Question

there are 32 teams in a traditional knockout tournament. each of these 32 teams are of a different quality and can be distinctly ranked from 1st to 32nd in terms of ability. assume that a higher-ability team always beats a lower-ability team. the teams are randomly seeded in the schedule of this knockout tournament. what is the probability that the second-best team meets the third best team in the semi-finals?

Answer 1

From the given 32 teams ,probability of teams in semi-finals  such that second best team meets the third best team is equal to 8/31.

As given in the question,

Total number of teams = 32

Condition given :

Second best team should meet the third best team in semi finals.

Round 1 probability where second best team meet with other 30 teams out of 31 apart from third best team = 30 /31

Round 2 probability where second best team meet with other 14 teams out of 15 apart from third best team = 14 /15

Round 3 probability where second best team meet with other 6 teams out of 7 apart from third best team = 6 /7

Round 4 probability where second best team meet with only third best team out of other 3 players = 1 /3

Required probability of teams in semi-finals  such that second best team meets the third best team

= (30/31) ×(14/15) × ( 6/7) ×(1/3)

= 8/31

Therefore, from the given 32 teams ,probability of teams in semi-finals  such that second best team meets the third best team is equal to 8/31.

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