TITLE: Maximum likelihood to throw exactly two 6s
QUESTION [3 upvotes]: One throws a dice $n$ times. For which value of $n$ is maximum the probability to obtain exactly two 6s?
I get $$n=11 \text{ or } n=12.$$

My solution:
the probability to obtain exactly two 6s in $n$ throws is (Bernoulli distribution)
$$ \binom{n}{2} \left( \dfrac{1}{6} \right)^2 \left(\dfrac{5}{6} \right) ^{n-2}.$$
To find its maximum consider
$$ f(x) = \dfrac{x(x-1)}{2} \dfrac{1}{36} \left(\dfrac{5}{6} \right) ^{x-2}.$$
Searching for its stationary points, differentiating one obtains
$$ x^2 \log \dfrac 5 6 + x (2- \log \dfrac 5 6)  -1 = 0, $$
whose solutions $x_1,x_2$ are s.t. $\lfloor x_1 \rfloor = 0, \lfloor x_2 \rfloor = 11.$
Substituing back in $f$ one finds that $$f(11) = f(12) > f(0) = f(1),$$ and therefore $n=11$ or $12.$
Is there another (faster/more interesting) way to obtain this result?

REPLY [0 votes]: You do not need calculus, since you can consider
$$\dfrac{f(n+1)}{f(n)} = \dfrac{P(\text{sixes}=2|\text{dice}=n+1)}{P(\text{sixes}=2|\text{dice}=n)} = \dfrac{n+1}{n-1}\times \dfrac56$$ 
which is greater than $1$ when $2 \le n \lt 11$, is less than $1$ when $n \gt 11$, and is equal to $1$ when $n=11$.  
So $11$ and $12$ give the maximum value of $f(n)=P(\text{sixes}=2|\text{dice}=n)$.
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