University of Kansas, Fall 2005
Philosophy 666: Rational Choice Theory
Ben Eggleston—eggleston@ku.edu

Solutions to problems from Resnik’s Choices

section 1-1:

1. characterizations (note: these five scenarios are open to interpretation):
   1. individual, game
   2. group
   3. individual, game
   4. group
   5. individual, game
2. presumably not epistemically rational, but possibly instrumentally rational (which is our concern)
3. No—even the goal of ceasing to be rational could be a rational goal (at least, an instrumentally rational one).

section 1-2:

1. table:

2. table:

The dominance principle supports Pascal’s reasoning because leading the life of a religious Christian is better in some states that being a pagan, and worse in none.

section 1-3:

1. characterizations (again, these scenarios are open to some interpretation):
   1. risk
   2. ignorance
   3. certainty (about immediate result); or ignorance (long-term result)
   4. certainty
   5. risk (statistical expectation); or ignorance (particular case)
2. table:

The outcome description you do not get sick is in more than one square because there is more than one act-state pair leading to it. The total probability that you will get an outcome so described (assuming you do not ensure it altogether by not eating there) is 65%.

section 1-4:

1. decision tree (should be represented graphically):
   1. accept $10,000 with no lawyer, etc.
   2. hire lawyer and demand $50,000
      1. get $25,000 offer and take it
      2. get $10,000 offer again
         1. accept $10,000 without suing
         2. sue
            1. win $50,000
            2. end up with nothing
2. strategies and table:
   1. S₁: do not hire lawyer; accept $10,000
   2. S₂: hire lawyer and demand $50,000; then accept either other ($25,000 or $10,000)
   3. S₃: hire lawyer and demand $50,000; then sue unless offered $25,000

section 2-1:

1. three scales
   1. first scale—several possible answers
      1. equally spaced: 20, 21.25, 22.5, 23.75, 25
      2. also fine: 20, 21, 22, 23, 25
      3. It would have been nice (but maybe too easy) for the new range to have been specified as 20–24 rather than 20–25, leading to the obvious answer of 20, 21, 22, 23, 24.
   2. second scale—again, several possible answers
      1. equally spaced: –5, –3.75, –2.5, –1.25, 0
      2. also fine: –5, –4, –3, –1, –0
      3. It would have been nice (but maybe too easy) for the new range to have been specified as –4 to 0 rather than –5 to 0, leading to the obvious answer of –4, –3, –2, –1, 0.
   3. third scale—again, several possible answers; one is 0.1, 0.2, 0.3, 0.4, 0.5
2. When each number is multiplied by –1, the scale gets reversed; this is unacceptable. When each number is multiplied by 0, everything is mergedd into one big indifference class; this, too, is unacceptable.
3. proof that if t satisfies condition c, and if u is an ordinal utility scale, then t satisfies conditions a and b:
   1. proof that t satisfies condition a—that t(u(x)) > t(u(y)) if and only if xPy:
      1. proof that if t(u(x)) > t(u(y)), then xPy:

2. proof that
if xPy, then t(u(x)) > t(u(y)):

Mark Wine came up with a much shorter proof with some ingenious moves in it. (This doesn’t just shorten each of the 16-line proofs from above, one by one; it replaces both of them—32 lines cut down to 6!!) Here it is:

2. proof that t satisfies condition b—that t(u(x)) = t(u(y)) if and only if xIy:
   1. proof that if t(u(x)) = t(u(y)), then xIy:

2. proof that if xIy, then t(u(x)) = t(u(y)):

4. To show that a function is an ordinal transformation, we must show that w ≥ v if and only if t(w) ≥ t(v).
   1. First, we consider t(x) = x – 2. We prove each half of the biconditional (the one stated above) in turn.
      1. proof that if w ≥ v, then t(w) ≥ t(v):

2. proof that if t(w) ≥ t(v), then w ≥ v:

2. Second, we consider t(x) = 3x + 5. We prove each half of the biconditional (the one stated above) in turn.
   1. proof that if w ≥ v, then t(w) ≥ t(v):

2. proof that if t(w) ≥ t(v), then w ≥ v:

5. Is t(x) = x² an ordinal transformation when applied to a scale whose numbers are greater than or equal to 0? Yes, Proof:
   1. proof that if w ≥ v, then t(w) ≥ t(v):

2. proof that t(w) ≥ t(v), then w ≥ v:

If some of the numbers are negative, then t(x) = x² is no longer an ordinal transformation. For example, if w = –3 and v = –4, then w ≥ v, but it is not the case that w² = 9 is greater than v² = 16, and so it is not the case that t(w) ≥ t(v).

6. It is impossible to use an ordinary finite number scale to represent the agent’s preferences because there are an infinite number of indifference classes.

If we could use the real numbers and concern ourselves with only moments of life or moments of death throes, then we could represent the agent’s preferences by assigning to any outcome the number of years (or days, or seconds) it lasts. For life, more is better; for death throes, subtract the time they last from the maximum of two weeks and, again, more is better (once you’ve done the subtraction).

section 2-2:

1. two tables
   1. table 2-5: act A₂
   2. table 2-6: acts A₁ and A₃
2. No, because the worst possible outcome from doing a facelift is worse than the worst possible outcome from not doing a facelift.
3. We must also know that the odds against S₂ occurring are not too long—that S₂ is not so unlikely that the $1.75-vs.-$10,000 comparison is essentially moot.
4. Here is a table showing that the maximin rule does not exclude a dominated act:

(A₁ is dominated, but not excluded by the maximin rule.)

Here is a table—based on the work of Mark Wine and Nathan Grubb—showing not only that the maximin rule does not necessarily exclude dominated acts, but also that it can exclude every act except a dominated one:

	S1	S2	S3
A1	0	1	3
A2	0	0	2

(A₂ is dominated, but is selected by the maximin rule, since that rule excludes A₁.)

Here is a table showing that the lexical maximin rule does not exclude a dominated act:

	S1	S2	S3
A1	6	6	7
A2	6	7	7

(A₁ is dominated, but not excluded by the lexical maximin rule.)

The table based on the work of Mark Wine and Nathan Grubb, above, also shows that the lexical maximin rule can exclude every act except a dominated one (excluding A₁ and uniquely selecting A₂, even though A₁ dominates A₂.)

section 2-3:

1. For decision table 2-19, the regret table is

The minimax-regret act is A₂, since its maximum regret value (5) is less than that of A₁ (21) and less than that of A₃ (26).

For decision table 2-20, the regret table is

A1	0	0	4
A2	8	12	0
A3	1	15	1

The minimax-regret act is A₁, since its maximum regret value (4) is less than that of A₂ (12) and less than that of A₃ (15).

2. Consider the following decision table:

Then the corresponding regret table is

	S1	S2	S3
A1	4	2	0
A2	4	1	0
A3	0	0	6

So neither A₁ nor A₂ is excluded (only A₃ is, with it maximum regret value of 6); but A₁ is dominated by A₂, so A₁ is a dominated act, and yet it is not excluded by the minimax regret rule.

3. When we apply a nonpositive linear transformation (i.e., u’ = au + b for a ≤ 0), then either the values of the act-state pairs get reversed (if a < 0) or the values of the act-state pairs all become the same (if a = 0).
4. Suppose, as stipulated, that we can convert one decision table into another by means of a positive linear transformation. To prove the desired result, we can proceed in either of (at least) two ways. First, we can do things the super-explicit, but super-tedious, way.

A. We begin by representing the first decision table as follows:

Thanks to the stipulated assumption, we can represent the second decision table as follows:

Now, if we let max (u_{_,x}) refer to the maximum value attained by u in any row of column x (whatever x may be), then we can represent the regret table for the first table as follows:

And we can represent the regret table for the second table as follows:

This equals the following:

This, in turn, can be rewritten as follows:

And this, in turn, can be simplified to the following:

Since a > 0, this, in turn, can be rewritten as follows:

And this, in turn, can be simplified to the following:

And this represents the regret table for the second decision table as desired, namely, with each element being a multiple of the corresponding elements in the regret table for the first decision table (table 3).

So that’s the very explicit, but tedious way of doing things. A shorter approach, but one demanding a little more care and precision along the way, follows. (It’s based on proofs I saw from Wes Smith and Kara Tan Bhala.)

B. Suppose we can convert one decision table into another by means of a positive linear transformation. How we can show that the regret table for the first decision table can be converted into the regret table for the second decision table by means of a transformation of the form u’ = au, with a ≥ 0? What we want to show, in effect, is that for every act-state pair, R’ = aR, for some a ≥ 0. Here is how we can establish this:

no.	claim	justification
1	For every act-state pair (or cell) in the second decision table, R’ = u’ – max’, where (1) u’ is the value of the act-state pair in question in the second decision table and (2) max’ is the maximum possible value associated with that state in the second decision table	definition of R’
2	For every act-state pair (or cell) in the second decision table, R’ = (au + b) – (a*max + b), where (1) a > 0, (2) u is the value of the act-state pair in question in the first decision table, and (3) max is the maximum possible value associated with that state in the first decision table	1, positive linear transformation converting first decision table into second decision table
3	For every act-state pair (or cell) in the second decision table, R’ = au + b – a*max – b, where the conditions stated in line 2 continue to hold.	2, simplification of equation
4	For every act-state pair (or cell) in the second decision table, R’ = au – a*max, where the conditions stated in line 2 continue to hold.	3, simplification of equation
5	For every act-state pair (or cell) in the second decision table, R’ = a(u – max), where the conditions stated in line 2 continue to hold.	4, rewriting of equation
6	For every act-state pair (or cell) in the second decision table, R’ = aR, where (1) a > 0 and (2) R is the regret score for the act-state pair in question in the first decision table.	5, definition of regret score
7	For every act-state pair (or cell) in the second decision table, R’ = aR, where (1) a ≥ 0 and (2) R is the regret score for the act-state pair in question in the first decision table.	6, plus the fact that if a > 0, then a ≥ 0

So that’s a somewhat more precise way of proceeding, but demands considerable precision in the way that the conditions are stated. In particular, you have to be careful about how you define max’ and max, since each of these is just the maximum value for its column (the maximum value associated with a particular state), not the maximum value for its whole table.

5. proof that if the scale u can be converted into the scale u’ by means of a positive linear transformation, and ditto for u’ and u’’, then ditto for u and u’’:

6. This problem involves finding a linear transformation that converts the values of any decision table into one whose largest value is 1 and whose smallest value is 0. Here’s one answer:

A. To convert any decision table into one whose largest value is 1 and whose smallest value is 0, follow these two steps. First make the lowest value 0, by subtracting the smallest entry from all the entries. Then divide all the (adjusted) entries by the (new value of the) largest entry, which will be the difference between the original maximum and the original minimum. In other words, if max is the largest value and min is the smallest, transform each entry in the table using the following formula:

u’ = (u – min)/(max – min).

To see that this is a positive linear transformation, notice that it can be written as u’ = u/(max – min) – min/(max – min), and think of this as an instance of the formula u’ = au + b. In order for our linear transformation to be a positive one, we need what corresponds to a—namely, 1/(max – min)—to be positive. And all this requires is that the denominator (max – min) be positive, which is assured as long as not all the values of the original table are the very same number.

So that’s one answer. But a more explicit way of arriving at the same result was presented by Mark Wine. Here is his approach:

B. To derive the transformation we need, we will start by defining some variables:

max =_df the maximum value of the original table (as above)

min =_df the minimum value of the original table (as above)

max’ =_df the maximum value of the transformed table (we know this will turn out to be 1)

min’ =_df the minimum value of the transformed table (we know this will turn out to be 0)

Now, since we are going to derive a positive linear transformation that takes us from the original table to the transformed table, we know that max’ and min’ are related to max and min by equations of the form

(1)    u’ = au + b

So, max’ and min’ are related to max and min by these equations:

(2) max’ = a(max) + b, and

(3) min’ = a(min) + b, for some a > 0

Note that a will be the same for both equations, and so will b: we’re dealing with a single transformation that will convert max into max’, and min into min’. Thus, the stipulation that a > 0 applies to both equations, and it’s there because we are deriving a positive linear transformation.

Since the largest value in the transformed table is going to be 1, and the smallest is going to be 0, we know these equations:

(4) max’ = 1

(5) min’ = 0

By combining equations 1 and 3, we get equation 5; and by combining equations 2 and 4, we get equation 6:

(6) a(max) + b = 1, and

(7) a(min) + b = 0, for a > 0

Now we can solve for the variables a and b in terms of the values max and min. (Remember, in a given situation, you would know the actual values of max and min. So don’t think of them as variables to be solved for; think of them as parameters you would know in any particular situation. Think of them as knowns, not unknowns.) Equation 7 implies

(8) b = –a(min)

Substituting this into equation 6, we get

(9) a(max) – a(min) = 1

This means that

(10) a(max – min) = 1

or

(11) a = 1/(max – min)

So now we have solved for a in terms of max and min. To solve for b, let’s substitute our formula for a (stated in equation 11) into equation 8:

(12) b = –[1/(max – min)]*(min)

This yields

(13) b = –min/(max – min)

Substituting our formulas for a and b (equations 11 and 13) into our original formula for our transformation as expressed in equation 1, we have

(14) u’ = [1/(max – min)]*u + [–min/(max – min)]

Through a couple of simplifying steps, we have the following equations:

(15) u’ = u/(max – min) – min/(max – min)

(16) u’ = (u – min)/(max – min)

Note that 16 is the transformation mentioned in solution A to this problem. As in that solution, we know that this linear transformation is a positive linear one because the coefficient of u—namely, 1/(max – min)—is positive as long as max > min, which we are assured of as long as there are two unequal numbers in the entire original decision table.

section 2-4:

1. a-indexes
   1. table 2-23
      1. For act A₁, the a-index is (1/2)*4 + (1/2)*0 = 2 + 0 = 2.
      2. For act A₂, the a-index is (1/2)*2 + (1/2)*1 = 1 + .5 = 1.5.
   2. table 2-24
      1. For act A₁, the a-index is (1/2)*10 + (1/2)*0 = 5 + 0 = 5.
      2. For act A2, the a-index is (1/2)*8 + (1/2)*7 = 4 + 3.5 = 7.5.
2. proof that the transformation of table 2-23 into table 2-24 is ordinal but not positive linear
   1. proof that the transformation of table 2-23 into table 2-24 is ordinal: The 2-23 numbers (0, 4, 2, and 1) have the same relative order (4th, 1st, 2nd, and 3rd) as the 2-23 numbers (0, 10, 8, and 7). This shows that, if the transformation from table 2-23 to table 2-24 is characterized as a function t(x), then for all w and v among the 2-23 numbers, and for all t(w) and t(v) among the 2-24 numbers, w ≥ v if and only t(w) ≥ t(v). And this means that the transformation is an ordinal one.
   2. proof that the transformation of table 2-23 into table 2-24 is not positive linear: Suppose there were a positive linear transformation from table 2-23 to table 2-24. Then for every u’ in table 2-24, u’ = au + b, with a > 0, for some corresponding u in table 2-23. Now suppose we solve for a and b using the first row of table 2-23 and the first row of table 2-24. (We need to use two elements from each table because we need two equations in order to solve for our two variables, a and b. And it just so happens that each row contains two elements.) Then we have 0 = a*0 + b, and we have 10 = a*4 + b. The first equation implies that b = 0; substituting this into the second equation, we can derive a = 2.5. So we have u’ = 2.5u + 0, or u’ = 2.5u; this equation successfully transforms the first two elements of table 2-23 into the first two elements of table 2-24, and is the only linear one that does so. But a moment’s investigation reveals that this equation does not successfully transform the third element of table 2-23 (i.e., 2) into the third element of table 2-23 (i.e., 8). This is sufficient to show that our sole possible positive linear transformation from table 2-23 into table 2-24 does not work; but for good measure, we could show the mismatch for the last elements of the tables, too (i.e., you do not get 7 by multiplying 1 by 2.5). Since the sole possible positive linear transformation turns out to fail upon further inspection, there is no positive linear transformation that converts table 2-23 into table 2-24.
3. an example of a decision table for which the optimism-pessimism rule fails to exclude a dominated act no matter what the agent’s optimism index is:

(The trick is to notice that the optimism-pessimism rule ignores intermediate values—i.e., every value that is not the minimum or the maximum—so two acts can be equally good, from the point of view of the optimism-pessimism rule, even if one act dominates the other in virtue of differences among intermediate values.)

4. proof that if the a-index for an act A is greater than or equal to that for an act B, it remains so when the utilities for the acts A and B are subjected to the same positive linear transformation:

5. [many reasonable answers]

section 2-5:

1. To show that the principle of insufficient reason never recommends a dominated act, we’ll prove the following: if A₁ is a dominated act, then it’s not recommended by the principle of insufficient reason.

2. There are infinitely many correct answers to this question; here is one of them.

Consider this decision table:

The principle of insufficient reason selects act A₁, since its expected utility of 3 is greater than A₂’s expected utility of 3.

Now consider this decision table:

	S1	S2
A1	2	4
A2	0	100

Clearly it is an ordinal transformation of the previous decision table—only the highest utility has been changed. But now the principle of insufficient reason selects act A₂, since its expected utility of 50 is greater than A₁’s expected utility of (as seen above) 3.

3. To show that the shortcut version of the principle of insufficient reason (i.e., summing across the rows) is equivalent to the original formulation, we’ll prove that, for any n (where n is the number of states),

(1/n)*u_1,1 + (1/n)*u_1,2 + (1/n)*u_1,3 + . . . + (1/n)*u_1,n ≥ (1/n)*u_2,1 + (1/n)*u_2,2 + (1/n)*u_2,3 + . . . + (1/n)*u_2,n

if and only if

u_1,1 + u_1,2 + u_1,3 + . . . + u_1,n ≥ u_2,1 + u_2,2 + u_2,3 + . . . + u_2,n

And this can be proved by noticing that the second inequality can be derived from the first by multiplying each side by n; and that the first inequality can be derived from the second by dividing each side by n.

section 2-6

1. We need to show that if the optimism-pessimism rule is indifferent between A₁, A₂, and A₃, then the agent’s optimism index is 1/2. Here is a proof:

2. We need to show two things: (1) if the agent’s optimism index is greater than 1/2, then A₁ and A₂ are both picked by the optimism-pessimism rule; and (2) if the agent’s optimism index is less than 1/2, then A₃ is picked.

Here is a proof of the first claim:

no.	claim	justification
1	The agent’s optimism index (a) is greater than 1/2.	assumption for proving conditional
2	The a-indexes for the acts are as follows: A1: a*1 + (1 – a)*0 A2: a*1 + (1 – a)*0 A3: a*(1/2) + (1 – a)*(1/2)	definition of a-index
3	The a-indexes for the acts are as follows: A1: a A2: a A3: a/2 + 1/2 – a/2	2, algebra
4	The a-indexes for the acts are as follows: A1: a A2: a A3: 1/2	3, algebra
5	The a-indexes for A1 and A2 are equal, and are greater than the a-index for A3.	4, with information from 1
6	The optimism-pessimism rule selects A1 and A2.	5, definition of optimism-pessimism rule

Here is a proof of the second claim:

3. See section 2-2, problem 4.
4. [many reasonable answers]

section 2-7

1. Here is a table for picking societies behind the veil of ignorance (simplified to consider just the Rawls and Harsanyi options and just three states in which a person might end up).

2. [see text]
3. [many reasonable answers]
4. [several reasonable answers]

section 3-2

1. one card drawn
   1. P(ace) = 4/52 = 1/13
   2. P(ace of hearts) = 1/52
   3. P(ace of hearts or king of hearts) = P(ace of hearts) + P(king of hearts) – P(ace of hearts and king of hearts) = 1/52 + 1/52 – 0 = 2/52 = 1/26
   4. P(ace or heart) = P(ace) + P(heart) + P(ace and heart) = 4/52 + 13/52 – 1/52 = 16/52 = 4/13
2. two cards drawn (first one not replaced)
   1. P(ace of hearts, then king of hearts) = P(ace of hearts) * P(king of hearts / ace of hearts) = 1/52 * 1/51 = 1/2,652
   2. P(two aces) = P(ace, then ace) = P(ace) * P(ace / ace) = 4/52 * 3/51 = 12/2,652 = 1/221
   3. P(no ace on either draw) = P(no ace, then no ace) = P(no ace) * P(no ace / no ace) = 48/52 *47/51 = 2,256/2652 = 188/221
   4. P(at least one ace and at least one heart): two approaches
      1. One kind of approach is to try to divide the ways the specified outcome might happen into several mutually exclusive events whose probabilities are easier to ascertain, and add them up. Here is an approach of this kind:
         P(at least one ace and at least one heart)
         = P(ace of hearts, then anything else) + P(anything else, then ace of hearts) +
         P(non-hearts ace, then non-ace heart) + P(non-ace heart, then non-hearts ace)
         = P(ace of hearts) * P(anything else / ace of hearts) + P(anything else) * P(ace of hearts / anything else) +
         P(non-hearts ace) * P(non-ace heart / non-hearts ace) +  P(non-ace heart) * P(non-hearts ace / non-ace heart)
         = (1/52 * 51/51) + (51/52 * 1/51) + (3/52 * 12/51) + (12/52 * 3/51)
         = 51/2,652 + 51/2,652 + 36/2,652 + 36/2,652
         = 174/2,652
         = 29/442
      2. Another kind of approach is to try to use formulas pertaining to probabilities to simplify the resulting computations. Here is an approach of this second kind:
         P(at least one ace and at least one heart)
         = 1 – P(no aces or no hearts)
         = 1 – [P(no aces) + P(no hearts) – P(no aces and no hearts)]
         = 1 – [P(no ace, then no ace) + P(no heart, then no heart) – P(neither an ace nor a heart, then neither an ace nor a heart)]
         = 1 – [P(no ace) * P(no ace / no ace) + P(no heart) * P(no heart / no heart)
         – P(neither an ace nor a heart) * P(neither an ace nor a heart / neither an ace nor a heart)
         = 1 – [48/52 * 47/51 + 39/52 * 38/51 – 36/52 * 35/51]
         = 1 – [2,256/2,652 + 1,482/2,652 – 1,260/2,652]
         = 1 – 2,478/2,652
         = 174/2,652
         = 29/442
3. In order for p and q to be mutually exclusive, it would have to be impossible for p and q to occur; that is, it would have to be the case that P(p & q) = 0. But P(p & q) = 1/4 > 0, so p and q are not mutually exclusive. We can compute P(p or q) as follows: P(p or q) = P(p) + P(q) – P(p & q) = 1/2 + 1/2 – 1/4 = 3/4.
4. Let p be the probability of rolling a 1. Then the other probabilities are 2p, 3p, 4p, 5p, and 6p. These must sum to 1, so we have 21p = 1, or p = 1/21. Then we have P(rolling an odd number) = P(rolling a 1, 3, or 5) = P(rolling a 1) + P(rolling a 3) + P(rolling a 5) = 1/21 + 3/21 + 5/21 = 9/21 = 3/7.
5. Here is a proof of the desired result:

6. Here is a proof of the desired result:

7. Taking the approach of Bayes’s theorem (but deriving it intuitively rather than jumping straight to the formula), we have the following:
   P(p/q)
   = P(both) / P(q)
   = P(p & q) / P(p & q or not p & q)
   = P(p & q) / [P(p & q) + P(not p & q)]
   = P(p) * P(q/p) / [P(p) * P(q/p) + P(not p) * P(q / not p)]
   = 1/4 * 1 / [1/4 * 1 + 3/4 * 1/5]
   = 1/4 / [1/4 + 3/20]
   = 1/4 / 8/20
   = 1/4 * 20/8
   = 20/32
   = 5/8
8. 800 type I urns with 6 blue and 4 red; 200 type II urns with 1 blue and 9 red
   1. P(type I) = 800 / (800 + 200) = 800/1,000 = 4/5
   2. P(red) = (800 * 4 + 200 * 9) / (800 * 10 + 200 * 10) = 5,000/10,000 = 1/2
   3. P(blue) = 1 – P(red) = 1 – 1/2 = 1/2
   4. P(type I / blue): two approaches
      1. One approach is to start with the expression to be evaluated and to manipulate it until we have expressions all of whose values we know:
         P(type I / blue)
         = P(both) / P(blue)
         = P(type I & blue) / P(blue)
         = P(type I) * P(blue / type I) / P(blue)
         = 4/5 * 6/10 / 1/2
         = 4/5 * 3/5 / 1/2
         = 12/25 / 1/2
         = 12/25 * 2
         = 24/25
      2. Another approach is to get some extra practice with Bayes’s theorem by not plugging in any values until we derive that formula:
         P(type I / blue)
         = P(both) / P(blue)
         = P(type I & blue) / P(type I & blue or not type I & blue)
         = P(type I & blue) / [P(type I & blue) + P(not type I & blue)]
         = P(type I) * P(blue / type I) / [P(type I) * P(blue / type I) + P(not type I) * P(blue / not type I)
         = 4/5 * 6/10 / [4/5 * 6/10 + 1/5 * 1/10]
         = 4/5 * 3/5 / [4/5 * 3/5 + 1/5 * 1/10]
         = 12/25 / [12/25 + 1/50]
         = 12/25 / [24/50 + 1/50]
         = 12/25 / [25/50]
         = 12/25 / 1/2
         = 12/25 * 2
         = 24/25
   5. P(type I / red): two approaches
      1. Again, one approach is to start with the expression to be evaluated and to manipulate it just until we have expressions all of whose values we know:
         P(type I / red)
         = P(both) / P(red)
         = P(type I & red) / P(red)
         = P(type I) * P(red / type I) / P(red)
         = 4/5 * 4/10 / 1/2
         = 4/5 * 2/5 / 1/2
         = 8/25 * 1/2
         = 8/25 *2
         = 16/25
      2. And, as before, another approach is to derive Bayes’s theorem and then plug values into that formula:
         P(type I / red)
         = P(both) / P(red)
         = P(type I & red) / P(type I & red or not type I & red))
         = P(type I & red) / [P(type I & red) + P(not type I & red)]
         = P(type I) * P(red / type I) / [P(type I) * P(red / type I) + P(not type I) * P(red / not type I)]
         = 4/5 * 4/10 / [4/5 * 4/10 + 1/5 * 9/10]
         = 4/5 * 2/5 / [4/5 * 2/5 + 1/5 * 9/10]
         = 8/25 / [8/25 + 9/50]
         = 8/25 / [16/50 + 9/50]
         = 8/25 / [25/50]
         = 8/25 / 1/2
         = 8/25 *2
         = 16/25
9. “Suppose you could be certain that the urn in the last example is of type II. Explain why seeing a blue ball drawn from the urn would not produce a lower posterior probability for the urn being of type II.” There are at least a couple of approaches to this problem. To see why there is some sort of problem here, notice that, intuitively, if you draw a blue ball, then you might be tempted to think that your urn is of type I, since blue balls are found much more frequently in urns of type I than in urns of type II, and your posterior probability that the urn is of type II might diminish accordingly. So we want to see what sort of posterior probability you get for the urn being of type II if you are certain, in advance, that it is of type II—that is, if your prior probability is that it is of type II is 1. As I said, there are at least a couple of approaches to this problem.
   1. One approach is to look at our formulas (inverse square law) and Bayes’s theorem one by one and notice that if P(type II) = 1, then P(type II / blue) = 1, too. In each case, we’ll derive the rule in question before applying it, just for practice.
      1. derivation and application of inverse square law
         P(type II / blue)
         = P(both) / P(blue)
         = P(type II and blue) / P(blue)
         = P(type II) * P(blue / type II) / P(blue)
         = 1 * 1/10 / 1/10
         = 1/10 / 1/10
         = 1
         Notice that when we substitute 1/10 for P(blue), this is because we know the urn is of type II—if we didn’t know this, we would have to substitute 1/2 for P(blue), as found in 8c, above.
      2. derivation and application of Bayes’s theorem
         P(type II / blue)
         = P(both) / P(blue)
         = P(type II & blue) / P(type II & blue or not type II & blue)
         = P(type II & blue) / [P(type II & blue) + P(not type II & blue)]
         = P(type II) * P(blue / type II) / [P(type II) * P(blue / type II) + P(not type II) * P(blue / not type II)]
         = 1 * 1/10 / [1 * 1/10 + 0 * 6/10]
         = 1/10 / [1/10 + 0]
         = 1/10 / 1/10
         = 1
         Again, we use 1/10 rather than 1/2 for P(blue), since we know the urn is of type II. For similar reasons, we use 0 for P(not type II)—we know it’s an urn of type II, so the probability that it’s not an urn of type II is just 0.
   2. Another approach is to focus on the probability that the urn is of type I rather than the probability that the urn is of type II. (This helps us think about the probability that the urn is of type II, albeit indirectly, because type I and type II exhaust the possibilities.) As mentioned above, both the inverse square law and Bayes’s theorem represent any posterior probability as proportional to the corresponding prior probability. So, according to either rule, the posterior probability P(type I / [some information]) is proportional to the prior probability P(type I). That is, the posterior probability is P(type I) times some other stuff. If we know that P(type I) is 0, then P(type I) times some other stuff—whatever that other stuff may be—is also going to be 0. So, according to either rule, the posterior probability that the urn is of type I is 0 times some other stuff, or 0, regardless of what “that other stuff” may be—regardless of what color ball is drawn, for example. Since the probability that the urn is of type I stays at 0 (posterior same as prior), so correspondingly the probability that the urn is of type II stays at 1 (again, posterior same as prior).

section 3-2a

1. Let q be the new information: eight tails and then two heads. Then, deriving Bayes’s theorem so that we know what information we also need to come up with, we have
   P(B/q)
   = P(both)/P(q)
   = P(B & q) / P(B & q or not B & q)
   = P(B & q) / [P(B & q) + P(not B & q)]
   = P(B) * P(q/B)  / [P(B) * P(q/B) + P(not B) * P(q / not B)]
   So, we need the following:
   P(B) = 0.01
   P(q/B) = (3/4)⁸ * (1/4)² = 3⁸/4⁸ * 1/4² = 3⁸/4¹⁰ = 3⁸/2²⁰
   P(not B) = 0.99
   P(q / not B) = (1/2)¹⁰ = 1/2¹⁰
   Substituting these values, we have
   0.01 * 3⁸/2²⁰ / [0.01 * 3⁸/2²⁰ + 0.99 * 1/2¹⁰]
   = 3⁸/(100 * 2²⁰) / [3⁸/(100 * 2²⁰) + 99/(100*2¹⁰)]
   = 3⁸/(100 * 2²⁰) / [3⁸/(100 * 2²⁰) + 99 * 2¹⁰/(100 * 2²⁰)]
   = 3⁸/(100 * 2²⁰) / [3⁸ + 99 * 2¹⁰ / (100 * 2²⁰)]
   = 3⁸ / (3⁸ + 99 * 2¹⁰)
   = 3⁸ / (3⁸ + 99 * 1,024)
   = 3⁸ / (3⁸ + 101,376)
   = 6,561 / (6,561 + 101,376)
   = 6,561 / 107,937
   = 0.06
   So, if you got those results in ten flips of the coin, your belief in the coin’s being biased would go up from 0.01 to 0.06.
2. Let q be the new information: some combination of eight tails and two heads. Again, we begin by deriving Bayes’s theorem:
   P(B/q)
   = P(both)/P(q)
   = P(B & q) / P(B & q or not B & q)
   = P(B & q) / [P(B & q) + P(not B & q)]
   = P(B) * P(q/B)  / [P(B) * P(q/B) + P(not B) * P(q / not B)]
   So, we need the following:
   P(B) = 0.01
   P(q/B) is 45 times the value for P(q/B) we found above, since there are 45 different ways in which you can get eight tails and two heads, and each of them has the same probability as the ways examined above. So P(q/B) = 45 * 3⁸/2²⁰.
   P(not B) = 0.99
   P(q / not B) is 45 times the value for P(q/B) we found above. So P(q / not B) = 45 * 1/2¹⁰.
   Substituting these values, we have
   0.01 * 45 * 3⁸/2²⁰ / [0.01 * 45 * 3⁸/2²⁰ + 0.99 * 45 * 1/2¹⁰]
   = 45 * (0.01 * 3⁸/2²⁰) / 45 * [0.01 * 3⁸/2²⁰ + 0.99 * 1/2¹⁰]
   = 0.01 * 3⁸/2²⁰ / [0.01 * 3⁸/2²⁰ + 0.99 * 1/2¹⁰]
   and we just saw, above, that this simplifies to 0.06.
   So the fact that the eight tails and two heads and tails can occur in any order, rather than a particular specified one, is irrelevant.
3. If you assigned a probability of 0 to the coin’s being biased, then no number of consecutive tails could cause you, through the employment of the inverse probability law or Bayes’s theorem, to arrive at a nonzero posterior probability for the coin’s being biased. This is because, according to both rules, P(p/q) = kP(p), where k is some other stuff (different stuff for one rule versus the other). If p is the coin’s being biased, then P(p) = 0, and so P(p/q) must also equal 0, regardless of what k is (that is, regardless of how large a number k is, based on the occurrence of some events that would be much more likely if the coin were biased than if it were not biased).

section 3-2b

1. We start with the following decision table:

Now, computing the expected monetary values of the four acts, we have the following:

EMV(A₁): Since A₁ is the act of investing in Daltex regardless of what the information source says, its EMV is just the EMV of investing in Daltex, which we know (from the text) is $76,250.

EMV(A₂): Since A₂ is the act of buying a savings certificate regardless of what the information source says, its EMV is just the EMV of buying a savings certificate, which we know (from the text) is $55,000.

EMV(A₃) = (1/200)($100,000) + (99/200)($55,000) + (1/4)($52,500) + (1/4)($55,000) = $500 + $27,225 + $13,125 + $13,750 = $54,600.

EMV(A₄) = (1/200)($55,000) + (99/200)($100,000) + (1/4)($55,000) + (1/4)($52,500) = $275 + $49,500 + $13,750 + $13,125 = $76,650.

The information-dependent strategies are A₃ and A₄; the one with the highest EMV is A₄, which has an EMV of $76,650. The non-information-dependent strategies are A₁ and A₂; the one with the highest EMV is A₁, which has an EMV of $76,250. So Clark’s best strategy is an information-dependent one, and it has an EMV that is $400 higher than the EMV of his best non-information-dependent one. So he should be willing to pay up to $400 for that information source. Note, however, that Clark wants access to the information source so that he can do the opposite of what it recommends—if it says the company is going public, he’ll put his money in savings; and if it says the company is not going public, he’ll invest in the hope that it actually is!

2. 800 type I urns with 6 blue and 4 red; 200 type II urns with 1 blue and 9 red
   bet on type I goes +$20/–$10; bet on type II goes +$80/–$10
   1. decision table and EMVs of the two bets:

EMV(A₁) = (4/5)($20) + (1/5)(–$10) = $16 – $2 = $14

EMV(A₂) = (4/5)(–$10) + (1/5)($80) = –$8 + $16 = $8

So A₁, or betting on type I, is the one to choose.

2. value of completely reliable information:

Now, computing the expected monetary values of the four acts, we have the following:

EMV(A₁): Since A₁ is the act of betting on type I regardless of what the information source says, its EMV is just the EMV of betting on type I, which we know (from part a) is $14.

EMV(A₂): Since A₂ is the act of betting on type II regardless of what the information source says, its EMV is just the EMV of betting on type II, which we know (from part a) is $8.

EMV(A₃) = (4/5)($20) + (0)($___) + (0)($___) + (1/5)($80) = $16 + $16 + $32

EMV(A₄) = (4/5)(–$10) + (0)($___) + (0)($___) + (1/5)(–$10) = –$8 – $2 = –$10

The information-dependent strategies are A₃ and A₄; the one with the highest EMV is A₃, which has an EMV of $32. The non-information-dependent strategies are A₁ and A₂; the one with the highest EMV is A₁, which has an EMV of $14. That’s a difference of $18; you should be willing to pay that much to learn what type the urn is.

3. revised probabilities and EMVs, given blue ball drawn

P(type I / blue) = 24/25 (from section 3-2, problem 8, part d)
P(type II / blue) = 1/25 (since this must be 1 – P(type I / blue))

	type I (24/25)	type II (1/25)
A1: bet on type I	$20	–$10
A2: bet on type II	–$10	$80

EMV(A₁) = (24/25)($20) + (1/25)(–$10) = $19.20 – $0.40 = $18.80

EMV(A₂) = (24/25)(–$10) + (1/25)($80) = –$9.60 + $3.20 = –$6.40

So A₁, or betting on type I, is the one to choose.

4. revised probabilities and EMVs, given red ball drawn

P(type I / red) = 16/25 (from section 3-2, problem 8, part e)
P(type II / red) = 9/25 (since this must be 1 – P(type I / red))

	type I (16/25)	type II (9/25)
A1: bet on type I	$20	–$10
A2: bet on type II	–$10	$80

EMV(A₁) = (16/25)($20) + (9/25)(–$10) = $12.80 – $3.60 = $9.20

EMV(A₂) = (16/25)(–$10) + (9/25)($80) = –$6.40 + $28.80 = –$22.40

So A₁, or betting on type I, is the one to choose.

5. value of color of ball drawn:

Now, computing the expected monetary values of the four acts, we have the following:

EMV(A₁): Since A₁ is the act of betting on type I regardless of what the information source says, its EMV is just the EMV of betting on type I, which we know (from part a) is $14.

EMV(A₂): Since A₂ is the act of betting on type II regardless of what the information source says, its EMV is just the EMV of betting on type II, which we know (from part a) is $8.

EMV(A₃) = (12/25)($20) + (8/25)(–$10) + (1/50)(–$10) + (9/50)($80)
= $9.60 – $3.20 – $0.20 + $14.40
= $20.60

EMV(A₄) = (12/25)(–$10) + (8/25)($20) + (1/50)($80) + (9/50)(–$10)
= –$4.80 + $6.40 + $1.60 – $1.80
= $1.40

The information-dependent strategies are A₃ and A₄; the one with the highest EMV is A₃, which has an EMV of $20.60. The non-information-dependent strategies are A₁ and A₂; the one with the highest EMV is A₁, which has an EMV of $14. That’s a difference of $6.60; you should be willing to pay that much to learn what color the ball is.

section 3-2c

1. The other two strategies are the following:
   3. withhold if fewer than ten die; market if ten or more die
   4. withhold if fewer than ten die; withhold if ten or more die
2. two tables of action probabilities (one for defective, one for fine), given P(10+/defective) = 1/2 and P(10+/fine) = 1/100
   1. batch defective:

2. batch fine:

3. completion of table 3-5:

If you use the maximin rule, you notice that the minima for the four strategies are –1,000, –500, –98, and –100. The maximum minimum is –98, which belongs to the third strategy.

[many reasonable answers to the question whether this is the only reasonable choice to make in the situation]

section 3-3a

1. proof that theorem 1 (P(p) + P(not p) = 1) holds under the classical interpretation of probability:

2. proof that theorem 2 (if p and q are equivalent, then P(p) = P(q)) holds under the classical interpretation of probability:

section 3-3b

1. proof that axiom 4 (P(p & q) = P(p) * P(q/p)) holds under the relative-frequency interpretation of probability:

First, we have to consider the case where nothing is both a P and an R:

no.	claim	justification
1	P(p & q) = PR(P & Q)	notational convention of relative-frequency interpretation
2	PR(P & Q) = #(P & Q & Rs)/#(Rs)	definition of relative-frequency interpretation
3	#(P & Q & Rs)/#(Rs) = 0/#(Rs)	Since nothing is both a P and an R, it follows that nothing is a P and a Q and an R.
4	0/#(Rs) = 0	arithmetic (since #(Rs) ≠ 0)
5	P(p & q) = 0	1–4, transitivity of =
6	P(p) = PR(P)	notational convention of relative-frequency interpretation
7	PR(P) = #(P and Rs)/#(Rs)	definition of relative-frequency interpretation
8	#(P and Rs)/#(Rs) = 0/#(Rs)	Nothing is both a P and an R.
9	0/#(Rs)	4, reiterated
10	P(p) = 0	6–9, transitivity of =
11	P(q/p) = PR(Q/P)	notational convention of relative-frequency interpretation
12	PR(Q/P) = 0	definition of relative-frequency interpretation
13	P(q/p) = 0	11 and 12, transitivity of =
14	0 = 0 * 0	arithmetic
15	P(p & q) = P(p) * P(q/p)	14, with substitution from 5, 10, and 13

Second, we have to consider the case where at least one thing is both a P and an R:

2. proof that theorem 2 (if p and q are equivalent, then P(p) = P(q)) holds under the relative-frequency interpretation of probability:

section 3-3c

1. completion of proof of axiom 3—proof for the case in which c > a + b:

2. Dutch Book proof that 0 ≤ a ≤ 1, where a is your betting quotient for “q given p”:

1. proof that a < 0 leads to a Dutch Book:

no.	claim	justification
1	a < 0	given
2	–a > 0	1, multiply each side by –1
3	Let strategy A1 be the strategy of betting for “q given p”.	just defining strategy A1
4	Rows 1 and 2 of table 3-9 describe all the genuine possibilities in which the bet is on.	The bet is off if p is false.
5	In row 1, strategy A1 pays (1 – a) = 1 – a = 1 + (–a) = 1 + (some positive number), which is positive.	reading table, with information from 2
6	In row 2, strategy A1 pays (–a) = –a, which is positive.	reading table, with information from 2
7	In each row describing a genuine possibility in which the bet is on, strategy A1 has a positive result.	4–6
8	A Dutch Book can be made against anyone who holds that a < 0.	7

2. proof that a > 1 leads to a Dutch Book:

no.	claim	justification
1	a > 1	given
2	a – 1 > 0	1, subtract 1 from each side
3	Let strategy A1 be the strategy of betting against “q given p”.	just defining strategy A1
4	Rows 1 and 2 of table 3-9 describe all the genuine possibilities in which the bet is on.	The bet is off if p is false.
5	In row 1, strategy A1 pays –(1 – a) = –1 + a = a – 1, which is positive.	reading table, with information from 2
6	In row 2, strategy A1 pays a, which is positive.	reading table, with information from 1
7	In each row describing a genuine possibility in which the bet is on, strategy A1 has a positive result.	4–6
8	A Dutch Book can be made against anyone who holds that a > 1.	7

3. proof that in the partially provided proof of axiom 4, the bookie’s bet (against p, against “q given p”, and for “p & q”) pays ab – c no matter what:

4. completion of proof of axiom 4—proof for the case in which c > ab:

5. Dutch Book arguments for the following:
   1. If p is impossible, P(p) = 0:
      1. proof that if p is impossible, then P(p) < 0 leads to a a Dutch Book:

2. proof that if p is impossible, then P(p) > 0 leads to a Dutch Book:

2. P(p) + P(not p) = 1:

We’ll need the following table:

		p		not p
p	not p	for	against	for	against
T	T	1 – a	–(1 – a)	1 – b	–(1 – b)
T	F	1 – a	–(1 – a)	–b	b
F	T	–a	a	1 – b	–(1 – b)
F	F	–a	a	–b	b

1. proof that P(p) + P(not p) < 1 leads to a Dutch Book:

2. proof that P(p) + P(not p) > 1 leads to a Dutch Book:

3. If p logically implies q, then P(p) ≤ P(q):

We’ll need the following table:

		p		q
p	q	for	against	for	against
T	T	1 – a	–(1 – a)	1 – b	–(1 –b)
T	F	1 – a	–(1 – a)	–b	b
F	T	–a	a	1 – b	–(1 – b)
F	F	–a	a	–b	b

proof that if p logically implies q, then P(p) > P(q) leads to a Dutch Book:

section 3-3d

[skip the problems]

section 4-1

1. We have the following situation:

The wining bet on Jack must be large enough for the expected monetary value of betting on Jack to equal the expected monetary value of betting on Ace. So we have the following equation (and equations derived from it):

(1/2)($5) + (1/2)(–$2) = (1/2)(–$10) + (1/2)(x)
$2.50 – $1 = –$5 + x/2
$1.50 = –$5 + x/2
$6.50 = x/2
$13 = x

So the payoff from Jack’s winning must be at least $13 in order for you to be willing to risk the $10 on his losing.

2. Let u’ = au + b, with a > 0. Then we have the following:
   au = u’ – b
   u = u’/a – b/a
   u = (1/a)u’ – (b/a)
   and this is the transformation that converts u’ back into u.
3. Let u’’ = cu’ + d, with c > 0. Then, with u’ defined as in problem 2, we can express u’’ in terms of u as follows:
   u’’ = cu’ + d
   = c(au + b) + d
   = acu + bc + d
   = (ac)u + (bc + d)
4. proof that if s and s’ are equivalent ratio scales, then if s(x) = 2s(y), then s’(x) = 2s’(y):

5. proof that adding a certain number b_i to each utility in column i (with possibly different b's in different columns) does not change the ordering of the acts (assuming, presumably, that we are concerned with expected utility):

We’ll begin with the following table:

	S1 (p1)	S2 (p2)	. . .	Sn (pn)
Ai	u1	u2	. . .	un
Aj	v1	v2	. . .	vn

And we’ll show the transformed values in the following table:

We just need to prove that if EU(A_i) ≥ EU(A_j) based on the utilities in the first table, then EU(A_i) ≥ EU(A_j) based on the utilities in the second table. Here’s a proof:

section 4-2:

1. The rule for maximizing expected utilities can be reformulated as a rule for minimizing expected disutilities.
2. proof that the expected utility of an act is equal to –1 times its expected utility:

3. proof that the rule mentioned in the answer to problem 1 and the use of a disutility scale yields the same ranking of acts as maximizing expected utility does:

We need to prove that if EU(A_i) ≥ EU(A_j), then EDU(A_i) ≤ EDU(A_j), where EU refers to expected utility and EDU refers to expected disutility.

4. For the modified game, we have the following situation:

The EMV is (1/2)($2) + (1/4)($4) + (1/4)($0)
= $1 + $1 + $0
= $2

If the game will be stopped after the n^th toss of the coin, we have the following situation:

	heads on toss 1 (1/2)	heads on (and not until) toss 2 (1/4)	. . .	heads on (and not until) toss n (1/2n)	heads on no toss through toss n (1/2n)
play the game	$2	$4		$2n	$0

The EMV is (1/2)($2) + (1/4)($4) + . . . + (1/2ⁿ)($2ⁿ) + (1/2ⁿ)($0)
= $1 + $1 + . . . + $1 + $0
= n($1) + $0
= $n

An EMVer should be willing to pay any amount to play the unrestricted St. Petersburg game because its EMV is infinite: it is just $1 + $1 + . . ., without end. Thus the EMV exceeds any price that may be asked in order to play the game.

5. No, the stated approach is not an adequate solution to the problem of relating EMVs to cash values, because it depends on stipulating that cash value will be identified with average winnings in the specified hypothetical cases. If someone is concerned that an act’s EMV may not be equivalent to any of its possible cash values, then it cannot be expected that such a person will be placated merely by any response that involves stipulating such an equivalence. That would be like responding to someone who does not believe it’s irrational to hold beliefs that violate the axioms of the probability calculus by just stipulating that irrationality will be defined as violation of those very axioms.

section 4-3 (p. 91)

1. We cannot assume there is a single best prize because an agent can satisfy the preference conditions and still rank two or more prizes in first place, so that two or more prizes tie for best.
2. It is desirable to assume that there is more than one basic prize, and to assume that the agent is not indifferent between all the prizes, so that the resulting utility function has more than one value. Is these assumptions were not true, the expected utility theorem would not be false; rather, any utility function with just one value would show it to be true, but only in that special case.
3. three lotteries’ chances at B:
   1. L(1, B, W): chance at B = 1
   2. L(1/2, L(1, W, B), L(1/2, B, W)): chance at B = (1/2)*(1 – 1) + (1/2)*(1/2) = (1/2)*(0) + 1/4 = 0 + 1/4 = 1/4
   3. L(a, B, B): chance at B = a + (1 – a) = a + 1 – a = 1
4. An agent should be indifferent between L(a, B, W) and L(1 – a, W, B) because each yields B with probability a and yields W with probability 1 – a. The difference between how they’re written is no more significant than the difference between the way the sum aB + (1 – a)W is written and the way the sum (1 – a)W + B is written.
5. A nested lottery (such as the one in problem 3b, above) can accommodate more than two prizes. You have a 50-percent chance of getting A, a 25-percent chance of getting B, and a 25-percent chance of getting C with any of the following six lotteries:
   1. L(1/2, A, L(1/2, B, C))
   2. L(1/2, A, L(1/2, C, B))
   3. L(1/4, B, L(2/3, A, C))
   4. L(1/4, B, L(1/3, C, A))
   5. L(1/4, C, L(2/3, A, B))
   6. L(1/4, C, L(1/3, B, A))

section 4-3 (p. 96)

1. proofs of claims about lotteries, using the existence part of the expected utility theorem (i.e., using conditions 1–3 on p. 90):
   1. L(1, x, y) I x:

2. L(0, x, y) I y:

3. L(a, x, y) I L(1 – a, y, x):

4. L(a, x, x) I x:

2. proof of part b of the substitution-of-lotteries condition (that if x I L(a, y, z), then L(c, v, x) I L(c, v, L(a, y, z)):

3. proof, using just the ordering condition, that if xIy and zIw, then xPz if and only if yPw:

4. two impossibility proofs, assuming rationality conditions of expected utility theorem
   1. proof that there is no number a or basic prize x distinct from B for which L(a, x, B) P L(a, B, B) or L(a, B, x) P L(a, B, B):
      1. proof that there is no number a or basic prize x distinct from B for which L(a, x, B) P L(a, B, B):

2. proof that there is no number a or basic prize x distinct from B for which L(a, B, x) P L(a, B, B):

2. proof that there is no number a or basic prizes x and y distinct from B for which L(a, x, y) P L(a, B, B):

Following the strategy suggested by Mark Wine, we’ll do this in two steps. First, we’ll establish this helpful lemma: if x I B, then L(a, x, y) I L (a, B, y) and L(a, y, x) I L(a, y, B).

This lemma is essentially the conjunction of two conditionals:
(1) if x I B, then L(a, x, y) I L (a, B, y)
(2) if x I B, then L(a, y, x) I L(a, y, B)

We’ll prove the first of these; the second can be proved analogously.

This proves that if x I B, then L(a, x, y) I L(a, B, y). Let us take this as sufficient to prove the lemma mentioned above, and use that lemma in the following proof of the main claim of this problem, namely, that there is no number a or basic prizes x and y distinct from B for which L(a, x, y) P L(a, B, B):

5. proofs about lotteries’ degrees
   1. proof that if no lottery of degree less than n is preferred to B, then there is no number a and lottery L of degree less than n for which L(a, B, L) P L(a, B, B) or L(a, L, B) P L(a, B, B):
      1. proof that if no lottery of degree less than n is preferred to B, then there is no number a and lottery L of degree less than n for which L(a, B, L) P L(a, B, B):

2. proof that if no lottery of degree less than n is preferred to B, then there is no number a and lottery L of degree less than n for which L(a, L, B) P L(a, B, B):

no.	claim	justification
1	No lottery of degree less than n is preferred to B.	assumption for proving conditional
2	Suppose there were a number a and lottery L of degree less than n for which L(a, L, B) P L(a, B, B).	assumption for indirect proof
3	L P B	2, via better-prizes condition
4	3 is impossible.	1
5	There is no number a and lottery L of degree less than n for which L(a, L, B) P L(a, B, B).	2–4

2. proof that if no lottery of degree less than n is preferred to B, then there is no number a and no lotteries L₁ and L₂ of degree less than n for which L(a, L₁, L₂) P L(a, B, B):

This can be proved using a proof just like that used in problem 4b, with three changes:

1. every reference to basic prize x must be replaced with a reference to lottery L₁
2. every reference to basic prize y must be replaced with a reference to lottery L₂
3. the justifying statement “B is defined as a basic prize that is at least as good as any other basic prize” must be replaced with the justifying statement “It is given that no lottery of degree less than n (such as L₁ or L₂) is preferred to B.”

6. proof that for no number a, L(a, B, B) P B, applying the conclusion of exercise 5 (that no lottery of degree greater than 0 is preferred to L(a, B, B) for any number a) to L(a, L(a, B, B), L(a, B, B)):

In class, Joe Morgan suggested a different strategy, which is used in the following proof:

After class, Courtney Gustafson suggested the following, also different, approach:

7. proof that for no number a, B P L(a, B, B):

In class, Courtney Gustafson suggested this approach:

no.	claim	justification
1	Suppose, for some number a, B P L(a, B, B).	assumption for indirect proof
2	L(a, B, B) P L(a, B, L(a, B, B))	1, via better-prizes condition
3	L(a, L(a, B, B), L(a, B, B)) I L(a, B, B)	reduction-of-compound-lotteries condition
4	L(a, L(a, B, B), L(a, B, B)) P L(a, B, L(a, B, B))	2 and 3, via ordering condition, part O6
5	L(a, B, B) P B	4, via better-prizes condition
6	It is not the case that B P L(a, B, B).	5, via ordering condition, part O1
7	Line 1 leads to a contradiction.	1 and 6
8	Line 1 is false.	7
9	For no number a, B P L(a, B, B).	8

8. proof that if BPx and xPW, then L(a, x, x) I x for any number a:

(Note that this proof does not use the given assumption—that B P x and x P W. This makes sense, since the result should hold even for prizes and lotteries that are as good as B or as bad as W, and not just for ones in between.)

section 4-3 (p. 98)

1. proof that c, as defined in statement k, is greater than 0:

2. proof that given any number k on the u-scale, there is some lottery x for which u(x) = k:

3. This problem involves reasoning much like that used in problem 6 of section 2-3.

To transform a 0-to-1 scale into a 1-to-100 scale, we just need to find an a and a b such that 1 = a(0) + b and 100 = a(1) + b. The first equation yields 1 = b; substituting this into the second equation, we have 100 = a + 1, meaning that a = 99. So we have the transformation u’ = 99u + 1.

To transform a –5-to-5 scale into a 0-to-1 scale, we just need to find an a and a b such that 0 = a(–5) + b and 1 = a(5) + b. The first equation yields 0 = –5a + b, or b = 5a; substituting this into the second equation, we have 1 = 5a + 5a, or 1 = 10a, or a = 1/10. Substituting this into the equation b = 5a, we have b = 5*(1/10), or b = 1/2. So we have the transformation u’ = (1/10)u + 1/2.

4. No, it does not make sense to say, of an agent whose preferences are measured on a von Neumann-Morgenstern utility scale, that he or she prefers a given prize twice as much as another. This is because a von Neumann-Morgenstern utility scale is unique (i.e., appropriate to an agent) only up to positive linear transformations. So, a scale might suggest that Ralph prefers a trip to Rome twice as much as a trip to London by assigning utilities of 20 and 10 to these two prizes, but  this scale could legitimately be transformed, using the function u’ = (1/10)u + 155, into one that assigns utilities of 157 and 156 to these two prizes. And no one would look at those two utilities—157 and 156—and think that they were attached to prizes one of which was twice as preferred as the other.

One thing it is helpful to do, in order to avoid the temptation to think that scores of 20 and 10 might reflect one prize being twice as preferred as another, is to notice that such scores might themselves be the result of a linear transformation of some other utility scale that might not suggest, at all, that one prize is twice as preferred as another. For example, instead of just noticing that the 20 and the 10 can be transformed into 157 and 156, notice also that maybe 157 and 156 were the “original” utilities, and that the 20 and 10 arose from a transformation (u’ = 10u – 1550) designed to make the first prize look twice as preferred as the second prize.

You could infer that one prize was twice as preferred, or twice as desired, as another if we could represent an agent’s preferences using a ratio scale. But von Neumann-Morgenstern utility scale are only interval scales, which fall short of ratio scales in that they lack natural zero points (which are essential to ratio scales).

section 4-3a:

1. No, it would not be correct to say that Miami was twice as hot as New York, since the Fahrenheit scale is only an interval scale, not a ratio scale. Using temperatures measured in degrees Fahrenheit, you can say things like “Miami was hotter than New York by twice the margin by which New York was hotter than Boston,” but you cannot say things like “Miami was twice as hot as New York.” Likewise for the Celsius scale. (See the answer to problem 4 on p. 98 of section 4-3 for more discussion of this point.)

Incidentally, the problem with saying things like “Miami was twice as hot as New York,” as just described, is not due to the fact that it’s temperature that’s under discussion; the problem is that the temperature scale being used is just an interval scale, not a ratio scale. There is a ratio scale for temperature, the kelvin scale, and if two things have kelvin temperatures such that one is twice the other (e.g., the temperature of one thing is 400 kelvins and the temperature of the other is 200 kelvins), then you can say that the first thing is twice as hot as the second. This is because the temperature of 0 kelvins signifies the (only hypothetically attainable) state of absolute zero, or the complete absence of heat, giving the kelvin scale a meaningful zero point. The Fahrenheit and Celsius scales are kept from being ratio scales by the fact that their zero points do not signify the utter absence of the things being measured, as do the zero points on ratio scales such as the kelvin temperature scale (as just explained), the miles-per-hour scale for measuring speed, the feet-and-inches scale for representing length, etc.

2. All the other utility functions of someone for whom u(x) = x is a utility function are positive linear transformations of the given utility function. So, they have the form u(x) = ax + b, where a > 0. (Note that when a = 1 and b = 0, then we just have our untransformed utility function u(x) = x.) Since they have the form u(x) = ax + b, they resemble the graphs of u(x) = x in being straight lines with positive slops, but they may differ in either or both of two ways:
   1. First, since a needn't be 1, their slopes needn’t be 1, or 45 degrees; when a is smaller than 1, they rise more gently (slopes of less than 1, or less than 45 degrees); when a is greater than 1, they are steeper (slopes of greater than 1, or more than 45 degrees).
   2. Second, since b needn’t be 0, they do not necessarily pass through the origin. When b is positive, they cross the vertical axis above the origin, and when b is negative, the cross the vertical axis below the origin.
3. If you are risk averse, then your utility curve is similar to the line y = x in increasing from left to right; but it differs in being concave down, whereas y = x has no concavity, because it is flat.
4. If you are risk seeking, then your utility curve is increasing from left to right, but is concave up.
5. If someone whose utility function for money was given by u(x) = x² up to $100,000 and by u(x) = (100,000)² after that, then we could conclude that the person was risk seeking for amounts up to $100,000, but then was so risk averse beyond that that there was no amount greater than $100,000 that he or she would prefer to $100,000 itself.

section 4-4c:

1. proof that the choices most people make in the situation described contradict the combined theories of utility and subjective probability:

2. If you are a fan of utility theory but have reservations about subjective probabilities, then the Ellsberg paradoxes confirms your suspicions by the following reasoning. The Ellsberg paradox contradicts the conjunction of utility theory and subjective-probability theory. Put this way, it portrays (1) utility theory and (2) subjective-probability theory as being on one side, and (3) the choices most agents make in Ellsberg situations as being on the opposing side. But all the contradiction really means is that 1, 2, and 3 are not all consistent—any two are consistent with each other, but not all three are consistent with each other. Put this way, it is clear that all three items are on a par; none has priority or less of a burden of proof than any other. So, confronted with the inconsistent triad of 1, 2, and 3, you can reject any one of them in order to eliminate the inconsistency. if you are a fan of 1 and are suspicious of 2, then instead of rejecting 3, you can accept 3 and claim that 2 is the one that must be rejected.
3. As a theory of ideally rational beings, whether decision theory is affected by paradoxes involving “what most people think” is debatable. On the one hand, since most people are not ideally rational, the paradoxes involving most people may seem to have no bearing on a theory of ideally rational beings. On the other, it might be maintained that the way most people think cannot, as a matter of principle, be very far short of genuine rationality. On this view, paradoxes drawing on what most people think would have more relevance.
4. Even if monetary values are replaced with their logarithms, then the St. Petersburg paradox can be reinstated by increasing the payoffs from {2, 4, 8, 16, . . .}, to {10², 10⁴, 10⁸, 10¹⁶, . . .}. Then the expected value is the following:
   (1/2)log₁₀(10²) + (1/4)log₁₀(10⁴) + (1/8)log₁₀(10⁸) + (1/16)log₁₀(10¹⁶) + . . .
   = (1/2)*2 + (1/4)*4 + (1/8)*8 + (1/16)*16 + . . .
   = 1 + 1 + 1 + 1 + . . .,
   which is the expected value that gives rise to the paradox as originally formulated.

section 4-5:

1. We need to use EU(blue) > EU(both) to solve for p, where p is the probability of the correct predictions—assuming, of course, that we can represent EU(blue) or EU(both) as a function of p. As it turns out, each of them is a function of p. We have the following:
   EU(blue) > EU(both)
   (1 – p)(0) + (1,000,000)p > (1,000)(p) + (1,001,000)(1 – p)
   1,000,000p > 1,000p + 1,001,000 – 1,001,000p
   2,000,000p > 1,001,000
   p > 1,001,000/2,000,000
   p > 1,001/2,000
   p > 50.05 percent
   So as long as p at least a little more than 50 percent, the paradox is preserved.
2. Here is a decision table for the Calvinist example that establishes it as a variant of the Predictor paradox:

3. Here is a decision table for the smoking example that establishes it as a variant of the Predictor paradox:

section 4-6:

1. To ascertain whether the physician should use the penicillin, we compute the expected utilities for the two acts. For ‘Give Penicillin’, we have (0.375)(0.5) + (0.125)(0) + (0.375)(1) + (0.125)(0.25)
   = (3/8)(1/2) + 0 + 3/8 + (1/8)(1/4)
   = 3/16 + 6/16 + 1/32
   = 6/32 + 12/32 + 1/32
   = 19/32

For ‘Do not’, we have (0.15)(1) + (0.35)(0.25) + (0.15)(1) + (0.35)(0.25)
= 3/20 + (7/20)(1/4) + 3/20 + (7/20)(1/4)
= 3/20 + 7/80 + 3/20 + 7/80
= 12/80 + 7/80 + 12/80 + 7/80
= 38/80
= 19/40

Clearly 19/32 is greater than 19/40, so the first act has the higher expected utility, and the physician should use the penicillin.

2. If the physician is uncertain about the effectiveness of antibiotic X, as well as about the effectiveness of penicillin, then we need a table twice as big, as follows: