Two players alternate flipping a coin until the result is head. How to derive that the probability for the first player to win is $2/3$? 
Two players, $A$ and $B$, alternately and independently flip a coin and
  the first player to obtain a head wins. Player $A$ flips first. What is
  the probability that $A$ wins?
Official answer: $2/3$, but I cannot arrive at it.

Thought process: Find the probability that a head comes on the $n$th trial. That's easy to do, it's just a Geometric random variable. Then find the probability that $n$th turn is player's A turn. Finally, multiply both probabilities.
When I came up with each probability, both of them depended on the amount of trials $n$, so my answer was a non-constant function of $n$.
However, what I find quite fantastic is that the answer is a constant, so the amount of tries until a head comes doesn't seem to matter.
 A: 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.
A: Let $p$ denote the probability that $A$ wins. There is a $.5$ probability $A$ wins on the first flip. If he flips tails, then in order to win $B$ must then flip tails. This situation occurs with probability $.25$ ($1/2 * 1/2$ for both tails) and in this case we are back to where we started. So
$$p = .5+ .25p$$
which yields $p=2/3$.
A: 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.$$p=\frac12+\frac12(1-p)\implies p=\frac23$$
A: 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 $$P= {2\over 3}$$
A: P(A wins) can be thought of a sum of infinite number of cases, starting with  >A wins the first round(0.5)

A loses first round, B loses second round, A wins the third round (0.5)^3

And so on. This forms an infinite gp with first term 0.5 and common product 0.5^2

P(A winning) = 0.5/(1-0.25)

