<s>[INST] 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?
 [/INST] 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.
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