Chiangmai
5 Answers
Clonge
Bob/PKB
AAABBB
AABBBA
ABBBAA
BBBAAA
BBAAAB
BAAABB
ABBABB
ABABAB
BABABA
BAABAA
ABBABA
BAABAB
I'm sure there are more, but you asked the OUTCOME, and it's going to be either A or B. Each team will win 3 (in whatever order), and the 7th game will determine the winner of the series. That is the outcome of a 7 game series. Had you said "Best of 7", the possibilities increase more (4 game sweep for either team, combinations of 5 games, 6 games).
| 11 years ago. Rating: 2 | |
Chiangmai
Permutations and Combinations
In the World Series, one National League team and one American League team compete for the title, which is awarded to the first team to win 4 games. In how many ways can a series of seven games be completed? (Thank you, Gerry, for helping with the answer):
Let’s call the teams A and B. I’ll compute the number of ways that A can win and and then double that to take B victories into account.
There is 1 way to win in 4 games.
For five games the fifth win must be by A and the first 4 games must have 3 wins by A and one by B.
There are (5)! / ((2!*(3!)) = (120) / (2 * 6) = (120 / 12) = 10 ways for A to win in 5
AAABB
AABAB
ABAAB
AABBA
AAABB
AABAB
BABAA
BAAAB
BBAAA
BAABA
So A can win in 4 ways.
For 6 games, the same combinatorics apply 6! / ((3!) * (3!)) = (6 * 5 * 4) / 6 = 20 ways
For 7 games the 7th game must be won by A and the other split 3-3 as in the 6 game problem.
6! / 3!)) * (3!)) = 20
Adding these up, A can win in 1 + 10 + 20 + 20 = 51 ways
Likewise, B can also win in 51 ways
Therefore your answer is : 2 * 51 = 102 ways ç========== Answer
| 13 years ago. Rating: 2 | |
First, there is only one way to win 4 straight.
Second, to win in 5 the last game is won by team A so 3 out of the first four need to be won by team A which is 4C3 which is 4 possible outcomes.
Third, to win in 6 the last game is won by team A so 3 out of the 5 need to be won which is 5C3 which is 10 possible outcomes.
Lastly, to win in 7 the last game is won by team A so 3 out of the 6 need to be won which is 6C3 which is 20 possible outcomes.
Total possible outcomes to win for team A is
= 1 + 4 + 10 + 20 = 35
Due to symmetry, the same can be said for team B and therefore the total possible outcomes for a best of 7 series is 70.
gingerboysmashers
Another way (perhaps simpler) to answer this would be the following.
Assume that there are exhibition games totaling 8 complete games played after the series in which the loser of the series must always win. Thus if the games went ABBABB with B winning the series, the extended version would be ABBABBAA, where A must win the last two games. This would be the case regardless of how quickly the series ended or who won the series (it could be AAAABBBB or ABABABAB).
To calculate the amount of combinations in which either A or B will win four of those eight games, we can use a simple combination, 8C4.
8!/(4!)(4!) = 8*7*6*5/4*3*2*1= 70 possible combinations.
| 11 years ago. Rating: 0 | |
