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This question already has an answer here:

Say I have a table of numbers 1-6. I throw a 6 sided die a number of times. Each time I get a number I have not already had, I mark it in the table. What is the expected number of times to throw the dice to fill out the table?

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marked as duplicate by Did probability Jun 21 '15 at 16:40

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Hint:

If $X$ denotes the number of throws that are needed then you can write: $$X=1+X_{1}+X_{2}+X_{3}+X_{4}+X_{5}$$ where $X_{i}$ has geometric distribution with parameter $\frac{6-i}{6}$.

Try to understand why this is the case.

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  • $\begingroup$ This is what I was thinking about, but I was not confident that I could show or prove it. My way of reasoning was: Given that we have k numbers checked in the table, we have 6-k left to find. Then we can just treat these as independent geometric distributions and add them together. $\endgroup$ – mathreadler Jun 21 '15 at 16:54

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