I am reviewing some probability puzzles, and trying to solve them under a standard timed duration. But, I think could be completely wrong in formulating the below exercise problem. So, I'd like someone to verify if my solution attempt is correct and if I've counted correctly.
[Blitzstein 1.37] A deck of cards is shuffled well. The cards are dealt one-by-one, until the first time an ace appears.
(a) Find the probability that no kings, queens or jacks appear before the first ace.
Solution.
Firstly, we are sampling without replacement. Moreover, I think that, since we are not dealing hands, the order in which the cards appear is important.
We are interested to form an $r$-tuple, $(x_1,x_2,\ldots,x_r)$ from the population $\{\heartsuit,\clubsuit,\spadesuit,\diamondsuit \} \times \{1,2,\ldots,13\}$, where all $x_j$'s are distinct.
Let $A_r$ be the event that the $r$'th draw is an ace. This last ace can be chosen in $4$ ways. Should there be no kings, queens and jacks before the first ace, we have $36$ cards to choose from. Thus, there are $(36)_{r-1}=\frac{36!}{(36-r+1)!}$ distinguishable choices for the first $(r-1)$ cards. Obviously, $r=1,2,\ldots,36$.
The desired probability is: $$P(\cup_{r=1}^{36} A_r) = \sum_{r=1}^{36} \frac{4 \cdot (36)_{r-1}}{(52)_r}$$
(b) Find the probability that exactly one king, exactly one queen and exactly one jack (appear in any order) before the first ace.
The desired probability is:
$$4^3 \sum_{r=1}^{36} \frac{4 \cdot (36)_{r-4}}{(52)_r}$$
Best Answer
You do not need to use $52$ or $36$ or your $r$ in the calculations for (a) or (b) as the positions of the other $36$ cards are irrelevant and you would get the same result if they were not there.
So for (a): the probability that at the first Ace appears before the first Jack, Queen or King is (by exchangeability) $\frac{4}{16}=\frac14$; one of the $16$ interesting cards must appear earlier than the others, each equally likely to do so, and $4$ of them are Aces;
and similarly for (b): the probability that exactly one king, exactly one queen and exactly one jack appear in any order before the first ace is $\frac{12}{16}\times\frac{8}{15}\times \frac{4}{14}\times \frac{4}{13} = \frac{16}{455}$.
You might want to check whether these correspond to your expressions.