You are rolling a fair 30-sided die. (a) What is the expected number of rolls required to get a 27?
(b) What is the expected number of rolls required to get a 27, followed later by a 28? For example, 9,28,1,27,27,3,28 would be an example of requiring 7 rolls and 9,3,27,28 would be 4 rolls.
(c) What is the expected number of rolls required to get a 27, and a 28 in either order? For example, 9,27,3,28 would be an example of 4 rolls and 1,28,14,28,5,27 would be an example of 6 rolls. [Hint: Think of the number of rolls to get 27 or 28, and then the number of rolls to get the remaining value.]
(d) What is the expected number of rolls to get all 30 possible values (in any order)?
I am thinking for (a) it would be 30 for (b) and (c) it is 60? But I am not sure and for (d) it is 30+30/2+30/3+30/4+...30/30?
Any hint or feedback would be helpful!
Your answers, including the newly added one for d), are, with one exception, right.
a) If we are rolling until we get a $27$, then the number of rolls has geometric distribution, parameter $1/30$. The mean number of rolls is $1/(1/30)$, that is, $30$.
b) The mean number of rolls until we get $27$ is $30$. Then the mean number of additional rolls until we get a $28$ is $30$, for a total mean of $60$.
c) The probability of getting a $27$ or $28$ is $1/15$, so the mean number of rolls until that happens is $15$. Then we have a mean waiting time of $30$ until the other number occurs, for a total of $45$.
d) This is the famous Coupon Collector's Problem (please see Wikipedia). The first roll for sure produces a new number. Then the mean waiting time until we get a second new number is $30/29$. After that, the mean waiting time until we get a new number is $\frac{30}{28}$, and so on up to $30/1$. So the mean number of rolls is $1+30/29+3028+30/27+\cdots +30/1$. I would rather write that backwards as $$30\left(1+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{30}\right).$$ *