Probability of team winning in a single-elimination tournament 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?
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.
@sds's answer is not correct. Use an all equal skilled 4 team tournament for example. Any team has a win chance of 0.25. But his formula gives 0.75.

Should be:

$$
P_2(i) = \sum_{j!=i}\frac{1}{2^{n}-1} P_v(i,j)
$$

$$
P_{r+1}(i) = \sum_{j!=i}\frac{2^{r-1}}{2^{n}-1} P_r(i,!j)P_v(i,j)P_r(j)
$$

$P_r(i,!j)$ is the probability that $i$ reached round $r$ without meeting $j$ in the previous matches.

To calculate $P_r(i,!j)$, use recursion again:

$$
P_2(i,!j) = \frac{2^{n}-2}{2^{n}-1}\sum_{k!=i,k!=j}\frac{1}{2^{n}-2} P_v(i,k)
$$

$$
P_{r+1}(i,!j) = \frac{2^{n}-2^{r}}{2^{n}-1}\sum_{k!=i, k!=j}\frac{2^{r-1}}{2^{n}-2} P_r(i,!j)P_v(i,k)P_r(k)
$$