A and B throw a pair of dice. A wins if he throws 6 before B throws 7 and B wins if he throws 7 before A throws 6. If A begins, his chance of winning is

a pair of dice is thrown

A=getting sum 6

B=getting sum 7

C=getting sum neither 6 nor 7

$$\begin{gathered} \mathrm{P}\left(\mathrm{A}\right)=\frac{5}{36} \mathrm{P}\left(\mathrm{B}\right)=\frac{6}{36}=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right. \\ \mathrm{P}\left(\mathrm{C}\right)=\frac{31}{36}\times \frac{5}{6}=\frac{155}{216} \end{gathered}$$

A begins the game

$$\begin{gathered} \mathrm{P}\left(\mathrm{Awins}\right)=\mathrm{P}\left(\mathrm{A}\right)+\mathrm{P}\left(\mathrm{C}\right) \mathrm{P}\left(\mathrm{A}\right)+(\mathrm{P}{\left(\mathrm{C}\right))}^2 \mathrm{P}(\mathrm{A})..... \\ =\frac{5}{36}+\frac{155}{216}·\frac{5}{36}+{\left(\frac{155}{216}\right)}^2\frac{5}{36}..... \\ =\frac{{\displaystyle \frac{5}{36}}}{1-{\displaystyle \frac{155}{216}}}=\frac{30}{61} \end{gathered}$$