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Eight players P1, P2, P3, ...........P8...

Eight players `P_1, P_2, P_3, ...........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 lt j`. Assuming that the players are paired at random in each round, what is the probability that the players `P_4`, reaches the final ?

Text Solution

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`P_(1) P_(2) P_(3) P_(4) P_(5) P_(6) P_(7) P_(8)`
Given that if `P_(i), P_(j)` play with `i lt j`, then `P_(i)` will win. For the first round, `P_(4)` should be paired with any one from `P_(5)` to `P_(8)`. It can be done in `.^(4)C_(1)` ways. Then `P_(4)` to be the finalist, at least one player from `P_(5)` to `P_(8)` should reach in the second round. Therefore, one pair should be from remaining 3 from `P_(5)` to `P_(8)` in `.^(3)C_(2)`. Then round, we have four players. Favorable ways is 1.
Now, total possible pairings is
`(.^(8)C_(2) xx .^(6)C_(2) xx .^(4)C_(2)xx .^(2)C_(2))/(4!) xx (.^(4)C_(2)xx .^(2)C_(2))/(2!)`
Therefore, the probability is
`(.^(4)C_(1).^(3)C_(2).^(3)C_(2)4!2!)/(.^(8)C_(2).^(6)C_(2).^(4)C_(2).^(2)C_(2).^(4)C_(2).^(2)C_(2))=(4)/(35)`
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