Use Chebyshev's inequality to calculate probability

Question

Use the Chebyshev's inequality to show that the probability that in $n$ throws of a fair die the number of sixes lies between $\frac{1}{6}n-\sqrt{n}$ and $\frac{1}{6}n+\sqrt{n}$ is at least $\frac{31}{36}$.

The Chebyshev's inequality goes as follows:

If $Y$ is a random variable and $\mathbb E(Y^2)<\infty$, then $$ \mathbb P(|Y|\geq t)\leq\frac{1}{t^2}\mathbb E(Y^2)\quad\text{for }t>0. $$

My problem here is that I don't know what $t$ to choose. My first guess was choosing $t=\frac{1}{6}n+\sqrt{n}$ and taking the complement. But then we get

$1-\mathbb P(|Y|\geq\frac{1}{6}n+\sqrt{n}).$

This gives us the interval $(-\frac{1}{6}n-\sqrt{n},\frac{1}{6}n+\sqrt{n})$, which is not what I need.

There must be some trick I guess to do this. Could someone help me out?

Answer 1

Let $X$ be a random variable that equals $1$ with probability $p=\frac{1}{6}$ and $0$ with probability $q=\frac{5}{6}$.
We have $\mathbb{E}[X]=\frac{1}{6}$ and $\text{Var}[X]=\frac{5}{36}.$ $Y=X+\ldots+X$ ($n$ times) is a random variable accounting for the number of sixes, with mean value $\mu=\mathbb{E}[Y]=\frac{n}{6}$ and $\sigma^2=\text{Var}[Y]=\frac{5n}{36}$.

By Chebyshev's inequality, $$ \mathbb{P}[|Y-\mu|>k\sigma]\leq \frac{1}{k^2} $$ hence by choosing $k$ as $\sqrt{\frac{36}{5}}$ we get: $$ \mathbb{P}\left[\left|Y-\frac{n}{6}\right|>\sqrt{n}\right]\leq \frac{5}{36} $$ so: $$ \mathbb{P}\left[Y\in\left(\frac{n}{6}-\sqrt{n},\frac{n}{6}+\sqrt{n}\right)\right]\geq\frac{31}{36} $$ as wanted.

---

If $n$ is pretty large ($n\geq 72$) a better approximation of the last probability is given by the central limit theorem:

$$ \sqrt{\frac{18}{5\pi n}}\int_{-\sqrt{n}}^{\sqrt{n}}\exp\left(-\frac{18x^2}{5n}\right)\,dx = \text{Erf}\left(3\sqrt{\frac{2}{5}}\right)\approx \color{red}{99.27\%}.$$

Answer 2

Hint: Let $Y=X-\frac 16n$, where $X$ is the number of sixes in $n$ throws and $t=\sqrt n$.

Answer 3

Rephrasing the Chebyshev's inequality, $$\mathbb P(|Y-\bar{Y}|\geq t\sigma_Y)\leq\frac{1}{t^2}\quad\text{for }t>0$$

If $X$ is the number of sixes in $n$ throws, then $\mathbb{E}(X)=\frac{n}{6}$, and $\sigma_X^2 = \frac{1}{6}\frac{5}{6}n=\frac{5}{36}n.$

Now pick $t = \frac{6}{\sqrt{5}}$ and you're done.

Note:

You're implicitly assuming that $\mathbb{E}(Y)=0$ in the inequality, which ins't necessarily true.

Answer 4

For $i=1,\ldots,n$, let $X_{i}$ be the random variable which equals 1 if roll $i$ comes up as 6, and 0 otherwise. Then the $X_{i}\overset{i.i.d}{\sim}\text{Bernoulli}(1/6),$ so the total number of sixes rolled is $$X=\sum_{i=1}^{n}X_{i}\sim\text{Binomial}(n,1/6).$$ $X$ has mean $n/6$ and variance $5n/36$. Then Chebyshev's inequality says that $$\mathbb{P}(|X-n/6|> \sqrt{n})\leq \frac{5n/36}{(\sqrt{n})^{2}},$$ or equivalently, $$\mathbb{P}(|X-n/6|\leq\sqrt{n})\geq 1-\frac{5n/36}{n}=\frac{31}{36}.$$