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Baseball's World Series matches 2 teams against each other

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Baseball's World Series matches 2 teams against each other [#permalink] New post 07 Jan 2004, 00:09
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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%

OPEN DISCUSSION OF THIS QUESTION IS HERE: baseball-s-world-series-matches-2-teams-against-each-other-3120.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 10 May 2014, 04:25, edited 1 time in total.
Edited the question and added the OA.
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Re: PS : World Series [#permalink] New post 07 Jan 2004, 05:57
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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 [#permalink] New post 07 Jan 2004, 06:54
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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 [#permalink] New post 07 Jan 2004, 08:33
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. :P
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 [#permalink] New post 07 Jan 2004, 08:57
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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 [#permalink] New post 07 Jan 2004, 09:03
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, 09:06, edited 1 time in total.
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 [#permalink] New post 07 Jan 2004, 09:06
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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 [#permalink] New post 07 Jan 2004, 09:14
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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 [#permalink] New post 07 Jan 2004, 18:19
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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 [#permalink] New post 07 Jan 2004, 19:31
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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 [#permalink] New post 07 Jan 2004, 19:34
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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 [#permalink] New post 07 Jan 2004, 22:26
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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 [#permalink] New post 07 Jan 2004, 23:44
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! :drunk 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?!

:shocked
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 [#permalink] New post 08 Jan 2004, 05:13
Thanks a lot fellas It was helpfull I see my mistake now.
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Re: Baseball's World Series matches 2 teams against each other [#permalink] New post 09 May 2014, 23:37
Praetorian 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


We are calculating the total number of outcome as 2 x 2 x 2 x 2 x 2 x 2 x 2 = 2^7
But here we assuming that W W W W W W W is a possibility
I don't think the series would continue after 4 wins and hence this is not an outcome at all..
Should we not calculate the total outcomes as : No of ways we can arrange WWWWLLL (Even this is not a possible case but this is what I can think of right now)
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Re: Baseball's World Series matches 2 teams against each other [#permalink] New post 10 May 2014, 04:37
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JusTLucK04 wrote:
Praetorian 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


We are calculating the total number of outcome as 2 x 2 x 2 x 2 x 2 x 2 x 2 = 2^7
But here we assuming that W W W W W W W is a possibility
I don't think the series would continue after 4 wins and hence this is not an outcome at all..
Should we not calculate the total outcomes as : No of ways we can arrange WWWWLLL (Even this is not a possible case but this is what I can think of right now)


We do want to include WWWWWWW case, when we do 1 - P(seven games) approach. This case would be one of the cases attributed to WWWW. The probability of this case is (1/2)^4. If you continue this to 7 games you get:
WWWW WWW --> P=(1/2)^7
WWWW WWL --> P=(1/2)^7
WWWW WLW --> P=(1/2)^7
WWWW LWW --> P=(1/2)^7
WWWW WLL --> P=(1/2)^7
WWWW LWL --> P=(1/2)^7
WWWW LLW --> P=(1/2)^7
WWWW LLL --> P=(1/2)^7
Sum = 8*(1/2)^7 = (1/2)^4. The same probability as we got for WWWW. So, 1 - P(seven games) would still give the correct answer.


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%

Say the teams are A and B. In order the series to consist of seven games each team has to win 3 games in the first 6 games: {A, A, A, B, B, B}. The probability that A wins is 1/2 and the probability that B wins is also 1/2, so the probability of {A, A, A, B, B, B} is (1/2)^6*6!/(3!3!).

Therefore the probability that the series will consist of fewer than 7 games is 1 - (\frac{1}{2})^6*\frac{6!}{3!3!} = 0.6875.

Answer: D.

OPEN DISCUSSION OF THIS QUESTION IS HERE: baseball-s-world-series-matches-2-teams-against-each-other-3120.html
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Re: Baseball's World Series matches 2 teams against each other   [#permalink] 10 May 2014, 04:37
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