How to find expectation of geometric distribution 2? Alec and Bill take alternate turns at kicking a football at a goal, and their probabilities of scoring a goal on each kick are $p_1$ and $p_2$ respectively, independently of previous outcomes. The first person to score allows the other person one more kick. If the other then scores, the game is drawn. If the other then misses, the first has won the game. Alec begins a game.
Given that $p_1 = p_2 = 1/3$, find the expected total number of kicks in the game.
 A: Let the expected total number of kicks be $c$. It is reasonably clear that $c$ exists. We condition on the results of the first two kicks.
If Alec hits on the first kick (probability $\frac{1}{3}$), then the number of kicks, and therefore the expectation is $2$, since Bill gets a turn.
If Alec misses and Bill hits (probability $\frac{2}{3}\cdot \frac{1}{3}$), then the total number of kicks, and therefore the expectation, is $3$.
If they both miss on their first kicks, then $2$ kicks have been wasted, and in essence the game begins again. Thus the conditional expectation of the number of kicks, given they both missed, is $2+c$. 
It follows that
$$c=\frac{1}{3}\cdot 2+\frac{2}{3}\cdot\frac{1}{3}\cdot 3+\frac{2}{3}\cdot\frac{2}{3}\cdot(2+c).$$
Solve this linear equation for $c$.  
Another way: The game in essence consists of tossing a coin that has probability $\frac{1}{3}$ of landing heads until we get a head, and then (because of the "last chance" rule) tossing once more. 
By the standard result for the expectation of a geometric random variable, the number of tosses until the first head has expectation $\frac{1}{1/3}=3$. The "last chance" kick makes the expectation $4$.
Remark: The conditioning approach, with minor changes, can be used in situations where Alec's probability $p_1$ is different from Bill's probability $p_2$. 
Alternately, we can find explicit expressions for the probability that the number $X$ of kicks is equal to $n$. Then we can write down an infinite series for the expectation, and sum the series. 
