imc

We split this problem into $2$ cases. First, we calculate the probability that all four are the same. After the first dice, all the numbers must be equal to that roll, giving a probability of $1\cdot \frac{1}{6}\cdot \frac{1}{6}\cdot \frac{1}{6}=\frac{1}{216}$. Second, we calculate the probability that three are the same and one is different. After the first dice, the next two must be equal and the third different. There are $4$ orders to roll the different dice, giving $4\cdot 1\cdot \frac{1}{6}\cdot \frac{1}{6}\cdot \frac{5}{6}=\frac{5}{54}$. Adding these up, we get $\frac{7}{72}$, or $\boxed{\frac{7}{72}}$.