probability in game

We know that either A or B can win .

Therefore $P(A)+P(B)=1$

But it is given that $P(A)=2P(B)$ Hence $2P(B)+P(B)=1$ $\implies P(B)=\dfrac{1}{3}$ Hence $P(A)=\dfrac{2}{3}$ There are infinite ways in which A can win with different probabilities.

$P(A) =$ A wins in first trial +A does not win in first trial and B does not win in second trial and A wins in third trial+ $\cdots$ .

This can be represented as $P(A)=P(W)+P(BWW)+P(BWBWW)+\cdots$ $P(A)=\dfrac{m}{m+n}+\dfrac{n}{m+n}\times\dfrac{m}{m+n}\times\dfrac{m}{m+n}+\cdots$ This we can see that it is an infinite $GP$ with

First term , $a = \dfrac{m}{m+n}$

Common Ratio = $r = \dfrac{mn}{(m+n)^{2}}$ Sum of infinite GP is $P(A) = \dfrac{a}{1-r}$ $P(A) = \dfrac{\dfrac{m}{m+n}}{1-\dfrac{mn}{(m+n)^{2}}}$ $\implies P(A) = \dfrac{mn+m^{2}}{m^{2}+mn+n^{2}}$ But $P(A) = \dfrac{2}{3}$ Therefore $\dfrac{2}{3}=\dfrac{mn+m^{2}}{m^{2}+mn+n^{2}}$ $\implies m=n$ Hence the required is $\boxed{\dfrac{m}{n}=1}$