1
$\begingroup$

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!!

$\endgroup$
  • 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
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.