Expected number of coin tosses untill number of head exceeds number of tails? This problem is basically from MIT OCW probability section.
A fair coin is flipped until the number of heads exceeds the number of tails. What is
the expected number of flips?
I tried this question and landed with 25/12 as the expected number of flips required. I want to check if this is correct.
 A: Replying to @Tavish, there is no way to get a sequence of even tosses where heads just became greater than tails (if heads are greater the difference is greater than two, so they where greater even before the last toss). Conversely for $2n + 1$ the sequences such that heads just became greater than head are such that the last toss is a head, before there were an equal number of heads and tails and tails were never strictly less than heads (that is $C_n$). So the probability that $X = 2n + 1$ should be $\frac{C_n}{2^{2n+1}}$. Hence
$$\mathbb{E}(X) = \sum_{n = 0}^\infty \frac{1}{n+1}  \binom{2n}{n} \frac{2n + 1}{2^{2n + 1}} = \sum_{n = 0}^\infty \binom{2n+1}{n+1} 2^{-(2n +1)}$$
Either I made a mistake somewhere or simply the expected value diverges, that anyway seems surprisingly
Edit: for completeness, the series diverges because the general term is asymptotically equivalent to $\binom{2n}{n} 2^{-2n}$ and by stirling approximation this is again asimptotically equivalent to $(\sqrt{\pi n})^{-1}$
