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The question I am looking over at right now is to use Chebyshev's inequality to compute an upper bound on the probability that my total earning is more than zero. The scenario is that I am playing a game (played $n$ times) where the chance of winning is $p<\frac{1}{2}$. If I win, I get a dollar, and if I lose, I have to pay a dollar.

Current Progress

I know that Chebyshev's inequality states that for $X$ as any random variable then for any $t>0$, then $P(|X-E(X)|\geq t)\leq \frac{Var(x)}{t^2}$. However, I am stuck as in finding the $E(X)$ and modifying the equation such that I can find the upper bound.

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$\frac{X+n}2$ has a binomial distribution with mean $np$ and variance $np(1-p)$

so $\mathbb E[X] = n(2p-1)$ and $\text{Var}(X) = \frac14 np(1-p)$

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