Two players alternate flipping a coin until the result is head. How to derive that the probability for the first player to win is $2/3$?

Let $p$ be the probability that the first player wins. A wins half the time when he tosses a head. The other half the time, B becomes the first tosser, and A wins if the first player doesn't win.$$p=\frac12+\frac12(1-p)\implies p=\frac23$$

So $A$ wins if head comes first time in odd number of tosses. Say $P_k$ is probability that head comes first time in $k$-th toss then $P_k = q^{k-1}p$ and

\begin{eqnarray} P &=& P_1+P_3+P_5+... \\ \\ &=&p+q^2p+q^4p+q^6p+....\\ \\ &=& {p\over 1-q^2}\\ \\& =& {1\over 1+q} \end{eqnarray}

where $p$ is probability that head comes in one toss and $q=1-p$.

So if the coin is fair, then $p=1/2=q$, so $$P= {2\over 3}$$

Here's another approach.

Consider a round of this game as A flipping and then B flipping. Of course the game may end in the middle of a round if A flips heads.

In the first round A has twice as great a chance to win as B. (A wins half the time by flipping heads); and B wins with tails-heads which occurs one quarter of the time).

But this two-to-one ratio of A winning to B winning holds in every round. So overall, A is twice as likely to win as B.

This gives A a $\frac23$ chance of winning to $\frac13$ for B.