I need a math/stats geek momentarily
 

There are four 7 game baseball series going on at once. Once a team wins 4 games the series is over. So a series can go either 4, 5, 6, or 7 games.

What are the chances that 3 of the series go 7 games, but not all 4 do it?

Ignore player abilities and all that. I just want a straight math percentage.

exactly 50%

They can do it or they cannot.

:1orglaugh

1 in 64 i think

Half of all the possibilities can be shown as (swap X for O for the other half):

XXXX - 2

XXXOX - 8
XXOXX
XOXXX
OXXXX

XXXOOX - 20
XXOXOX
XOXXOX
OXXXOX
XXOOXX
XOXOXX
OXXOXX
XOOXXX
OXOXXX
OOXXXX

XXXOOOX - 22
XXOXOOX
XOXXOOX
OXXXOOX
XXOOOXX
XOXOOXX
OXXOOXX
XOOOXXX
OOXOXXX
OXOOXXX
OOOXXXX

So, we have 52 possible outcomes. There's a 22/52 chance that any one series will go 7 games and 30/52 chance it won't. We have four series and for three, and only three, of those series to be 7 games long ->

number of ways A does happen
_____________________________ ..... OR .... P(A)/ (1 -P(A))
number of ways A does not happen

Where:
P(A) = 22/52 * 22/52 * 22/52 * 30/52 * 4 = 1277760/7311616

1 - P(A) = (22/52 * 22/52 * 22/52 *22/52) + (30/52 * 30/52 * 22/52 * 22/52 * 6) + (30/52 *30/52 * 30/52 * 22/52 * 4) + (30/52 * 30/52 * 30/52 * 30/52) =
(234256 + 2613600 + 2376000 + 810000)/7311616 = 7077360/7311616

(1277760/7311616)
__________________ = 1277760/7077360 =
(7077360/7311616)

5324/29489 or about an 18% chance.

22/52*22/52*22/52*30/52

4.36%

Edited: to remove geeky exposition

What you're showing is the possibilty of any one <B>specific</B> match of a series being the only one not 7 games long.

Look at it this way- There are four series to be played, and three of them must be 7 games and one of the others is of 4-6 games.

OK, label each of the series A-D, but use lower case to designate the 4-6 game set. You'll then see there are 4 possible permutations:
a-B-C-D = 30/52*22/52*22/52*22/52
A-b-C-D = 22/52*30/52*22/52*22/52
A-B-c-D = 22/52*22/52*30/52*22/52
A-B-C-d = 22/52*22/52*22/52*30/52

This is how P(A), the number of possible ways, is derived. So we have <B>22/52*22/52*22/52*30/52*4</B>. But that has to be divided by 1 - P(A), the number of ways it doesn't happen, to give the odds.

Got an e-mail that shows a bit of lack of understanding on what "odds" are, so one last time in...

See http://www.saintjoe.edu/~karend/m111/m111c7.html for the difference between "probability" (P(A)), and "odds" (P(A)/P(not A)). Also check on the definition of "sample space".

Breakdown of the <B>complete</B> sample space, using UPPER CASE to show a seven game match:

a-B-C-D = 30/52*22/52*22/52*22/52 | These
A-b-C-D = 22/52*30/52*22/52*22/52 | meet
A-B-c-D = 22/52*22/52*30/52*22/52 | the
A-B-C-d = 22/52*22/52*22/52*30/52 | criteria.
__________________________________________
a-b-c-d | No 7 game matches. (30/52*30/52*30/52*30/52) +

A-b-c-d | Just
a-B-c-d | one match
a-b-C-d | of 7
a-b-c-D | games. (30/52*30/52*30/52*22/52*4) +

a-b-C-D | Two
a-B-c-D | matches
A-b-c-D | of
a-B-C-d | 7
A-b-C-d | games
A-B-c-d | each. (30/52*30/52*22/52*22/52*6) +

A-B-C-D | All 7 game matches. (22/52*22/52*22/52*22/52)

~~~~~~~

So, the odds are app. 3/20, 3:20, 3 to 20 or, as a percentage, 15%.

answer is :

12/256

Peace

Thanks, folks. :)