Baseball's World Series matches 2 teams against each other

The correct answer is 75%.

You have a total of eight different possibilities:

4-0 in favor of Team A
4-1 in favor of Team A
4-2 in favor of Team A
4-3 in favor of Team A
4-0 in favor of Team B
4-1 in favor of Team B
4-2 in favor of Team B
4-3 in favor of Team B

So the total number of choices for the game to end with less than 7 games = 6 different possibilities

Therefore % less than 7 games = 6/8 = 75%

In order to determine the probability that the World Series will last less than 7 games, we can first determine the probability that the World Series WILL last exactly 7 games and then subtract this value from 1.

In order for the World Series to last exactly 7 games, the first 6 games of the series must results in 3 wins and 3 losses for each team.

Let's analyze one way this could happen for Team 1:

Game 1 Game 2 Game 3 Game 4 Game 5 Game 6
T1 Wins T1 Wins T1 Wins T1 Loses T1 Loses T1 Loses


There are many other ways this could happen for Team 1. Using the permutation formula, there are 6!/(3!)(3!) = 20 ways for Team 1 to win 3 games and lose 3 games in the first 6 games.

Logically, there are also 20 ways for Team 2 to win 3 games and lose 3 games in the first 6 games.

Thus, there are a total of 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:



Thus the probability that the World Series will last less than 7 games is 100% - 31.25% = 68.75%.

The correct answer is D.

I have answer 42.86% but not matching the option.

Explanation:
Let us find probability that game will consist of exactly 7 games.
So there are 4 wins & 3 losses, but out of these wins, one win has to be last
Ex: WWWLLL W
possibilities = 6!/(3!*3!) = 20 (dividing so as to eliminate repetitions)

Now for all possible events, there are only four sub-possibilities : 1) exactly 4 games played, 2) exactly 5 games played, 3) exactly 6 games played, 4) exactly 7 games played
1) 4 games = total possibilities = 1 (WWWW)
2) 5 games = total possibilities = 4!/(3!*1!) = 4 (for ex: WWWL W)
3) 6 games = total possibilities = 5!/(3!*2!) = 10 (for ex: WWWLL W)
7) 7 games = as explained above = 20
So total possible outcomes = 1 + 4 + 10 + 20 = 35
prob of exact 7 games = 20/35 = 57.14%

prob of fewer than 7 games = 100 - 57.14 = 42.86%

Here's how I did it, but I don't think its correct

So to finish in less than 7 games it will finish in either 4,5 or 6 games.

4 games will be (1/2) ^ 4 = 1/16, one team winning the four games
5 games will be (5C4)(1/2^5) = 5/32
6 games will be (6C4)(1/2^6)=15/64

Adding em up, 29/64 close enough to the correct answer choice

Now, i'm not too sure whether I should have considered that for instance in the case of 4 games we would have 2 cases, 1/16 if one team wins and 1/16 if another team wins, so total probability would be 2/16 = 1/8?

Same with the other scenarios

Could anybody please clarify this? Appreciate it!

Thanks!
Cheers
J

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.

We do want to include such cases as WWWWWWW, 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.

Bunuel..I have no clue about what you are trying to convey here...
I will show you the calculations which seem logical to me

Ways in which series can occur-->
1. No L: WWWW = 1
2. 1 L and W at the end= WWWL-W = 4!/3! = 4
3. 2 L and W at the end= WWWLL-W = 5!/(3!*2!) = 10
4. 3 L and W at the end= WWWLLL-W = 6!(3!*3!) =20

Total number of outcomes: 35
Favorable outcomes is case 4: 20

P for Seven game series is 20/35 = 57.14%

4 games:

WWWW --> (1/2)^4. Since either team can win, then for this case the probability is 2*(1/2)^4 = 1/8.

5 games:

LWWWW, WLWWW, WWLWW, WWWLW --> 4*(1/2)^5. Since either team can win, then for this case the probability is 2*4*(1/2)^5 = 1/4.

6 games:

WWWLLW, WWLWLW, WLWWLW, LWWWLW, WWLLWW, WLWLWW, LWWLWW, WLLWWW, LWLWWW, LLWWWW --> 10*(1/2)^6. Since either team can win, then for this case the probability is 2*10*(1/2)^6 = 5/16.

Overall the probability = 1/8 + 1/4 + 5/16 = 11/16 = 06875. The same answer as we got above.

Hope it's clear now.

Responding to a pm:

There are two teams A and B. We assume there are no draws (though it is usually mentioned in the question). They play against each other in a best of 7 series. That is, the one team which wins more matches out of the 7 played, wins. Also, if a team wins 4 matches say team A wins first 4 matches played, team A is announced the winner and other 3 matches do not take place. Say team A wins 3 matches and team B wins 2 matches out of 5. Then, if team A wins the 6th match, team A is the winner and the 7th match is not played.

So they need to play a minimum of 4 matches (all 4 won by the same team) and a maximum of 7 matches (4 won by 1 team and 3 won by the other)

We need to find
P(Less than 7 games) = 1 - P(7 games)

7 games will be played when the win matrix of 6 games looks something like this {A, A, A, B, B, B} or {A, A, B, A, B, B} or {B, A, B, A, B, A} etc...
That is each team would have won exactly 3 games. You can arrange 3 As and 3 Bs in 6!/(3!*3!) ways.

What is the probability that 7 games will be played? A wins 3 and B wins 3. Each team can win a match with a probability of (1/2).

Probability of getting {A, A, A, B, B, B} = (1/2)*(1/2)*(1/2)*(1/2)*(1/2)*(1/2)

Probability that 7 games will be played = \((1/2)^6 * 6!/(3!*3!)\)

Probability that less than 7 games will be played = \(1 - (1/2)^6 * 6!/(3!*3!)\)

First note that every game has a Win and a Loss. One team wins and the other loses. So I am not sure what you mean by W W W W W W. I am assuming that this is the set for a particular team. If so, note that if one team wins 4 matches, no more matches are played. So this is not valid. If your point is just whether the probability of W W W is the same as that of W W L then yes, it is. After all the probability of winning a match for a team is 1/2. So in either case, the probability of this will be (1/2)^3. It is similar to the Heads and Tails of a coin.




I think Bunuel has already pointed out the flaw in this approach. You need to consider that either team could get this result and hence adjust accordingly.

Case 1: Only 4 games are played: WWWW - team A could win all 4 or team B could win all 4. Total 2 cases. Probability is 2*(1/2)^4 = 1/8

Case 2: 5 games are played: LWWWW, WLWWW, WWLWW, WWWLW - Again, this set can belong to either team so 2 cases. Probability is 2*4*(1/2)^5 = 1/4.

Case 3: 6 games are played: 4 wins and 2 losses arranged in 6!/(4!*2!) = 15. But we need to exclude the 5 ways discussed above in cases 1 and 2: WWWWLL, LWWWWL, WLWWWL, WWLWWL, WWWLWL (note that we added Ls at the end to make the 2 loss scenario). The reason we exclude these cases is that in such cases there will be only 4 or 5 matches. 6 matches will not be played. Again, this set can belong to either team so 2 cases. Probability is 2*10*(1/2)^6 = 5/16.

Overall the probability = 1/8 + 1/4 + 5/16 = 11/16 = 0.6875.

Even if I consider either cases...the no of events will be twice in each case and the probability will remain the same..
I think the issue is in the basic understanding that the probability of happening of W W W W is not the same as W W W L W and hence I cannot just add the cases to find the probability...AM I RIGHT in my understanding here
What I assumed is ..I am considering each individual case..so the probability of happening of each case is the same..and hence I can add all the cases and divide them by the total number of possible cases to get the probability...
In your approach we calculate the probability of happening of each type of case and then add all the probabilities as they are mutually exclusive..So the different answers..

Please correct me if I am wrong

Note that the 2 is immaterial if you are calculating favorable cases and total cases. In case you are working with probability, (1/2)^4 = 1/16 needs to be multiplied by 2 to get 1/8.

You use favorable cases/total cases only when the probability of each case is the same. The probability of WWWW is the same as the probability of WWWL but the probability of WWWW (1/16) is not the same as the probability of LLWWWW (1/64)

4 games = 2 cases
5 games = 8 cases
6 games = 20 cases
7 games = 40 cases.
Less than 7 cases = 30/70 = 42.8

We are asked the prob., please see what's wrong with that?

Bunuel
KarishmaB

I have similar query: 
Total cases are ''70 and favorable are 30 out of them. Hence should be 3/7.

4 games = 2 cases
5 games = 8 cases
6 games = 20 cases
7 games = 40 cases.
Less than 7 cases = 30/70 = 42.8­

The point is that these cases are not equally weighted. There are four times as many 5-game cases as there are 4-game cases. However, the combined probability of 4-game cases is 1/8, while the probability of 5-game cases is 1/4, which is only twice as many. So, we cannot simply add favorable cases and divide by the total number of cases.­