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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

Samar Imam Zaidi
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    Does "P2 and P4 were in different groups" mean that they were in different starting groups, or that they were never in the same group (including the final group)? I believe the current answers assume the later. – Cain Aug 04 '21 at 16:50

4 Answers4

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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 $\frac{3}{4}\times\frac{2}{3}=\frac{1}{2}$

acat3
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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

Henry
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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 $\frac 48=\frac 12$.
dan_fulea
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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 $\frac{6}{8} = \frac{3}{4}$.

Cain
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