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This question already has an answer here:

I've been thinking about this for a while but don't know how to go about obtaining the answer. Find the expectation of the times of rolls if u want to get all 6 numbers when rolling a 6 sided die.

Thanks

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marked as duplicate by Michael Joyce, jameselmore, wythagoras, kjetil b halvorsen, Marconius Oct 12 '15 at 19:14

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Intuitively

You first roll a die. You get some value. Now, the probability of getting a different value on the next one is $\frac{5}{6}$ . Hence the expected value for different values in two rolls is $\frac{6}{5}$. You continue this and get

$\frac{6}{6} + \frac{6}{5} + \frac{6}{4} + \frac{6}{3} + \frac{6}{2} + \frac{6}{1} = \color{blue}{14.7}$ rolls

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  • $\begingroup$ Thanks. Is the expected value of $6/5$ that you obtained as this is effectively a geometric distribution of the number of tries before we get a success(i.e. a different number) and hence E[.] = 1/p? $\endgroup$ – Jojo Oct 1 '15 at 1:55
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    $\begingroup$ @Jojo Yes. Look at what happens at the end. There is only one possibility, so probability is 1/6 and that one possibility, the expected rolls is 6. Now continue this process backwards. That is more convincing intuitively. $\endgroup$ – Shailesh Oct 1 '15 at 1:57

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