$3$ Bracket Knockout Tournament Probability $12$ people in $3$ brackets ($A,B$, and $C$) compete against each other in a knockout tournament. The final round has $4$ contestants, the $3$ winners from brackets $A,B$ and $C$, plus $1$ of the dropouts from brackets $A$ and $B$ brought back into the game at random. What is the probability of winning for a player in group $A$ or $B$ (assuming equal chance to advance to the next stage of the tournament and equal chance to be brought back in).
The way I though of this was that there are two ways that a person in Group $A$ or $B$ can win. They can win the tournament by simply winning all their matches, or by losing, being brought back and winning the final round. Adding these two probabilities should give the answer.
I calculated their chance of winning by simply winning all their matches as $\frac1{12}$, since each contestant has equal chance to advance and there are $12$.
For the other scenario, I said that there is a $\frac12$ chance they lose their first match (since all participants have equal chance of advancing) , multiplied by a $\frac16$ chance they get voted back in, (since of the original $8$ people from $A$ and $B$ only $2$ will advance and there will be $6$ dropouts and all dropouts have equal probability of being brought back). Then multiplied by $\frac14$ since there are $4$ participants in the final round and each has equal chance of winning, So $\frac12 \cdot \frac16 \cdot \frac14$.
Overall I got $\frac1{12} + \frac12 \cdot \frac16 \cdot \frac14=\frac5{48}$, which is not the answer.
The options for the answer are:

*

*$\frac3{16}$


*$\frac3{32}$


*$\frac1{16}$


*$\frac18$.
Thanks
 A: *

*A random player's chance of winning four games in a row is $\left(\frac{1}{2}\right)^4 = \frac{1}{16} \neq \frac{1}{12}$.

*You haven't counted the players who win their first match and lose their second (and are awarded the final random spot).


Its hard to count all the possibilities, but here we go:
Win all games: $\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2} = \frac{1}{16}$
Lose, Lottery Spot, Win, Win: $\frac{1}{2}\cdot \frac{1}{6}\cdot \frac{1}{2}\cdot \frac{1}{2} = \frac{1}{48}$
Win, Lose, Lottery Spot, Win, Win: $\frac{1}{2}\cdot\frac{1}{2}\cdot \frac{1}{6}\cdot \frac{1}{2}\cdot \frac{1}{2} = \frac{1}{96}$
Add these up to get $\frac{9}{96} = \frac{3}{32}$.

Alternatively,
$$\underbrace{\frac{3}{8}}_{\text{get to the final from A/B}} \cdot \underbrace{\frac{1}{4}}_{\text{win the final}} = \frac{3}{32}$$

Edit (to address OP comment):
Checking our answer against the Law of Total Probability:
The lottery mechanism makes it more difficult to check each player's chance of winning. The C players must win all their games. Each of the four C players has a one-sixteenth chance. The A and B players have better chances. Eight players must first earn one of three finals spots and then win the finals:
$$\underbrace{4\cdot\frac{1}{16}}_{C} + \underbrace{8 \cdot \frac{3}{32}}_{A/B} = \frac{1}{4} + \frac{3}{4} = 1$$
A: Two ways to advance to final round:
$$P(\text{win bracket})=\frac{1}{4}\\P(\text{wild card}) = \frac{3}{4}\times\frac{1}{6}=\frac{1}{8}$$
Total probability to advance:
$$P(\text{advance}) = \frac{1}{4}+\frac{1}{8}=\frac{3}{8}$$
Total probability to win:
$$P(\text{win tournament}) = \frac{3}{8}\times\frac{1}{4}=\frac{3}{32}$$
