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# Baseball's World Series matches 2 teams against each other

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Re: Baseball's World Series matches 2 teams against each other  [#permalink]

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11 May 2014, 00:57
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%
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Re: Baseball's World Series matches 2 teams against each other  [#permalink]

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11 May 2014, 04:55
2
2
JusTLucK04 wrote:
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.
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Re: Baseball's World Series matches 2 teams against each other  [#permalink]

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12 May 2014, 22:30
Titleist 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%

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 = $$1 - (1/2)^6 * 6!/(3!*3!)$$
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Re: Baseball's World Series matches 2 teams against each other  [#permalink]

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13 May 2014, 04:56
JusTLucK04 wrote:
VeritasPrepKarishma wrote:
Titleist 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%

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 = $$1 - (1/2)^6 * 6!/(3!*3!)$$

1.Do you mean to imply that the probability of getting a W W W W W W is the same as getting W W W L L L..Similar to H and T case in coin

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.

Quote:
2.
Whats wrong in this approach?
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%

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.
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Re: Baseball's World Series matches 2 teams against each other  [#permalink]

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13 May 2014, 06:38
VeritasPrepKarishma wrote:

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
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Re: Baseball's World Series matches 2 teams against each other  [#permalink]

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14 May 2014, 01:08
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JusTLucK04 wrote:
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)
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Baseball's World Series matches 2 teams against each other  [#permalink]

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23 Aug 2015, 21:46
I did this using the following method

4 W #ways = 1 prob = (1/2)^4
4 W + 1 L #ways = 5!/4! = 5 (exclude 1 way which is WWWWL) prob =4* (1/2)^5
4 W + 2 L #ways = 6!/4!2! = 15 (exclude 5 ways same as previous count) prob = 10*(1/2)^6

Total prob = (1/2)^4 + 4*(1/2)^5 + 10*(1/2)^6
= (1/2)^4 [ 1 + 4/2 + 10/4]
= (1/16) *[22/4]
= 22/64 = 11/32

now this can happen for either teams so net prob = 2x 11/32 = 11/16 = 68.75%
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Re: Baseball's World Series matches 2 teams against each other  [#permalink]

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19 Jul 2016, 20:18
4 games have to be played anyhow to determine if a team wins or lose as per the question.So minimum of 4 games or maximum of 6 games(fewer than 7).
4 games:-A team either A or B wins 3 games and the fourth game has to be won by either of the two teams to conclude the game within 4 games.
{WWW}W=the first 3 games can be arranged in any order=3!/3! Followed by the 4th game that has to be won and its position is fixed.
So,3!/3!*1/2^4=1/16,since the game can be won by either of the teams=2*1/16=1/8.
5 games-This is only possible in case of 3 wins and 1 loses,and the following game will be won by either of the teams.
3 wins and 1 los can be arranged in different manner,followed by 5th game-win for either A or B in which its position will remain fixed.
{WWWL}W=4!/3!1!*1/2^5=4/32=1/8*2=1/4(for either of the two teams winning the game).
6 games-This is only possible when there is a win of 3 games and 2 losses by either of the teams,followed by the final game which has to be won by either of the teams.
{WWWLL}W=5!/3!2!*1/2^6=10/64*2(either of the 2 teams winning the game)=5/16.
So total probability=1/8+1/4+5/16=11/16=0.6875.

OR

P(event happens) + P(event doesn't happen) = 1.

In a best-of-seven series, either 4, 5, 6, or 7 games are played.

P(event happens) = P(fewer than 7 games) = P(4, 5 or 6 games)
P(event doesn't happen) = P(not fewer than 7 games) = P(exactly 7 games)

Since either fewer than 7 games are played (meaning 4, 5, or 6 games) or exactly 7 games are played, the sum of the two probabilities above must be 1
P(fewer than 7 games) + P(exactly 7 games) = 1

So P(fewer than 7 games) = 1 - P(exactly 7 games).

To determine P(exactly 7 games), let's say that the teams playing are team A and team B. The problem implies that each team has an equal chance of winning a game. The only way that 7 games will be played is if A wins 3 of the first 6 games and B wins the other 3.

P(AAABBB) = 1/2 * 1/2 * 1/2 * 1/2 * 1/2 * 1/2 = 1/64
Since any arrangement of AAABBB will yield a good outcome, we have to account for all the different ways we could arrange AAABBB: 6!/(3!*3!) = 20
Multiplying the results above, we see that:
P(each team wins 3 of the first 6 games) = 1/64 * 20 = 5/16.
Thus, P(exactly 7 games) = 5/16.

So P(fewer than 7 games) = 1 - 5/16 = 11/16 = 68.75%.
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Re: Baseball's World Series matches 2 teams against each other  [#permalink]

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17 Jan 2018, 07:42
Bunuel wrote:
riskietech 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%

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%

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 $$1 - (\frac{1}{2})^6*\frac{6!}{3!3!} = 0.6875$$.

I almost get the right answer, I think of 1 - (1/2)^7 * (7!) / [(4!)(3!)]
now, I understand that there is no need for the result of the last game.
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Re: Baseball's World Series matches 2 teams against each other  [#permalink]

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18 Jan 2018, 16:11
Hi - I did the following but I'm not sure if it's correct.

Total possible ways is 2^7 = 128

Now, total ways to choose 4 wins out of 7 is 7C4 = 7!/(4!*3!) = 35

Probability of at least 6 wins = 1- P(win all 7) = 1 - (35/128) = 73%.
Re: Baseball's World Series matches 2 teams against each other &nbs [#permalink] 18 Jan 2018, 16:11

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