Your bullet points amount to saying that you're going to flip the coin until the number of heads exceeds the number of tails. Suppose that this happens on the $n$-th flip; then after $n-1$ flips you must have had equal numbers of heads and tails, so $n=2m+1$ for some $m$, you now have $m+1$ heads, and the ratio of heads to flips is $\frac{m+1}{2m+1}$. If $p_n$ is the probability of stopping after the $(2n+1)$-st flip, the expected ratio of heads to flips is $$\sum_{n\ge 0}\frac{p_n(n+1)}{2n+1}\;.$$ Thus, the first step is to determine the $p_n$.
Clearly $p_0=\frac12$: we stop after $1$ toss if and only if we get a head. If we stop after $2n+1$ tosses, where $n>0$, the last toss must be a head, half of the first $2n$ tosses must be heads, and for $k=1,2,\dots,2n$ the first $k$ tosses must not include more heads than tails. The problem of counting such sequences is well-known: these are Dyck words of length $2n$, and there are $C_n$ of them, where $$C_n=\frac1{n+1}\binom{2n}n$$ is the $n$-th Catalan number. Each of those $C_n$ sequences occurs with probability $\left(\frac12\right)^{2n}$, and each is followed by a head with probability $\frac12$, so $$p_n=C_n\left(\frac12\right)^{2n}\cdot\frac12\;,$$ and the expected ratio is $$\frac12\sum_{n\ge 0}C_n\left(\frac12\right)^{2n}\frac{n+1}{2n+1}=\frac12\sum_{n\ge 0}\frac1{4^n(2n+1)}\binom{2n}n\;.$$
Very conveniently, the Taylor series for $\arcsin x$ is $$\arcsin x=\sum_{n\ge 0}\frac1{4^n(2n+1)}\binom{2n}nx^{n+1}\;,$$ valid for $|x|\le 1$, so the expected ratio is $$\frac12\sum_{n\ge 0}\frac1{4^n(2n+1)}\binom{2n}n=\frac12\arcsin 1\approx 0.7854\;.$$
Added: I should emphasize that this calculation applies only to the stated strategy. As others have noted, that strategy is known not to be optimal, though it's quite a good one, especially for being so simple.
The trick is to write $X = X_1 + X_2 + \dots + X_k$, where $X_i$ is "the expected number of flips to get the $i$th head after getting the $(i-1)$th head".
Now it is not too hard to see that each $X_i$ is identically distributed and that the probability that $X_i = j$ (for $j \geq 1$) is equal to $p(1-p)^{j-1}$. (In other words, each $X_i$ is a geometric random variable). A simple calculation shows that $\mathbb{E}[X_i] = \frac{1}{p}$, so by linearity of expectation $\mathbb{E}[X] = \frac{k}{p}$.
Best Answer
Let
Xbe the random variable which counts the number of flips till we get aT(tails). Then: $$E[X] = \frac{1}{2}\cdot1 + \frac{1}{2}(1+E[X])$$ $$\implies E[X] = 2$$ The first term on LHS is if we get aTon a our first flip (with probability $\frac{1}{2}$). The second term is if we get aH(heads) on our first flip (with probability $\frac{1}{2}$), which means that we are effectively starting over with1extra flip.Now let
Ybe the random variable which counts the number of flips till we get aHTsequence. Then: $$E[Y] = \frac{1}{2}(1+E[Y])+\frac{1}{2}(1+E[X])$$ $$\implies E[Y] = 4$$ The first term on LHS is if we get aTon our first flip (with probability $\frac{1}{2}$), which means we are effectively starting over with1extra flip. The second term if we get aHon our first flip (with probability $\frac{1}{2}$), then all we want is aT, which means we want to count 1 + expected number of flips to get aT.