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In a knockout tournament of 2n equally skilled players, S₁,,S₂, S₃,… are participating. In each round, players are divided in pairs at random and winner from each pair moves in the next round. If ,S₂ reaches the semi-final, then the probability that $S_{1}$ wins the tournament is 1/84. The value of n equals

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answer is 6.

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Detailed Solution

Given S, reaches the semifinals.
Since all other players are equally likely to win the finals with probability p. We have

$\begin{array}{l} or (2^{n} - 1) p + \frac{1}{4} = 1 \\ or (2^{n} - 1) p = \frac{3}{4} \\ or p = \frac{3}{4 (2^{n} - 1)} \\ If p = \frac{1}{84}, then \frac{1}{84} = \frac{3}{4 (2^{n} - 1)} \\ or 2^{n} - 1 = 63 \\ or 2^{n} = 64 \\ n = 6 \end{array}$

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