Probability of winning a turn-based game with a random element
Let us denote by $\Pr(w, b)$ the probability of P2 winning given its their turn and there are $w$ white stones and $b$ black stones remaining and similarly we write $\overline\Pr(w, b)$ for the probability of P2 winning given it is P1's turn.
From the rules of the game we get that any of the following cases can occur in P2's turn (assuming that enough stones are left):

White is drawn and P2 loses. This happens with probability $w / (w + b)$.
Black is drawn and a white stone vanishes. The probability for this is $b / (w + b) \cdot w / (w + b - 1)$.
Black is drawn and a black stone vanishes. We find the probability for this to be $b / (w + b) \cdot (b - 1) / ( w + b - 1)$.

If it is P1's turn to play we find similar cases:

White is drawn and P2 wins. This happens with probability $w / (w + b)$.
Black is drawn and a white stone vanishes. The probability for this is $b / (w + b) \cdot w / (w + b - 1)$.
Black is drawn and a black stone vanishes. We find the probability for this to be $b / (w + b) \cdot (b - 1) / ( w + b - 1)$.

We can now factor $\Pr$ and $\overline\Pr$ using our case distinctions and some simple probability theory:
$$
\begin{align*}
\Pr(w, b) &amp;= \Pr(\text{Case 1.2}) \cdot \overline\Pr(w - 1, b - 1) + \Pr(\text{Case 1.3}) \cdot \overline\Pr(w, b - 2), \\
\overline\Pr(w, b) &amp;= \Pr(\text{Case 2.1}) + \Pr(\text{Case 2.2}) \cdot\Pr(w - 1, b - 1) + \Pr(\text{Case 2.3}) \cdot \Pr(w, b - 2)
\end{align*}
$$
Note that we can combine these 2 equations to find a recursive formula for $\Pr$ (which I will not type out here for brevity).
Coupling these with the initial values $\Pr(w, b)$ where $w + b \leq 4$ (which can be precomputed) will give us an algorithmic way to find the desired probability.