Titleist wrote:
Baseball's World Series matches 2 teams against each other in a best-of-seven series. The first team to win four games wins the series and no subsequent games are played. If you have no special information about either of the teams, what is the probability that the World Series will consist of fewer than 7 games?
(A) 12.5%
(B) 25%
(C) 31.25%
(D) 68.75%
(E) 75%
Responding to a pm:
There are two teams A and B. We assume there are no draws (though it is usually mentioned in the question). They play against each other in a best of 7 series. That is, the one team which wins more matches out of the 7 played, wins. Also, if a team wins 4 matches say team A wins first 4 matches played, team A is announced the winner and other 3 matches do not take place. Say team A wins 3 matches and team B wins 2 matches out of 5. Then, if team A wins the 6th match, team A is the winner and the 7th match is not played.
So they need to play a minimum of 4 matches (all 4 won by the same team) and a maximum of 7 matches (4 won by 1 team and 3 won by the other)
We need to find
P(Less than 7 games) = 1 - P(7 games)
7 games will be played when the win matrix of 6 games looks something like this {A, A, A, B, B, B} or {A, A, B, A, B, B} or {B, A, B, A, B, A} etc...
That is each team would have won exactly 3 games. You can arrange 3 As and 3 Bs in 6!/(3!*3!) ways.
What is the probability that 7 games will be played? A wins 3 and B wins 3. Each team can win a match with a probability of (1/2).
Probability of getting {A, A, A, B, B, B} = (1/2)*(1/2)*(1/2)*(1/2)*(1/2)*(1/2)
Probability that 7 games will be played = \((1/2)^6 * 6!/(3!*3!)\)
Probability that less than 7 games will be played = \(1 - (1/2)^6 * 6!/(3!*3!)\)
1.Do you mean to imply that the probability of getting a W W W W W W is the same as getting W W W L L L..Similar to H and T case in coin