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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 $\alpha=1$).

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    $\begingroup$ This is the coupon collecting problem with unequal probabilities. Von Schelling, Coupon collecting for unequal probabilities, AMM 61.5 (1954) pp306-311 apparently answers it, although I don't have a subscription to T&F and cannot verify this. $\endgroup$ Commented Oct 31, 2022 at 17:01
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    $\begingroup$ Thank you @PeterTaylor $\endgroup$ Commented Oct 31, 2022 at 19:00
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    $\begingroup$ I don't think "as $α$ grows, the required expected number of balls $m$ cannot increase" can be correct. As $\alpha$ increases, $p_1$ necessarily decreases. In the large-$\alpha$ limit, $m \to p_1^{-1} = 1 + \alpha (n(n+1)/2 - 1)$ Perhaps you meant to say that $mp_1$ cannot increase. $\endgroup$ Commented Nov 1, 2022 at 13:10
  • $\begingroup$ Thank you @PeterTaylor . Yes, corrected. $\endgroup$ Commented Nov 1, 2022 at 18:56

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From the referenced paper, I am writing in terms of their variables, $k$ is the number of bins or type of coupons:

Let $n_1$ be the time where the last of the missing events is observed. Let $n_2$ be the time where the second last of the missing events is observed, etc. until $n_k$:

Define $$ S_kf(p_1,\ldots,p_m)=\sum_{1\leq i_1<i_2<\cdots<i_k\leq m}f(p_{i_1},\ldots,p_{i_m}), $$ which means that the function is $f$ is to be formed for all $\binom{k}{m}$ combinations of the $k$ indices indicated in the subscript of $S$ and that all these terms are to be added. For example $$ S_3 \frac{1}{p_1+p_2}= \frac{1}{p_1+p_2}+\frac{1}{p_2+p_3}+\frac{1}{p_1+p_3} $$ The general equation given below is still not really clean, but you can simplify using your constraint of linear probabilities. The paper has the distribution and the cumulative distribution as well. However we have the expectation which follows in inclusion-exclusion manner thus $$ \mathbb{E}(n_m)=S_k \left\{ \binom{m-1}{m-1} \frac{1}{p_1+p_2+\cdots+p_m}+\right. $$ $$ -\binom{m}{m-1}\frac{1}{p_1+p_2+\cdots+p_{m+1}} +\cdots $$ $$ \left. +(-1)^{k-m} \binom{k-1}{m-1}\frac{1}{p_1+p_2+\cdots+p_k}. \right\} $$

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    $\begingroup$ Thank you for your answer. It seems still very difficult to manage it in contrast with the type of probability definition $p_i=\alpha i p_1$ which is very simple, i.e., it seems a result so general that could be significantly simplified by considering the above probability definition. Do you think it is possible to find an elementary function $f(n,\alpha)$ such that the expected number of balls in the question is, for $n\to\infty$, asymptotic to (or has the same order of) $f(n,\alpha)$ - at least when $\alpha=1$, in which case we are just looking for a function $f(n)$? $\endgroup$ Commented Oct 31, 2022 at 19:24
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    $\begingroup$ You can take a look at my edit in the question. $\endgroup$ Commented Nov 1, 2022 at 9:55

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