What is the probability of getting at least one six with three dice. Problem:
A person rolls $3$ six sided dice. What is the probability that the largest value will be $6$?
Answer:
Let $p$ be the probability that we seek. The probability that the first die comes up $6$ is $1/6$. The probability that the second die comes up $6$ is $1/6$. The same for the third die. As such, I might think the answer is:
$$ p = \dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{6}  = \dfrac{1}{2} $$
but that is wrong because I am counting the roll of all sixes $3$ times. Hence I claim that the
correct answer is less than $\dfrac{1}{2}$.
\begin{align*}
p &= \dfrac{ 3(5)^2 + 3(5)+1}{6^3} = \dfrac{75+15+1}{216} \\
p &= \dfrac{91}{216}
\end{align*}
To check the answer, I compute the probability that all the dice will be $5$ or less. Call this
probability $p_5$. Note that $p + p_5 = 1$.
\begin{align*}
p_5 &= \dfrac{ 5^3 }{6^3}  \\
p &= 1 - p_5 = 1 - \dfrac{ 5^3 }{6^3} = \dfrac{ 6^3 - 5^3 }{6^3 } \\
p &= \dfrac{ 91}{216}
\end{align*}
Is my solution right? Is my first method correct?

It is correct, and this can be generalized by the Binomial Theorem, since the numbers in the numerator of the first method are precisely the binomial coefficients.
