Conditional expectation of throwing a die. 
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 
  
  
*
  
*$9$
  
*$15$
  
*$16$
  
*$17$
  

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.
 A: 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
