This formula says:
$I.$ either the starting player (call him $F$) wins in first move $(\frac{w}{b+w})$
$II.$ or $F$ draws a black ball $\frac{b}{b+w}$
AND the second player (label him $S$) has to lose.
In other words, $P(b,w) =$ Pr($F$ picks a white ball) + Pr($F$ picks a black ball)$\times$Pr($S$ loses)
Now, need Pr($S$ loses). Since the question provides only a probability for the winner, and imparts no other information on the loser, thus consider Pr($S$ loses) = $1 - Pr(S$ wins). What's $Pr(S$ wins)?
In case II, $F$ went first and drew a black ball.
Subsequently, at $S$'s turn, $b - 1$ & $w$ remain.
In other words, $S$ (and NOT $F$) is now the starting player in a game that begins with $b-1$ black balls and $w$ white ones !
Thus, $Pr(S$ wins) is exactly $P(b-1, w)$,
because $S$ can now be considered as the starting player in a game that begins with $b-1$ black balls and $w$ white ones.
Thus, Pr($S$ loses) = $1 - P(b-1, w)$.
As an alternative approach, we can model this as an infinite series.
Starting with player A:
$$
{P(A\ wins)\ =\ P(A\ wins\ 1^{st}\ turn)\ +\ P(A\ wins\ 2^{nd}\ turn)\ +\ P(A\ wins\ 3^{rd}\ turn)\ +\ ...}
$$
First looking at the individual cases for player A:
$$
P(A\ wins\ 1^{st}\ turn)\ =\ (A\ chooses\ white\ ball)\
$$
$$
=\ \frac{4}{12}\ =\ \frac{1}{3}
$$
$$
P(A\ wins\ 2^{nd}\ turn)\ =\ (A,B,C\ all\ choose\ nonwhite\ balls)(A\ chooses\ white\ ball)\
$$
$$
=\ \bigg(\frac{8}{12}*\frac{8}{12}*\frac{8}{12}\bigg)\frac{1}{3}\
=\ \bigg(\frac{2}{3}\bigg)^3\frac{1}{3}
$$
$$
P(A\ wins\ 3^{rd}\ turn)\ =\ (A,B,C\ all\ choose\ nonwhite\ balls)(A,B,C\ all\ choose\ nonwhite\ balls)(A\ chooses\ white\ ball)\
$$
$$
=\ \bigg(\frac{8}{12}*\frac{8}{12}*\frac{8}{12}\bigg)\bigg(\frac{8}{12}*\frac{8}{12}*\frac{8}{12}\bigg)\frac{1}{3}\ =\ \bigg(\frac{2}{3}\bigg)^6\frac{1}{3}
$$
Combining these cases back into the total probability:
$$
P(A\ wins)\ =\ \frac{1}{3}\ +\ \bigg(\frac{2}{3}\bigg)^3\frac{1}{3}\ +\ \bigg(\frac{2}{3}\bigg)^6\frac{1}{3}\ +\ ...
$$
This series can be modeled as an infinite sum:
$$
P(A\ wins)\ = \frac{1}{3} \sum\limits_{n=0}^{\infty}\bigg[\bigg(\frac{2}{3}\bigg)^3\bigg]^n
$$
This is a infinite sum is a geometric series and can be represented as:
$$
P(A\ wins)\ = \frac{1}{3} \frac{1}{1-\big(\frac{2}{3}\big)^3} = \frac{9}{19}
$$
The same geometric series holds for players B and C and we only need to multiply by appropriate probabilities for their cases:
$$
P(B\ wins)\ = \frac{2}{3}\frac{1}{3} \frac{1}{1-\big(\frac{2}{3}\big)^3} = \frac{6}{19}
$$
$$
P(C\ wins)\ = \bigg(\frac{2}{3}\bigg)^2\frac{1}{3} \frac{1}{1-\big(\frac{2}{3}\big)^3} = \frac{4}{19}
$$
Best Answer
I preassume that the $2$ players draw a ball turn by turn.
Let $E$ denote the event that the first ball drawn is a white one. Then:
$$P(\text{A wins})=P(\text{A wins}|E)P(E)+P(\text{A wins}|E^c)P(E^c)$$
Now realize that: $$P(\text{A wins}|E^c)=1-P(\text{A wins})$$
This because under condition $E^c$ the game somehow starts over, but now with player $B$ as the one who draws the first ball.
Defining $p:=P(\text{A wins})$ we come to the following equation in $p$:
$$p=1\times\frac15+(1-p)\times\frac45=1-\frac45p$$
and find that: $$p=\frac59=0.5555\dots$$