A card draw problem concerning Coupon collector's problem

Question

I got a problem from my probability test. It is:
There are 21 cards labeled with each of A, B, C, 1, 2, . . . , 9. For each of A, B, C, there is one card labeled with it, and for each of 1, 2, . . . , 9, there are two cards labeled with it. Consider to repeat trials each of which is “to draw one from 21 cards uniformly at random.”
Answer the following questions.
(1) Let X denote the number of trials repeated until any one of A, B, C appears. Find the expectation of X.
(2) Let Y denote the number of trials repeated until each of A, B, C appears at least once. Find the expectation of Y.
(3) Let Z denote the number of trials repeated until each of 1, 2, . . . , 9 appears at least once. Please tell which is greater, the expectation of Z or the expectation of Y, and also explain your reason.
(4) Let P denote the probability that each of A, B, C appears at least once after n trials. Describe P using n.

I know situation (1) is a geometric distribution, through it is a little complex to calculate E(X), I could do that. But from (2), I cannot image how to write its probability formula, I searched the Wiki and know it seems like a variation of Coupon collector's problem. By intuition, I think it should be 21+21+21=63 times? and (3) should be (21/2)*9=94.5?
Could you give me some hints and your thoughts on this problem? Thank you!

Answer 1

$(1)$ Indeed geometric distribution where the number of trials needed to arrive at a success are counted. Here the parameter is $p=\frac3{21}$ so that $\mathbb EX=\frac1p=\frac{21}3=7$.

$(2)$ Here we could write $Y=Y_1+Y_2+Y_3$ where $Y_1$ is the same as $X$ in $(1)$, $Y_2$ has geometric distribution with parameter $p=\frac2{21}$ (if e.g. the first success is a draw of card $A$ then for reaching the second success we need to draw $B$ or $C$ because drawing $A$ is in that case not a success anymore) and $Y_3$ has geometric distribution with parameter $p=\frac1{21}$. With linearity of expectation we find: $$\mathbb EY=\mathbb EY_1+\mathbb EY_2+\mathbb EY_3=\frac{21}3+\frac{21}2+\frac{21}1$$

$(3)$ As in $(2)$ you could write here $Z=Z_1+\cdots+Z_9$ (caution: the index does not correspond with the label but with chronology of successes) and calculate $\mathbb EZ$ the same way as done in $(2)$. Then you can of course compare this with $\mathbb EY$ and answer the question. You will find that $\mathbb EY>\mathbb EZ$ and actually you are not asked explicitly to find $\mathbb EZ$ but to just to answer and "explain your reason" (I leave that upto you).

$(4)$ This can expressed as $P(Y\leq n)$.

Answer 2

1. The probability $p$ of any one of $A$, $B$ or $C$ appearing equals $p = \frac{3}{21} = \frac{1}{7}$. This is indeed a geometric distribution, so we find: $$E(X) = \frac{1}{p} = 7$$

2. The probability of $A$ occurring equals $\frac{1}{21}$, and the same is true for $B$ and $C$. We cannot simply multiply 21 by 3, though, because we could get a $B$ or $C$ while looking for $A$. Rather, look at it this way. First, we need to find one of the letters, with probability $\frac{3}{21} = \frac{1}{7}$. As soon as we have found one, we need to get one of the two remaining letters with probability $\frac{2}{21}$. Then, finally, we need to get the last remaining letter with probability $\frac{1}{21}$. We thus find: $$E(Y) = \frac{21}{3} + \frac{21}{2} + \frac{21}{1} = 38.5$$

Can you apply a similar reasoning to questions $3$ and $4$?