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Old 07-03-2003, 11:51 AM   #1
FATPad
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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.
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Old 07-03-2003, 11:57 AM   #2
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exactly 50%

They can do it or they cannot.

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Old 07-03-2003, 12:00 PM   #3
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1 in 64 i think
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Old 07-03-2003, 01:16 PM   #4
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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.
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Last edited by fiveyes; 07-03-2003 at 01:18 PM..
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Old 07-03-2003, 02:00 PM   #5
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22/52*22/52*22/52*30/52

4.36%

Edited: to remove geeky exposition

Last edited by Penthouse_mike; 07-03-2003 at 02:05 PM..
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Old 07-03-2003, 02:45 PM   #6
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Quote:
Originally posted by Penthouse_mike
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.
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Old 07-03-2003, 04:43 PM   #7
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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%.
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Last edited by fiveyes; 07-03-2003 at 04:45 PM..
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Old 07-03-2003, 04:44 PM   #8
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Old 07-03-2003, 06:16 PM   #9
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answer is :

12/256

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Old 07-04-2003, 06:58 AM   #10
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Thanks, folks.
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