Some years ago, I found a paper with all the formulas for the balls into bins problem when the "areas" (i.e., probabilities to capture a ball) of the bins are all different. However, the formulas looked quite involved and cumbersome in the general case. Now, I am instead trying to solve an elementary version of the balls into bins problem with a non-uniform probability of capturing balls, which I firmly believe has a simple and clean answer.
We are given $n$ bins $b_1, b_2, \ldots, b_n$. In a sequential fashion, at each time step, one ball is placed into bin $b_i$ with probability $p_i$, where $\sum_{i=1}^{n} p_i=1$, and $p_i=\alpha i p_1$ for a given constant $\alpha\ge 1$ for all integer $i\in\{2,3\ldots,n\}$.
Question: What is expected number $m$ of balls that we need to throw to have that all $n$ bins contains at least one ball?
Edit: For any given fixed value of $n\in\mathbb{N}$, as $\alpha$ grows, the required expected number of balls $m p_1$ cannot increase. For the minimum value of $\alpha$ in the problem, which is $1$ ($\alpha\ge 1$), it seems that $m=\frac{\beta}{p_1}$ for some constant $\beta$. Since $\frac{1}{p_1}$ balls are always necessary to make bin $b_1$ non-empty, I guess that there is a constant $\gamma(\alpha)\in [1,\beta]$ depending on $\alpha$ such that $\frac{\gamma(\alpha)}{p_1}$ is the expected number of balls required, but I do not know how $\gamma$ varies with $\alpha$. Anyway for $n\to\infty$ we always have $m\in\Theta\left(\frac{1}{p_1}\right)$ (which is equal to $\Theta(n^2)$ for $\alpha=1$).