World SeriesProbability puzzles require you to weigh all the possibilities and pick the most likely outcome.
The world series consists of up to seven games, the first team to win four wins the series. Team A will play in their home field for games 1, 2, and 6 and 7 if necessary. Games 3, 4, and 5 (if necessary) will be played at team B's home field. Assuming each team has a 50% chance of winning every game, which team is more likely to have the home field advantage the majority of the time? For extra credit what is the expected number of games played in each home field?
AnswerIn general the probability of either team winning exactly 3 out of n (n>3) games is (n-1)*(n-2)*(n-3)/(3*2^n).
For a series to last exactly n games at least one team must have won exactly 3 out of the last (n-1). Thus:
The probability of the series lasting 4 games is 1/8.
The probability of the series lasting 5 games is 1/4.
The probability of the series lasting 6 games is 5/16.
The probability of the series lasting 7 games is 5/16.
The expected number of games played in team A's home field is: 2*(1/8) + 2*(1/4) + 3*(5/16) + 4*(5/16) = 188/64.
The expected number of games played in team B's home field is: 2*(1/8) + 3*(1/4) + 3*(5/16) + 3*(5/16) = 184/64.
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