2014 AMC 10B Problem 16

There are $6^4$ equally likely ordered outcomes. If exactly three dice show the same value, choose the repeated value in $6$ ways, the different value in $5$ ways, and the position of the different die in $4$ ways. This gives $6\cdot5\cdot4=120$ outcomes. If all four dice match, there are $6$ outcomes. The probability is $\frac{120+6}{6^4}=\frac{126}{1296}=\frac7{72}$. Thus, the correct answer is B .