There are infinitely many correct answers. Here are a couple, copied or adapted from responses turned in:
- I can get to Lawrence via I-70 or 24/40. I-70 is fast, but has a toll; 24/40 is quicker, but has a toll. 24/40 is free, but relatively slow.
- On Saturday, I can either do my recycling or do food shopping. Recycling will be quick and easy, but food shopping will take a long time because the grocery store will be crowded and there will be long lines.
Again, here are a couple of correct answers copied or adapted from responses turned in:
- I can either get up when the alarm goes off, or hit the snooze button. If I get up, I will be on time. If I don’t, I might be on time, but I might be late.
- When Jeanette gets dressed for school this morning she comes to the problem of footwear. Knowing she has to work in the shop on a project, she can either wear closed-toed sneakers that afford comfort and protection for her feet, or she can wear her sandals that, although comfortable, do not protect her feet at all. She just has some finish painting to do on the model she has built, so it is unlikely that she will drop anything heavy on her feet, but nonetheless the danger is still present in the environment.
Again, here are a couple of correct answers copied or adapted from responses turned in:
- I can either bet that the card I draw will be higher than my three friends’ cards, or not. If I don’t bet, I neither win nor lose. If I do bet, I believe that I have a 25-percent chance of winning and a 75-percent chance of losing.
- You’re at El Mezcal on a Friday night and it is packed. If you stay, there is a 10-percent chance that you will be seated within 30 minutes and a 90-percent chance that you won’t. If you leave and go to La Familia, there is a 40-percent chance that you will be seated faster and a 60-percent chance that you won’t.
There are infinitely many correct answers. Here are a couple, copied or adapted from responses turned in:
- Suppose you prefer Honda to Toyota, prefer Ford to Chevy, are indifferent between Mercedes and BMW, and are indifferent between Toyota and Chevy. Using the letters ‘P’ and ’I’, along with letters or other short names for the mentioned options (you don’t have to use the letters a, b, c, . . .), concisely represent these preferences.
- Suppose you prefer shoes to a nice dinner, prefer DVDs to shoes, are indifferent between a nice dinner and a night at the bar, are indifferent between an Xbox game and DVDs, and prefer any previous option to buying a Michael Moore statue. Using the letters ‘P’ and ’I’, along with letters or other short names for the mentioned options (you don’t have to use the letters a, b, c, . . .), concisely represent these preferences.
Here’s the most commonly submitted sound proof. There are other sound ones, too.
# claim justification 1 j P k given 2 k P m given 3 m P n given 4 n P j given 5 j P m 1 and 2, transitivity 6 j P n 5 and 3, transitivity 7 contradiction 4 and 6, completeness
Again, here are a couple of correct answers copied or adapted from responses turned in:
- I prefer a 40-minute nap to a 20-minute one, a 60-minute nap to a 40-minute one, a 90-minute nap to a 60-minute one, a 120-minute nap to a 90-minute nap, and no nap to a 120-minute one.
- I wish to place a bet on which of the following three baseball teams will miss the playoffs this year: Chicago, Detroit, and Minnesota. I want to place the bet based on these teams’ records versus each other. Chicago is 10–5 versus Detroit, Detroit is 10–5 versus Minnesota, and Minnesota is 10–7 versus Chicago.
There are infinitely many correct answers. Here is one adapted from one that was turned in:
Suppose we have eight items that we’ve labeled a, b, c, . . . h. Arrange these items in an ordered list that represents the preferences stated below. Put any tied items in alphabetical order. State which character or typographical features (e.g., comma, semicolon, line break, etc.) you are using as your separator for ‘P’, and which for ‘I’.
e P d
c P b
a P d
e I g
f I a
b P e
h I c
b P fThese preferences can be represented by the following ordered list, where a line break is the separator for ‘P’ and the comma is the separator for ‘I’:
c, h
b
a, e, f, g
d(Correction posted September 22: Actually, these preferences cannot be represented by an ordered list, because they are not complete, because they do not indicate the relative standing of a and f relative to e and g. Some relation between a or f, and e or g, would need to be added in order to make the preferences complete. For example, a I e would suffice to correct the problem, and then the answer given above would be correct. Or a P e would suffice to correct the problem, but then we would have an ordered list of five rows rather than four.)
Again, there are infinitely many correct answers. Here is one adapted from one that was turned in:
Here is a list of preferences. For ease of reference, they are numbered. What is the largest number n that makes the following sentence true? “The preferences numbered 1 through n can all be represented by an ordered list.”
1. × P ■
2. ▲ P ☼
3. × P ☼
4. ● P ▲
5. ☼ P ■
6. ▲ I ×n = 5
(Correction posted September 22: Actually, all of the stated preferences can be represented by an ordered list, so the correct answer is 6. Line 2 was supposed to say ▲ I ☼, and in that case 5 would be the correct answer.)
There are infinitely many correct answers. Here is one adapted from one that was turned in:
x u(x) Boondock Saints 6 Never Ending Story 5 Full Metal Jacket 4 Legends of the Fall 3 Ferris Bueller’s Day Off 2 Apocalypse Now Redux 1
Again, there are infinitely many correct answers. Here is one that was turned in:
x u(x) Boondock Saints 10 Never Ending Story 9.9996 Full Metal Jacket 9.9995 Legends of the Fall 9.9994 Ferris Bueller’s Day Off 9.979 Apocalypse Now Redux 9.978
The answer is 2.
The answer is 4.
Here is an answer adapted from one that was turned in:
You can either go to the gym or go swim at the lake. If you go to the gym and it’s not crowded, you’ll have a good time, whereas if it’s crowded, you’ll have a bad time. If you go swim at the lake, then you’ll have a great time if it doesn’t rain and a terrible time if it does. Here’s the matrix for this situation:
gym not crowded, and no rain gym not crowded, and rain gym crowded, and no rain gym crowded, and rain go to gym good good bad bad go swim at lake great terrible great terrible
S_{1} S_{2} S_{3} S_{4} A_{1} 3 9 4 4 A_{2} 4 5 8 4 A_{3} 2 4 9 6
- maximin
- A_{1}’s minimum = 3
- A_{2}’s minimum = 4
- A_{3}’s minimum = 2
- So, A_{2} would be selected by this rule.
- maximax
- A_{1}’s maximum = 9
- A_{2}’s maximum = 8
- A_{3}’s maximum = 9
- So, A_{1} and A_{3} would be selected by this rule.
- optimism/pessimism, with an optimism index of 3/5
- α-index for A_{1} = (3/5)(9) + (1 – 3/5)(3) = (3/5)(9) + (2/5)(3) = 27/5 + 6/5 = 33/5
- α-index for A_{2} = (3/5)(8) + (1 – 3/5)(4) = (3/5)(8) + (2/5)(4) = 24/5 + 8/5 = 32/5
- α-index for A_{3} = (3/5)(9) + (1 – 3/5)(2) = (3/5)(9) + (2/5)(2) = 27/5 + 4/5 = 31/5
- So, A_{1} would be selected by this rule.
- maximizing expected utility using the principle of insufficient reason
- EU(A_{1}) = (1/4)(3) + (1/4)(9) + (1/4)(4) + (1/4)(4) = 3/4 + 9/4 + 4/4 + 4/4 = 20/4
- EU(A_{2}) = (1/4)(4) + (1/4)(5) + (1/4)(8) + (1/4)(4) = 4/4 + 5/4 + 8/4 + 4/4 = 21/4
- EU(A_{3}) = (1/4)(2) + (1/4)(4) + (1/4)(9) + (1/4)(6) = 2/4 + 4/4 + 9/4 + 6/4 = 21/4
- So, A_{2} and A_{3} would be selected by this rule.
S_{1} S_{2 }(1/5) S_{3} A_{1} 3 9 2 A_{2} 4 2 7 A_{3} 4 5 4 A_{4} 3 7 3
- maximin
- A_{1}’s minimum = 2
- A_{2}’s minimum = 2
- A_{3}’s minimum = 4
- A_{4}’s minimum = 3
- So, A_{3} would be selected by this rule.
- maximax
- A_{1}’s maximum = 9
- A_{2}’s maximum = 7
- A_{3}’s maximum = 5
- A_{4}’s maximum = 7
- So, A_{1} would be selected by this rule.
- optimism/pessimism, with an optimism index of 1/2
- α-index for A_{1} = (1/2)(9) + (1 – 1/2)(2) = (1/2)(9) + (1/2)(2) = 9/2 + 2/2 = 11/2
- α-index for A_{2} = (1/2)(7) + (1 – 1/2)(2) = (1/2)(7) + (1/2)(2) = 7/2 + 2/2 = 9/2
- α-index for A_{3} = (1/2)(5) + (1 – 1/2)(4) = (1/2)(5) + (1/2)(4) = 5/2 + 4/2 = 9/2
- α-index for A_{4} = (1/2)(7) + (1 – 1/2)(3) = (1/2)(7) + (1/2)(3) = 7/2 + 3/2 = 10/2
- So, A_{1} would be selected by this rule.
- maximizing expected utility using the principle of insufficient reason
- EU(A_{1}) = (2/5)(3) + (1/5)(9) + (2/5)(2) = 6/5 + 9/5 + 4/5 = 19/5
- EU(A_{2}) = (2/5)(4) + (1/5)(2) + (2/5)(7) = 8/5 + 2/5 + 14/5 = 24/5
- EU(A_{3}) = (2/5)(4) + (1/5)(5) + (2/5)(4) = 8/5 + 5/5 + 8/5 = 18/5
- EU(A_{4}) = (2/5)(3) + (1/5)(7) + (2/5)(3) = 6/5 + 7/5 + 6/5 = 19/5
- So, A_{2} would be selected by this rule.
EU($300) > EU(30-percent chance at $950 and 70-percent chance at $0)
u($300) > (30/100)u($950) + (70/100)u($0)
100u($300) > 30u($950) + 70u($0)
100[u($0) + x] > 30[u($0) + x + y] + 70u($0)
100u($0) + 100x > 30u($0) + 30x + 30y + 70u($0)
100u($0) + 100x > 100u($0) + 30x + 30y
100x > 30x + 30y
70x > 30y
(7/3)x > y
y < (7/3)x
u($0) = 0
u($300) = 10
u($950) = 11
EU(60-percent chance at $900 and 40-percent chance at $0) > EU($550)
(60/100)u($900) + (40/100)u($0) > u($550)
60u($900) + 40u($0) > 100u($550)
60[u($0) + x + y] + 40u($0) > 100[u($0) + x]
60u($0) + 60x + 60y + 40u($0) > 100u($0) + 100x
100u($0) + 60x + 60y > 100u($0) + 100x
60x + 60y > 100x
60y > 40x
y > (2/3)x
u($0) = 0
u($550) = 1
u($900) = 11
u($500) = u($100) + x
u($800) = u($500) + y_{1} = u($100) + x + y_{1}
EU($500) > EU(60-percent chance at $800 and 40-percent chance at $100)
u($500) > (60/100)u($800) + (40/100)u($100)
u($100) + x > (60/100)[u($100) + x + y_{1}] + (40/100)u($100)
100[u($100) + x] > 60[u($100) + x + y_{1}] + 40[u($100)]
100u($100) + 100x > 60u($100) + 60x + 60y_{1} + 40u($100)
100u($100) + 100x > 100u($100) + 60x + 60y_{1}
100x > 60x + 60y_{1}
40x > 60y_{1}
x > (3/2)y_{1}
y_{1} < (2/3)x
u($500) = u($100) + x
u($700) = u($500) + y_{2} = u($100) + x + y_{2}
EU(50-percent chance at $700 and 50-percent chance at $100) > EU($500)
(50/100)u($700) + (50/100)u($100) > u($500)
(50/100)[u($100) + x + y_{2}] + (50/100)u($100) > u($100) + x
50[u($100) + x + y_{2}] + 50[u($100)] > 100[u($100) + x]
50u($100) + 50x + 50y_{2} + 50u($100) > 100u($100) + 100x
100u($100) + 50x + 50y_{2} > 100u($100) + 100x
50x + 50y_{2} > 100x
50y_{2} > 50x
y_{2} > x
If you prefer more money to less, then y_{2} should be less than y_{1} because y_{2} is the utility interval between $500 and $700, whereas y_{1} is the utility interval between $500 and $800. The utility interval from $500 to $700 should be less than the utility interval from $500 to $800. The inequalities derived in the answers to problems 20 and 21 are not consistent with this.
Here’s a good answer:
C_{1} | C_{2} | |
R_{1} | 6 | 7 |
R_{2} | 4 | 3 |
Here’s a good answer:
C_{1} | C_{2} | C_{3} | |
R_{1} | 6 | 2 | 3 |
R_{2} | 8 | 2 | 1 |
R_{3} | 4 | 1 | 7 |
C_{1} | C_{2} | |
R_{1} | 6 | 4 |
R_{2} | 3 | 6 |
solving for p:
EU(C_{1}) = EU(C_{2})
p(6) + (1 – p)(3) = p(4) + (1 – p)(6)
6p + 3 – 3p = 4p + 6 – 6p
3p + 3 = –2p + 6
5p = 3
p = 3/5solving for q:
EU(R_{1}) = EU(R_{2})
q(6) + (1 – q)(4) = q(3) + (1 – q)(6)
6q + 4 – 4q = 3q + 6 – 6q
2q + 4 = –3q + 6
5q = 2
q = 2/5
C_{1} | C_{2} | |
R_{1} | 6 | 7 |
R_{2} | 9 | 4 |
solving for p:
EU(C_{1}) = EU(C_{2})
p(6) + (1 – p)(9) = p(7) + (1 – p)(4)
6p + 9 – 9p = 7p + 4 – 4p
–3p + 9 = 3p + 4
–6p = –5
p = 5/6solving for q:
EU(R_{1}) = EU(R_{2})
q(6) + (1 – q)(7) = q(9) + (1 – q)(4)
6q + 7 – 7q = 9q + 4 – 4q
–q + 7 = 5q + 4
–6q = –3
q = 1/2
Game A | C_{1} | C_{2} |
R_{1} | 5, 6 | 7, 2 |
R_{2} | 9, 3 | 4, 1 |
Game B | C_{1} | C_{2} |
R_{1} | 5, 6 | 3, 9 |
R_{2} | 7, 1 | 4, 2 |
Game C | C_{1} | C_{2} |
R_{1} | 5, 6 | 8, 3 |
R_{2} | 4, 2 | 9, 4 |
Game C is a battle of wills.
Game B is a prisoner’s dilemma.
The answers to these problems are in a separate PDF file. Its URL is http://web.ku.edu/~utile/courses/rct2/homework_2_problems_for_credit_29-31.pdf.
It is (16 – 7)/(18 – 7), or 9/11.
- What is the row player’s relative concession when u_{1} = 14?
Well, when u_{1} = 14 the row player’s relative benefit is (14 – 7)/(18 – 7), or 7/11, so the row player’s relative concession is 1 – 7/11, or 4/11.
It is (12 – 8)/(14 – 8), or 4/6, or 2/3.
- what is the column player’s relative benefit?
It is (12 – 6)/(13 – 6), or 6/7.
It is (7 – 3)/(10 – 3), or 4/7.
- what is the column player’s relative benefit?
It is (14 – 12)/(21 – 12), or 2/9.
The following problems refer to the following situation. Randy, Paula, and Simon are important figures on a highly, somewhat inexplicably, successful television program. When their contracts are being renewed for the next season, Randy can command a salary of $1,000,000, Paula a salary of $2,000,000, and Simon a salary of $1,500,000. Following the example of the stars of Friends, they find that if they band together in pairs, they can get the producers of the show to pay them even more, and that if they negotiate as a trio, they do even better. It also happens that for them, utility is directly proportional to dollar amounts. Specifically, their situation has the following characteristic function:
Finally, assume that imputations list Randy’s utility first, then Paula’s, then Simon’s.
To show that the {Randy, Paula, Simon} coalition satisfies the superadditivity condition, we have to show that the value of the {Randy, Paula, Simon} coalition is greater than the sum of the values of any two coalitions that could form it. There are three pairs of coalitions that could form the {Randy, Paula, Simon} one. Let’s consider them each in turn.
First, the {Randy, Paula, Simon} coalition has a value of 6,000,000, which is greater than the sum of Randy’s value of 1,000,000 and the {Paula, Simon} coalition’s value of 3,800,000.
Second, the {Randy, Paula, Simon} coalition has a value of 6,000,000, which is greater than the sum of Paula’s value of 2,000,000 and the {Randy, Simon} coalition’s value of 3,700,000.
Third, the {Randy, Paula, Simon} coalition has a value of 6,000,000, which is greater than the sum of Simon’s value of 1,500,000 and the {Randy, Paula} coalition’s value of 3,600,000.
From the answer to problem 35, we can see that pairs of coalitions that can form the {Randy, Paula, Simon} coalition have total values of 4,800,000, 5,700,000, and 5,100,000. The largest of these sums of 5,700,000. So, assuming that the second producer said something true, the first producer must have mentioned a dollar amount less than $5,700,000. That is, x < 5,700,000.
Randy is complaining about a violation of the individual-rationality condition.
Here’s one correct answer: (1,300,000, 2,300,000, 1,500,000).
profile 1 | profile 2 | |||||||
Larry | Moe | Curly | society | Larry | Moe | Curly | society | |
a | b | c | b | b | d | c | d | |
b | c | d | c | a | b | d | c | |
c | a | b | c | a | b | b | ||
d | d | a | b | c | a | a |
correction (sent to e-mail distribution list on Friday, November 3): In profile 2, Larry’s last-place alternative should be d, not b.
answer: b
answer: c
answer: b
Here’s one. (Keep an eye on alternatives a and b to see the violation of condition I.)
profile 1 | profile 2 | |||||||||
Jerry | George | Elaine | Kramer | society | Jerry | George | Elaine | Kramer | society | |
a | d | c | b | a, b, c, d | a | d | c | c | c, d | |
b | a | d | c | b | a | d | d | a | ||
c | b | a | d | c | b | a | b | b | ||
d | c | b | a | d | c | b | a |
Yes, we can know that it does, because it associates, with each profile, a social ordering that—because it is identical with some individual’s preference ordering—is complete and transitive.
- Can we know, in advance of the random selection of the individuals and the specification of the resulting social ordering, whether this social welfare function satisfies condition CS? Why or why not?
No, we cannot know that. It might satisfy CS—it’s possible that, among the r social orderings, there will be one in which every strict-preference paid occurs (a P b, b P a, a P c, c P a, etc.). But it is also possible that some strict-preference pair won’t occur in any of the r social orderings, in which case CS will be violated.
- Can we know, in advance of the random selection of the individuals and the specification of the resulting social ordering, whether this social welfare function satisfies condition I? Why or why not?
No, we cannot know that. It might satisfy I—it’s possible that, for every pair of profiles in which there is consistency among the citizens’ rankings of any two alternatives, the ranking of those two alternatives will be consistent in the social orderings, too. But it is also possible that there will be some pair of profiles, and some pair of alternatives, such that the citizens’ ranking of those two alternatives is consistent from one profile to the next, but the corresponding social orderings do not rank them consistently.
profile 1 | profile 2 | |||||||
Larry | Moe | Curly | society | Larry | Moe | Curly | society | |
a | b | c | c | a | b | c | c | |
b | c | d | b | b | a | d | b | |
c | a | b | a | c | c | b | d | |
d | d | a | d | d | d | a | a |
answer: b
answer: a
answer: c