What is the probability that the highest of the three numbers obtained from three dice will be exactly a $4$? 
Question: You are going to toll three dice. What is the probability that the highest of the three numbers will be exactly a $4$?

My attempt: 
The first die obtains number $4$ with probability $\frac{1}{6}.$
Then the remaining two dice can take numbers $1,2,3,4$ only. So each of them has probability $\frac{4}{6}.$
In total, the required probability is 
$$\frac{1}{6}\times \frac{4}{6} \times \frac{4}{6} \times 3 = \frac{2}{9}.$$
But apparently this is not the answer. 
It is $\frac{37}{216}.$
May I know where is the flaw in my reasoning?
 A: You've summed the probabilities of the events that one particular die shows a $4$ and the other two show not more than $4$. Unfortunately, those events are not disjoint, so you've double-counted the cases where several $4$s appear. To fix that, it helps to separate your computations by the number of $4$s rolled:
The probability that exactly one $4$ is rolled, and the other dice show less, is $\frac 16 \times \frac 36 \times \frac 36 \times 3 = \frac{27}{216}$.
The probability that exactly two $4$s are rolled, and the third die shows less, is $\frac 16 \times \frac 16 \times \frac 36 \times 3 = \frac{9}{216}$.
The probability that all three dice show a $4$ is $\frac 16 \times \frac 16 \times \frac 16 = \frac{1}{216}$.
Together, this makes $\frac{37}{216}$.
A: P(highest of the three numbers will be exactly a $4$)
=P(highest of the three numbers will be at most a $4$)-P(highest of the three numbers will be at most a $3$)
=${\Big(\frac46\Big)}^3-{\Big(\frac36\Big)}^3$
=$\frac{64}{216}-\frac{27}{216}$
=$\frac{37}{216}$
