2014
Problem - 1320
Four fair six-sided dice are rolled. What is the probability that at least three of the four dice show the same value?
There are two qualified cases: all have the same value and only three have the same value.
There are $C_6^1\times\left(\frac{1}{6}\right)^4=\frac{1}{216}$ chances that four have the same value.
The probability that only three of them have the same value equals $C^1_6\times C_4^1\times\left(\frac{1}{6}\right)^3\times\frac{5}{6}=\frac{20}{216}$.
Therefore the final result is $$\frac{1}{216}+\frac{20}{216}=\boxed{\frac{7}{72}}$$