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CEO
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Baseball's World Series matches 2 teams against each other [#permalink]
 07 Jan 2004, 01:09
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0% (00:00) correct
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Baseball's World Series matches 2 teams against each other in a best-of-seven series. The first team to win four games wins the series and no subsequent games are played. If you have no special information about either of the teams, what is the probability that the World Series will consist of fewer than 7 games?
(A) 12.5%
(B) 25%
(C) 31.25%
(D) 68.75%
(E) 75%
please explain
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Director
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Re: PS : World Series [#permalink]
07 Jan 2004, 06:57
praetorian123 wrote: Baseball's World Series matches 2 teams against each other in a best-of-seven series. The first team to win four games wins the series and no subsequent games are played. If you have no special information about either of the teams, what is the probability that the World Series will consist of fewer than 7 games?
(A) 12.5%
(B) 25%
(C) 31.25%
(D) 68.75%
(E) 75%
please explain
Isn't this a previous Manhattan GMAT question of the week?
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Probability that league is finished in 6 games = 1 - P
where P = probability that league is finished in 7 games.
In order to go for seven games each team has to win 3 each in the first 6 games
Total combinations are 6! / ( 3! * 3! ) = 20
The last game has two combinations ( win or lose )
so total combinations = 20 * 2 = 40
Since each game has two possible values 7 games will have 2^7 = 128
P = 40/128
Probability that league is finished in 6 games = 1-40/128 = 88/128
= 68.75
Please post the correct answer.
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Director
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anandnk wrote: Probability that league is finished in 6 games = 1 - P
where P = probability that league is finished in 7 games.
In order to go for seven games each team has to win 3 each in the first 6 games
Total combinations are 6! / ( 3! * 3! ) = 20
The last game has two combinations ( win or lose )
so total combinations = 20 * 2 = 40
Since each game has two possible values 7 games will have 2^7 = 128
P = 40/128
Probability that league is finished in 6 games = 1-40/128 = 88/128
= 68.75
Please post the correct answer.
That is the correct answer - i don't think you needed anyone to tell you that. 
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Hmm. I am not very confident is solving probability questions. I do solve permutation/combination problems well. I even dont know baseball or playing cards( unsual right - I dont attempt probs with deck of cards, I need to know about cards ). SO I definitely needed the answer.
Thanks
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Director
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anandnk wrote: Hmm. I am not very confident is solving probability questions. I do solve permutation/combination problems well. I even dont know baseball or playing cards( unsual right - I dont attempt probs with deck of cards, I need to know about cards ). SO I definitely needed the answer.
Thanks
Your explanation is very close to Manhattan GMAT's official answer - it's quite uncanny:
"There are many other ways this could happen. Using the permutation formula, there are 6!/(3!)(3!) = 20 ways for the two teams to split the first 6 games (3 wins for each).
There are then 2 possible outcomes to break the tie in Game 7. Thus, there are a total of 20 ├Ч 2 = 40 ways for the World Series to last the full 7 games.
The probability that any one of these 40 ways occurs can be calculated from the fact that the probability of a team winning a game equals the probability of a team losing a game = 1/2.
Given that 7 distinct events must happen in any 7 game series, and that each of these events has a probability of 1/2, the probability that any one particular 7 game series occurs is.
Since there are 40 possible different 7 game series, the probability that the World Series will last exactly 7 games is: (1/2)^7 = 1/128
40*1/128 = 40/128 = 31.25%
Thus the probability that the World Series will last less than 7 games is 100% - 31.25% = 68.75%.
The correct answer is D."
Last edited by Titleist on 07 Jan 2004, 10:06, edited 1 time in total.
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This is unbelievable. Truely I came up with it on my own. I did try combination of 4,5 and 6 games.
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Director
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anandnk wrote: This is unbelievable. Truely I came up with it on my own. I did try combination of 4,5 and 6 games.
I believe you! No innuendos here. Just fascination.
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I have a problem with this question let me know where I am wrong.
If number of games played is to be less than 7 then a team can win
either in first 4 games or 4 games in 5 or 4 games in 6
so desired combinations are
4C4 + 5C4+6C4 = 1+5+15 = 21
Total combinations = 7C4 = 35
so Proability that seven games are not played = 21/35 = 3/5 = 0.6
How is this possible ?
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Director
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anandnk wrote: I have a problem with this question let me know where I am wrong.
If number of games played is to be less than 7 then a team can win
either in first 4 games or 4 games in 5 or 4 games in 6
so desired combinations are
4C4 + 5C4+6C4 = 1+5+15 = 21
Total combinations = 7C4 = 35
so Proability that seven games are not played = 21/35 = 3/5 = 0.6
How is this possible ?
I don't know what it is you're exactly trying to show here. Although I have an idea - but in any case it's not making much sense to me.
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I used reverse probability to get the correct answer. However the other method should also give me the same result right.
I would like you to tell me what is wrong with my recent aproach. I am getting 60% instead of 68.75%
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Director
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When you use, for example, 6c4, you ignore the possibility that some of those iterations are AAAABB, or the like-- in other words, you're counting a series where one team loses the first four games and somehow wins the rest.
Since after losing 4 games a team is eliminated, some of those scenarios are impossible.
On the other hand, I'm drunk right now, so I might have misunderstood you...
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Director
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stoolfi wrote: When you use, for example, 6c4, you ignore the possibility that some of those iterations are AAAABB, or the like-- in other words, you're counting a series where one team loses the first four games and somehow wins the rest.
Since after losing 4 games a team is eliminated, some of those scenarios are impossible.
On the other hand, I'm drunk right now, so I might have misunderstood you...
stoolfi - you kill me!  i'm quite drunk myself. however, mr. anandank, i'm giving up 4 minutes of sleep just to show you the error in your method.
there are 4c4 ways of winning 4 out of 4 games. and the probability is 1/2^4 since you're playing 4 games.
secondly, 4 out of 5 games...5c4 right??? WRONG!!!
write them down
LWWWW
WLWWW
WWLWW
WWWLW
WWWWL exclude
so you have 4 secnarios (x2) multiply by 1/2^5
thus 2/(2^4)
Lastly you have 4 out of 6 games
Scenarios
LLWWWW
WLLWWW
WWLLWW
WWWLLW
LWLWWW
WLWLWW
WWLWLW
LWWLWW
WLWWLW
LWWWLW
so you have 10 scenarios x 2 = 20 scenarios multiplied by 1/2^6
thus you have 20/64+8/32+2/16=44/64=0.6875 or 68.75%
But why the heck would you ever want to do it this way?
I'm going to bed! It's freak'n close to 3 am?!
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Thanks a lot fellas It was helpfull I see my mistake now.
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