Find the probability of winning a football tournament 
There is a football tournament between teams A,B and C. Teams who have lost two games in total will be eliminated. Team A and B play each other first and C progresses to the next round. The winner of each game will play the next game with the player who didn't play, and the loser will have the next bye until one team is eliminated. When one team is eliminated, the remaining two continue to play until one of them is eliminated, and the other finally wins, and the game ends. Let the probability of both sides winning each game be $1/2$.


What is the probability that C will eventually win the tournament?

We need to look at all scenarios that C will win:

*

*A beats B, C beats A, C beats B, C beats A : Probability = $(1/2)^4=1/16$.

*B beats A, C beats B, C beats A, C beats B : Probability = $(1/2)^4 = 1/16$.

*A beats B, C beats A, B beats C, B beats A, C beats B: Probability = $(1/2)^5 = 1/32$

*B beats A, C beats B, A beats C, A beats B, C beats A: Probability = $(1/2)^5 = 1/32$

*A beats B, A beats C, A beats B, C beats A, C beats A Probability = $(1/2)^5 = 1/32$
6.B beats A, B beats C, B beats A, C beats B, C beats B Probability = $(1/2)^5 = 1/32$
So $P(C~wins)$ = 1/4. However the solution is given by:
$$ P(C~wins) = 1 - 2((1/2)^4 + 7(1/2)^5) = 7/16.$$
I'm not sure how this is calculated.
 A: Addendum added, to provide alternative approach, that has no shortcut.

Shortcut.
Since you are only interested in C's win factor, and since A and B play each other in game 1, without loss of generality, A wins game 1, and B loses game 1.
This implies that A and C are about to play each other, in game 2, with neither of them having a loss.
This implies that at this point A and C have the same win factor.  So, if you calculate B's win factor as $p(B)$, then C's win factor must be
$[1 - p(B)]/2.$
Further, B must win out, to win the tournament.
So, at this point, just before game 2, the only thing that is relevant is determining $p(B).$
Therefore, without loss of generality, A wins game 2 and plays B in game 3.
Since B has to win out, B will have to win game 3.
The only question is how many more games will he have to play.
So, B has to beat A in game 3.  Then, B has to win out, eliminating C in game 4 and A in game 5.
Therefore, under the assumption that B lost game 1,
$p(B)$ must be $(1/8)$, because B must win games 3, 4, and 5.  And if does so, he wins the tournament.
Therefore C's win factor must be $[1 - (1/8)]/2$.

Addendum
Alternative approach that uses no shortcuts.
Without loss of generality A beats B in game 1.
Either C wins game 2, against A or he doesn't.
Separate cases will be examined, and within these cases, sub-cases.

$\underline{\text{Case 1: C loses to A in game 2}}$
Probability of this scenario is $(1/2)$. 
Then, either A beats B in game 3 or he doesn't.
$\underline{\text{Case 1.1: A beats B in game 3}}$ 
Probability of this scenario is $(1/2) \times (1/2) = (1/4)$. 
B is now toast, so C must beat A in both games 4 and 5. 
Overall probability of this scenario is $(1/4) \times (1/4) = (1/16).$
$\underline{\text{Case 1.2: B beats A in game 3}}$ 
Probability of this scenario is $(1/2) \times (1/2) = (1/4)$. 
At this point, each of A,B,C have 1 loss.  C must defeat B in game 4 and defeat A in game 5.
Overall probability of this scenario is $(1/4) \times (1/4) = (1/16).$

$\underline{\text{Case 2: C beats A in game 2}}$
Probability of this scenario is $(1/2)$. 
Then, either C beats B in game 3 or he doesn't.
$\underline{\text{Case 2.1: C beats B in game 3}}$ 
Probability of this scenario is $(1/2) \times (1/2) = (1/4)$. 
B is now toast, C has no losses, and A has one loss.  At this point, C is a $3$ to $1$ favorite. Assuming that C then wins, the overall probability of this scenario is $(1/4) \times (3/4) = (3/16).$
$\underline{\text{Case 2.2: C loses to B in game 3}}$ 
Probability of this scenario is $(1/2) \times (1/2) = (1/4)$. 
At this point, all $3$ players have one loss, and A will play B in game 4.  In this scenario, C wins the tournament if and only he wins game 5.
Overall probability of this scenario is $(1/4) \times (1/2) = (1/8).$

So, in Cases 1.1, 1.2, 2.1, and 2.2 respectively, C's probabilities are $(1/16), (1/16), (3/16), (1/8).$
Therefore, C's overall probability of winning is $(7/16).$
A: The answer given is $1$ minus the probability that $C$ loses.
You can calculate the winning lines for $C$ by using a set of rules to emulate the games.
Consider a string built from $A,B,C$. The string must start with an $A$ or a $B$. No letter can appear more than twice. No letter can appear consecutively, unless some other letter has appeared twice before. If both $A$ and $B$ have appeared twice, the string is terminated.
$C$ is only allowed to appear at most once. The appearance of a letter means that team lost in the previous round.
The valid strings then are:

*

*ABAB

*BABA

*ACABB

*ACBAB

*ABCAB

*ABCBA

*ABACB

The last five are repeated with $B$ swapped with $A$.
This gives $C$ $14$ winning positions, and a probability of winning of $\frac{14}{32}=\frac{7}{16}$.
