Football game, penalty kicks, maximizing winning chances Two teams are shooting penalty kicks. Firstly, both teams are performing a series of 5 penatly kicks (first teams A then B then A B A B A B A B). If the score of the series is decided we stop the series. If the series ends in a draw, then the teams shoot single penatly kicks until one team score and the other doesn't. First shoots a player from team $A$ and then from team $B$.
In a series of 5 penalty kicks, the first 6 shots ended in a goal, that is $|A_G \ B_G \ | \ A_G \ B_G \ | \ A_G \ B_G | -- \ | -- |$
The probability of scoring = $p=0,8$ and it is the same for every player. Which is more probable - finishing shooting penalty kicks in the series of 5 shots or starting the series of single shots? What is the expected number of penalty kicks until the end of the football game?
I think that the first one is more probable.
In order for the game to end during the five shots we need to have following scenarios:
(the first three shots are as above) $A_G, \ B_G$ - A, B scored a goal, $A_N, B_N$- A, B didn't score, $-$ - nothing happended
$  A_G \ B_N \ | \ -- $
$ A_N \ B_G \ | \ -- $
$ \ A_G \ B_G \ | \ A_G \ B_N $ 
$ \ A_G \ B_G \ | A_N \ B_G $
$ A_N \ B_N \ | \ A_G \ B_N $  
$ \ A_N \ B_N \ | \ A_N \ B_G $
So if $E$ = the game finished during the five shot series then $P(W) = p (1-p) + (1-p)p + p^2 p (1-p) \cdot 2 + (1-p)^2 p (1-p) \cdot 2 = 0,5376$.
While in the second case (single shots are needed) we must have $A_G B_G A_G B_G $ or $A_G B_G A_N B_N$ or $A_N B_N A_G B_G$ or $A_N B_N A_N B_N$
And here $P(W^c) = 0,4624$
Is that correct? 
Could you help me with the rest?
 A: Getting the probability of finishing the game during the 5 penalty kicks is easier than getting the expected number of kicks to end the game:
Find the probability that the penalty kicks will end in a draw. That means the number of goals will be the same each side, and all 5 kicks will
be used by each side. 
\begin{align*}
  \mathbb{P}\left(\text{Ends in draw}\right) &=  \sum_{k=0}^5\binom{5}{k}^2\cdot \left(\frac{4}{5}\right)^{2k} \cdot \left(\frac{1}{5}\right)^{10-2k} \\
  &= \frac{3122577}{9765625}\\
  \implies \mathbb{P}\left(\text{Game ends in penalty kicks}\right) &= 1-\mathbb{P}\left(\text{Ends in draw}\right) = 0.6802481152
\end{align*}
Let $E$ be the number of kicks (after a draw) till one of the teams scores and other does not, then if $p=0.8$ and $q=0.2$,
\begin{align*}
  E &= \left(p^2+q^2\right)\left(E+2\right)+2pq\left(2\right) \\
  E &= \frac{25}{4}
\end{align*}
Hoping my understanding about the problem is correct, I have only an answer from computer enumeration, where the expected number of 
goals during a draw will be $10+E$:
\begin{align*}
  \mathbb{E}\left(\text{total no. of kicks from both teams}\right) &= \frac{18187961}{1562500} = 11.64029504
\end{align*}
