Consider a box containing one black ball and one white ball. Every time a black ball is drawn, a die is rolled and a number of black balls equal to the result of the die are added to the box. We want to determine $P(X=k)$ where $X$ is the random variable that represents the number of draws required to obtain a white ball.
Examples I found: $$\begin{split} P(X=1)&=\frac 12\\ P(X=2)&=\frac 1{24}\\ P(X=3)&=\dfrac 1{2.6^2}\displaystyle \sum_{k=1}^6\dfrac {k+1}{k+2}\sum_{i=1}^{6}\dfrac 1{i+k+1}= 0.0472490588115\dots$ \end{split}$$
it appears that there is no elegant formula for $P(X=k)$.
Note: the OP is asking for the computation of the probability of ending on a specified round. I expect that's hard to get, and here I just consider the expected value. Too long for a comment.
Consider the simpler problem for which you just add a single black ball each time you draw a black. Thus, at the $n^{th}$ round, there would be $n$ black balls and a single white one. This is the classical case of Polya Urns.
In that case, the probability of a win on round $k$ would have probability $\frac 1{k+1}$ so the odds of failing for $n-1$ consecutive rounds would be $$\frac 12\times \frac 23\times \frac 34\times \cdots \times \frac {n-1}{n}=\frac 1n$$
It follows that the probability of drawing the white ball for the first time on the $n^{th}$ turn is $$\frac 1n\times \frac 1{n+1}$$
Thus the expected number of rounds you need in order to get that white ball is given by $$\sum_{n=1}^{\infty} n\times \frac 1n\times \frac 1{n+1}=\infty$$.
As the expected number is infinite for this game, it is clearly infinite for yours.