Say I start with a box of $n$ balls. At each step, I remove some $k$ balls, where $k$ is from the discrete uniform distribution ${\mathcal {U}}\{1,b\} $ where $b$ is the number of balls present in the box for that step.
I'm interested in the distribution of the number of draws required to empty the box.
For example, with $n=5$, we can find that the PMF of the distribution for $1,2,3,4,5$ draws required is $1/5,5/12,7/24,1/12,1/120$.
I've done the small cases using a simple Markov chain and getting the distribution of times to absorption, but this becomes computationally unwieldy for $n$ much more than a hundred or so.
I'd like to be able to get the distribution for $n$ in the thousands.
Is there a more direct way to achieve this?
As an aside, I found that the expected number of draws is a pretty result of $H_n$, which is rigorously shown in the related question here.
EDIT: A clarifying example.
Let us use the case of $n$ of 5. We want to get the probability the box is emptied in exactly three draws.
Let us represent a sequence of draws (steps) as $\{d_1,d_2,d_3\}$ where each $d_x$ represents the number of balls removed from the box at step $x$.
Then, the possible draw sequences for emptying the box in precisely three draws (steps) are:
$\{\{3,1,1\},\{1,3,1\},\{1,1,3\},\{2,2,1\},\{2,1,2\},\{1,2,2\}\}$
These have the corresponding within-step probabilities of:
$\{\{1/5,1/2,1\},\{1/5,1/4,1\},\{1/5,1/4,1/3\},\{1/5,1/3,1\},\{1/5,1/3,1/2\},\{1/5,1/4,1/2\}\}$
Multiplying these to get the probability of each possible step sequence gives:
$\{1/10,1/20,1/60,1/15,1/30,1/40\}$
Which when totaled gives the desired $7/24$ total probability of depleting the box of 5 balls in precisely 3 drawing steps.
For reference of readers, this turns out to be $\frac{\left| S_n^{(r)}\right| }{n!}$ for $n$ balls and $r$ draws, where $| S_n^{(r)}|$ is the unsigned Stirling number of the first kind.