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For no reason other than my own curiosity, I have been trying to figure out the solution to the following question:

Given a bag of x marbles, what is the probability of having seen every different marble at least once after y draws with replacement?

I am interested in making different graphs for specific values of x just to know what it would look like but I can only figure it out when x=y. The moment y>x, I do not know how to approach it.

I have scribbled on more pages than is reasonable, so I hope I can find an answer here.

Thanks!

RobPratt
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user2696806
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    This is a rephrasing of the usual Coupon Collector Problem – lulu Sep 28 '23 at 10:13
  • The probability is $\dfrac{x!, S_2(y,x)}{x^y}$ where $S_2(y,x)$ is a Stirling number of the second kind. For a large number of urns $x$, an approximation based on Erdős and Rényi is $\exp(-x\exp(-\frac yx))$, though this is not so good for a small number of urns. – Henry Sep 28 '23 at 11:14
  • Here's a way without appealing to Stirling numbers. Enumerate your marbles as $C_1, \dots,C_x$. Let $E_j$ be the event that marble $C_j$ wasn't observed at all among these $y$ sampled marbles. You need to find $$\mathbb{P}(E_1^c\cap \dots \cap E_x^c)$$ If you write this as $$1-\mathbb{P}(E_1\cup \dots \cup E_x)$$ you can use inclusion/exclusion. Can you finish? –  Sep 28 '23 at 19:13
  • @MatthewH. This way of seeing it helped me find it, thanks a lot! :) – user2696806 Sep 30 '23 at 17:37

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