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#1 |
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Confirmed User
Join Date: Oct 2001
Posts: 6,693
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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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#2 |
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CjOverkill
Industry Role:
Join Date: Apr 2003
Location: Woldwide
Posts: 1,328
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exactly 50%
They can do it or they cannot. ![]()
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#3 |
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Confirmed User
Industry Role:
Join Date: Feb 2002
Location: The 510
Posts: 4,545
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1 in 64 i think
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#4 |
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Confirmed User
Join Date: Aug 2001
Location: New Orleans
Posts: 1,680
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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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#5 |
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Confirmed User
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Join Date: Jun 2003
Location: Silicon Valley
Posts: 317
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22/52*22/52*22/52*30/52
4.36% Edited: to remove geeky exposition |
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#6 | |
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Confirmed User
Join Date: Aug 2001
Location: New Orleans
Posts: 1,680
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Quote:
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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#7 |
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Confirmed User
Join Date: Aug 2001
Location: New Orleans
Posts: 1,680
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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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#8 |
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There can be only one
Industry Role:
Join Date: Aug 2001
Location: Somewhere else
Posts: 39,075
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SIG TOO BIG |
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#9 |
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Confirmed User
Join Date: Mar 2003
Posts: 400
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answer is :
12/256 Peace |
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#10 |
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Confirmed User
Join Date: Oct 2001
Posts: 6,693
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Thanks, folks.
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