In our homework we had the following problem:

Let's consider $n$– many particles. Each of them has the probability $\frac{1}{N}$ to land in one of $N$-many numbered boxes, where $N>n$. We select $n$-many boxes. What is the probability that in each of those $n$ boxes is exactly one particle?

Assume that (like Bose-Einstein statistic) the cases which can be generated by simply swapping the particles are identical. It only matters how many particles are in a box and not which particles.

**My approach:**

It is clear that the elements of the sample space $\Omega$ can be counted by the combinatorial model of "indistinguishable objects and distinguishable boxes" hence we define $\Omega:=\{\omega\in\mathbb{N}^n\mid1\leq\omega_1\leq\dots\leq \omega_n\leq N\}$, where $|\Omega|={N+n-1\choose n}$.

The problem is that we don't know the probability distribution/the probability of each element of $\Omega$. The information that each particle lands with probability $\frac{1}{N}$ in one box doesn't help. Sure it makes sense to assume that the probability is equally distributed among the elements of $\Omega$ (which would turn this into a pretty easy problem) but this information can't be elicited from the question.

So in my opinion we can't solve this question without any further information on the probability distribution.

Am I missing something? Or do you agree?

## Best Answer

AssumingBose-Einstein statistics, the answer to the quoted question is of course just ${1\over\binom{N+n-1}{n}}$, because whichever might be the selected $n$ boxes, having exactly one particle in each of them corresponds to exactly one of the equally-likely outcomes in the sample space.The issue, it seems to me, is whether assuming something "

likeBose-Einstein statistics" is the same as assuming Bose-Einstein statistics. (If it is not the same, then I agree the question hasn't provided enough information to be answered.) The second quoted paragraph saysApparently this is intended to specify that Bose-Einstein statistics are to be assumed; but, saying "

likeBose-Einstein" -- and only mentioning the "indistinguishable particles" part of that model -- doesn't make it very clear that we are to also assume the other essential part of that model, namely that the $\binom{N+n-1}{n}$ outcomes in the sample space are equally-likely.Also, nothing else in the quoted problem clarifies this issue. In particular, the statement ...

... implies that the particle locations are uniformly distributed among the boxes, although it does

notimply that these are also mutuallyindependent. (So we also cannot assume a sample space of $N^n$ equally-likely outcomes.) In fact,if we assume this independence, then the $\binom{N+n-1}{n}$ outcomes in the sample space are not equally likely-- contradicting the Bose-Einstein model.To see this, just consider the case of $n=2, N=3$, assuming independent locations of the particles: $$\begin{align}P(\text{both particles land in box$_1$})=\underbrace{P(\text{particle$_1$ lands in box$_1$})}_{1\over 3}\cdot \underbrace{P(\text{particle$_2$ lands in box$_1$})}_{1\over 3} ={1\over 9}\end{align}$$ whereas in the Bose-Einstein model we have $$P(\text{both particles land in box$_1$})={1\over\binom{3+2-1}{2}}={1\over 6}.$$

As an aside, one might wonder: If the particles are allocated one-at-a-time -- and their locations must be

uniformlybut not independentlydistributed among the $N$ boxes -- then how exactly can this be done to produce Bose-Einstein statistics? E.g., by what pseudorandom algorithm can this be simulated?Such an algorithm is implicit in the following article (p. 1655): Ijiri Y, Simon H A. Some distributions associated with Bose-Einstein statistics. Proc Natl Acad Sci U S A. 1975;72(5):1654-1657. Namely:

Allocate the particles one-at-a-time to $N$ boxes numbered $1,...,N,$ by repeatedly applying the following rule, starting with no particles yet allocated: