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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.

louie mcconnell
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  • Interesting. Well, as a crude first shot I'd note that it has to happen at tosses divisible by 3, and the probability that it happens at toss 3n (possibly not for the first time) is $\frac{1}{2^{3n}} \binom{3n}{n}$. Then I'd add these up. Granted, we are double counting (and triple and so on), but at least this gives a start. – lulu Jul 15 '15 at 21:36
  • @lulu: the sum of the first six terms is more than $1$ so that is indeed an upper bound. – Henry Jul 15 '15 at 22:06
  • @Henry Ha! So it is. Funny...I'd have thought it was such an unlikely event that I could ignore most cross terms. Apparently not. – lulu Jul 15 '15 at 23:02

2 Answers2

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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.

user2566092
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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 $\frac{3}{4}(3-\sqrt{5})$.

Henry
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