Uniform balls-in-bins allocation ($(B_1,\dots ,B_m)\sim \mathrm{Multinomial}(n;1/m,\dots ,1/m)$)

Place $n$ labelled balls independently into $m$ labelled bins, each with probability $1/m$. The vector of bin counts has a multinomial distribution, and each count separately has the binomial distribution $\mathrm{Bin}(n,1/m)$. Distinct bin counts have covariance $-n/m^2$, so they are dependent for $n>0$ and $m>1$. Counts of bins of a specified size are sums of indicator variables; their expectations follow from the linearity of expectation, without any independence between bins.

For $n$ balls in $n$ uniformly and independently chosen boxes, put $p_n=(1-1/n{)}^n$. The empty-box indicators in uniform allocation have negative pairwise covariance, so $\mathrm{var}(K_n)\le n{p}_n(1-p_n)\le n/4$. Chebyshev's inequality givesSince $p_n\to e^{-1}$, this proves the displayed convergence in probability. The centring and the negative covariance are essential to this short proof; an expectation limit alone would not establish concentration.

Place $n\ge 1$ balls independently and uniformly into $m\ge 2$ boxes. If $I_i$ indicates that box $i$ is empty, then $\mathbb{E}{I}_i=p$, $\mathrm{var}(I_i)=p(1-p)$, and, for $i\ne j$,Indeed all balls avoid one specified box with probability $p$ and two specified boxes with probability $q$. The indicator random variable identities give the expectations, variances, and covariance. Finally $1-2/m<(1-1/m{)}^2$ proves strict negativity. Thus the number $K$ of empty boxes has $\mathbb{E}K=mp$ and $\mathrm{var}(K)=mp(1-p)+m(m-1)(q-p^2)$ by the variance of a sum.

In a uniform balls-in-bins allocation with $m\to \infty$ and $n/m\to \lambda \in (0,\infty )$, a fixed bin count has convergence in distribution to a Poisson distribution variable of mean $\lambda$ by the Poisson limit theorem. If $N_j$ counts bins containing exactly $j$ balls, symmetry and the linearity of expectation give $\mathbb{E}[N_j]/m=\mathbb{P}(B_1=j)\to e^{-\lambda }{\lambda }^j/j!$ for each fixed $j$. In particular the expected empty fraction tends to $e^{-\lambda }$, and the expected fraction with at least two balls tends to $1-(1+\lambda )e^{-\lambda }$. These are statements about expectations; convergence of the random fractions requires an additional argument.