Two players A and B throw alternately a pair of dice. A wins if he throws 8 before B throws 7 and B wins if he throws 7 before A throws 8. If A starts the game, then the ratio P(A) : P(B) =

Let a = P(A throw sum 8 in one trial)$=\frac{5}{36}$
Let b= P(A do not throw sum 8 and B do not throw
sum 7 successively).
Then 

$\begin{array}{l}b= [1 – P(sum8\left)\right][1-\mathrm{P}(\mathrm{sum}7\left)\right]=\left(1-\frac{5}{36}\right)\left(1-\frac{6}{36}\right)=\left(\frac{31}{36}\right)\left(\frac{5}{6}\right)=\frac{155}{216}\\ \mathrm{P}\left(\mathrm{A}\text{ wins }\right)=\mathrm{a}+\mathrm{ba}+{\mathrm{b}}^2\mathrm{a}+{\mathrm{b}}^3\mathrm{a}+\dots \dots \dots \mathrm{\infty }\\ \Rightarrow P\left(A\right)=\frac{\mathrm{a}}{1-\mathrm{b}}=\frac{\left(\frac{5}{36}\right)}{\left(\frac{216-155}{216}\right)}=\frac{30}{61}\\ \\ \Rightarrow \mathrm{P}\left(\mathrm{B}\right)=1-\mathrm{P}\left(\mathrm{A}\right)=\frac{31}{61}\\ P\left(A\right):P\left(B\right)=\frac{30}{61}:\frac{31}{61}=30:31\end{array}$