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  $P\left(A\right):P\left(B\right)=$

Let a=P(A throw sum 8 in onetrail) $=\frac{5}{36}$
Let b=P( A do not throw sum 8 and B do ot throw sum 7 successively).
Then b=[1-P(sum8)]
[1-P(sum7)]= $(1-\frac{5}{36})(1-\frac{6}{36})=\left(\frac{31}{36}\right)\left(\frac{5}{6}\right)=\frac{155}{216}$
P(A wins) $=a+ba+b^{2}a+b^{2}a+\mathrm{.........}\infty$ 
$=\frac{a}{1-b}=\frac{\left(\frac{5}{36}\right)}{\left(\frac{216-155}{216}\right)}=\frac{30}{61}\Rightarrow P\left(B\right)$$=1-P\left(A\right)=\frac{31}{61}$