# Inifinte coins flipping

Suppose there are infinitely many coins labled $1,2,3...$ and define a R.V. $X_n=1$ when the coin $n$ is a head and $X_n=0$ when the coin n is a tail. Also we assume the for any integer m, it is not possible that $X_i=1$ for all $i\ge m$. Now define another new R.V $Z=\sum_{n\ge 1}\frac{X_n}{2^n}$. So question i) What is $P(Z\le0.625)$ and ii)In general, for all $x\in (0,1)$, what is the probability $Pr(Z\le x)$

I have pondered this question quite a long time, but have no idea how to do it. I found that $0.625=\frac{1}{2}+\frac{1}{8}$ and it seems quite like some terms of $Z$ but still not quite work to solve the question.

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The terms are such that if all terms after the $n$-th coin are 1, their total contribution to $Z$ is exactly the contribution of the $n$-th coin. You assume that this event has probability $0$ (which is true for independent coin flips), so their contribution will be strictly less than the contribution of the $n$-th coin.
So for the first question: compute first $P(Z > \frac{1}{2} + \frac{1}{8})$: we need $X_1$ to be 1 (or all other terms sum to less than $\frac{1}{2}$, and either $X_2$ or $X_3$ or both to be 1 too: if they are both $0$ the other terms after $X_3$ sum to less than $\frac{1}{8}$. So the probability $P(Z > \frac{1}{2} + \frac{1}{8}) = \frac{1}{2}\cdot(1-\frac{1}{2}\cdot\frac{1}{2})$, so the asked for probability is its complement.