Probability of two teams meet up in FIFA tounament **Second round is single elimination round.
**Tournament is from 16 teams elimination follower by quarter finals,semi-finals and a final.
**The losing semi-finalist contest a third place match.
In world cup 2018,if 16 teams in the second round are evenly matched, and what is the probability that Brazil n Russia will meet in a match during tournament.
Found a lot of example that is don't have the third place match meet.
 A: Simpler version: I'm assuming that for this question, the tournament starts from the second phase, and that the position of one team in the knockout draw is independent on the position of other team (other than two teams cannot be at the same slot in the draw, of course). 
Since two teams meet exactly at most once, we know that 
$$
\mathcal{P}(\text{$A$ and $B$ meet}) = \mathbb{E}[\text{number of matches between $A$ and $B$}],
$$ 
where $\mathbb{E}$ denotes the expectation value. We note that probability that A and B meet is the same for each A and B, if the knockout draw is completely random. Let $p$ be that probability. 
Let the teams be numbered from $1$ to $16$.
\begin{align*}
\sum_{A=1}^{16} \sum_{B=A+1}^{16} \mathcal{P}(\text{$A$ and $B$ meet}) &= \sum_{A=1}^{16} \sum_{B=A+1}^{16}  \mathbb{E}[\text{number of matches between $A$ and $B$}] \\ &= \mathbb{E}[\text{number of matches in the knockout phase}]
\end{align*}
The LHS is $p$ times the number of pairs of teams $\binom{16}{2}$, and the RHS is the number of matches in the knockout phase, which is $16$. Thus
\begin{align*}
\binom{16}{2}p &= 16 \\
p &= 16 \big/ \binom{16}{2} = \frac{2}{15}.
\end{align*}

Complicated version:
In a real FIFA world cup, the probability distribution of the places in the knockout draw depend on which teams are from the same preliminary group: the teams are placed such that teams from the same preliminary group can only play in the final or in the third place match in the knockout phase. the probability that A and B meet in the whole tournament is:
$$
\mathcal{P}(\text{$A$ and $B$ meet in the tournament}) = \mathcal{P}(\text{$A$ and $B$ are in the same preliminary group}) + \mathcal{P}(\text{$A$ and $B$ are not in the same preliminary group}) \times
 \mathcal{P}(\text{$A$ and $B$ proceed to the second round} | \text{$A$ and $B$ are not in the same preliminary group}) \times
 \mathcal{P}(\text{$A$ and $B$ are play against each other in the knockout phase} | \text{($A$ and $B$ are not in the same preliminary group) and ($A$ and $B$ proceed to the second round)} ).
$$
Here, $\mathcal{P}(\text{$A$ and $B$ are in the same preliminary group})$ depends on the positions of the teams in the FIFA ranking, due to the way the groups are determined. Also, 
$$
\mathcal{P}(\text{$A$ and $B$ are play against each other in the knockout phase} | \text{($A$ and $B$ are not in the same preliminary group) and ($A$ and $B$ proceed to the second round)} )
$$
is more difficult than the previous one: Let's say that $p_1$ is the probability that teams from the same preliminary group meet and $p_2$ is the probability that teams from different preliminary groups meet. 
Again, we have
$$
\sum_{A=1}^{16} \sum_{B=A+1}^{16} \mathcal{P}(\text{$A$ and $B$ meet}) = \mathbb{E}[\text{number of matches in the knockout phase}] = 16.
$$
Howere, here $ \mathcal{P}(\text{$A$ and $B$ meet})$ is not constant. Rather, we have $8$ pairs of teams from the same group, and $\binom{16}{2}-8 = 112$ pairs of teams from different groups. Thus the LHS is $8p_1 + 112p_2$.
In a FIFA tournament, two teams from the same group meet in the group phase if and only if they both proceed to the final or they both proceed to the 3rd place match. If the teams are evenly matched, probability that a team proceeds to the final is $1/8$ because $1$ out of $8$ teams from that side of the draw proceed to the final. The same for 3rd place match. Thus, 
$$
p_1 = (1/8)^2 + (1/8)^2 = 1/32.
$$
We get
$$8 \times (1/32) + 112 p_2 = 16,$$
which gives $p_2= 63/448$. 
Finally, we get
$$
\mathcal{P}(\text{$A$ and $B$ meet in the tournament}) = \mathcal{P}(\text{$A$ and $B$ are in the same preliminary group}) + \mathcal{P}(\text{$A$ and $B$ are not in the same preliminary group}) \times
 \mathcal{P}(\text{$A$ and $B$ proceed to the second round} | \text{$A$ and $B$ are not in the same preliminary group}) \times
 \frac{63}{448}.
$$
With the assumption that Brazil and Russia always proceed to the next round, we get
$$
\mathcal{P}(\text{BRA and RUS meet in the tournament}) =  P + (1-P) \times 1\times
 \frac{63}{448},
$$
 where
$$P=\mathcal{P}(\text{BRA and RUS are in the same preliminary group}),$$
which, as I explained, is not trivial to know.

With answer this long, I almost certainly made mistakes in the calculations and numbers. Please comment if you find any.
