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!!

  • 3
    $\begingroup$ 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). $\endgroup$ – 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.


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