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.
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?