# Optimal times of Rolls to get exactly 2 six

I'm encountered with an interesting question, that is, we have a fair die, and we got to choose how many time to roll this die, let's say $$n$$ rolls. We then log the result, which is an $$n$$-tuple. If the result contains exactly 2 six, we win the game. how do we choose $$n$$. The way I want to do is the use binomial distribution to count exactly 2 successes in $$n$$ rolls. That is: $$P(win) = {n\choose 2}{\frac{1}{6}}^2{\frac{5}{6}}^{n-2}.$$ With $$\frac{1}{6}$$ is the probability of rolling 6 in one roll and $$\frac{5}{6}$$ is the chance not to roll 6 in one roll. we then set: $$f(n) = {n\choose 2}{\frac{1}{6}}^2{\frac{5}{6}}^{n-2}$$ Differentiate the function, set the derivative to $$0$$ and find the value of $$n$$ that maximize the probability of winning. Conceptually, this method make sense. But as you can see, differentiating this thing is a tedious thing to do.

I'm pretty sure there's an easier way to solve this problem other than using binomial random variables, but I just cannot find them, everything I think of eventually comes back to binomial and I'm pretty upset. could anyone please help me and show me an easier way to attack this problem, thank you very much!!

• Differentiating is tough when you have a binomial symbol, but you could just replace that with $\frac {n(n-1)}2$. Still, I expect it's a lot easier to just do it numerically (the derivative is a pretty nasty function and you'd still need to find its root(s) numerically). – lulu Feb 15 at 2:02

Just consider the ratio of the successive term

$$\frac {f(n+1)} {f(n)} = \frac {5(n+1)} {6(n-1)}$$

The above ratio is greater than $$1$$ if and only if

$$5(n+1) > 6(n - 1) \iff n < 11$$

Thus we can conclude that

$$\begin{cases} f(n+1) > f(n) & \text{when} & n < 11 \\ f(n+1) = f(n) & \text{when} & n = 11 \\ f(n+1) < f(n) & \text{when} & n > 11 \\ \end{cases}$$

So from this we can conclude that $$f(n)$$ is maximized when $$n = 11, 12$$, for any $$n \in \mathbb{N}$$

In fact the mode of a binomial distribution is around the mean, so one intuitive guess is we pick $$np = 2$$ which yield $$n = 12$$ as well.