Which outcome is more likely, getting four kings within 2x cards or a specific pair of triples (from a group) within x cards?
Here's a partial answer giving the expected number of cards drawn to get 4 Kings for the first time, and a conceptual answer to the two triplets question.
Notably, the "first time" requirement makes this problem trickier than it looks at first, and is the main complication blocking me from presenting a complete answer with the two triplets.
For 4 Kings, the probability that round $n$ yields the four kings is the probability of having 3 kings at round $n-1$ and drawing the last king at round $n$.
$$p(n) = \frac{1}{52-(n-1)} \times \frac{{48 \choose n-4}}{{52 \choose n-1}}$$
Summing this from $n = 4$ to $52$ yields 0.25, which we use to normalize and find the expectation:
$$\sum_{n=4}^{52} \frac{n \times p(n)}{0.25} = 42.4$$
I confirmed this answer using computer simulations. Incidentally, the expected number of rounds to achieve two triplets is roughly 21.2 by simulation and is the long term winner (barely).
When considering the expected round to achieve two triplets, it's natural to notice that by the pigeonhole principle, it must be less than 37, since it's impossible to not have two triplets by the time the 37th card is drawn. We get this number from 16 (4 cards x 4 face ranks) + 8*2 (8 suits between 2 and 9 with 2 cards each) + 4 (our first triplet with the maximal 4 cards allowed); drawing the 37th card must necessarily complete a second triplet.
I recommend using the above approach, that is, finding $p(n)$ by the product of the probabilities for
the event that at round $n-1$ we don't have two triplets, but it's possible to draw a single card and complete our second triplet
the event that at round $n$ we draw a single card that completes the second triplet
I'll describe the mathematical concepts to answer this problem.
The minimal situation at round $n-1$ can be described using:
$$\binom{9}{1}\binom{4}{3}\binom{8}{1}\binom{4}{2} \times ...$$
This counts the number of ways you can have a "full house" within the 9 non-face ranks. This is the minimal condition at round $n-1$ for it to be possible to complete two triplets by round $n$.
$$\frac{2}{52 - (n-1)}$$
This expression counts the probability of drawing a single card that completes the second triplet, under our minimal condition.
For all the remaining cards, we need to count the number of ways to place them into 4 face ranks and 7 non-face ranks without inducing more non-face triplets. This can be expressed as counting the number of ways you can place $n$ indistinguishable balls into $k$ bins, with a max of $m$ balls per bin (see http://math.stackexchange.com/questions/89417/number-of-ways-to-put-n-unlabeled-balls-in-k-bins-with-a-max-of-m-balls-in for a good reference).
Lastly, for all non-face ranks that we place 2 balls into (draw 2 cards in), we need to adjust the number of winning draws at round $n$ that will complete a second triplet.
Conceptually I believe this should work, but this is a partial answer because I'm having difficulties expressing it all into a single equation that yields the expectation occurring by simulation, 21.2. If I make more progress on this, I'll update this answer.