jmc

Consider the people around the table, sequentially, to be A, B, C and D. Now, with probability $\frac{1}{6}$, persons A and C, seated opposite one another, will roll the same number. In that case, each of B and D may roll any of 5 numbers not equal to the number rolled by both A and C. So the probability of no two consecutive persons rolling the same number in the case where A and C roll the same is $\frac{1}{6}\cdot \frac{5}{6}\cdot \frac{5}{6}$. A and C roll differently with probability $\frac{5}{6}$, in which case each of B and D must choose from only 4 numbers, because A and C offer different numbers to be avoided. So, the probability then is $\frac{5}{6}\cdot \frac{4}{6}\cdot \frac{4}{6}$. Adding the two cases gives $\frac{5(5+4\cdot 4)}{6^3}=\frac{3\cdot 5\cdot 7}{3\cdot 2\cdot 36}=\boxed{\frac{35}{72}}$.