# Solve the problem using Chebyshev inequality

The problem is the following:

The symmetric coin is tossed 1600 times. What is the probability that the head will be shown up more than 1200 times?

Attempt.

Using the formula $\mathbb{P}(|X-MX|)>e)≤ DX/e^2$ I put the numbers in it

$$\mathbb{P}(|X-800|>1200)\le 400/1200^2$$

But do not get the answer which is $\le 1/800$.

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You did everything right, except that to find when it is greater than $1200$ you need $|X-800|>400$ not $|X-800|>1200$. Also note, that this will include cases $X>1200$ as well as $X<400$, and since the distribution is symmetric around $800$, all you need to do is divide $400/400^2$ by $2$.

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Why should I put 400 instead of 1200? –  saakyan Feb 16 at 18:10
Because $|X-800|>z$ is $X-800>z$ or $X-800<-z$, i.e. $X>800+z$ or $X<800-z$, and you need $X>1200$. Think about it differently: suppose you have median of $X$ equal $8$ and you are asked when $X$ is greater than $12$. Well, when $X>12$ or $X-8>4$, not $X-8>12$. –  Vadim Feb 16 at 18:14
Let $X$ be the number of heads. Then $X$ has mean $\mu=800$ and variance $\sigma^2=(1600)(1/2)(1/2)=400$. The Chebyshev Inequality says that $$\Pr(|X-\mu|\ge k\sigma)\le \frac{1}{k^2}.$$ In our case, we are interested in something related to the probability that $X\ge 1200$. Now $1200=\mu +k\sigma$, where $k=20$. So by the Chebyshev Inequality we have $$\Pr(|X-800|\ge 400)\le \frac{1}{20^2}.$$ But the distribution of $X$ is symmetric about $800$, since the coin is fair. It follows that $$\Pr(X\le 1200)\le \frac{1}{2}\cdot\frac{1}{20^2}=\frac{1}{800}.$$
We have shown that the probability that $X\ge 1200$ is $\le \frac{1}{800}$. So in fact $\Pr(X\gt 1200)\lt \frac{1}{800}$.
We have $\sigma=20$ and $\mu=800$. to use the Chebyshev Inequality, we need to find the appropriate $k$. We want $1200=800+(k)(20)$, so $k=20$. –  André Nicolas Feb 16 at 18:24
(more) Basically we are interested in the right tail, from $1200$ up. Actually, the problem asks about $1201$ or more. So we could replace the $k=20$ by solving $1201=800+(k)(20)$, and get an improved (slightly!) estimate probability is $\le \frac{1}{2}\cdot\frac{1}{(20.05)^2}$. But both the earlier bound and this one are quite poor, Central Limit Theorem gives better estimates. –  André Nicolas Feb 16 at 18:31