Nine person of equal strength playing in tournament Nine persons $P_i$ , i = 1, 2, ...., 9 of equal strength are playing a tournament such that they are first grouped in to three groups A, B, C each containing three persons at random and a winner from each group is selected and. a new group is formed and finally from this group the winner of the tournament is decided.
Probability that $P_2$  and  $P_4$    were in different groups given that $P_4 $ is the winner of the tournament is   equal to
(A) $\frac{2}{3}$
(B) $\frac{1}{2}$
(C) $\frac{1}{4}$
(D) $\frac{3}{4}$
Let us treat it a case of 9 distinct balls (each player) and three sets of three identical boxes though three identical boxes are put in three distinct boxes
Let us put ball representing $P_2$ & $P_4$ in different boxes and then make the number of combinati0ns
${}^7{C_3} \times {}^4{C_2} = 210$
Now the arrangement of three groups is $210\times3!=1260$
Now total number of cases without restriction = ${}^9{C_3} \times {}^6{C_3} = 1680$
Now the arrangement of three groups is $1680\times3!=10080$
Probability that $P_2$ and $P_4$ are in different group = $\frac{1260}{10080}=\frac{1}{8}$
I am not sure about my approach
 A: Who wins has nothing to do with initial group arrangement so probability of $P_{2}$ not in the same initial group with the winner is $\frac{6}{8}=\frac{3}{4}$.
Probability of $P_{2}$ not making to the next round is $\frac{2}{3}$.
Probability that $P_{2}$ and $P_{4}$ never meet given that $P_{4}$ win is $\frac{3}{4}\times\frac{2}{3}=\frac{1}{2}$
A: That is too low.
$P_4$ is in an initial group with two of the other eight, and in the final group with two others
so the probability $P_2$ was in an initial group with $P_4$ is $\frac28=\frac14$
and (since $P_4$ won and so must have been in the final group) was in either  group with $P_4$ is $\frac28+\frac28=\frac{1}{2}$
meaning the probability $P_2$ and $P_4$ were in different initial groups is $1-\frac14=\frac34$
while the probability $P_2$ and $P_4$ were not both in the same initial group or final group, given that $P_4$ won is $1-\frac12=\frac12$.
Of your ${9 \choose 3}{6 \choose 3}{3 \choose 3}=1680$ ways of distributing the three initial groups, you will find ${3 \choose 1}{7 \choose 1}{6 \choose 3}{3 \choose 3}=420$ have $P_2$ and $P_4$ in the same initial group and ${3 \choose 1}{7 \choose 2}{6 \choose 3}{3 \choose 3}=1260$ have them in different initial groups
A: Here is a modelling situation that considers the fact that the players had to play and win. There are three steps in the given history, so we model by a tree with three steps. The solution uses the "symmetries" of the modelling space. (Given by the action of some permutation group.)

I will denote the (total) set of the players  by $T\{\ 1,2,3,4,5,6,7,8,9\ \}$, and my $1$ is the winner $P_4$, my $2$ is the player $P_2$ from the OP.

*

*At the first step we are taking a random partition of the $9$ players in $3$ ordered groups, there are
$$
\binom 9{3\ 3\ 3}
=\frac{9!}{3!\; 3!\; 3!}
=1680
$$
chances for this. Let $(S_1,S_2,S_3)$ be this partition, $S_1,S_2,S_3$ being sets with three elements and $T=S_1\sqcup S_2\sqcup S_3$.

*At the second step we are picking a random element (the winner) from each of these sets, $w_1\in S_1$,  $w_2\in S_2$,  $w_3\in S_3$, getting the set $S=\{w_1,w_2,w_3\}$.

*At the final step we pick a winner $w\in S$, i.e. $w$ is among $w_1,w_2,w_3$.

So the modelling probability space $\Omega$ is the space of all data $\omega$ of the shape
$$
\omega = (\ (S_1,S_2,S_3)\ ,\ S=(w_1,w_2,w_3)\ ,\ w\ )
$$
with the above properties.
(So $\Omega$ has $\binom 9{3\ 3\ 3}\cdot 3^3\cdot 3$ elements.)
Each element is equally probable.
We restrict to the subspace $\Omega_1$ where $w=1$ is the winner.

So far we have covered only the modelling part, things have been fixed in a precise way. Now we let some group of permutations act on the space $\Omega_1$. Instead of taking the whole group of permutations fixing the winner $1$ let us consider the small group with eight elements of cyclic permuations of $2,3,4,5,6,7,8,9$, taken in this order.
Then for each $\omega\in\Omega_1$, in the "orbit" of all states obtained from $\omega$ by applying one by one these eight permutations...

*

*there are exactly two where $2$ belongs to the initial group ($S_1$ or $S_2$ or $S_3$,) where the non-permuted $1$ lives in, and occupies one of the three places,

*there are exactly two where $2$ belongs to the final group $S$ where the non-permuted $1$ lives in, and occupies one of the three places,

*and exactly four where $2$ belongs to either the initial group or the final group of the winner $w=1$, since the initial group of the winner and $S$ have only $1$ in common.


The OP is unclear about the needed probability, so let us make clear statements. The conditional probability, conditioned by $\Omega_1$ (i.e. $w=1$ is the winner)...

*

*that $1,2$ are in the same initial group is $\frac 28=\frac 14$,

*that $1,2$ are in the same final group is $\frac 28=\frac 14$,

*that $1,2$ are in the same either initial or final group is $\frac 48=\frac 12$.

A: This answer assumes that the question is only referring to starting group.
Without loss of generality, lets name the starting groups A, B, C such that P4 $\in$ A.
By symmetry, $P(P_2 \in A) = P(P_i \in A)$ for every $i \ne 4$, since there is nothing unique about $P_2$. Out of 8 balls, 2 will be in A and 6 will be not in A. Therefore we get the total chance is simply $\frac{6}{8} = \frac{3}{4}$.
