Suppose in the Minor Hockey League Championship play off series between Nepean P
ID: 3679654 • Letter: S
Question
Suppose in the Minor Hockey League Championship play off series between Nepean Pirates and Kanata Thunders, there are three possible playoff games planned. A team that wins two games is declared the champion. Outcome of each game is either a win or a loss - there are no ties! The first game is played in the Pirates Arena, the Second game is played in the Thunder's arena, and if required the third game will be played in the Pirates arena. A team wins with a probability of 2/3rd in its home arena, and with a probability of l/3rd in the opposition's arena. What is the probability that Nepean Pirates will be declared the Champion?Explanation / Answer
As the game begins playing in the pirates field, the winning probability for the first game for pirates will be
2/3 and , the second game is played in thunder, so winning probability for pirates is 1/3, The third set of the game is played in the pirates arena, where again the winnning probability is 2/3. since the pirates get a chance to play on it home land twice, The winning probability is 2/3
The probability of team pirates losing (which is the same as team thunder winning) a single game is 1/3 = 1 2/3.
Best-of-Three
A best-of-one is a rather simple case. The probability that team pirates wins is just 2/3.
Therefore, the first "non-trivial" case is the best-of-three playoff series. A tree for this looks like:
game 1 game 2 game 3 outcome probability pirates wins2/3 pirate wins
2/3 — team pirates
wins series (2/3) * (2/3) = 4/9 thunder wins
1/3 pirate wins
2/3 team pirate
wins series (2/3) *(1/3)* (2/3) = 4/27 thunder wins
1/3 team thunder
wins series 2/3 * 1/3 * 1/3 = 2/27 thunder wins
1/3 pirate wins
2/3 pirate wins
2/3 team pirate
wins series 2/3 * 2/3 * 1/3 = 2/27 thunder wins
1/3 team thunder
wins series 2/3 * 1/3 * 1/3 = 2/27 thunder wins
1/3 — team thunder
wins series 1/3 * 1/3 = 1/9