6. Two people take turns rolling dice, each rolling two at a time. The first person to roll a sum greater than 6 on the two dice wins, otherwise the turn passes to the other person. The probability of the first person to roll winning is $\qquad$
The probability of winning in a single throw is $\frac{7}{12}$. Let $A=$ "the first thrower wins", so $P(A)=\frac{7}{12}+\left(\frac{5}{12}\right)^{2} \cdot \frac{7}{12}+\left(\frac{5}{12}\right)^{4} \cdot \frac{7}{12}+\cdots=\frac{\frac{7}{12}}{1-\left(\frac{5}{12}\right)^{2}}=\frac{84}{119}=\frac{12}{17}$.