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A total of $n$ balls, numbered $1,\ldots,n$, are put into $n$ urns, also numbered $1,\ldots,n$, in such a way that the $i$th ball is equally likely to go into any of the urns $1,\ldots,i$. What is the expected number of empty urns? What is the probability that none of the urns are empty?

Let $X_i=1$ if the $i$th urn is empty, and $X_i=0$ otherwise. Then $X=\sum_{i=1}^nX_i$ counts the number of empty urns. Since the $j$th ball must choose any of its possible urns but the empty one,

$$P[X_i=1]=\prod_{j=i}^n\frac{j-1}{j}$$

hence,

$$E[X]=\sum_{i=1}^nE[X_i]=\sum_{i=1}^n\prod_{j=i}^n\frac{j-1}{j}$$

Is this work for the first part correct? If no, then where did I go wrong? If yes, then does the expression above converge to any elementary closed-form algebraic expression?

Then, how should I approach the second part of the question, since I'm lost on approach?

whorl
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  • First part is incorrect. Think about how many options each ball has. For 1 empty box, you first choose a box that will receive two balls($\binom{n-1}{1}$), then choose which 2 balls go in the box($\binom{n}{2}$) and the remaining $n-2$ balls need to be put into $n-2$ boxes.... – Saket Choudhary May 05 '15 at 03:36

2 Answers2

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(1) Your first part is correct. To simplify your result, your product equals $\dfrac{i-1}{n}$, and summing that from $1$ to $n$ yields $\dfrac{n-1}{2}$.

(2) For part two: The only way to get no empty urns is for the $i^{th}$ ball to go into the $i^{th}$ urn, for all $i = 1$ to $n$. The probability of this, therefore, is: $\prod_{i=1}^n \dfrac{1}{i} = \dfrac{1}{n!}$.

Mick A
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  • The statement, "for each ball there is a choice for $n$ urns" isn't correct. The $i^\text{th}$ ball only has $i$ possible urns for it can enter. –  May 19 '23 at 11:35
  • @MatthewH. Yes, you're absolutely right. I must have mis-read the question. I've changed my answer now. Thanks for letting me know. – Mick A May 20 '23 at 07:18
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I believe your work for the first part is correct (contrary to what another answerer stated) since the $i$th ball can only go into urns $1,\dots i$. So the $i$th urn can only be filled by balls $i, i+1, \dots n$. To simplify your product into a closed form solution, you can treat it as a telescoping series that simplifies to $\frac{i-1}{n}$. The $E[X]$ then becomes $$\frac1n\sum^{n}_{i=1}(i-1)$$

Тyma Gaidash
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Aditi Misra
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