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?

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- Mar 13th 2011, 04:10 PMjellyksong(Simple) World series probability question?
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? - Mar 13th 2011, 04:41 PMPlato
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. - Mar 13th 2011, 04:49 PMRandom Variable
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. - Mar 13th 2011, 06:10 PMPlato
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. - Mar 13th 2011, 07:02 PMRandom Variable
Well, the answer you posted makes no sense. Team A wins in 35/16 ways?

EDIT: Did you mean $\displaystyle \binom{7}{4} $ ways?