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If $n$ balls are thrown into $k$ bins (uniformly at random and independently), what is the probability that every bin gets at least one ball? i.e. If we write $X$ for the number of empty bins, what is $P(X=0)$?

I was able to calculate the $E(X)$ and thus bound with Markov's inequality $P(X>=1) \le E(X)$ but I don't how to work out an exact answer.

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up vote 13 down vote accepted

What is the chance that all $k$ bins are occupied?

For $1\leq i\leq k$, define $A_i$ to be the event that the $i$th bin stays empty. These are exchangeable events with $P(A_1\cdots A_j)=(1-{j\over k})^n$ and so by inclusion-exclusion, the probability that there are no empty bins is $$P(X=0)=\sum_{j=0}^k (-1)^j {k\choose j}\left(1-{j\over k}\right)^n.$$

Stirling numbers of the second kind can be used to give an alternative solution to the occupancy problem. We can fill all $k$ bins as follows: partition the balls $\{1,2,\dots, n\}$ into $k$ non-empty sets, then assign the bin values $1,2,\dots, k$ to these sets. There are ${n\brace k}$ partitions, and for each partition $k!$ ways to assign the bin values. Thus, $$P(X=0)={{n\brace k}\,k!\over k^n}.$$

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@Colonel: But note this approximation to the result. – joriki Oct 11 '12 at 23:27

I propose to use combinatorics. Namely, stars and bars formulae.

  1. Number of outcomes with at least one ball in every bin is $\tbinom{n - 1}{k-1}.$
  2. Number of outcomes with any number of balls in every bin is $\tbinom{n + k - 1}{n}.$

Now just divide.

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To count the outcomes for this question, using inclusion-exclusion formula is correct, but $n$ choose $k$ and then multiplied by $k!$ is only correct if the question is asking that "each bin has only one ball". If n balls are thrown into $n$ bins, then, a simple answer is $n!$. So the question asking at least one ball is complicated. We have to write down each possible way and sum up each combination, just like another question asking about each day per week has at least one call. If we have exact numbers, we can count the numbers of outcomes, but here, it is better to ask at least one ball with the setting of $n$ balls thrown into $n$ bins.

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Set $X_i = 1$ if there is at least 1 ball in the $\textit{i}$th bin; and $0$ otherwise [$\textit{i}$ goes from 1 to k]. Then this question is asking what is $P[\sum\limits_{i=1}^k X_i = k]$. Given that each $X_i$s are independent of each other, $P[\sum\limits_{i=1}^k X_i=k]) = (P[X_i=1])^k =(1-P[X_i=0])^k =(1-(\frac{k-1}{k})^n)^k$

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