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Suppose I do $n$ coinflips with an unfair coin that shows head with $p=\frac{5}{12}$. Now I define the random variable $X_i \in \{-1, 1\}$ of the $i$-th flip with $-1, 1$ describing tail and head respectively. Because my coin is unfair I am more likly to end up with $X_i = -1$.

Looking at $Z_n = \sum_{i = 1}^n X_i$ it seems intuitive, that $Z_n \to -\infty$ or $P(Z_n > k) \to 0$ for any $k \in \mathbb{Z}$ as $n \to \infty$.

I struggle however to show this formally. I was trying to transform $Z_n$ such that it is binomially distributed and work with that but didn't get anywhere really.

Any help is appreciated!

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    Keyword: Law of large numbers. // For quantitative estimates, use $$P(Z_n>c)\leqslant E(e^{tZ_n})e^{-tc}=E(e^{tX})^ne^{-tc}$$ for every $t>0$ and choose $t$ such that $$E(e^{tX})=pe^t+(1-p)e^{-t}<1$$ to conclude that $$P(Z_n>c)\leqslant Cr^n$$ for some finite $C$ and some $r<1$, hence the conclusion that $$P(Z_n>c)\to0$$ for every fixed $c$. All this works with $$r=\sqrt{4p(1-p)}$$ – 2017-01-26
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    Thanks a lot! Can you explain how you get from $E(e^{tZ_n})$ to $E(e^{tX}$)? – 2017-01-26
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    Sum of $n$ i.i.d. random variables distributed like $X$. – 2017-01-26
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    ah of course! Thanks again. – 2017-01-26

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