A bag contains a white and b black balls. Two players A and B alternately draw a ball from the bag, replacing  the ball each time after the draw till one of them draws a white ball and wins the game. A begins the game. If the probability of A winning the game is three times that of B, the ratio a:b is

Let W denote the event of drawing a white ball at any draw and B that for a black ball. Then

$P\left(W\right)=\frac{a}{a+b}\text{ and }P\left(B\right)=\frac{b}{a+b}$

P(A wins the game)

$\begin{array}{l}=\mathrm{P}(\mathrm{W}\text{ or }\mathrm{BBW}\text{ or }\mathrm{BBBBW}\text{ or }\dots )\\ =\mathrm{P}\left(\mathrm{W}\right)+\mathrm{P}\left(\mathrm{BBW}\right)+\mathrm{P}\left(\mathrm{BBBBW}\right)+\cdots \\ =\mathrm{P}\left(\mathrm{W}\right)+\mathrm{P}\left(\mathrm{B}\right)\mathrm{P}\left(\mathrm{B}\right)\mathrm{P}\left(\mathrm{W}\right)+\mathrm{P}\left(\mathrm{B}\right)\mathrm{P}\left(\mathrm{B}\right)\mathrm{P}\left(\mathrm{B}\right)\mathrm{P}\left(\mathrm{B}\right)\mathrm{P}\left(\mathrm{W}\right)+\cdots \\ =\mathrm{P}\left(\mathrm{W}\right)+\mathrm{P}\left(\mathrm{W}\right)\mathrm{P}(\mathrm{B}{)}^2+\mathrm{P}(\mathrm{W}\left)\mathrm{P}\right(\mathrm{B}{)}^4+\cdots \\ =\frac{\mathrm{P}\left(\mathrm{W}\right)}{1-\mathrm{P}(\mathrm{B}{)}^2}=\frac{\mathrm{a}(\mathrm{a}+\mathrm{b})}{{\mathrm{a}}^2+2\mathrm{ab}}=\frac{\mathrm{a}+\mathrm{b}}{\mathrm{a}+2\mathrm{b}}\end{array}$

$\text{ Also }P\left(B\text{ wins the game }\right)=1-\frac{a+b}{a+2b}=\frac{a}{a+2b}$

According to the given condition,

$\frac{a+b}{a+2b}=3\frac{b}{a+2b}\Rightarrow a=2b\Rightarrow a:b=2:1$