If I'm flipping through a deck of cards one-by-one and pulling each one out, what's the chance I pull all 4 jacks, all 4 queens, or all 4 aces before I pull 3 kings?
One idea I had was to compute the answer as $P(\text{draw 3 of a kind of J, Q, A before 3 of a kind of K}) \times P(\text{draw last J, Q, or A before 3rd K})$, somehow using the Negative Hypergeometric Distribution, but I don't think it works because, for example, a 3 of a kind of 2's can't be classified as a "success" or "failure" in this problem.
Best Answer
By inclusion-exclusion:
The probability all four Jacks occur before a third king is the same as the probability that the final two cards of a reduced eight-card deck (containing only jack and kings) are both kings is $\frac{4}{8}\cdot\frac{3}{7}$
The probability that all four Jacks and all four Queens occur before a third king is the same as the probability that the final two cards of a reduced twelve-card deck (containing only jacks, queens, and kings) are both kings is $\frac{4}{12}\cdot\frac{3}{11}$.
Finally, all four Jacks, Queens, and Aces occur before the third king is the same as the probability that the final two cards of a reduced sixteen-card deck (containing only jacks, queens, kings, and aces) are both kings is $\frac{4}{16}\cdot\frac{3}{15}$.
By symmetry, all other relevant probabilities can be found. Letting $J$ be the event that all four Jacks occurred before the third king, and similarly defining $Q$ and $A$, we have:
$$\Pr(J\cup Q\cup A) = \Pr(J)+\Pr(Q)+\Pr(A)-\Pr(J\cap Q)-\Pr(J\cap A)-\Pr(Q\cap A)+\Pr(J\cap Q\cap A)$$
$$= 3\cdot \dfrac{4\cdot 3}{8\cdot 7} - 3\cdot\dfrac{4\cdot 3}{12\cdot 11}+\dfrac{4\cdot 3}{16\cdot 15}=\dfrac{647}{1540}\approx 0.42013$$