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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.

Write $p_n=(1-1/n{)}^n$, the probability that all balls avoid a specified box. Each of the empty-box indicators in uniform allocation is an indicator random variable with success probability $p_n$. Since $I_i^2=I_i$, its expectation and variance are

Let $q_n=(1-2/n{)}^n$. The product of two distinct empty-box indicators in uniform allocation is the indicator random variable of both boxes being empty, so $\mathbb{E}(I_i{I}_j)=q_n$. ThusIndeed $0\le 1-2/n<(1-1/n{)}^2$ for $n\ge 2$, and taking the positive integer power $n$ preserves this strict inequality. This proves the negative covariance directly.

By linearity of expectation, $\mathbb{E}{N}_n=n{p}_n$. The variance of a sum and the negative covariance just proved giveApply Chebyshev's inequality to the empty-box indicators in uniform allocation:This proves the required concentration of the empty-box fraction. Since $p_n\to e^{-1}$, it also yields convergence in probability of $N_n/n$ to $e^{-1}$.