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In a knockout tournament $2^{𝑛}$ equally skilled players; $𝑆_{1}, 𝑆_{2}, . . . . . . . . 𝑆_{2^{𝑛}}$ are participating. In each round players are divided in pair at random and winner from each pair moves in the next round. If $𝑆_{2}$ reaches the semifinal then the probability that $𝑆_{1}$ wins the tournament is 1/20. Find the value of $\frac{n}{4}$ .

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Detailed Solution

No. of ways that 𝑆2 reaches semi final $= (2^{n} - 1) c_{3}$
No. of ways that both 𝑆2 & 𝑆1 reaches semifinal $= (2^{n} - 1) c_{2}$
Prob. that 𝑆1 wins the tournament
$\frac{(2^{n} - 2) c_{2} . (\frac{1}{4})}{(2^{n} - 1) c_{3}}$

Given $\frac{(2^{n} - 2) c_{2}}{4 \times (2^{n} - 1) c_{3}} = \frac{1}{20} \Rightarrow n = 4$

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