Let $X$ be the # of tosses till first H. Then, $(X = 1) \cap (X > 1) = \phi$. Using Total Expectation Theorem \begin{align} E[X] &= P(X = 1)E[X|X = 1] + P(X > 1)E[X|X > 1] \newline &= 0.5 * 1 + 0.5 E[X] \newline \Rightarrow E[X] &= 2 \end{align} $P(X = 1) = 0.5$ because then we get the head in the first toss itself. Since $P(X = 1) + P(X > 1) = 1$, we have $P(X > 1) = 0.5$. $E[X] = E[X|X > 1]$ because the tosses are independent and thus memoryless.