Probability question about coin flipping If I flip an unbiased coin an infinite number of times, what is the probability that, at some point, the number of heads will be twice the number of tails?
I have already tried making a branching probability tree, but I just get overwhelmed and it seems to go on forever... I don't know how to reconcile it. 
 A: A closed form solution would be perhaps hard (too hard?) to derive, but you can at least come up with a straightforward upper bound by summing up the probability that this happens for $3n$ flips over all values of $n$. For given $3n$, the probability that you get twice as many heads as tails is ${{3n} \choose n}2^{-3n}$. So your upper bound is $\sum_n {{3n} \choose n} 2^{-3n}$. The terms very quickly go to zero as $n$ increases so you could even use inclusion-exclusion on the first few terms (giving a more complicated expression) and then just sum the rest of the terms (giving an over-estimate) that is still very close to the true answer.
A: Empirically it seems to be $$\dfrac{3}{8^1} + \dfrac{6}{8^2} + \dfrac{21}{8^3} + \dfrac{90}{8^4} + \dfrac{429}{8^5} + \dfrac{2184}{8^6} + \dfrac{11628}{8^7} + \dfrac{63954}{8^8} + \dfrac{360525}{8^9} + \dfrac{2072070}{8^{10}} + \dfrac{12096045}{8^{11}} + \dfrac{71524440}{8^{12}} + \dfrac{427496076}{8^{13}} + \dfrac{2578547760}{8^{14}} + \dfrac{15675792072}{8^{15}} + \dfrac{95951017602}{8^{16}} + \cdots $$
which seems to be about $0.573$.
Added: As a sum it seems to be $\displaystyle \sum_{n=1}^{\infty} \dfrac{2}{8^n(3n-1)} {3n \choose n}$ which is $\frac{3}{4}(3-\sqrt{5})$.
