A bag contains ‘W’ white balls and ‘R’ red balls. Two players P1 and…

Probability of drawing a white ball at any draw

$=\frac{W}{W+R}$

Now, P₂ can be the winner in general on 2r^th, $r \geq 1$, draw.

That means in the first (2r - 1) draws no player draws a white ball and 2r^th draw results in a white ball for P₂.

If the corresponding probability is p_r, then

$p_r=\left(\frac{R}{W+R}\right)^r . \left(\frac{R}{W+R}\right)^{r-1} . \left(\frac{W}{W+R}\right)$

$=\left(\frac{R}{W+R}\right)^{2 r} . \frac{W}{R}$

Hence, required probability

$=\sum\limits_{r=1}^{\infty} p_r=\frac{W}{R} \sum\limits_{r=1}^{\infty}\left(\frac{R}{W+R}\right)^{2 r}$

$=\frac{R}{(W+2 R)}$