Probability of first complete set from a standard 52 card deck Suppose we have a standard, $52$-card deck that's been shuffled using a strong RNG and an unbiased shuffling algorithm. We draw one card at a time, without replacement, and stop as soon as we observe all four Aces.
Given $k$, what is the probability that we observe all four Aces before we observe four of any other rank on the $k$th card, where $k = 4, 5, \ldots, 40$?
In other words, and for example, if $k = 20$, what is the probability that the first $20$ cards contain exactly four Aces, with the fourth Ace being drawn on the $20th$ card and that the remaining sixteen cards are not four of any other rank?
For $k = 4, 5, 6, 7$ this looks to be straightforward. But for $k \geq 8$ this feels increasingly headache-y.
Thoughts on an approach
For the $k = 20$ case from above, it seems like I have to, among other things, count the number of ways to partition $16$ [the remaining cards] with $12$ objects [the remaining ranks] where each object can get no more than three of a copy and then weighting each of these partitions. For example, I could observe a partition like $3 + 3 + 3 + 3 + 3 + 1$ as permutations of KKKQQQJJJTTT9998AAAA where one of the As is always at the end. Then I would weight each partition of $16$ from $12$ by the number of permutations and number of ways of choosing similar ranks.
I can always simulate this, but for fun, I'm trying to see if there is a practical, exact approach.
EDIT 6/9/2017
I seem to have caused some confusion with my example and original wording. Here is (hopefully) better wording of what I'm after.
 Fix $k$. The sample space is the set of permutations in which the fourth ace is drawn on the $k$th card. What proportion of those permutations contain only three or fewer occurrences of all other ranks in the first $k$ cards?
 A: For $k=20$, say, you need the $32$ remaining cards to contain no aces, and at least one of every other rank. Let's choose those $32$ cards first. If our choice is successful, then we also need the $20$th card to be an ace; conditional on what we already know this is a $4/20=1/5$ chance, since there are $4$ aces in the first $20$ cards.
The total number of choices for the remaining $32$ cards is $\binom{52}{32}$. The number of those choices which have no aces is $\binom{48}{32}$. Now we have to subtract the number of choices with no aces which are also missing some other rank; we can do this using inclusion-exclusion.
There are $\binom{44}{32}$ combinations missing one specified other rank, $\binom{40}{32}$ combinations missing two specified other ranks, and so on. So the total number of combinations missing aces and at least one other rank is
$$\binom{12}1\binom{44}{32}-\binom{12}2\binom{40}{32}+\binom{12}3\binom{36}{32}-\binom{12}4\binom{32}{32}.$$
For general $k$, keep taking terms in the pattern $(-1)^{i+1}\binom{12}i\binom{48-4i}{52-k}$, but cut off any terms where $48-4i<52-k$.
Overall, then, your probability is 
$$\frac{\binom{48}{32}-\binom{12}1\binom{44}{32}+\binom{12}2\binom{40}{32}-\binom{12}3\binom{36}{32}+\binom{12}4\binom{32}{32}}{\binom{52}{32}}\times\frac{4}{20}\approx 0.003.$$
A: Math Misery has clarified the question further in the comments below. It's now apparent that I misinterpreted the question. The answer I wrote addresses what I thought the question was, not what it is.
First, we say what the probability is that you'll draw your fourth ace on the $k$th card. There $\binom{52}{4}$ possible ways that the four aces could be positioned in the 52 cards, and exactly $\binom{k-1}{3}$ of them satisfy the condition. So the probability for this is $\binom{k-1}{3}\big/\binom{52}{4}$.
Now we look at the remaining 48 cards and ignore aces. The question is, what is the probability $p$ that of the 12 card values, none will appear four times in the first $k-4$ cards of 48? 
If we let $A_K, A_Q, A_J, A_{10},\dots$, be the event that all four kings, queens, jacks, 10s, etc., appear in the first $k-4$ cards, then applying the inclusion-exclusion formula, we find that 
$$
\begin{align}
p &= 1 - 12P(A_K) + \binom{12}{2}P(A_K \cap A_Q) -  \binom{12}{3}P(A_K \cap A_Q \cap A_J) + \dots \\
&= \sum_{i=0}^{12} (-1)^i \binom{12}{i}\frac{\binom{48-4i}{52 - k}}{\binom{48}{52-k}}.
\end{align}
$$
For example, when $i = 3$, the expression $\binom{48-4i}{52 - k}\big/\binom{48}{52-k}$ represents the probability that none of the $12 = 4i$ kings, queens and jacks appear in the final $52-k$ cards of the $48$ that are left.
So the overall probability is
$$
\frac{\binom{k-1}{3}}{\binom{52}{4}}\sum_{i=0}^{12} (-1)^i \binom{12}{i}\frac{\binom{48-4i}{52 - k}}{\binom{48}{52-k}}.
$$
