Eight players $P_1, P_2, \ldots P_8$ play a knock out tournament. It is known that whenever the players $P_i$ and $P_j$ play, the player $P_i$ will win if $i < j$. Assuming that the players are paired at random in each round, then the probability that the player $P_4$ reaches the final is ______.
Step-by-Step Solution
Key Concept: In this deterministic tournament where P_i beats P_j iff i < j, P_4 reaches the final only if it avoids players P_1, P_2, P_3 in rounds 1 and 2. The probability depends on pairing configurations: P_4 must face exactly one of {P_5, P_6, P_7, P_8} in round 1, then one of the survivors from {P_5, P_6, P_7, P_8} in round 2, while P_1, P_2, P_3 must eliminate each other.
Eight players are paired in four pairs in $\frac{1}{4!}\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2} = 105$ ways. For at least two players to reach the second round, exactly two players must play against each other in the first round from the top four seeds, with the remaining player playing one of the bottom four. This occurs in $\binom{4}{2} \times 4\binom{3}{1} \times \binom{2}{1} = 36$ ways. The required probability is $\frac{4}{35}$.
Correct Answer: 0.1142