What is the probability that the player $P_4$ reaches the semi final $16$ players $P_1,P_2,....,P_{16}$ play a knockout tournament.
It is known that whenever the players $P_i$ and $P_j$ play, the player $P_i$ will win if $i<j$.
Assuming the players are paired at random in each round, what is the probability that the player $P_4$ reaches the semi final?

I know that $P_1$ will anyhow reach final and $P_{16}$ will not clear the first round.
I dont know how to solve further.
 A: $P_4$ will reach the semifinal if and only if none of $P_1,P_2,P_3$ is in the same quarter of the final draw as $P_4$.

It follows that the probability that $P_4$ reaches the semifinal is 
$${\small{\frac{\binom{12}{3}}{\binom{15}{3}}}}={\small{\frac{44}{91}}}$$
$$
\overline{
\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\;\;\;
}$$
The above solution assumes the distribution of the final draw, assuming random repairings, is the same as the distribution of a random draw, with no repairings. That seemed intuitively clear (by symmetry), but based on lulu's comment, I began to worry about whether the random repairings were perhaps relevant, hence I tried a different approach, as shown below, which explicitly assumes random repairings. 

With some corrections applied, based as on comments by Henry and Mees de Vries, here is a revised version of that approach . . .

The probability that $P_4$ gets to round two is ${\large{\frac{12}{15}}}={\large{\frac{4}{5}}}$, since in round one, $P_4$ has to avoid playing any of $P_1,P_2,P_3$.

Consider two cases . . .

Case $(1)$:$\;$In the first round, none of $P_1,P_2,P_3$ plays $P_4$, and two of $P_1,P_2,P_3$ play each other.

The probability that this case occurs is
$$
\left({\small{\frac{4}{5}}}\right)\left({\small{\frac{3}{13}}}\right)
=
{\small{\frac{12}{65}}}
$$
Given that this case occurs, the probability that $P_4$ wins the next match is ${\large{\frac{5}{7}}}$, since in round two, $P_4$ needs to avoid playing any of the two remaining stronger players.

Case $(2)$:$\;$In the first round, none of $P_1,P_2,P_3$ plays $P_4$, and no two of $P_1,P_2,P_3$ play each other.

The probability that this case occurs is
$$
\left({\small{\frac{4}{5}}}\right)\left({\small{\frac{10}{13}}}\right)
=
{\small{\frac{8}{13}}}
$$
Given that this case occurs, the probability that $P_4$ wins the next match is ${\large{\frac{4}{7}}}$, since in round two, $P_4$ needs to avoid playing any of $P_1,P_2,P_3$.

Summing the results for the two cases, it follows that the probability that $P_4$ reaches the semifinal is
$$
\left({\small{\frac{12}{65}}}\right)\left({\small{\frac{5}{7}}}\right)
+
\left({\small{\frac{8}{13}}}\right)\left({\small{\frac{4}{7}}}\right)
=
{\small{\frac{44}{91}}}
$$
Note:$\;$The two approaches yield the same result, as I initially expected, but wasn't sure of.
A: In the first round $P_4$ plays with equal probability against any other team. The probability that $P_4$ wins this round is therefore given by ${12\over15}$. We now condition on this event, i.e.,  that in the first round $P_4$ has played against a $P_k$ with $k>4$.
In the second round there are $7$ adversaries left for $P_4$, all of them equiprobable. We have now have to see how many of them are better  than $P_4$. This depends on whether in the first round there was a match among $1$–$3$. Denote the probability for this to have happened by $p$. Given that $P_4$ in the first round has not played against one of $1$–$3$ we obtain
$$p={2\over13}+{11\over13}\cdot{1\over11}={3\over13}\ .\tag{*}$$
In the case covered by $p$ two of the adversaries of $P_4$ are better than $P_4$, in the case covered by $1-p$ there are three of them. It follows that the overall probability $p_*$ that $P_4$ wins in both rounds is given by
$$p_*={4\over5}\bigl(p\cdot{5\over7}+(1-p){4\over7}\bigr)={44\over91}=0.484\ .$$
$$$$
$(^*)$ The probability that $1$ plays against $2$ or $3$ is ${2\over13}$. If $1$ plays against a $P_k$ with $k>4$ the probability that $2$ plays against $3$ is ${1\over11}$. 
A: Pr (Player $P_4$ wins the Round of 16)= Pr (Player $P_4$ is not tied with players 1,2, or 3 in the first round)= $$\frac{\binom{12}{1}}{\binom{15}{1}}=\frac{12}{15}$$
Player $P_4$ will win the quarter finals only if he is not tied with players 1,2 or 3 in the quarters. 
The possible cases for Players $P_1$,$P_2$,$P_3$ are:


*

*$W$: Players $P_1$ and $P_3$ proceed to the quarterfinals. $P_2$ gets tied with $P_1$ in the round of 16 and is eliminated.

*$X$: Players $P_1$ and $P_2$ proceed to the quarterfinals. $P_3$ gets tied with $P_1$ in the round of 16 and is eliminated.

*$Y$: Players $P_1$ and $P_2$ proceed to the quarterfinals. $P_3$ gets tied with $P_2$ in the round of 16 and is eliminated.

*$Z$: All three players proceed to the quarterfinals. 
Player $P_1$ will always proceed to the quarterfinals. 
Let $A$ denote the event that player $P_4$ wins the quarterfinal.
Using the law of total (conditional) probability,
$$Pr (A)= Pr(A/W).Pr(W)+Pr(A/X).Pr(X)+Pr(A/Y).Pr(Y)+Pr(A/Y)+Pr(A/Z).Pr(Z)$$
$$=\frac{5}{7}\frac{1}{15}+\frac{5}{7}\frac{1}{15}+\frac{5}{7}\frac{1}{15}+\frac{4}{7}\left(1-\frac{3}{15}\right)$$
$$=\frac{1}{7}+\frac{48}{105}=\frac{63}{105}$$
Pr (Player $P_4$ plays semi final)= Pr (Player $P_4$ wins the Round of 16)Pr ($A$)
$$=\frac{12}{15}\frac{63}{105}=\frac{12}{25}=0.48$$
A: Player $P_4$ shares their semi-final group with three other players out of the remaining 15; they advance to the semi-final if and only if none of those three players are $P_1, P_2, P_3$. This means the probability of them advancing is
$$
\frac{\binom{12}{3}}{\binom{15}{3}} = \frac{44}{91} \approx 0.48.
$$
A: First Round = 16 men. Second Round = $\frac{16}{2} = 8 $ men. 3rd Round or Semi Final = $\frac{8}{2} = 4 $ men. Therefore, $P{_{4}}$ must defeat $3$ men to reach the Semi Final. Those 3 men must be from $P{_{5}}P{_{6}}P{_{7}}P{_{8}}P{_{9}}P{_{10}}P{_{11}}P{_{12}}P{_{13}}P{_{14}}P{_{15}}P{_{16}}  $ = 12 men in total. This can be done in $\binom{12}{3}$ ways.Let us call this a Narrow view. In Broad view, 3 opponents can be selected in $\binom{15}{3}$ ways.Then $\frac{Narrow View}{Broad View}$ = $\frac{\binom{12}{3}}{\binom{15}{3}}$ = $\frac{220}{455}$ = $0.48$
