7. Roll a die 6 times, let the number obtained on the $i$-th roll be $a_{i}$, if there exists a positive integer $k$ such that $\sum_{i=1}^{k} a_{i}=6$ the probability $p=\frac{n}{m}$, where $m, n$ are coprime positive integers, then $\log _{6} m-\log _{7} n=$ $\qquad$
7. 1 .
Solution: When $k=1$, the probability is $\frac{1}{6}$;
When $k=2$, $6=1+5=2+4=3+3$, the probability is $5 \cdot$ $\left(\frac{1}{6}\right)^{2}$;

When $k=3$, $6=1+1+4=1+2+3=2+2+2$, the probability is $(3+6+1) \cdot\left(\frac{1}{6}\right)^{3}=10 \cdot\left(\frac{1}{6}\right)^{3}$;

When $k=4$, $6=1+1+1+3=1+1+2+2$, the probability is $(4+6) \cdot\left(\frac{1}{6}\right)^{4}=10 \cdot\left(\frac{1}{6}\right)^{4}$;

When $k=5$, $6=1+1+1+1+2$, the probability is $5 \cdot$ $\left(\frac{1}{6}\right)^{5}$;
When $k=6$, the probability is $\left(\frac{1}{6}\right)^{6}$;
Therefore, $p=\frac{1}{6}+5 \cdot\left(\frac{1}{6}\right)^{2}+10 \cdot\left(\frac{1}{6}\right)^{3}+10 \cdot$ $\left(\frac{1}{6}\right)^{4}+5 \cdot\left(\frac{1}{6}\right)^{5}+\left(\frac{1}{6}\right)^{6}=\frac{1}{6} \times\left(1+\frac{1}{6}\right)^{5}=$ $\frac{7^{5}}{6^{6}}$,
Thus, $n=7^{5}, m=6^{6}$, hence $\log _{6} m-\log _{7} n=1$.