Probability of having always flipped more $H$ than $T$ in an infinite coin flip sequence A biased coin has probability $p \in [0,1]$ of landing heads ($H$) and hence probability $1-p$ of landing tails ($T$). We will flip this coin infinitely many times, obtaining a sequence $(x_i)_{i=1}^\infty$ of flips, where $x_i \in \{H, T\}$ for $i \in \mathbb{N}$. What is the probability that for all $N \in \mathbb{N}$, $(x_i)_{i=1}^N$ contains strictly more $H$s than $T$s?
I've tried using the Ballot Theorem and taking limits, but nothing has seemed to work so far. 
 A: The probability we're looking for is that the first flip is heads, and there is never a tie after that (hence never a first tie). This probability is 
$$A = p\left(1 - \sum_{n=1}^{+\infty} a_n\right), $$
where (conditionally on getting heads on the first toss) $a_n$ is the probability that heads is always ahead until the first tie occurs at $2n$ flips. 
We have $a_n = N_n p^{n-1}(1-p)^n$, where $N_n$ is an integer that can be described as follows. $N_n$ is the number of paths in the plane, in which a step consists of moving right by one unit or up by one unit, that go from $(0,1)$ to $(n-1,n)$ without ever touching the diagonal. The reflection method described here shows that this is the Catalan number $C_{n-1}$. Therefore, letting $c(x) = \sum_{n=0}^{+\infty} C_n x^n$, we have
$$A = p - p\sum_{n=1}^{+\infty} C_{n-1}p^{n-1}(1-p)^n = p-p(1-p)c[p(1-p)].$$
Using the formula $xc(x) = \frac{1}{2} - \frac{1}{2}\sqrt{1-4x}$ from the same Wikipedia source gives us
$$A = p - \frac{1}{2}\left(1 - \sqrt{1 - 4p + 4p^2}\right) = \frac{1}{2}(2p-1 + |2p-1|).$$
Put more simply, $A = 2p - 1$ if $p > 1/2$, and $A = 0$ otherwise.
A: An alternate method: A gambler wins $1$ unit with probability $p$ and loses 1 unit otherwise. Let $q=1-p.$ The gambler is playing against an infinitely rich casino. The question as asked corresponds to a gambler starting with 0 capital and who wins the first bet and then never goes broke (never returns to 0 capital.)
Assume the gambler starts with 0 capital and wins the first bet so his capital is now 1. The probability of the gambler ultimately going broke from this state is $q/p,$ if $p>0.5$ and 1 if $p\le 0.5$ by solving difference equations. (See Feller Vol. 1, Ch. XIV or this). Therefore the desired probability of winning the first bet and then never going broke is $$p(1-q/p)=2p-1,\text{ if }p>0.5 $$
and 0 if $p\le 0.5$
