A question about the expected number of games being played In this question this game is presented:

Suppose in an independent game which has 2 players, player 1 and
  player 2, the probability of player 1 to win each game is r. To be the
  overall winner of the game, one of the players needs to win 2 more
  games than the other.

Now, instead of asking for the probability of player 1 to win, I want to know what is the expected number played to end the game?
 A: Let $a$ be the expected number of additional games that need to be played if the two players are tied in wins.  In particular, $a$ is our required number, since the players start off tied. 
Let $b$ be the expected number of additional games if Player $1$ is leading by $1$. And let $c$ be expected number of additional games if Player $1$ is trailing by $1$. We have the equations
$$\begin{align}a&=1+rb+(1-r)c,\\
b&=1+(1-r)a, \\
c&=1+ra.\end{align}$$ 
Use the above system of linear equations to find $a$. 
Remarks: $1.$ To justify the first equation, note that for sure we will be playing one more  game. That's the $1$ in the equation. And we will not be finished. With probability $r$ our expected number of games beyond that  will be   $b$, and with probability $1-r$ our expected number of games beyond that will be $c$. That yields the first equation.  It may be prettier and clearer to write the  equation as $a=r(1+b)+(1-r)(1+c)$. 
The justifications for the other two equations are similar. Again, one might like to write the second equation as $b=r(1)+(1-r)(a+1)$, and do a similar rewrite of the third equation. 
$2.$ As pointed out by Marc van Leeuwen, the argument tacitly assumes that the expectations exist. To show that they do is not difficult. Whatever $r$ is, the probability of two opposite results in a row is $2r(1-r)$, which is $\le  1/2$. So the probability that a game length $2n$ or more is $\le (1/2)^n$, and therefore the expected length is finite.  
$3.$ We used a strategy broadly similar to the one used in the related question about the probability that Player $1$ wins. This kind of strategy tends to work more widely for expectations than for probabilities, because of the linearity of expectation.  
