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Let T denote the number of times we have to roll a fair dice before each face has appeared at least once and let N denote the number of different faces appearing in the first six rolls. Then

E(T|N=3) is?

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    Closely related to [this problem](http://math.stackexchange.com/a/1639566/173262). – 2017-01-28
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    Related to coupon-collecting problem. – 2017-01-28

1 Answers 1

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Take a look at the Coupon Collector's Problem, the idea is the same.

Given that you have obtained three faces in the first six rolls, the number of rolls $T_4, T_5$, and $T_6$ to obtain the remaining three faces is distributed as Geometric with parameter $3/6$, $2/6$, and $1/6$ respectively.

So:

$$E[T|N=3]=E[6+T_4+T_5+T_6]=6+6/3+6/2+6=17$$

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    the expectation of T4 T5 and T6 are 1, 2 and 5 respectively. So the answer should be 11? – 2017-01-28
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    Given that you already obtain three faces, the probability to obtain a new face in a roll is $1/2$ (you have a 50% chance to obtain a face that was already rolled, and a 50% chance to obtain a new face). So $E[T_4]=2$. After you have obtained four faces, the probability to obtain a new face in a roll changes to $1/3$, and so on. – 2017-01-28
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    Yes yes. I got it now. Thank-you so much – 2017-01-28
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    correction: "Given that you already obtained..." – 2017-01-28
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    You are welcome! – 2017-01-28