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Diamond Enthusiast

Picture of Peteeo
Posted
? What are the odds for tne number of World Series games?
I believe there to be 70 4 best of seven combinations.
2 , 4 games sweeps team a or b
8, 5 game combinations
20 ,6 game combinations
40 ,7 game combinations.

2 * n C r
2 * 7 C 4
2*35
Correct?
 
Posts: 211 | Location: Vadnais Heights MN. | Registered: 06-15-02Reply With QuoteEdit or Delete MessageReport This Post
Silver Enthusiast
Picture of Pin~Jinx
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Could you please be more explicit as to what you mean by ‘4best of 7 combos’ and what do 2,8,20…stand for?

Pin~Jinx/anarchist unclear
 
Posts: 629 | Location: Karachi | Registered: 06-27-02Reply With QuoteEdit or Delete MessageReport This Post
Diamond Enthusiast

Picture of Peteeo
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I'm probably making this harder in my head than it needs to be.
For a complete world series Team A can win in a sweep of 4 games or Team B can win in a sweep of 4 games. 2 ways to have a sweep.
2^7 is 128 combinations but not all are possible. Team A cannot win 5 out of 7 games as the series ends with the 4th win.
I count 20 ways that Team A or B can win in 6 games as an example.
With 70 possibleby my count most world series go the full 7 games.. 40/70. A sweep as happend this year is very rare. 2/70.
Main question..Did I miss any of the combinations?
Thanks
 
Posts: 211 | Location: Vadnais Heights MN. | Registered: 06-15-02Reply With QuoteEdit or Delete MessageReport This Post
Picture of qualserve
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There are only 2 combinations of each scenario you present: 4-0 x 2, 4-1 x 2, 4-2 x 2, 4-3 x 2.
For instance, you state there are 20 ways that A or B can win in 6 games?? Heck no, 4-2 is the only combination that makes a six game series, right?
 
Posts: 193 | Location: Merrimack, NH, United States | Registered: 06-03-02Reply With QuoteEdit or Delete MessageReport This Post
Diamond
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Qualserve-
Peteeo is considering the case where team A wins the first game, but then loses the remaining games to be different from the case where team A only wins the second game. The case where team A only wins the thirds game is considered yet another case, as is the case where team A only wins the fourth game. All of those give a total record of 4-1, but they are 4 separate cases. Then we have the 4 corresponding cases where team B only wins one game to round out the 8 possible 5 game situations.
 
Posts: 5891 | Location: Indiana | Registered: 06-13-02Reply With QuoteEdit or Delete MessageReport This Post
Diamond
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Peteeo - My probability is a little rusty, but your formula does look right as far as I can tell.

I decided to go through a brute force method (writing down all the possibilities) for 4, 5, and 6 game series. I got the same answers as you did. The 7 game possibilities are a bit much for a brute force method.
 
Posts: 5891 | Location: Indiana | Registered: 06-13-02Reply With QuoteEdit or Delete MessageReport This Post
Picture of qualserve
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Gotcha!! Seemed too easy!! Wink
 
Posts: 193 | Location: Merrimack, NH, United States | Registered: 06-03-02Reply With QuoteEdit or Delete MessageReport This Post
Diamond Enthusiast

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Thanks for the feedback.
It just seemed to happen too often that the series would go to 7 games.. There's a reason for it. Both teams are good to get there so they've got a nearly 50:50 chance to win and
the probabities of the mose permutations of the result..
 
Posts: 211 | Location: Vadnais Heights MN. | Registered: 06-15-02Reply With QuoteEdit or Delete MessageReport This Post
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