problem 37

(a): Ignore all the cards except $J, Q, K, A$ . There are $16$ of those, $4$ of which are aces. Each card has an equal chance of being first in the list, so the answer is $\frac{1}{4}$ .
Source: https://math.stackexchange.com/a/3726869/649082
(b): Ignore all the cards except $J, Q, K, A$ . There are $4$ choices for a king, $4$ choices for a queen and $4$ choices for a jack with $3!$ permutations of the cards. Then, there are $4$ choices for an ace. The remaining $12$ cards can be permuted in $12!$ ways, so the answer is $\frac{4^{3} \times 3! \times 4 \times 12!}{16!}$ .