Probability that total number of sixes in $n$ dice tosses

Question

Let X be the total count of fives in $n$ dice ups and downs s.t. it is in between $\frac 1 k n +/- \sqrt n$ with k=6. Calculate Pr(X).

I personally think that I should integrate the probability density function of n such that it is in between the two values? I dont know how to find that pdf but I know for n throws, you expect to get n/6 sixes. any hin would be helpful thanks!

Answer 1

$p= \frac{1}{6}$, $q=1-p$, $\mu = \frac{n}{6}$, $\sigma = \sqrt{npq}$. With Chebychev:

$$P(|X-\mu|\lt \sqrt{n}) \geq 1- \frac{\sigma^2}{n}= 1- \frac{npq}{n} = 1- \frac{5}{36}=\frac{31}{36}$$

Answer 2

Hint: $P(a< X\leq b) = P(X\leq b) - P(x\leq a) = F_X(b) - F_X(a)$

Answer 3

Hint:

• Let $Y \sim Bin(n, \frac16)$

• Use Chebyshev's inequality on $Pr \left(|Y-\mathbb{E}(Y)|\ge \sqrt{n} \right)$

Answer 4

This is a problem which can be solved using the concentration inequalities approach. Specifically, you use (a variant of) Markov's inequality to obtain $$\mathbb{P}(X-n/6>x)\le \inf_{\theta>0}e^{-\theta x}\mathbb{E}(e^{\theta (X-n/6)}),$$ which is called the Chernoff's bound in literature. Now, noting that $X=\sum_{k=0}^n I(X_k=1)$, where $X_k$ is the random variable which is $1$ is we get a $6$ in the $k$ th trial, and $0$ otherwise, so that $\mathbb{E}X_k = 1/6$. Furthermore, $\{X_k\}_{k=1}^n$ are all i.i.d. Then, $$\mathbb{E}(e^{\theta(X-n/6)})=\prod_{k=1}^n \mathbb{E}(e^{\theta(X_k-1/6)})=\left(1/6e^{5\theta/6}+5/6e^{-\theta/6}\right)^n$$ It is now a matter of minimizing the function $e^{-\theta x}\left(1/6e^{5\theta/6}+5/6e^{-\theta/6}\right)^n$ over all $\theta >0$. We can easily find that the minimum is obtained at the solution of $$-x + 5n\frac{e^{5\theta/6}-e^{-\theta/6}}{e^{5\theta/6}+5e^{-\theta/6}}=0\implies \theta = \ln \left(\frac{n+x}{n-x/5}\right)$$ Substituting this value of $\theta$ in the original expression finds a bound on the probability, from which the desired result can be obtained, maybe after a few manipulations.