Probability Questions: Sheldon Ross 7th ed. Baseball probability

[mattclgn]

Baseball probability question

On the morning of September 30, 1982 the won lost records of the three leading baseball teams were as follows
W L
braves 87 72
giants 86 73
dodgers 86 73

Each team had 3 remaining games to be played. 3 Giants Vs dodgers games 3 braves vs Padres. Suppose all outcomes of the games are independent and equally likely to be won by either participant,. what are the probabilities that each team wins the division.

In the event of a tie, same scenario in a playoff.

Okay, so the answers to this are 38/64, 13/64/, and 13/64

net victories are 15,14,14 ...

I am not really sure how to even approach? do I just do the ideal victory scenario for each one. Do I have to factor in a scenario such as Braves win all three Giants one Dodgers 2 + probability of Braves in all three Giants 3 Dodgers 0? I would imagine I have to use Bayes Theorem somehow before the end?

 

[haruspex]

I don't think it lends itself to any particularly neat methods. You can make use of the giants/dodgers symmetry. From the perspective of the braves, all that matters in their games is whether the same team wins all three, so only two cases to consider.
It's not stated, but I presume the padres are too far behind to have a chance. (On the info given they could have a net score of 13, then win all three.)

 

[WWGD]

Note that the Braves can win without a playoff only by either winning all of their games --guaranteed-- or by winning 2 , if the Giants and Dodgers split the wins. And , since wins by any team are equally-likely, the probability of Giants winning the pennant are the same as the Dodgers'.

Then you need to see the probability of Braves winning in a playoff, having won just one game against the Padres and with the other teams splitting. So, yes, you need to consider all scenarios where Braves win; I don't see where/why Bayes would be used here. And, in the case of a tie, how is the playoff done, is it just a tie-breaking game, or is there something else to be done?

 

[HallsofIvy]

Baseball probability question

On the morning of September 30, 1982 the won lost records of the three leading baseball teams were as follows
W L
braves 87 72
giants 86 73
dodgers 86 73

Each team had 3 remaining games to be played. 3 Giants Vs dodgers games 3 braves vs Padres. Suppose all outcomes of the games are independent and equally likely to be won by either participant,. what are the probabilities that each team wins the division.

Look at the possibilities for records at the end of the season.
If the braves win all three games, then their record will be 90, 72. That has probability (1/2)^3= 1/8.
If the braves win two and lose one, their record will be 89, 73. That has probability 3(1/2)^3= 3/8.
If the braves win one and lose two, their record will be 88, 74. That has probability 3/8 as well.
If the braves lose all three, their record will be 87, 75. That has probability 1/8.
(Of course, those add to 1.)

If the giants win all three games, then their record will be 89, 72 . That has probability (1/2)^3= 1/8. Since they are playing against the dodgers, this same situation leaves the dodgers at 86, 75.
If the giants win two and lose one, their record will be 88, 73 and the dodgers 87, 77. That has probability 3(1/2)^3= 3/8.
If the giants win one and lose two, their record will be 87, 74 and the dodgers 88, 76. That has probability 3/8 as well.
If the giants lose all three, their record will be 86, 75 and the dodgers 89, 75. That has probability 1/8.

The braves win, without a playoff, if
1) They end 90, 72 since the other teams cannot catch them in that case. The probability is 1/8.
2) They end 89, 73 and neither the giants nor the dodgers win all 3 of their games. That second condition has probability 3/8+ 3/8= 6/8= 3/4. The probability the braves with two and lose one was 3/8 so this case has probability (3/4)(3/8)= 9/32.
3) They end 88, 74. In every scenario, either the giants or the dodgers will also have at least 88 so the probability in this case is 0.
The probability the braves win without a playoff is 1/8+ 9/32= 4/32+ 9/32= 13/32.

Do the same analysis to determine the probability the giants and the dodgers will win without a play off.

In the event of a tie, same scenario in a playoff.

There will be a playoff if
1) Both the braves and the giants get 89 wins- probability (3/8)(9/32)= 27/256
2) Both the braves and the dodgers get 89 wins- probability (3/8)(1/8)= 3/64
3) Both the braves and the giants get 88 wins- probability (3/8)(3/8)= 9/64
(in this case the dodgers have 87 wins)
4) Both the braves and the dodgers get 88 wins- probability (3/8)(3/8)= 9/64
(in this case, the giants have 87 wins)
5) Both the braves get 87 wins and one of the other teams 87 then the third team will have more than 87 wins so this is not a possibility.
Further, the giants and dodgers cannot tie.

Apply the same analysis to the play off in cases 1 through 4.

Okay, so the answers to this are 38/64, 13/64/, and 13/64abil

net victories are 15,14,14 ...

I am not really sure how to even approach? do I just do the ideal victory scenario for each one. Do I have to factor in a scenario such as Braves win all three Giants one Dodgers 2 + probability of Braves in all three Giants 3 Dodgers 0? I would imagine I have to use Bayes Theorem somehow before the end?

 

[haruspex]

If the giants win all three games, then their record will be 89, 72 . That has probability (1/2)^3= 1/8. Since they are playing against the dodgers, this same situation leaves the dodgers at 86, 75.
If the giants win two and lose one, their record will be 88, 73 and the dodgers 87, 77. That has probability 3(1/2)^3= 3/8.
If the giants win one and lose two, their record will be 87, 74 and the dodgers 88, 76. That has probability 3/8 as well.
If the giants lose all three, their record will be 86, 75 and the dodgers 89, 75. That has probability 1/8.

As WWGD and I wrote, this part can be simplified. You don't care which of giants/dodgers come out ahead. Whichever, it's a 1/4 chance that the winner will have a 17 point net score, and 3/4 it will be 16 points.
Btw, this does look rather like a homework question. mattclgn, should this have been posted in a different forum?

 

[verty]

@mattclgn:

I have used this book as well. Do the earlier problems if you haven't, that book is very progressive and the difficulty ramps up quite quickly. If you can do all the earlier problems, it shouldn't be too difficult to do this question; IIRC it is similar but about three times longer than the immediately preceding question.

 

[Ray Vickson]

Baseball probability question

On the morning of September 30, 1982 the won lost records of the three leading baseball teams were as follows
W L
braves 87 72
giants 86 73
dodgers 86 73

Each team had 3 remaining games to be played. 3 Giants Vs dodgers games 3 braves vs Padres. Suppose all outcomes of the games are independent and equally likely to be won by either participant,. what are the probabilities that each team wins the division.

In the event of a tie, same scenario in a playoff.

Okay, so the answers to this are 38/64, 13/64/, and 13/64

net victories are 15,14,14 ...

I am not really sure how to even approach? do I just do the ideal victory scenario for each one. Do I have to factor in a scenario such as Braves win all three Giants one Dodgers 2 + probability of Braves in all three Giants 3 Dodgers 0? I would imagine I have to use Bayes Theorem somehow before the end?

I don't get the given answers. Let {Giants, Dodgers} be regarded as a "super" team whose divisional score A is the highest of the giants and dodgers divisional score. Let B be the divisional score of the Braves. We have P(A=88) = 6/8, P(A = 89) = 2/8, while B = 87, 88, 89, 90 with respective probabilities 1/8, 3/8, 3/8, 1/8. The random variables A and B are independent.

Due to Giants, Dodgers symmetry, we have:
P(Braves win) = P(B > A)
P(tie) = P(B = A); P(Braves-Giants tie) = P(Braves-Dodgers tie) = (1/2)P(tie)
P(Giants win) = P(Dodgers win) = (1/2)P(B < A).

These result in different numbers from your listed answers.

 

[PeroK]

I get the answers given as follows:

First, note that the probability the Giants-Dodgers games go 3-0 is 1/4 and 2-1 is 3/4.

The Padres can win in one of four ways:

a) Win all 3 games: P(a) = 1/8 = 8/64

b) Win 2 games and Giants-Dodgers is 2-1: P(b) = (3/8)(3/4) = 9/32 = 18/64

c) Win 2 games, Giants-Dodgers is 3-0 and win play-off: P(c) = (3/8)(1/4)(1/2) = 3/64

d) Win 1 game, Giants-Dodgers is 2-1 and win play-off: (3/8)(3/4)(1/2) = 9/64

So, total probability Padres win is 38/64.

 

[mattclgn]

As WWGD and I wrote, this part can be simplified. You don't care which of giants/dodgers come out ahead. Whichever, it's a 1/4 chance that the winner will have a 17 point net score, and 3/4 it will be 16 points.
Btw, this does look rather like a homework question. mattclgn, should this have been posted in a different forum?

Technically no...or maybe,...I'm independently studying for SOA exam P. Failed my first go, so I thought maybe the problem was that I needed a more difficult textbook. this was problem 73 in the Ross book. Hoping if I go through thoroughly I can get it this time.

 

[mattclgn]

Look at the possibilities for records at the end of the season.
If the braves win all three games, then their record will be 90, 72. That has probability (1/2)^3= 1/8.
If the braves win two and lose one, their record will be 89, 73. That has probability 3(1/2)^3= 3/8.
If the braves win one and lose two, their record will be 88, 74. That has probability 3/8 as well.
If the braves lose all three, their record will be 87, 75. That has probability 1/8.
(Of course, those add to 1.)

If the giants win all three games, then their record will be 89, 72 . That has probability (1/2)^3= 1/8. Since they are playing against the dodgers, this same situation leaves the dodgers at 86, 75.
If the giants win two and lose one, their record will be 88, 73 and the dodgers 87, 77. That has probability 3(1/2)^3= 3/8.
If the giants win one and lose two, their record will be 87, 74 and the dodgers 88, 76. That has probability 3/8 as well.
If the giants lose all three, their record will be 86, 75 and the dodgers 89, 75. That has probability 1/8.

The braves win, without a playoff, if
1) They end 90, 72 since the other teams cannot catch them in that case. The probability is 1/8.
2) They end 89, 73 and neither the giants nor the dodgers win all 3 of their games. That second condition has probability 3/8+ 3/8= 6/8= 3/4. The probability the braves with two and lose one was 3/8 so this case has probability (3/4)(3/8)= 9/32.
3) They end 88, 74. In every scenario, either the giants or the dodgers will also have at least 88 so the probability in this case is 0.
The probability the braves win without a playoff is 1/8+ 9/32= 4/32+ 9/32= 13/32.

Do the same analysis to determine the probability the giants and the dodgers will win without a play off.


There will be a playoff if
1) Both the braves and the giants get 89 wins- probability (3/8)(9/32)= 27/256
2) Both the braves and the dodgers get 89 wins- probability (3/8)(1/8)= 3/64
3) Both the braves and the giants get 88 wins- probability (3/8)(3/8)= 9/64
(in this case the dodgers have 87 wins)
4) Both the braves and the dodgers get 88 wins- probability (3/8)(3/8)= 9/64
(in this case, the giants have 87 wins)
5) Both the braves get 87 wins and one of the other teams 87 then the third team will have more than 87 wins so this is not a possibility.
Further, the giants and dodgers cannot tie.

Apply the same analysis to the play off in cases 1 through 4.

Thank You this was helpful