This is a variant of the Coupon Collector problem.
Suppose that you draw $X \sim \text{Poi}(n^2)$ coupons. How can we show, with high probability, that the same coupon was not drawn (i.e., repeated) more than $1.1n$ times? (assume that $n$ is very large).
My hunch is that we need to use some kind of bound (Chebyshev Inequality or Chernoff bound) but I'm struggling with how to get to that point.
Consider the $i^{\text{th}}$ coupon and let $X_i$ denote the (random) number of times the $i^{\text{th}}$ coupon is drawn. Since $X \sim \text{Poi}(n^2)$, $X_i$ is also a Poisson random variable with mean $n^2 \cdot \frac{1}{n} = n$. Using the concentration inequality for Poisson random variables which states that for $Y \sim \text{Poi}(\lambda)$, $\Pr(Y > \lambda + y) \leq \exp\left( -\frac{y^2}{2(\lambda + y)}\right)$, we obtain \begin{align*} \Pr(X_i > 1.1n) \leq \exp\left( -\frac{0.01n^2}{(2 \cdot 1.1n)}\right) = \exp\left(-\frac{n}{220}\right). \end{align*}
If you want to show that this holds for all coupons, note that $X_i$'s are independent and hence union bound gives us $\Pr( \max_{i} X_i > 1.1n) \leq n \exp\left(-\frac{n}{220}\right)$, which is very small for large $n$.