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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?

Bridget
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1 Answers1

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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:

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

Momo
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  • the expectation of T4 T5 and T6 are 1, 2 and 5 respectively. So the answer should be 11? – Bridget Jan 28 '17 at 17:30
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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. – Momo Jan 28 '17 at 17:46
  • Yes yes. I got it now. Thank-you so much – Bridget Jan 28 '17 at 17:58
  • correction: "Given that you already obtained..." – Momo Jan 28 '17 at 17:58
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    You are welcome! – Momo Jan 28 '17 at 17:58