The probability of meeting in a tournament, version 2: fixed ranking This is a variation on this question
$2^n, n\in\mathbf N$ tennis players compete in a tournament. In the first round, they partition into a set of $2^{n-1}$ disjoint pairs. The two players in each pair compete against each other. The $2^{n-1}$ winners form a set of $2^{n-2}$ disjoint pairs and compete in the next round, and so on. This competition lasts for $n$ rounds. The partition in each round is uniformly random. The players are strictly ranked and the higher ranked player always beats the lower one. Given a particular ranking, for two chosen players of $i$'th and $j$'th ranking, what is the probability that they will compete against each other in a pair? 
 A: Let $i\lt j$, and lower numbers are better players.
They meet in round $k$ if $j$ is the top of his branch of $2^{k-1}$ players, $i$ is top of the neighbouring branch of $2^{k-1}$ players, so they get paired up .  The odds of this happening is
$${2^n-j\choose 2^{k-1}-1}{2^n-i-2^{k-1}\choose 2^{k-1}-1}\over{ {2^n-1\choose 2^{k-1}-1}{2^n-2^{k-1}\choose 2^{k-1}}}$$
Then sum that from $k=1$ to $n$.  
As a sanity check, here are the probabilities for a tournament with eight players.
$$\frac{1}{105}\left[\begin{array}{cccccccc}
     0 &  105  &  70  &  47  &  33  &  25  &  20  &  15\\
   105 &    0  &  35  &  31  &  27  &  23  &  19  &  15\\
    70 &   35  &   0  &  27  &  24  &  21  &  18  &  15\\
    47 &   31  &  27  &   0  &  21  &  19  &  17  &  15\\
    33 &   27  &  24  &  21  &   0  &  17  &  16  &  15\\
    25 &   23  &  21  &  19  &  17  &   0  &  15  &  15\\
    20 &   19  &  18  &  17  &  16  &  15  &   0  &  15\\
    15 &   15  &  15  &  15  &  15  &  15  &  15  &   0
\end{array}\right]$$
A: Interesting problem. Just for a warm up first: 
Let $n = 4$, then the players are ranked from lowest to highest as $\{x_{1}, x_{2}, x_{3}, x_{4}\}$. For the 1st round, there will be 2 disjoint pairs. Pattern:
$$ p_{1}, p_{2} $$
with $p_{i}$ is pair $i$. So the number of disjoint pairs are:
$$ \frac{\binom{4}{2} \binom{2}{2}}{2} = 3  $$
Division by 2 because for example: $(x_{1}, x_{2}), (x_{3},x_{4})$ is same as $(x_{3}, x_{4}), (x_{1},x_{2})$. These are: 
$$(x_{1}, x_{2}), (x_{3}, x_{4}) \rightarrow (x_{2}, x_{4})$$
$$(x_{1}, x_{3}), (x_{2}, x_{4}) \rightarrow (x_{3}, x_{4})$$
$$(x_{1}, x_{4}), (x_{2}, x_{3}) \rightarrow (x_{3}, x_{4})$$
Righthand side is the final round.
Now, what is the probability that $(x_{1}, x_{2})$ will appear in the draw? it is $1/3$, probability of $(x_{1},x_{i>1})$ will always be $1/3$. 
Probability $(x_{2}, x_{3})$ is 1/3. Probability $(x_{2},x_{4})$ is 2/3. 
Probability $(x_{3}, x_{4})$ is 1..!
It seems to be quite difficult to answer your problem.
