Baseball's World Series matches

[rajeev.bajpai]

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%

Hi, This is my first post, hope this is the right forum for MGMAT challenging problems related Qs.

Please explain why the answer is not 30/70? (1/2)^7 as explained in the answer considers a different set. In my opinion, The total possibilities are:

4-0 (Combinations 2*1=2), 4-1(2*4=8), 4-2(2*5c2 = 20) and 4-3(2*20=40)

Fewer than 7 games would have probability 30/70.

The given answer considers even the results like 5-0,6-1 etc.

Thanks.

[rajeev.bajpai]

Can an instructor help me resolve this confusion please.

Thanks.

[shaji]

This is quite intriguing!!!. A choice between the devil and the deep blue sea!!!;because all the given answers are incorrect.
The closest of the choices to the correct answer is C.
What is the source of the problem and please re-check the answer choices and revert and I shall be happy to explain this matter further.

[shaji]

Can an instructor help me resolve this confusion please.

Thanks.

I am very sorry!!!. The correct answer indeed is D. Please ignore my earlier post.
Solution:
The quickest mode is reverse gear, meaning consider that 7 games need to be played to decide the championship. The Probabilty of any one team winning is (2*(20/128)) if seven games need to be played.
The required prob is 1-(2*(20/128)); which is D.

[rajeev.bajpai]

Shaji,
This is from Manhattan Challenging problems and I know they calculated the probability this way.

I agree that (1 - prob for rest of the events) is a good method in many occassions, but it does not seem to be the case here. By applying this logic in this case, we are dealing with the problem as if all seven games were played with any possible outcomes of seven games (6-1, ,7,0, 5-2 etc) - That is when you get 128 in the denominator (2^7).

The total possibilities set is restricted here by condition of winning 4 games. So your set changes to only those possibilities where winning team wins no more than 4 games. The set become 4-0,4-1,4-2,4-3 in favor of any team. Now all probabilities should be calculated by cosidering only this set.

Coming back to the the way answer D is calculated, what if I change the question slightly? What if I remove the 4 games condition so that all games are played? And I ask the probability of not having a 4-3 final score? You would still be applying the same logic and getting the same answer. But that would be correct because you dealt within the right set, considering all possibilities like 6-1,7-0,5-2 scores.

[shaji]

Rajeev;
"By applying this logic in this case, we are dealing with the problem as if all seven games were played with any possible outcomes of seven games (6-1, ,7,0, 5-2 etc) - That is when you get 128 in the denominator (2^7).". This is NOT true. notice the 20 possibilities(numerator) elliminates all those cases.

"The total possibilities set is restricted here by condition of winning 4 games. So your set changes to only those possibilities where winning team wins no more than 4 games. The set become 4-0,4-1,4-2,4-3 in favor of any team. Now all probabilities should be calculated by cosidering only this set". U excell yourself!!!. This is precisely crux of the matter. Please note that the sum of the 4 possibilities cited you is the required probability. I suggest the reverse gear since you will have to evaluate only one probability which would take just 1/4th the time, which is very crucial for tests like GMAT. Quick estimation is one of the importnat skills of management aptitude.

[rajeev.bajpai]

Shaji, Because you are interested in solving my problem, I would explain further to help you understand my point.

I added a question for you in my last post,
"What if I remove the 4 games condition so that all games are played? And I ask the probability of not having a 4-3 final score?"
Answering this would have hepled you understand my question.

Try a simple one,
In the original question (with the condition that the first team to win 4 games wins the tournament and no subsequent games are played), in how many ways can the series be played? (hint: 4-0 score can be achieved in 2 ways,4-1 can be achieved in 2*4=8 ways,4-2 in 20 ways, 4-3 in 40 ways).

Let me know if you find answer to the above question. If you find a total 70 possible ways, try to calculate the probability of a 4-0 score. If you come up with 2/70 as your answer, next think why it is not same as {2*(1/2)^4} that would help you understand what I am trying to ask.

Thanks.

[shaji]

"Because you are interested in solving my problem" Yes!!!; that's pretty much the size of it. The probability of a 4-0 result is NOT 2/70 but is indeed {2*(1/2)^4} .
I would recommend you write down all possibilities(sometimes known as the "brute force" method) and do a complete audit to satisy yourself that 2/70 is incorrect!!!. I stongly recommed the "brute force" technique to enable U to get a hang of probability principles.

"What if I remove the 4 games condition so that all games are played? And I ask the probability of not having a 4-3 final score?"Answering this would have hepled you understand my question. Yes!!!; again I am aware of 'your problem". Removing 4 games condition is IRELEVANT since it makes the problem very trivial as it removes the 'challenge' entirely. The required probability for the trivial situation is((total number of possibilities if all 7 games are played-((4-3))/total number of possibilities))=(128-70)/128.

[rajeev.bajpai]

I tried to simplify the problem as much as I could but it just doesnt work :)

Please do not post a response now because I really want an instructor to have a look. The discussion between us may lead them to think that I do not need their help anymore.

btw, Good luck with your preparation!! I would work on my probability knowedge and learn the principles as you advised.

Thanks for all your help!!

[RonPurewal]

hi all -

answer (d) is indeed correct. first i'll put forward a formal explanation; then i'll debunk the objection about invalid outcomes.

--

FORMAL EXPLANATION:

let's just go ahead and calculate, DIRECTLY, the probabilities of all of the possible outcomes. this way there can be no possible argument.

4 GAMES
the probability that one team wins the series in 4 games is (1/2)^4, or 1/16. since either team could pull this off, that's 2 * 1/16, or 1/8.

5 GAMES
for the series to last 5 games, the first four games must be 3-1. there are four ways in which this can happen, from the perspective of the winning team (note W = win, L = loss):
WWWL, WWLW, WLWW, LWWW. (each of these would be followed by the series-ending fourth W, in the fifth game)
the probability of each of these events is (1/2)^5, or 1/32, so the combined probability of this set of four is 4/32 = 1/8. since there are two teams, that's 2/8, or 1/4 chance that the series will last 5 games.

6 GAMES
for the series to last 6 games, the first five games must be 3-2. there are 5!/(3!2!) = 10 ways in which that can happen, from the perspective of the winning team...or you could just list them:
WWWLL
WWLWL
WLWWL
LWWWL
WWLLW
WLWLW
LWWLW
WLLWW
LWLWW
LLWWW
each of these is followed by another "W" for the sixth game.
so that's (1/2)^6 = 1/64 for each possibility; so 10/64 per team; so 20/64 = 5/16 overall for a length of 6 games.

add them: 1/8 + 1/4 + 5/16 = 11/16 = 0.6875, so (d) is correct.

--

if you do the 1 - prob(seven games) calculation, YES it is true that you are including series such as 6-1 and so on.
but this is actually ok.

here's the reason: (this is sort of subtle, but i'll give it a shot since this is the challenge-problem crowd)
when you do the 1 - x calculation, you're tacitly representing every possible outcome of the series as "7 games". but the thing is - this will mean that series turning out as 6-1, 5-2 etc. will actually be representations of series that would actually be truncated after fewer games. therefore, you DO want to include these series.
for instance:
one possible five-game series is WWLWW, whose probability is (1/2)^5, or 1/32.
if you run this out to seven games, there are four "possibilities": WWLWWLL, WWLWWLW, WWLWWWL, and WWLWWWW. if you add up the probabilities of these four "possibilities", then you get 4 * (1/2)^7, which is still 1/32.
so you're still good.

[700+]

4 GAMES
the probability that one team wins the series in 4 games is (1/2)^4, or 1/16. since either team could pull this off, that's 2 * 1/16, or 1/8.

5 GAMES
for the series to last 5 games, the first four games must be 3-1. there are four ways in which this can happen, from the perspective of the winning team (note W = win, L = loss):
WWWL, WWLW, WLWW, LWWW. (each of these would be followed by the series-ending fourth W, in the fifth game)
the probability of each of these events is (1/2)^5, or 1/32, so the combined probability of this set of four is 4/32 = 1/8. since there are two teams, that's 2/8, or 1/4 chance that the series will last 5 games.

I'm a bit confused. In a series of 4 games, for a team to win the series, the team would have to win 3 games. So there 4 ways (4!/3!) in which a team could win the series of 4 games. So should not the probability that 1 team wins the series be 4 x (1/16) = 1/4?

[shaji]

hi all -

answer (d) is indeed correct. first i'll put forward a formal explanation; then i'll debunk the objection about invalid outcomes.

--

FORMAL EXPLANATION:

let's just go ahead and calculate, DIRECTLY, the probabilities of all of the possible outcomes. this way there can be no possible argument.

4 GAMES
the probability that one team wins the series in 4 games is (1/2)^4, or 1/16. since either team could pull this off, that's 2 * 1/16, or 1/8.

5 GAMES
for the series to last 5 games, the first four games must be 3-1. there are four ways in which this can happen, from the perspective of the winning team (note W = win, L = loss):
WWWL, WWLW, WLWW, LWWW. (each of these would be followed by the series-ending fourth W, in the fifth game)
the probability of each of these events is (1/2)^5, or 1/32, so the combined probability of this set of four is 4/32 = 1/8. since there are two teams, that's 2/8, or 1/4 chance that the series will last 5 games.

6 GAMES
for the series to last 6 games, the first five games must be 3-2. there are 5!/(3!2!) = 10 ways in which that can happen, from the perspective of the winning team...or you could just list them:
WWWLL
WWLWL
WLWWL
LWWWL
WWLLW
WLWLW
LWWLW
WLLWW
LWLWW
LLWWW
each of these is followed by another "W" for the sixth game.
so that's (1/2)^6 = 1/64 for each possibility; so 10/64 per team; so 20/64 = 5/16 overall for a length of 6 games.

add them: 1/8 + 1/4 + 5/16 = 11/16 = 0.6875, so (d) is correct.

--
This is indeed the the forward gear approach including the 'brute force'

if you do the 1 - prob(seven games) calculation, YES it is true that you are including series such as 6-1 and so on. This is NOT true. 6-1 is NOT included!!. See the forward gear approach!. (6-1) connot occur as per the problem, since the games stop the moment any team wins 4 games.
but this is actually ok. This is the fallacy!!!

here's the reason: (this is sort of subtle, but i'll give it a shot since this is the challenge-problem crowd)
when you do the 1 - x calculation, you're tacitly representing every possible outcome of the series as "7 games". but the thing is - this will mean that series turning out as 6-1, 5-2 etc. will actually be representations of series that would actually be truncated after fewer games. therefore, you DO want to include these series.
for instance:
one possible five-game series is WWLWW, whose probability is (1/2)^5, or 1/32.
if you run this out to seven games, there are four "possibilities": WWLWWLL, WWLWWLW, WWLWWWL, and WWLWWWW. if you add up the probabilities of these four "possibilities", then you get 4 * (1/2)^7, which is still 1/32.
so you're still good.

I tried to simplify the problem as much as I could but it just doesnt work :)

Please do not post a response now because I really want an instructor to have a look. The discussion between us may lead them to think that I do not need their help anymore.

btw, Good luck with your preparation!! I would work on my probability knowedge and learn the principles as you advised.

Thanks for all your help!!

4 GAMES
the probability that one team wins the series in 4 games is (1/2)^4, or 1/16. since either team could pull this off, that's 2 * 1/16, or 1/8.

5 GAMES
for the series to last 5 games, the first four games must be 3-1. there are four ways in which this can happen, from the perspective of the winning team (note W = win, L = loss):
WWWL, WWLW, WLWW, LWWW. (each of these would be followed by the series-ending fourth W, in the fifth game)
the probability of each of these events is (1/2)^5, or 1/32, so the combined probability of this set of four is 4/32 = 1/8. since there are two teams, that's 2/8, or 1/4 chance that the series will last 5 games.

I'm a bit confused. In a series of 4 games, for a team to win the series, the team would have to win 3 games. So there 4 ways (4!/3!) in which a team could win the series of 4 games. So should not the probability that 1 team wins the series be 4 x (1/16) = 1/4?

[tim]

700+, this is not a "series of four games", but a series where the first team to win four games is the champion. So the only way the series can end after four games is if the same team wins all four of them..

Shaji, it looks like all you did was quote several people in your post without adding any content of your own. If you still have a question about this one, can you make it clear what your question is?