In a tournament of football (or whatever sport you prefer), let's say we have $2^n$ teams and for all $2^{n-1}(2^n-1)$ possible matches we know the probability of one team winning against the other. The tournament is a single-elimination tournament (therefore having $n$ rounds) and the teams are to be randomly allocated for the first round. How do we go about computing the probability of team $x$ winning the tournament?
[Math] Probability of team winning in a single-elimination tournament
probability
Related Solutions
Team $A$ wins the game if they win four matches times before $B$ wins four matches. Thus the first approach measures the probability of team A doing so when team $B$ wins zero, one, two, or three matches before $A$'s fourth victory.
$$P(A) = \left(\binom 30 p^3q^0+\binom 41 p^3q^1+\binom 52 p^3q^2+\binom 63 p^3q^3\right)p$$
Now imagine the teams kept playing for a full seven matches even after one of them wins the game. Team $A$ wins the game if they win at least four matches among those seven, since if they do so then team $B$ can win at most three matches before $A$ wins their fourth.
$$\begin{align}P(A) & = \binom 77 p^7q^0+\binom 76 p^6q^1+\binom 75 p^5q^2+\binom 74p^4q^3\end{align}$$
We can see these are equal by taking the terms of the first equation, and including the imagined games after victory. Then if you use the binomial theorem to expand...
$$\begin{align}P(A) &= \binom 30 p^4q^0(p+q)^3+\binom 41 p^4q^1(p+q)^2+\binom 52 p^4q^2(p+q)+\binom 63 p^4q^3 \\[1ex] &= \binom 30 p^4q^0(p+q)^3+\binom 41 p^4q^1(p+q)^2+\binom 52 p^5q^2+\left(\binom 52+\binom 63\right) p^4q^3\\[1ex] &\vdots \\[1ex] &= \binom 77 p^7q^0+\binom 76 p^6q^1+\binom 75 p^5q^2+\left(\binom 30+\binom 41+\binom 52+\binom 63\right) p^4q^3 \\[1ex]& = \binom 77 p^7q^0+\binom 76 p^6q^1+\binom 75 p^5q^2+\binom 74p^4q^3\end{align}$$
For the first round, since your team has a $0.9$ probability of winning, they have $0.1$ probability of being eliminated.
From the second round onwards, your team must win the previous rounds in order to progress (or, pessimistically, have a chance to be eliminated in subsequent rounds)
Your team has a $0.9$ probability of winning the first round, and a $1-0.8=0.2$ probability of losing the second. Hence the probability of your team being eliminated in the second round is $0.9 \times 0.2 = 0.18$. Similarly, they have a $0.9\times 0.8=0.72$ probability of progressing.
Using this logic, the remaining probabilities are (check them!):
Eliminated in the third round: $0.9\times 0.8\times 0.3=0.216$
Eliminated in the finals: $0.9\times 0.8\times 0.7\times 0.4=0.2016$
Winning the tournament: $0.9\times 0.8\times 0.7\times 0.6=0.3024$
Best Answer
For each team $i$ the probability of getting into the second round is $$P_2(i)=\frac{1}{2^n-1}\sum_{j\ne i}P_v(i,j)$$ where $P_v(i,j)$ is the probability that $i$ will defeat $j$.
Similarly, the probability that $i$ will get into the round $r+1$ is $$P_{r+1}(i)=P_r(i)\frac{1}{2^{n-r+1}-1}\sum_{j\ne i}P_v(i,j)P_r(j)$$
So, all you need to do is compute $P_n(i)$ recursively.