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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

  1. $9$
  2. $15$
  3. $16$
  4. $17$

First I didn't understand the problem fully. Conditional expectation is given by $$E(T=t|N=3)=\sum_{i} t_i\frac{P(T=t_i,N=3)}{P(N=3)}$$ Now how am I going to proceed from here. $N=3$ for first six throws so $P(N=3)=\frac{1}{2}$ right? Any hint or help would be helpful. Thank you.

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    $P(N=3)$ is not $\frac12$ – 2017-01-08
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    Ok then I must have mistaken. Can you give me a hint @MarkFischler – 2017-01-08
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    @KushalBhuyan Find the probability of throwing $6$ dice and having $3$ different numbers. – 2017-01-08
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    $P(N=3)=\frac{{6}\cdot{5}\cdot{4}\cdot{4}\cdot{4}\cdot{4}}{{6}\cdot{6}\cdot{6}\cdot{6}\cdot{6}\cdot{6}}$, is that correct? – 2017-01-08

1 Answers 1

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I wouldn't recommend using the formula, since then you need to calculate $P(T,N=3)$ which isn't a lot easier than the conditional.

Start by imagining N=3. This means that after the sixth roll, there have been three different faces seen. So the expected time to see all 6 conditional on this is $$ 6 + \mbox{(expected time it takes to see the last three faces)}.$$

So we need calculate the additioal time it takes to roll the last 3 faces. When you start attempting this, any one of the 3 faces you haven't seen yet will be a new face. Thus on each attempt, you have a 3/6 = 1/2 probability of getting a new face. The expected number of independent trials to see an event with probability $p$ is $1/p$. Thus, the expected time to roll the fourth new face starting from when you rolled the third is 1/(1/2) = 2. After you roll the fourth face, the probability of rolling a new face goes down to 2/6 = 1/3. So the expected time to roll the 5th face is 3. Finally, after the fifth is roll you only have a 1/6 chance of rolling that final face, so the expected time to do that is 6.

So the expected time to see the last three faces is 2 + 3 + 6 = 11. So the answer is 6 +11 = 17

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    "last three faces is 2 + 3/2 + 6 = 9.5" Shouldn't that be $2+3+6 = 11$? The answer would finally be $17$, which fits OP's options. – 2017-01-08
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    @O.VonSeckendorff Whoops, thank you! – 2017-01-08
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    I didn't understand this line --"Starting from roll 7 with 3 faces seen, you have a probability of 1/2 of rolling a new face on each attempt". Can you please elaborate? – 2017-01-08
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    @KushalBhuyan . Edited to elaborate that part. Does it make more sense now? – 2017-01-08
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    Yes it does now. Thank you very much. +1 – 2017-01-08