Assume four teams, the Antelopes, Bobcats, Cougars, and Donkeys, play a tourname

Question

Assume four teams, the Antelopes, Bobcats, Cougars, and Donkeys, play a tournament. In the first round, the Antelopes play the Bobcats, and the Cougars play the Donkeys. The two winners meet in the finals for the championship. The Antelopes are the best team. They win any game they play with probability 0.7. The other three teams are all equal. Each one can beat one of the other two with probability 0.5. Assume the games are independent. What is the probability that each team wins the tournament? 2.18

Explanation / Answer

1. Let the event of each team winning a game be given as follows :

P(A) = Probability that Antelopes win a game

P (B) = Probability that Bobcats win a game

P (C) = Probability that Cougars win a game

P (D) = Probability that Donkeys win a game

Then the probability that each team wins the tournament is calculated as given below :

1. P(A wins tournament) = (0.7*0.5*0.7)+(0.7*0.5*0.7)) = 0.49 : If they win the first and second round against the winner from C or D.

2. P( B wins tournament) = ((0.3*0.5*0.5)+(0.3*0.5*0.5)) = 0.15 : If A loses the first round and they win both rounds.

3. P (C wins tournament )

= P(B wins and C wins then C wins second round) + P(A wins and C wins and then C wins second round)

= (0.3*0.5*0.5)+(0.7*0.5*0.3) = 0.18

3. P (D wins tournament )

= P(B wins and D wins then D wins second round ) + P(A wins and D wins and then D wins second round)

= (0.3*0.5*0.5)+(0.7*0.5*0.3) = 0.18

Adding these, we get ,

P(A) + P(B) + P(C) + P(D) = 1

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