In the World Series, two teams play each other until one team wins 4 games. How many different series of games is possible? (i.e., Team A wins the series by winning games 1, 2, 4, and 6)
Would it just be 7 choose 4?
In the World Series, two teams play each other until one team wins 4 games. How many different series of games is possible? (i.e., Team A wins the series by winning games 1, 2, 4, and 6)
Would it just be 7 choose 4?
This is perhaps the most famous question in probability.
It may well date back to the eleventh centenary.
Playing any series of seven games will determine a winner. RIGHT?
“Who has four wins, there are no ties’s”.
$\displaystyle \sum\limits_{k = 4}^7 {\binom{7}{k}2^{ - k} } = 0.5$ for each team.
If Team A wins in 4 games, they must win the first three games. If Team A wins in 5 games, they must win 3 out of the first 4 games. If Team A wins in 6 games, they must win 3 out of first 5 games. And if team A wins in 7 games, they must win 3 out of the first 6 games.
So number of ways Team A wins is $\displaystyle \binom{3}{3} + \binom{4}{3} + \binom{5}{3} + \binom{6}{3} $
And the number of ways Team B can win is obviously the same.
Do you understand that none of that makes any difference?
Team A wins if any of the sequences $\displaystyle AAAABBB$ occurs?
Suppose that we require that seven games be played period no matter the outcomes. Do you see that any string of four A's and three B's mans that team A wins?
As I said above, this question has a long history.
I can trace it back ot at least the $\displaystyle 11^{th}$ century.