TITLE: Two players alternate flipping a coin until the result is head. How to derive that the probability for the first player to win is $2/3$?
QUESTION [5 upvotes]: 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.

REPLY [8 votes]: 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.