It's a KnockOut

In order to reach the final $p_4$ needs to face two of $p_5,p_6,p_7$ and $p_8$ in the first two rounds. His chance of facing one of them in the first round is $\frac{4}{7}$ .

Next, his chance of facing one more in the second round depends on the other matches of the first round. If each one of $p_1,p_2$ and $p_3$ faces one of $p_5,p_6$ and $p_7$ (supposing $p_4$ faced $p_8$ in the first round without loss of generality) then he can't reach the final. So two of $p_5,p_6$ and $p_7$ must have faced each other in the first round and the chance of this happenning is $\frac{9}{15}=\frac{3}{5}$ .

Finally once $p_4$ and one of $p_5,p_6$ and $p_7$ are on the second round the chance of them facing each other is $\frac{1}{3}$ .

Thus the chance of $p_4$ reaching the final is $\frac{4}{7}\cdot\frac{3}{5}\cdot\frac{1}{3}=\frac{4}{35}=0.114$

We can divide the tournament into two 4-man "conferences". $p_4$ will win his conference and reach the final if and only if the three better players are all in the other conference. We can place them in order. For $p_1$ , there are three spots in $p_4$ 's conference and four in the other, so he has a $\frac{4}{7}$ chance of being in the other. This takes up one spot, so $p_2$ has a $\frac{3}{6}$ chance. This leaves $\frac{2}{5}$ for $p_3$ . Multiplying all of these together gives $\frac{4}{35}$ .

Lets first find the total number of cases.

Apply the simple concept of distribution to divide the players in sets of two for round 1. This would give $\frac{8!}{2!^4.4!}$ number of ways for round one. Now for round 2, we would be having a total of 4 players left, again distributing them into 2 sets of two players $\frac{4!}{2!^2.2!}$ . So total number of cases are $\boxed {\frac{8!}{2!^4.4!}.\frac{4!}{2!^2.2!}}$

Now for the favourable cases. p4 must play match with either of p5,p6,p7 or p8. Now we have 3 players less than p4 and 3 players more than p4 in subscript. It can't be that the lesser players would play with higher players as then p4 won't win the semi final, hence 2 of the higher players must play a match among them. So we have to "select one from the 4 higher players to match against p4" * "select any 2 from the remaining 3 players to match among them"*"Distribute the 3 lower players and 1 higher player". In the semis, we have only one way with which p4 can reach the finals(i.e. it must play with the higher player remaining). Framing the above line Mathematically, we get $\boxed{4C_{1}.3C_{2}.\frac{4!}{2!^2.2!}.1}$

Divide them for the probability

The number of ways in which $p_1, p_2, ..., p_8$ can be paired is
= $\frac{1}{4!} {8 \choose 2} {6 \choose 2} {4 \choose 2} {2 \choose 2}$
= 105 ways

Now, at least two players certainly reach the second round in between $p_1, p_2,$ and $p_3$ .

$p_4$ can reach in final if in the first round, this happens $\rightarrow$

• Exactly two players play against each other in between $p_1, p_2, p_3$
• the remaining player plays against one of the players from $p_5, p_6, p_7, p_8$ .
• $p_4$ plays against one of the remaining three from $p_5, p_6, p_7, p_8.$

This can be possible in
= ${3 \choose 2} {4 \choose 1} {3 \choose 1}$
= 36 ways

$\Rightarrow$ Probability that $p_4$ and exactly one of $p_5, ... p_8$ reach second round is
= $\frac{36}{105}$ = $\frac{12}{35}$

If $p_1, p_i, p_4$ and $p_j$ where $i$ = 2 or 3 and $j$ = 5 or 6 or 7 reach the second round, then they can be paired in
= $\frac{1}{2!} {4 \choose 2}{2 \choose 2}$ = 3 ways

But $p_4$ will reach the final if $p_1$ plays against $p_i$ and $p_4$ plays against $p_j$ .
Hence, the probability that $p_4$ reaches the final round from the second round is $\frac{1}{3}$ .

Therefore, probability that $p_4$ reaches the final round is $\frac{12}{35} \times \frac{1}{3}$ = $\frac{4}{35}$

Let us set this up like a tourney. It is a row of 8 people and places 1, 2 ; 3, 4 ; 5, 6 ; 7, 8 face each other. If you split it into two groups by the half, you know in the half of $p_4$ , there can only be $p_5, p_6, p_7,$ and $p_8$ .

Therefore the numerator is : $8 \cdot 4 \cdot 3 \cdot 2 \cdot 4!$

because there are 8 ways to put the $p_4$ , $4\cdot 3 \cdot 2$ ways to put down $p_5 , p_6, p_7, p_8$ and $4!$ to be the rest.

Therefore the probability is $\frac{8 \cdot 4 \cdot 3 \cdot 2 \cdot 4!}{8!} \approx 0.114$