Question

Thirty-two players ranked 1 to 32 are playing in a knockout tournament. Assume that in every match between any two players, the better-ranked player wins, the probability that ranked 1 and ranked 2 players are winner and runner up respectively, is $p$the find the value of $\left[ \dfrac{2}{p} \right]$ where $\left[ . \right]$ is the greatest integer function. \[\]
A.2\[\]
B.3\[\]
C.4\[\]
D.1\[\]

Answer 1

Hint: We need to arrange the tournament in such a way that the top two ranked players do play before the final. The 32 player knockout tournament will have four rounds. Find the probabilities of two top ranked players not playing with each other in all the rounds before the final and multiply them . \[\]

Complete step by step answer:
As given in the question that thirty-two players ranked 1 to 32 are playing in a knockout tournament. Let us denote them as ${{A}_{1}},{{A}_{2}}...,{{A}_{32}}$ where the subscript represents their rank. \[\]
It has also been assumed in the question that in every match between any two players, the better-ranked player wins. It means any player ${{A}_{i}}$ is going to win against any player ${{A}_{j}}$ when $i>j$ where $i,j\in \left\{ 1,2,3,..32 \right\}$. We shall first find the probability of the first ranker ${{A}_{1}}$ and the second ranker ${{A}_{2}}$ to play the final match.\[\]
A knockout tournament with 32 players is going to take four rounds first, second, third, semi-final and final. We know that in every match a player is going to be knocked out of the tournament. So we have 32 players in the first round, 16 players in the second, 8 players in the third , 4 players in the semi-final and 2 players in the final. \[\]
We do not want the ${{A}_{1}}$ and ${{A}_{2}}$ to play in the first round because we want them to play in the final . If they play ${{A}_{1}}$ is going to win and ${{A}_{2}}$ is going to be knocked out as assumed. So ${{A}_{1}}$ can only with the rest 30 players out of 31(including ${{A}_{1}}$ ) players. So the probability of ${{A}_{1}}$ go to play the final with ${{A}_{2}}$ is ${{P}_{1}}=\dfrac{30}{31}$.\[\]
We also do not want players ${{A}_{1}}$ and ${{A}_{2}}$ to play in the second round. The number of players ${{A}_{1}}$ can play in the second round is 15 and excluding ${{A}_{2}}$ is 14. So the probability of ${{A}_{1}}$ will go to play the final with ${{A}_{2}}$ is ${{P}_{2}}=\dfrac{14}{15}$.\[\]
We see similarly in the third round that the number of players ${{A}_{1}}$ can play in the second round is 7 and excluding ${{A}_{2}}$ is 6. So the probability of ${{A}_{1}}$ playing the final with ${{A}_{2}}$ is ${{P}_{3}}=\dfrac{6}{7}$.\[\]
We see in the semi-final that ${{A}_{1}}$ can play in the second round is 3 and excluding ${{A}_{2}}$ is 3. So the probability of ${{A}_{1}}$ playing the final with ${{A}_{2}}$ is ${{P}_{4}}=\dfrac{2}{3}$.\[\]
So the probability of ${{A}_{1}}$ playing final with ${{A}_{2}}$ is
\[{{P}_{1}}{{P}_{2}}{{P}_{3}}{{P}_{4}}=\dfrac{30}{31}\times \dfrac{14}{15}\times \dfrac{6}{7}\times \dfrac{2}{3}=\dfrac{16}{31}\]
 The value of probability is denoted in the question as $p$. We are asked to find the greatest integer smaller than $\dfrac{2}{p}$. So
\[\left[ \dfrac{2}{p} \right]=\left[ \dfrac{2\times 31}{16} \right]=3\]

So, the correct answer is “Option B”.

Note: We need to take care of the fact that in each round we are finding the probability of top two players NOT playing before final and we need to take care that we are asked to find the value of $\left[ \dfrac{2}{p} \right]$ not $\left[ p \right]$ or $\left[ \dfrac{1}{p} \right]$.