6. Two people take turns throwing dice, each throwing two at a time. The first person to get a sum greater than 6 on the two dice wins; otherwise, the other person throws. What is the probability that the first person to throw wins? $\qquad$
6. $\frac{12}{17}$.

The probability of the sum of the points on two dice being greater than 6 is $\frac{21}{36}=\frac{7}{12}$, thus, the probability of the first player winning is
$$
\frac{7}{12}+\left(\frac{5}{12}\right)^{2} \times \frac{7}{12}+\left(\frac{5}{12}\right)^{4} \times \frac{7}{12}+\cdots=\frac{12}{17} .
$$