$ ,
$(2,1,0)$ , $(1,2,0)$ , $(1,1,1)$ ,
$(1,2,0)$ , $(0,3,0)$ , $(0,2,1)$ ,
$(1,1,1)$ , $(0,2,1)$ , $(0,1,2)$ ,
$(2,0,1)$ , $(1,1,1)$ , $(1,0,2)$ ,
$(1,1,1)$ , $(0,2,1)$ , $(0,1,2)$ ,
$(1,0,2)$ , $(0,1,2)$ , $(0,0,3)$

We can then see that there are $3^3=27$ End States.
Probabilities are not Equal : $P((3,0,0))=P((0,3,0))=P((0,0,3))=1/27$ , while $P((1,1,1))=6/27$ , while $P((0,1,2))=P((0,2,1))=P((1,0,2))=P((1,2,0))=P((2,0,1))=P((2,1,0))=3/27$

There are $4$ Cases where bin $1$ is Empty : Probability is $1/27+1/27+3/27+3/27=8/27$

In Other Experiment , we will have $4$ Cases where bin $1$ is Empty , though the Probability of those Cases will be Equal because we will have $10$ total Cases & it might not match $8/27$ : In that Experiment , Probability is $4/10$

Experiment matters , which in turn will make Order matter or not !

ADDENDUM 1 :

OP Query : (3) If we had thrown the balls at the same time, would we not care about order if they were unlabelled?

When we throw the balls 1 at a time , we can observe the Intermediate Stages.
Eventual End State can occur in many ways.
Probability will vary with the number of End States & the number of ways to arrive there.
With $n=2$ , we can see $(2,0)$ , $(1,1)$ , $(1,1)$ , $(0,2)$
Hence Probability that $1^{st}$ is Empty is $1/4$

When we throw the balls all together , we can not see the Intermediate States.
It is same as somebody else putting the balls 1 at a time , then choosing 1 End State to give to us to observe.
We can count the number of End States.
We can also count the number of Cases where $1^{st}$ bin is Empty.
We then get the Probability.
With $n=2$ , we can see $(2,0)$ , $(1,1)$ , $(0,2)$
Hence Probability that $1^{st}$ is Empty is $1/3$

ADDENDUM 2 :

I did not want to complicate the Issue hence I did not earlier mention that the Difference between the 2 "Experiments" is what control whether the Order matters or not.
I saw a comment by user "user" which hints at that , hence I decided to include it now.

At Atomic level or Quantum level , when we throw all balls (Particles) at once , whether the balls (Particles) are "Distinguishable" or not will decide what type of Particle we have.
When we have 2 Bosons in 2 bins , it can be $(2,0)$ , $(1,1)$ , $(0,2)$ & Probability of first bin Empty is $1/3$
When we have 2 Fermions in 2 bins , it can be $(2,0)$ , $(1,1)$ , $(1,1)$ , $(0,2)$ & Probability of first bin Empty is $1/4$

More here :
https://en.wikipedia.org/wiki/Bose%E2%80%93Einstein_statistics#History , the Bin Example is given there.
https://farside.ph.utexas.edu/teaching/sm1/lectures/node76.html : Nice tabular Example is given here

In Summary :
What we Observe at Quantum level [ Electrons & Energy States & ETC ] is controlled by the Nature of the Particles.
What we Observe at Macro level [ by throwing balls into bins ] is controlled by the Experimental Process.
In Current Case , we can calculate based on Order , to get the Probability.

Answer 3

Throwing in order (2 balls, 2 bins)
1st ball: (01), (10)
2nd ball: (02), (11), (20)
Probability that bin $1$ is empty after all the balls are thrown = $\frac{1}{3}$

Probability that bin $1$ is empty at any time during the "experiment" = $\frac{2}{5}$

Throwing all the balls together (2 balls, 2 bins)
(02), (11), (20).
Probability that bin $1$ is empty = $\frac{1}{3}$

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