Question

In a knockout tournament ${2^{nd}}$ equally skilled players ${S_1},{S_{2,..........,}}{S_n}$ are participating. In each round, players are divided in pairs at random and the winner from each pair moves in the next round. If ${S_2}$ reaches the semi-final round then the probability that ${S_1}$ wins the tournament is $\dfrac{1}{{84}}$. Then the value of n equals_________.

Answer 1

Hint: According to given in the question we have to determine the value of n when in a knockout tournaments ${2^{nd}}$ equally players ${S_1},{S_{2,..........,}}{S_n}$ are participating. In each round, players are divided in pairs at random and the winner from each pair moves in the next round. If ${S_2}$ reaches the semi-final round then the probability that ${S_1}$ wins the tournament is$\dfrac{1}{{84}}$. So, first of all we have to determine the probability of winning when there are ${2^{nd}}$ teams. A total of 4 players reached the semi-final.
Now, we have to determine the probability that if some-one else's team wins the match and then we have to determine the probability if no one is special.
Now, we have to determine the value of n with the help of the probability that ${S_1}$ wins the tournament which is $\dfrac{1}{{84}}$.

Complete step by step answer:
Step 1: First of all we have to determine the probability of winning when there are ${2^{nd}}$team total 4 players reach the semi-final as mentioned in the solution hint. Hence,
$ \Rightarrow $Probability of winning ${S_1} = \dfrac{1}{4}$
Step 2: Now, we have to determine the probability if some-one else wins as mentioned in the solution hint.
$ \Rightarrow $Probability of winning some-one else$ = \dfrac{3}{4}$
Step 3: Now, we have to determine the probability if no one is special so they will have the same chance of winning.
Step 4: Now, as we know that there are ${2^n} - 1$ of them, the probability that one of them ${S_1}$wins. Hence,
$ \Rightarrow \dfrac{3}{{4({2^n} - 1)}} = \dfrac{1}{{84}}$
Now, we have to solve the expression as obtained just above, by applying cross-multiplication. Hence,
$
   \Rightarrow \dfrac{3}{{({2^n} - 1)}} = \dfrac{1}{{21}} \\
   \Rightarrow ({2^n} - 1) = 21 \times 3 \\
   \Rightarrow {2^n} = 63 + 1 \\
   \Rightarrow {2^n} = 64...............(1) \\
 $
Step 5: Now, to obtain the value of n we have to compare the power of 2 before that we have to make the base common in the expression as obtained in the solution step 4. Hence,
$
   \Rightarrow {2^n} = {2^6} \\
   \Rightarrow n = 6 \\
 $

Hence, we have obtained the value of n which is 6 when in a knockout tournaments ${2^{nd}}$ equally players ${S_1},{S_{2,..........,}}{S_n}$ are participating. In each round, players are divided in pairs at random and the winner from each pair moves on to the next round. If ${S_2}$ reaches the semi-final round then the probability that ${S_1}$ wins the tournament is $\dfrac{1}{{84}}$.

Note: To determine the probability we have to divide the favourable event by the sample space or total number of outcomes.
If the probability of an event to occur is P(A) then the probability of an event that does not occur is 1-P(A).