Jill is willing in one year (assuming no marriage now)  Jill is not willing in one year (assuming no marriage now)  
marry now  marriage and no extra money  marriage and no extra money 
wait  marriage and extra million dollars  million dollars but no marriage 
God exists  God does not exist  
lead the life of a religious Christian  go to heaven  at least as good as pagan life 
be a pagan  go to hell  pagan life 
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.
the food is spoiled but you can handle it (35%)  the food is spoiled and you cannot handle it (35%)  the food is not spoiled (30%)  
eat there  you do not get sick  you get sick  you do not get sick 
do not eat there  you do not get sick  you do not get sick  you do not get sick 
The outcome description you do not get sick is in more than one square because there is more than one actstate 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%.
better offer following hiring of lawyer  no better offer following hiring of lawyer, but victory in court  no better offer following hiring of lawyer, and defeat in court  
S_{1}:  $10,000  $10,000  $10,000 
S_{2}:  $25,000  $10,000  $10,000 
S_{3}:  $25,000  $50,000  $10,000 
no.  claim  justification 
1  t(u(x)) > t(u(y))  assumption for proving conditional 
2  t(u(x)) > t(u(y)) or t(u(x)) = t(u(y))  1, ‘or’ introduction 
3  t(u(x)) ≥ t(u(y))  2, definition of ≥ 
4  u(x) ≥ u(y)  3, condition c 
5  u(x) > u(y) or u(x) = u(y)  4, definition of ≥ 
6  u(x) > u(y)  assumption—considering first possibility in line 5 
7  xPy  6, condition a (since u is an ordinal utility scale) 
8  u(x) = u(y)  assumption—considering second possibility in line 5 
9  u(x) = u(y) or u(y) > u(x)  8, ‘or’ introduction 
10  u(y) ≥ u(x)  9, definition of ≥ 
11  t(u(y) ≥ t(u(x))  10, condition c 
12  not [t(u(x)) > t(u(y))]  11, definition of ≥ 
13  contradiction  1 and 12 
14  line 8 is false  13 
15  line 6 must be true  lines 6 and 8 were the only possibilities from line 5 
16  xPy  15 (truth of line 6, xPy derived from it on line 7) 
 proof that
if xPy, then t(u(x)) > t(u(y)):
no.  claim  justification 
1  xPy  assumption for proving conditional 
2  u(x) > u(y)  1, condition a (since u is an ordinal utility scale) 
3  u(x) > u(y) or u(x) = u(y)  2, ‘or’ introduction 
4  u(x) ≥ u(y)  3, definition of ≥ 
5  t(u(x)) ≥ t(u(y))  4, condition c 
6  t(u((x)) > t(u(y)) or t(u(x)) = t(u(y))  5, definition of ≥ 
7  t(u(x)) > t(u(y))  assumption—considering first possibility in line 6 
8  t(u(x)) = t(u(y))  assumption—considering second possibility in line 6 
9  t(u(x)) ≥ t(u(y)) and t(u(y)) ≥ t(u(x))  8, definition of ≥ 
10  u(x) ≥ u(y) and u(y) ≥ u(x)  9, condition c 
11  u(x) = u(y)  10, definition of ≥ 
12  xIy  11, condition b (since u is an ordinal utility scale) 
13  contradiction  1 and 12 
14  line 8 is false  13 
15  line 7 must be true  lines 7 and 8 were the only possibilities from line 6 
16  t(u(x)) > t(u(y))  7, repeated 
Mark Wine came up with a much shorter proof with some ingenious moves in it. (This doesn’t just shorten each of the 16line proofs from above, one by one; it replaces both of them—32 lines cut down to 6!!) Here it is:
no.  claim  justification 
1  w ≥ v if and only if t(w) ≥ t(v)  statement of condition c, which we are allowed to assume 
2  not [w ≥ v] if and only if not [t(w) ≥ t(v)]  1, negating both sides of a biconditional keeps the whole thing true 
3  w < v if and only if t(w) < t(v)  2, combining the negations and weak inequalities on each side of the ‘if and only if’ yields reversed strong inequalities on each side 
4  u(y) < u(x) if and only if t(u(y)) < t(u(x))  3, substitute u(y) for w and u(x) for v 
5  u(y) < u(x) if and only if xPy  condition a (which we are entitled to assert since u is an ordinal utility scale) 
6  t(u(y)) < t(u(x)) if and only if xPy  4 and 5, you can go from ‘A if and only if B’ and ‘A if and only if C’ to ‘B if and only if C’ 
 proof that t satisfies condition b—that t(u(x)) = t(u(y)) if and only if xIy:
 proof that if t(u(x)) = t(u(y)), then xIy:
no.  claim  justification 
1  t(u(x)) = t(u(y))  assumption for proving conditional 
2  t(u(x)) ≥ t(u(y)) and t(u(y)) ≥ t(u(x))  1, definition of ≥ 
3  u(x) ≥ u(y) and u(y) ≥ u(x)  2, condition c 
4  u(x) = u(y)  3, definition of ≥ 
5  xIy  4, condition b (since u is an ordinal utility scale) 
 proof that if xIy, then t(u(x)) = t(u(y)):
no.  claim  justification 
1  xIy  assumption for proving conditional 
2  u(x) = u(y)  1, condition b (since u is an ordinal utility scale) 
3  u(x) ≥ u(y) and u(y) ≥ u(x)  2, definition of ≥ 
4  t(u(x)) ≥ t(u(y)) and t(u(y)) ≥ t(u(x))  3, condition c 
5  t(u(x)) = t(u(y))  4, definition of ≥ 
no.  claim  justification 
1  w ≥ v  assumption for proving conditional 
2  w – 2 ≥ v – 2  1, subtract 2 from each side 
3  t(w) ≥ t(v)  2, definition of t(x) 
 proof that if t(w) ≥ t(v), then w ≥ v:
no.  claim  justification 
1  t(w) ≥ t(v)  assumption for proving conditional 
2  w – 2 ≥ v – 2  1, definition of t(x) 
3  w ≥ v  2, add 2 to each side 
 Second, we consider t(x) = 3x + 5. We prove each half of the biconditional (the one stated above) in turn.
 proof that if w ≥ v, then t(w) ≥ t(v):
no.  claim  justification 
1  w ≥ v  assumption for proving conditional 
2  3w ≥ 3v  1, multiply each side by 3 
3  3w +5 ≥ 3v + 5  2, add 5 to each side 
4  t(w) ≥ t(v)  3, definition of t(x) 
 proof that if t(w) ≥ t(v), then w ≥ v:
no.  claim  justification 
1  t(w) ≥ t(v)  assumption for proving conditional 
2  3w +5 ≥ 3v + 5  1, definition of t(x) 
3  3w ≥ 3v  2, subtract 5 from each side 
4  w ≥ v  3, divide each side by 3 
no.  claim  justification 
1  w ≥ v  assumption for proving conditional 
2  w^{2} ≥ wv  1, multiply both sides by w 
3  wv ≥ v^{2}  1, multiply both sides by v 
4  w^{2} ≥ v^{2}  2 and 3, transitivity of ≥ 
5  t(w) ≥ t(v)  4, definition of t(x) 
 proof that t(w) ≥ t(v), then w ≥ v:
no.  claim  justification 
1  t(w) ≥ t(v)  assumption for proving conditional 
2  w^{2} ≥ v^{2}  1, definition of t(x) 
3  v^{2} ≤ w^{2}  2, reversed 
4  v^{2}/w^{2} ≤ 1  3, divide each side by w^{2} 
5  (v/w)^{2} ≤ 1  4, left side rewritten 
6  v/w ≤ 1  5, only numbers less than 1 have squares that are less than 1 
7  v ≤ w  6, multiply each side by w 
8  w ≥ v  7, reversed 
If some of the numbers are negative, then t(x) = x^{2} is no longer an ordinal transformation. For example, if w = –3 and v = –4, then w ≥ v, but it is not the case that w^{2} = 9 is greater than v^{2} = 16, and so it is not the case that t(w) ≥ t(v).
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).
S_{1}  S_{2}  
A_{1}  6  6 
A_{2}  6  7 
(A_{1} 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:
S_{1} S_{2} S_{3} A_{1} 0 1 3 A_{2} 0 0 2 (A_{2} is dominated, but is selected by the maximin rule, since that rule excludes A_{1}.)
Here is a table showing that the lexical maximin rule does not exclude a dominated act:
S_{1} S_{2} S_{3} A_{1} 6 6 7 A_{2} 6 7 7 (A_{1} 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 A_{1} and uniquely selecting A_{2}, even though A_{1} dominates A_{2}.)
A_{1}  3  21  7 
A_{2}  5  0  0 
A_{3}  0  26  12 
The minimaxregret act is A_{2}, since its maximum regret value (5) is less than that of A_{1} (21) and less than that of A_{3} (26).
For decision table 220, the regret table is
A_{1} 0 0 4 A_{2} 8 12 0 A_{3} 1 15 1 The minimaxregret act is A_{1}, since its maximum regret value (4) is less than that of A_{2} (12) and less than that of A_{3} (15).
S_{1}  S_{2}  S_{3}  
A_{1}  6  8  6 
A_{2}  6  9  6 
A_{3}  10  10  0 
Then the corresponding regret table is
S_{1} S_{2} S_{3} A_{1} 4 2 0 A_{2} 4 1 0 A_{3} 0 0 6 So neither A_{1} nor A_{2} is excluded (only A_{3} is, with it maximum regret value of 6); but A_{1} is dominated by A_{2}, so A_{1} is a dominated act, and yet it is not excluded by the minimax regret rule.
A. We begin by representing the first decision table as follows:
(table 1)  S_{1}  S_{2}  . . . 
A_{1}  u_{1,1}  u_{1,2}  . . . 
A_{2}  u_{2,1}  u_{2,2}  . . . 
. . .  . . .  . . .  . . . 
Thanks to the stipulated assumption, we can represent the second decision table as follows:
(table 2)  S_{1}  S_{2}  . . . 
A_{1}  au_{1,1} + b  au_{1,2} + b  . . . 
A_{2}  au_{2,1} + b  au_{2,2} + b  . . . 
. . .  . . .  . . .  . . . 
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:
(table 3)  S_{1}  S_{2}  . . . 
A_{1}  max (u_{_,1}) – u_{1,1}  max (u_{_,2}) – u_{1,2}  . . . 
A_{2}  max (u_{_,1}) – u_{2,1}  max (u_{_,2}) – u_{2,2}  . . . 
. . .  . . .  . . .  . . . 
And we can represent the regret table for the second table as follows:
(table 4)  S_{1}  S_{2}  . . . 
A_{1}  max (au_{_,1} + b) – (au_{1,1} + b)  max (au_{_,2} + b) – (au_{1,2} + b)  . . . 
A_{2}  max (au_{_,1} + b) – (au_{2,1} + b)  max (au_{_,2} + b) – (au_{2,2} + b)  . . . 
. . .  . . .  . . .  . . . 
This equals the following:
(table 5)  S_{1}  S_{2}  . . . 
A_{1}  max (au_{_,1} + b) – au_{1,1} – b  max (au_{_,2} + b) – au_{1,2} – b  . . . 
A_{2}  max (au_{_,1} + b) – au_{2,1} – b  max (au_{_,2} + b) – au_{2,2} – b  . . . 
. . .  . . .  . . .  . . . 
This, in turn, can be rewritten as follows:
(table 6)  S_{1}  S_{2}  . . . 
A_{1}  max (au_{_,1}) + b – au_{1,1} – b  max (au_{_,2}) + b – au_{1,2} – b  . . . 
A_{2}  max (au_{_,1}) + b – au_{2,1} – b  max (au_{_,2}) + b – au_{2,2} – b  . . . 
. . .  . . .  . . .  . . . 
And this, in turn, can be simplified to the following:
(table 7)  S_{1}  S_{2}  . . . 
A_{1}  max (au_{_,1}) – au_{1,1}  max (au_{_,2}) – au_{1,2}  . . . 
A_{2}  max (au_{_,1}) – au_{2,1}  max (au_{_,2}) – au_{2,2}  . . . 
. . .  . . .  . . .  . . . 
Since a > 0, this, in turn, can be rewritten as follows:
(table 8)  S_{1}  S_{2}  . . . 
A_{1}  a*max (u_{_,1}) – au_{1,1}  a*max (u_{_,2}) – au_{1,2}  . . . 
A_{2}  a*max (u_{_,1}) – au_{2,1}  a*max (u_{_,2}) – au_{2,2}  . . . 
. . .  . . .  . . .  . . . 
And this, in turn, can be simplified to the following:
(table 9)  S_{1}  S_{2}  . . . 
A_{1}  a*(max (u_{_,1}) – u_{1,1})  a*(max (u_{_,2}) – u_{1,2})  . . . 
A_{2}  a*(max (u_{_,1}) – u_{2,1})  a*(max (u_{_,2}) – u_{2,2})  . . . 
. . .  . . .  . . .  . . . 
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 actstate pair, R’ = aR, for some a ≥ 0. Here is how we can establish this:
no. claim justification 1 For every actstate pair (or cell) in the second decision table, R’ = u’ – max’, where
(1) u’ is the value of the actstate pair in question in the second decision table and
(2) max’ is the maximum possible value associated with that state in the second decision tabledefinition of R’ 2 For every actstate 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 actstate pair in question in the first decision table, and
(3) max is the maximum possible value associated with that state in the first decision table1, positive linear transformation converting first decision table into second decision table 3 For every actstate 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 actstate 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 actstate 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 actstate pair (or cell) in the second decision table, R’ = aR, where
(1) a > 0 and
(2) R is the regret score for the actstate pair in question in the first decision table.5, definition of regret score 7 For every actstate pair (or cell) in the second decision table, R’ = aR, where
(1) a ≥ 0 and
(2) R is the regret score for the actstate 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.
no.  claim  justification 
1  u’ is a positive linear transformation of u  given 
2  u’’ is a positive linear transformation of u’  given 
3  u’ = au + b, for some a > 0  1, definition of positive linear transformation 
4  u’’ = cu’ + d, for some c > 0  2, definition of positive linear transformation 
5  u’’ = c(au + b) + d, for some a, c > 0  4, with substitution from 3 
6  u’’ = cau + cb + d, for some a, c > 0  5, simplified 
7  u’’ = (ca)u + (cb + d), for some a, c > 0  6, with grouping 
8  u’’ = eu + f, for some e > 0  7, introducing e = ac and f = cb + d 
9  u’’ is a positive linear transformation of u  8, definition of positive linear transformation 
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.
S_{1}  S_{2}  S_{3}  
A_{1}  1  99  100 
A_{2}  1  2  100 
(The trick is to notice that the optimismpessimism 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 optimismpessimism rule, even if one act dominates the other in virtue of differences among intermediate values.)
no.  claim  justification 
1  the aindex for A > the aindex for B  given 
2  aMAX_{A} + (1 – a)min_{A} > aMAX_{B} + (1 – a)min_{B}  1, definition of aindex 
3  c(aMAX_{A} + (1 – a)min_{A}) > c(aMAX_{B} + (1 – a)min_{B}), for all c > 0  2, multiply each side by c 
4  c(aMAX_{A}) + c(1 – a)min_{A} > c(aMAX_{B}) + c(1 – a)min_{B}, for all c > 0  3, distribute c across sums 
5  (a)cMAX_{A} + (1 – a)cmin_{A} > (a)cMAX_{B} + (1 – a)cmin_{B}, for all c > 0  4, rearrange factors within each term 
6  (a)cMAX_{A} + ad + (1 – a)cmin_{A} > (a)cMAX_{B} + ad + (1 – a)cmin_{B}, for all c > 0  5, add d/a to each side 
7  (a)(cMAX_{A} + d) + (1 – a)cmin_{A} > (a)(cMAX_{B} + d) + (1 – a)cmin_{B}, for all c > 0  6, do some factoring 
8  (a)(cMAX_{A} + d) + (1 – a)cmin_{A} + (1 – a)d > (a)(cMAX_{B} + d) + (1 – a)cmin_{B} + (1 – a)d, for all c > 0  7, add (1 – a)d to each side 
9  (a)(cMAX_{A} + d) + (1 – a)(cmin_{A} + d) > (a)(cMAX_{B} + d) + (1 – a)(cmin_{B} + d), for all c > 0  8, do some factoring 
10  the aindex for cA + d > the aindex for cB + d, for all c > 0  9, definition of aindex 
11  the aindex for A, subjected to an unspecified positive linear transformation, it greater than the aindex for B, subjected to the same positive linear transformation  10, definition of positive linear transformation 
no.  claim  justification 
1 
A_{1} is a dominated act. 
assumption for proving conditional 
2 
There exists some act, which we’ll call A_{2},
such that 
1, definition of dominated act 
3 
There exists some act, which we’ll call A_{2},
such that if n is the number of states, 
2, multiply each side of each inequality by 1/n (a positive number) 
4  (1/n)*u_{2,1} + (1/n)*u_{2,2} (1/n)*u_{2,3} + . . . (1/n)*u_{2,n} > (1/n)*u_{1,1} + (1/n)*u_{1,2} (1/n)*u_{1,3} + . . . (1/n)*u_{1,n}  3, summing across states 
5  expected utility of A_{2} > expected utility of A_{1}  4, definition of expected utility 
6 
A_{1} is not recommended by the principle of insufficient reason. 
5, definition of principle of insufficient reason 
Consider this decision table:
S_{1}  S_{2}  
A_{1}  2  4 
A_{2}  0  6 
The principle of insufficient reason selects act A_{1}, since its expected utility of 3 is greater than A_{2}’s expected utility of 3.
Now consider this decision table:
S_{1} S_{2} A_{1} 2 4 A_{2} 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 A_{2}, since its expected utility of 50 is greater than A_{1}’s expected utility of (as seen above) 3.
(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.
no.  claim  justification 
1 
The optimismpessimism rule is indifferent between A_{1}, A_{2}, and A_{3}. 
assumption for proving conditional 
2  The optimismpessimism rule is indifferent between A_{1} and A_{3}.  1, nature of indifference 
3  The aindex for A_{1} is equal to the aindex for A_{3}.  2, definition of indifference for optimismpessimism rule 
4  The aindex for A_{1} is a*1 + (1 – a)*0, where a is the agent’s optimism index.  table, definition of aindex 
5  The aindex for A_{1} is a, where a is the agent’s optimism index.  4, algebra 
6  The aindex for A_{3} is a*(1/2) + (1 – a)*(1/2).  table, definition of aindex 
7 
The aindex for A_{3} is a/2 + 1/2 – a/2. 
6, algebra 
8 
The aindex for A_{3} is 1/2. 
7, algebra 
9 
a = 1/2, where a is the agent’s optimism index 
3, substituting results of 5 and 8 
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 aindexes for the acts are as follows:
A_{1}: a*1 + (1 – a)*0
A_{2}: a*1 + (1 – a)*0
A_{3}: a*(1/2) + (1 – a)*(1/2)definition of aindex 3 The aindexes for the acts are as follows:
A_{1}: a
A_{2}: a
A_{3}: a/2 + 1/2 – a/22, algebra 4 The aindexes for the acts are as follows:
A_{1}: a
A_{2}: a
A_{3}: 1/23, algebra 5 The aindexes for A_{1} and A_{2} are equal, and are greater than the aindex for A_{3}.
4, with information from 1 6 The optimismpessimism rule selects A_{1} and A_{2}.
5, definition of optimismpessimism rule Here is a proof of the second claim:
no.  claim  justification 
1 
The agent’s optimism index (a) is less than 1/2. 
assumption for proving conditional 
2 
The aindexes for the acts are as follows: 
definition of aindex 
3 
The aindexes for the acts are as follows: 
2, algebra 
4 
The aindexes for the acts are as follows: 
3, algebra 
5 
The aindexes for A_{1} and A_{2} are less than the aindex for A_{3}. 
4, with information from 1 
6 
The optimismpessimism rule selects A_{3}. 
5, definition of optimismpessimism rule 
poorer than most people  median income  richer than most people  
differenceprinciple society  bad  o.k.  quite good 
utilitarian society  very bad  a little better than o.k.  very good 
no.  claim  justification 
1 
p implies q 
given 
2 
P(p) ≠ 0 
given 
3 
P(q/p) = P(p & q)/P(p) 
theorem 4 
4 
p is equivalent to p & q 
1, plus hint given in problem 
5 
P(p) = P(p & q) 
4 and theorem 2 
6 
P(q/p) = P(p)/P(p) 
3, with substitution from 5 
7 
P(q/p) = 1 
6, simplified 
no.  claim  justification 
1 
P(p & q) = P(p) * P(q/p) 
axiom 4 
2 
0 ≤ P(q/p) ≤ 1 
axiom 1, part b 
3 
0 * P(p) ≤ P(q/p) * P(p) ≤ P(p) 
2, each part multiplied by P(p) 
4 
0 ≤ P(p) * P(q/p) ≤ P(p) 
3, with simplification and rearrangement 
5 
0 ≤ P(p & q) ≤ P(p) 
4, with substitution from 1 
6 
[0 ≤ P(p & q)] and [P(p & q) ≤ P(p)] 
5, spelled out 
7 
P(p & q) ≤ P(p) 
6, with first conjunct dropped 
public and “public” (1/2)*(1/100) = 1/200 
public and “not” (1/2)*(99/100) = 99/200 
not and “public” (1/2)*(1/2) = 1/4 
not and “not” (1/2)*(1/2) = 1/4 

A_{1}: invest, regardless of advice  $100,000  $100,000  $52,500  $52,500 
A_{2}: savings, regardless of advice  $55,000  $55,000  $55,000  $55,000 
A_{3}: invest if “public” and savings if “not”  $100,000  $55,000  $52,500  $55,000 
A_{4}: savings if “public” and invest if “not”  $55,000  $100,000  $55,000  $52,500 
Now, computing the expected monetary values of the four acts, we have the following:
EMV(A_{1}): Since A_{1} 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(A_{2}): Since A_{2} 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(A_{3}) = (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(A_{4}) = (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 informationdependent strategies are A_{3} and A_{4}; the one with the highest EMV is A_{4}, which has an EMV of $76,650. The noninformationdependent strategies are A_{1} and A_{2}; the one with the highest EMV is A_{1}, which has an EMV of $76,250. So Clark’s best strategy is an informationdependent one, and it has an EMV that is $400 higher than the EMV of his best noninformationdependent 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!
type I (4/5) 
type II (1/5) 

A_{1}: bet on type I  $20  –$10 
A_{2}: bet on type II  –$10  $80 
EMV(A_{1}) = (4/5)($20) + (1/5)(–$10) = $16 – $2 = $14
EMV(A_{2}) = (4/5)(–$10) + (1/5)($80) = –$8 + $16 = $8
So A_{1}, or betting on type I, is the one to choose.
 value of completely reliable information:
I and “I” (4/5)*(1) = 4/5 
I and “II” (4/5)*(0) = 0 
II and “I” (1/5)*(0) = 0 
II and “II” (1/5)*(1) = 1/5 

A_{1}: bet on type I  $20  $20  –$10  –$10 
A_{2}: bet on type II  –$10  –$10  $80  $80 
A_{3}: bet on I if “I” and II if “II”  $20  –$10  –$10  $80 
A_{4}: bet on II if “I” and I if “II”  –$10  $20  $80  –$10 
Now, computing the expected monetary values of the four acts, we have the following:
EMV(A_{1}): Since A_{1} 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(A_{2}): Since A_{2} 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(A_{3}) = (4/5)($20) + (0)($___) + (0)($___) + (1/5)($80) = $16 + $16 + $32
EMV(A_{4}) = (4/5)(–$10) + (0)($___) + (0)($___) + (1/5)(–$10) = –$8 – $2 = –$10
The informationdependent strategies are A_{3} and A_{4}; the one with the highest EMV is A_{3}, which has an EMV of $32. The noninformationdependent strategies are A_{1} and A_{2}; the one with the highest EMV is A_{1}, 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.
 revised probabilities and EMVs, given blue ball drawn
P(type I / blue) = 24/25 (from section 32, 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)A_{1}: bet on type I $20 –$10 A_{2}: bet on type II –$10 $80 EMV(A_{1}) = (24/25)($20) + (1/25)(–$10) = $19.20 – $0.40 = $18.80
EMV(A_{2}) = (24/25)(–$10) + (1/25)($80) = –$9.60 + $3.20 = –$6.40
So A_{1}, or betting on type I, is the one to choose.
 revised probabilities and EMVs, given red ball drawn
P(type I / red) = 16/25 (from section 32, 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)A_{1}: bet on type I $20 –$10 A_{2}: bet on type II –$10 $80 EMV(A_{1}) = (16/25)($20) + (9/25)(–$10) = $12.80 – $3.60 = $9.20
EMV(A_{2}) = (16/25)(–$10) + (9/25)($80) = –$6.40 + $28.80 = –$22.40
So A_{1}, or betting on type I, is the one to choose.
 value of color of ball drawn:
I and blue (4/5)*(3/5) = 12/25 
I and red (4/5)*(2/5) = 8/25 
II and blue (1/5)*(1/10) = 1/50 
II and red (1/5)*(9/10) = 9/50 

A_{1}: bet on type I  $20  $20  –$10  –$10 
A_{2}: bet on type II  –$10  –$10  $80  $80 
A_{3}: bet on I if “I” and II if “II”  $20  –$10  –$10  $80 
A_{4}: bet on II if “I” and I if “II”  –$10  $20  $80  –$10 
Now, computing the expected monetary values of the four acts, we have the following:
EMV(A_{1}): Since A_{1} 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(A_{2}): Since A_{2} 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(A_{3}) = (12/25)($20) + (8/25)(–$10) + (1/50)(–$10) + (9/50)($80)
= $9.60 – $3.20 – $0.20 + $14.40
= $20.60EMV(A_{4}) = (12/25)(–$10) + (8/25)($20) + (1/50)($80) + (9/50)(–$10)
= –$4.80 + $6.40 + $1.60 – $1.80
= $1.40The informationdependent strategies are A_{3} and A_{4}; the one with the highest EMV is A_{3}, which has an EMV of $20.60. The noninformationdependent strategies are A_{1} and A_{2}; the one with the highest EMV is A_{1}, 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.
market  withhold  
1: market if fewer than ten die; market if ten or more die  1  0 
2: market if fewer than ten die; withhold if ten or more die  1/2  1/2 
3: withhold if fewer than ten die; market if ten or more die  1/2  1/2 
4: withhold if fewer than ten die; withhold if ten or more die  0  1 
 batch fine:
market  withhold  
1: market if fewer than ten die; market if ten or more die  1  0 
2: market if fewer than ten die; withhold if ten or more die  99/100  1/100 
3: withhold if fewer than ten die; market if ten or more die  1/100  99/100 
4: withhold if fewer than ten die; withhold if ten or more die  0  1 
batch defective  batch fine  
1: market if fewer than ten die; market if ten or more die  (1)(–1,000) + (0)(0) = –1,000  (1)(100) + (0)(–100) = 100 
2: market if fewer than ten die; withhold if ten or more die  (1/2)(–1,000) + (1/2)(0) = –500  (99/100)(100) + (1/100)(–100) = 98 
3: withhold if fewer than ten die; market if ten or more die  (1/2)(–1,000) + (1/2)(0) = –500  (1/100)(100) + (99/100)(–100) = –98 
4: withhold if fewer than ten die; withhold if ten or more die  (0)(–1,000) + (1)(0) = 0  (0)(100) + (1)(–100) = –100 
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]
no.  claim  justification 
1 
P(p) = #(pcases)/#(total possibilities) 
definition of classical interpretation 
2 
P(not p) = #(notpcases)/#(total possibilities) 
definition of classical interpretation 
3 
P(p) + P(not p) = #(pcases)/#(total possibilities) + #(notpcases)/#(total possibilities) 
1 and 2 
4 
#(pcases)/#(total possibilities) + #(notpcases)/#(total possibilities) = [#(pcases) + #(notpcases)]/#(total possibilities) 
3, combining common denominator 
5 
[#(pcases) + #(notpcases)]/#(total possibilities) = #(total possibilities)/#(total possibilities) 
The set of pcases and set of notpcases are mutually exclusive and exhaustive subsets of the total possibilities. 
6 
#(total possibilities)/#(total possibilities) = 1 
same numerator and denominator on left side 
7 
P(p) + P(not p) = 1 
3–6, transitivity of = 
no.  claim  justification 
1 
p and q are equivalent 
assumption for proving conditional 
2 
P(p) = #(pcases)/#(total possibilities) 
definition of classical interpretation 
3 
#(pcases) = #(qcases) 
1 
4 
#(pcases)/#(total possibilities) = #(qcases) /#(total possibilities) 
3 
5 
#(qcases) /#(total possibilities) = P(q) 
definition of classical interpretation 
6 
P(p) = P(q) 
2, 4, and 5, transitivity of = 
First, we have to consider the case where nothing is both a P and an R:
no. claim justification 1 P(p & q) = P_{R}(P & Q)
notational convention of relativefrequency interpretation 2 P_{R}(P & Q) = #(P & Q & Rs)/#(Rs)
definition of relativefrequency 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) = P_{R}(P) notational convention of relativefrequency interpretation 7 P_{R}(P) = #(P and Rs)/#(Rs)
definition of relativefrequency 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) = P_{R}(Q/P) notational convention of relativefrequency interpretation 12 P_{R}(Q/P) = 0 definition of relativefrequency 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:
no.  claim  justification 
1 
P(p & q) = P_{R}(P & Q) 
notational convention of relativefrequency interpretation 
2 
P_{R}(P & Q) = #(P & Q & Rs)/#(Rs) 
definition of relativefrequency interpretation 
3 
#(P & Q & Rs)/#(Rs) = #(P & Q & Rs)/#(Rs) * #(P & Rs)/#(P & Rs) 
right side is same as left side, times 1 
4 
#(P & Q & Rs)/#(Rs) * #(P & Rs)/#(P & Rs) = #(P & Rs)/#(Rs) * #(P & Q & Rs)/#(P & Rs) 
right side is same as left side, with numerators switched 
5 
#(P & Rs)/#(Rs) * #(P & Q & Rs)/#(P & Rs) = P_{R}(P) * P_{R}(Q/P) 
definition of relativefrequency interpretation 
6 
P_{R}(P) * P_{R}(Q/P) = P(p) * P(q/p) 
notational convention of relativefrequency interpretation 
7 
P(p & q) = P(p) * P(q/p) 
1–6, transitivity of = 
no.  claim  justification 
1 
p and q are equivalent 
assumption for proving conditional 
2 
P(p) = P_{R}(P) 
notational convention of relativefrequency interpretation 
3 
P_{R}(P) = #(P & Rs)/#(Rs) 
definition of relativefrequency interpretation 
4 
#(P & Rs) = #(Q & Rs) 
1 
5 
#(P & Rs)/#(Rs) = #(Q & Rs)/#(Rs) 
4, with both sides divided by same thing 
6 
#(Q & Rs)/#(Rs) = P_{R}(Q) 
definition of relativefrequency interpretation 
7 
P_{R}(Q) = P(q) 
notational convention of relativefrequency interpretation 
8 
P(p) = P(q) 
2, 3, and 5–7, transitivity of = 
section 33c
no.  claim  justification 
1 
p and q are mutually exclusive. 
given 
2 
c > a + b 
given 
3 
c – a – b > 0 
2, subtract a and b from each side 
4 
Let strategy A_{1} be the strategy of betting for p, betting for q, and betting against the disjunction p or q. 
just defining strategy A_{1} 
5 
Rows 2–4 of table 38 describe all the genuine possibilities. 
1 (mutually exclusive statements can’t both be true at once) 
6 
In row 2, strategy A_{1} pays (1 – a) +
(–b) + –(1 – c) 
reading table, with information from 2, above 
7 
In row 3, strategy A_{1} pays (–a) + (1
– b) + –(1 – c) 
reading table, with information from 2, above 
8 
In row 4, strategy A_{1} pays (–a) + (–b) + c

reading table, with information from 2, above 
9 
In each row describing a genuine possibility, strategy A_{1} has a positive result. 
5–8 
10 
A Dutch Book can be made against anyone who holds that c > a + b. 
9 
 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 A_{1} be the strategy of betting for “q given p”.
just defining strategy A_{1} 4 Rows 1 and 2 of table 39 describe all the genuine possibilities in which the bet is on.
The bet is off if p is false. 5 In row 1, strategy A_{1} 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 A_{1} 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 A_{1} has a positive result.
4–6 8 A Dutch Book can be made against anyone who holds that a < 0.
7
 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 A_{1} be the strategy of betting against “q given p”.
just defining strategy A_{1} 4 Rows 1 and 2 of table 39 describe all the genuine possibilities in which the bet is on.
The bet is off if p is false. 5 In row 1, strategy A_{1} pays –(1 – a)
= –1 + a
= a – 1,
which is positive.reading table, with information from 2 6 In row 2, strategy A_{1} 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 A_{1} has a positive result.
4–6 8 A Dutch Book can be made against anyone who holds that a > 1.
7
no.  claim  justification 
1 
Rows 1–4 of table 310 represent all the possibilities. 
2 truth values for each of 2 statements, 2^{2} = 4 
2 
In row 1, the bookie’s bet pays –(1 – a)b + –(1
– b) + (1 – c) 
reading table (as amended) 
3 
In row 2, the bookie’s bet pays –(1 – a)b + (b)
+ (–c) 
reading table (as amended) 
4 
In row 3, the bookie’s bet pays (ab) + (0) +
(–c) 
reading table (as amended) 
5 
In row 4, the bookie’s bet pays (ab) + (0) +
(–c) 
reading table (as amended) 
6 
In each row describing a genuine possibility, the bookie’s bet pays ab – c. 
1–5 
no.  claim  justification 
1 
c > ab 
given 
2 
c – ab > 0 
1, subtract ab from each side 
3 
Let the bookie’s bet be [for p, for ”q given p”, and against “p and q”]. 
just defining the bookie’s bet 
4 
Rows 1–4 of table 310 describe all the possibilities. 
2 truth values for each of 2 statements, 2^{2} = 4 
5 
In row 1, the bookie’s bet pays (1 – a)b + (1 –
b) + –(1 – c) 
reading table (as amended) 
6 
In row 2, the bookie’s bet pays (1 – a)b + (–b)
+ (c) 
reading table (as amended) 
7 
In row 3, the bookie’s bet pays (–ab) + (0) +
(c) 
reading table (as amended) 
8 
In row 4, the bookie’s bet pays (–ab) + (0) +
(c) 
reading table (as amended) 
9 
In each row describing a genuine possibility, the bookie’s bet pays c – ab. 
4–8 
10 
In each row describing a genuine possibility, the bookie’s bet has a positive result. 
9, with information from 2 
no.  claim  justification 
1 
p is impossible. 
assumption for proving conditional 
2 
P(p) < 0 
given 
3  P(p) = a  definition of having a betting quotient of a 
4  a < 0  2, with substitution from 3 
5 
–a > 0 
4, multiply each side by –1 
6 
Let strategy A_{1} be the strategy of betting for p. 
just defining strategy A_{1} 
7 
Let the stakes on p be S. 
specifying the stakes 
8 
Row 2 of table 36 describes the only genuine possibility. 
1 (an impossible statement cannot be true) 
9 
In row 2, strategy A_{1} pays –aS 
reading table, with information from 5 
10 
In the only row describing a genuine possibility, strategy A_{1} has a positive result. 
9–9 
11 
A Dutch Book can be made against anyone who holds that P(p) < 0. 
10 
 proof that if p is impossible, then P(p) > 0 leads to a Dutch Book:
no.  claim  justification 
1 
p is impossible. 
assumption for proving conditional 
2 
P(p) > 0 
given 
3  P(p) = a  definition of having a betting quotient of a 
4  a > 0  2, with substitution from 3 
5 
Let strategy A_{1} be the strategy of betting against p. 
just defining strategy A_{1} 
6 
Let the stakes on p be S. 
specifying the stakes 
7 
Row 2 of table 36 describes the only genuine possibility. 
1 (an impossible statement cannot be true) 
8 
In row 2, strategy A_{1} pays aS, 
reading table, with information from 4 
9 
In the only row describing a genuine possibility, strategy A_{1} has a positive result. 
7–8 
10 
A Dutch Book can be made against anyone who holds that P(p) > 0. 
9 
 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
 proof that P(p) + P(not p) < 1 leads to a Dutch Book:
no.  claim  justification 
1 
P(p) + P(not p) < 1 
given 
2  P(p) = a  definition of having a betting quotient of a 
3  P(not p) = b  definition of having a betting quotient of b 
4  a + b < 1  1, with substitution from 2 and 3 
5 
1 – a – b > 0 
4, subtract a and b from each side 
6 
Let strategy A_{1} be the strategy of betting for p and for not p. 
just defining strategy A_{1} 
7 
Let the stakes on p be 1, and let the stakes on not p be 1 as well. 
specifying the stakes 
8 
Rows 2 and 3 of the foregoing table describe all the genuine possibilities. 
p and not p cannot be both true (row 1) or both false (row 4). 
9 
In row 2, strategy A_{1} pays (1 – a) +
(–b) 
reading table, with information from 5 
10 
In row 3, strategy A_{1} pays (–a) + (1
– b) 
reading table, with information from 5 
11 
In the rows describing all the genuine possibilities, strategy A_{1} has a positive result. 
8–10 
12 
A Dutch book can be made against anyone who holds that P(p) + P(not p) < 1. 
9 
 proof that P(p) + P(not p) > 1 leads to a Dutch Book:
no.  claim  justification 
1 
P(p) + P(not p) > 1 
given 
2  P(p) = a  definition of having a betting quotient of a 
3  P(not p) = b  definition of having a betting quotient of b 
4  a + b > 1  1, with substitution from 2 and 3 
5 
a + b – 1 > 0 
4, subtract 1 from each side 
6 
Let strategy A_{1} be the strategy of betting against p and against not p. 
just defining strategy A_{1} 
7 
Let the stakes on p be 1, and let the stakes on not p be 1 as well. 
specifying the stakes 
8 
Rows 2 and 3 of the foregoing table describe all the genuine possibilities. 
p and not p cannot be both true (row 1) or both false (row 4). 
9 
In row 2, strategy A_{1} pays –(1 – a) +
(b) 
reading table, with information from 5 
10 
In row 3, strategy A_{1} pays (a) + –(1
– b) 
reading table, with information from 5 
11 
In the rows describing all the genuine possibilities, strategy A_{1} has a positive result. 
8–10 
12 
A Dutch book can be made against anyone who holds that P(p) + P(not p) > 1. 
9 
 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:
no.  claim  justification 
1 
p logically implies q 
assumption for proving conditional 
2 
P(p) > P(q) 
given 
3  P(p) = a  definition of having a betting quotient of a 
4  P(q) = b  definition of having a betting quotient of b 
5  a > b  2, with substitution from 3 and 4 
6 
a – b > 0 
5, subtract b from each side 
7 
Let strategy A_{1} be the strategy of betting against p and for q. 
just defining strategy A_{1} 
8 
Let the stakes on p be 1, and let the stakes on q be 1 as well. 
specifying the stakes 
9 
Rows 1, 3, and 4 of the foregoing table describe all the genuine possibilities. 
1 (p cannot be true while q is false—so row 2 is not a genuine possibility). 
10 
In row 1, strategy A_{1} pays –(1 – a) +
(1 – b) 
reading table, with information from 6 
11 
In row 3, strategy A_{1} pays (a) + (1 –
b) 
reading table, with information from 6 
12 
In row 4, strategy A_{1} pays (a) + (–
b) 
reading table, with information from 6 
13 
In the rows describing all the genuine possibilities, strategy A_{1} has a positive result. 
9–12 
14 
If p logically implies q, a Dutch book can be made against anyone who holds that P(p) > P(q). 
13 
section 33d
[skip the problems]
section 41
Ace wins (1/2)  Jack wins (1/2)  
bet on Ace  $5  –$2 
bet on Jack  –$10  x 
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 = xSo 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.
no.  claim  justification 
1  s and s’ are equivalent ratio scales.  assumption for proving conditional 
2 
s(x) = 2s(y) 
assumption for proving conditional 
3 
s’(z) = as(z) for all z, for some a > 0 
1, definition of equivalent ratio scales 
4 
s’(x) = as(x) 
3, applied to x 
5 
as(x) = a2s(y) 
2, multiply each side by a 
6 
a2s(y) = 2as(y) 
commutativity of multiplication 
7 
s’(y) = as(y) 
3, applied to y 
8 
as(y) = s’(y) 
7, symmetry of = 
9 
2as(y) = 2s’(y) 
8, multiply each side by 2 
10 
s’(x) = 2s’(y) 
4–6, 9, transtivity of = 
We’ll begin with the following table:
S_{1} (p_{1}) S_{2} (p_{2}) . . . S_{n} (p_{n}) A_{i} u_{1} u_{2} . . . u_{n} A_{j} v_{1} v_{2} . . . v_{n} And we’ll show the transformed values in the following table:
S_{1} (p_{1})  S_{2} (p_{2})  . . .  S_{n} (p_{n})  
A_{i}  u_{1} + b_{1}  u_{2} + b_{2}  . . .  u_{n} + b_{n} 
A_{j}  v_{1} + b_{1}  v_{2} + b_{2}  . . .  v_{n} + b_{n} 
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:
no.  claim  justification 
1  EU(A_{i}) ≥ EU(A_{j}), based on the utilities in the first table  assumption for proving conditional (where EU refers to expected utility) 
2  p_{1}u_{1} + p_{2}u_{2} + . . . p_{n}u_{n} ≥ p_{1}v_{1} + p_{2}v_{2} + . . . p_{n}v_{n}  1, definition of EU 
3 
p_{1}u_{1} + p_{2}u_{2}
+ . . . p_{n}u_{n} + p_{1}b_{1} + p_{2}b_{2}
+ . . . + p_{n}b_{n} ≥ 
2, add [p_{1}b_{1} + p_{2}b_{2} + . . . + p_{n}b_{n}] to each side 
4 
p_{1}u_{1} + p_{1}b_{1}
+ p_{2}u_{2} + p_{2}b_{2} + . . . + p_{n}u_{n}
+ p_{n}b_{n} ≥ 
3, rearranging terms 
5 
p_{1}(u_{1} + b_{1}) + p_{2}(u_{2} +
b_{2}) + . . . + p_{n}(u_{n} + b_{n}) ≥ p_{1}(v_{1} + b_{1}) + p_{2}(v_{2} + b_{2}) + . . . + p_{n}(v_{n} + b_{n}) 
4, with grouping 
6  EU(A_{i}) ≥ EU(A_{j}), based on the utilities in the second table  5, definition of EU 
no.  claim  justification 
1  Let the expected utility of some act A_{1} be u.  just defining variables 
2 
u can be represented as p_{1}u_{1} + p_{2}u_{2}
+ . . . + p_{n}u_{n}, where u_{i} is the utility associated with A_{1} if state i occurs, and where p_{i} is the probability that state i occurs 
definition of expected utility 
3 
The expected disutility of A_{1} can be represented as p_{1}(–u_{1}) + p_{2}(–u_{2}) + . . . p_{n}(–u_{n}), where u_{i} and p_{i} are defined as above. 
definition of expected disutility 
4 
p_{1}(–u_{1}) + p_{2}(–u_{2}) + . . . p_{n}(–u_{n}) = –p_{1}u_{1} – p_{2}u_{2} – . . . – p_{n}u_{n} 
simplification 
5  –p_{1}u_{1} – p_{2}u_{2} – . . . – p_{n}u_{n} = –(p_{1}u_{1} + p_{2}u_{2} + . . . + p_{n}u_{n})  factoring 
6  –(p_{1}u_{1} + p_{2}u_{2} + . . . + p_{n}u_{n}) = –u  2, multiply each side by –1 and reverse 
7 
The expected disutility of A_{1} can be represented as –u. 
3–6, transitivity of = 
8 
The expected disutility of A_{1} can be represented as –1 times its expected utility 
7, with substitution from 1 
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.
no.  claim  justification 
1 
EU(Ai) ≥ EU(Aj) 
assumption for proving conditional 
2 
–EU(Aj) ≤ –EU(Aj) 
1, multiplying each side by –1 
3 
EDU(A_{i}) = –EU(A_{i}) 
proved in solution to problem 2 
4 
EDU(A_{j}) = –EU(A_{j}) 
proved in solution to problem 2 
5  EDU(Ai) ≤ EDU(Aj)  2, with substitution from 3 and 4 
heads on toss 1 (1/2)  heads on (and not until) toss 2 (1/4)  heads on no toss through toss 2 (1/4)  
play the game  $2  $4  $0 
The EMV is (1/2)($2) + (1/4)($4) + (1/4)($0)
= $1 + $1 + $0
= $2If 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 (1/2^{n}) heads on no toss through toss n (1/2^{n}) play the game $2 $4 $2^{n} $0 The EMV is (1/2)($2) + (1/4)($4) + . . . + (1/2^{n})($2^{n}) + (1/2^{n})($0)
= $1 + $1 + . . . + $1 + $0
= n($1) + $0
= $nAn 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.
no.  claim  justification 
1 
u[L(a, x, y)] = au(x) + (1 – a)u(y) 
expected utility property (condition 3 on p. 90) 
2 
u[L(1, x, y)] = 1u(x) + (1 – 1)u(y) 
1, substituting 1 for a 
3 
u[L(1, x, y)] = u(x) + 0u(y) 
2, simplifying right side 
4  u[L(1, x, y)] = u(x)  3, simplifying right side 
5  L(1, x, y) I x  4, via condition 2 on p. 90 
 L(0, x, y) I y:
no.  claim  justification 
1 
u[L(a, x, y)] = au(x) + (1 – a)u(y) 
expected utility property (condition 3 on p. 90) 
2 
u[L(0, x, y)] = 0u(x) + (1 – 0)u(y) 
1, substituting 0 for a 
3 
u[L(0, x, y)] = 0u(x) + 1u(y) 
2, simplifying right side 
4  u[L(0, x, y)] = u(y)  3, simplifying right side 
5  L(0, x, y) I y  4, via condition 2 on p. 90 
 L(a, x, y) I L(1 – a, y, x):
no.  claim  justification 
1 
u[L(a, x, y)] = au(x) + (1 – a)u(y) 
expected utility property (condition 3 on p. 90) 
2 
u[L(1 – a, y, x)] = (1 – a)u(y) + (1 – (1 – a))u(x) 
1, substituting 1 – a for a, y for x, and x for y 
3 
u[L(1 – a, y, x)] = (1 – a)u(y) + (1 – 1 + a))u(x) 
2, simplifying right side 
4  u[L(1 – a, y, x)] = (1 – a)u(y) + au(x)  3, simplifying right side 
5  u[L(1 – a, y, x)] = au(x) + (1 – a)u(y)  4, rearranging right side 
6  u[L(a, x, y)] = u[L(1 – a, y, x)]  5, transitivity of = 
7  L(a, x, y) I L(1 – a, y, x)  6, via condition 2 on p. 90 
 L(a, x, x) I x:
no.  claim  justification 
1 
u[L(a, x, y)] = au(x) + (1 – a)u(y) 
expected utility property (condition 3 on p. 90) 
2 
u[L(a, x, x)] = au(x) + (1 – a))u(x) 
1, substituting x for y 
3 
u[L(a, x, x)] = au(x) + u(x) – aux 
2, simplifying right side 
4  u[L(a, x, x)] = u(x)  3, simplifying right side 
5  L(a, x, x) I x  4, via condition 2 on p. 90 
no.  claim  justification 
1 
x I L(a, y, z) 
assumption for proving conditional 
2 
Exactly one of the following is true: 
ordering condition, parts O1–O4 
3  Suppose possibility 1 is true: L(c, v, x) P L(c, v, L(a, y, z))  considering one possibility 
4 
x P L(a, y, z) 
3, via betterprizes condition 
5 
4 is precluded by 1. 
ordering condition, part O3 
6 
Possibility 1 leads to a contradiction. 
3–5 
7 
Suppose possibility 3 is true: L(c, v, L(a, y, z)) P L(c, v, x) 
considering another possibility 
8 
L(a, y, z) P x 
7, via betterprizes condition 
9 
8 is precluded by 1. 
ordering condition, part O3 
10 
Possibility 3 leads to a contradiction. 
7–9 
11 
Possibility 2 must be true. 
2, 6, 10 
12 
L(c, v, x) I L(c, v, L(a, y, z)). 
11, 2 
no.  claim  justification 
1 
xIy and zIw 
assumption for proving conditional 
2  xIy  1, conjunction elimination 
3  zIw  2, conjunction elimination 
4 
xPz 
assumption for proving first half of biconditional 
5 
yPz 
4 and 2, via ordering condition, part O6 
6 
yPw 
5 and 3, via ordering condition, part O7 
7 
if xPz, then yPw 
4–6 
8 
yPw 
assumption for proving second half of biconditional 
9 
xPw 
8 and 2, via ordering condition, part O6 
10 
xPz 
9 and 3, via ordering condition, part O7 
11 
if yPw, then xPz 
8–10 
12 
xPz if and only if yPw 
7 and 11 
no.  claim  justification 
1 
Suppose there were a number a or basic prize x distinct from B for which L(a, x, B) P L(a, B, B). 
assumption for indirect proof 
2  x P B  1, via betterprizes condition 
3  2 is impossible.  B is defined as a basic prize that is at least as good as any other basic prize. 
4 
There is no number a or basic prize x for which L(a, x, B) P L(a, B, B). 
1–3 
5 
There is no number a or basic prize x distinct from B for which L(a, x, B) P L(a, B, B). 
4 
 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):
no.  claim  justification 
1 
Suppose there were a number a or basic prize x distinct from B for which L(a, B, x) P L(a, B, B). 
assumption for indirect proof 
2  x P B  1, via betterprizes condition 
3  2 is impossible.  B is defined as a basic prize that is at least as good as any other basic prize. 
4 
There is no number a or basic prize x for which L(a, B, x) P L(a, B, B). 
1–3 
5 
There is no number a or basic prize x distinct from B for which L(a, B, x) P L(a, B, B). 
4 
 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.
no.  claim  justification 
1  x I B  assumption for proving conditional 
2 
L(a, x, y) P L(a, B, y) or L(a, x, y) I L(a, B, y) or L(a, B, y) P L(a, x, y) 
ordering condition, part O4 
3  Suppose L(a, x, y) P (a, B, y).  considering first possibility 
4  x P B  3, via betterprizes condition 
5 
It is not true that x P B. 
1, via ordering condition, part O3 
6 
Line 3 leads to a contradiction. 
4 and 5 
7 
Line 3 is false. 
6 
8 
It is not true that L(a, x, y) P (a, B, y). 
7 
9 
Suppose L(a, B, y) P L(a, x, y). 
considering third possibility 
10 
B P x 
9, via betterprizes condition 
11 
It is not true that B P x. 
1, via ordering condition, part O3 
12 
Line 9 leads to a contradiction. 
10 and 11 
13 
Line 9 is false. 
12 
14 
It is not true that L(a, B, y) P L(a, x, y). 
13 
15 
L(a, x, y) I L(a, B, y) 
2, 8, and 14 
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):
no.  claim  justification 
1  Suppose there were a number a or basic prizes x and y distinct from B for which L(a, x, y) P L(a, B, B).  assumption for indirect proof 
2  It is not true that x P B.  B is defined as a basic prize that is at least as good as any other basic prize. 
3  It is not true that y P B.  B is defined as a basic prize that is at least as good as any other basic prize. 
4  x P B or x I B or B P x  ordering condition, part O4 
5  x I B  considering middle possibility 
6  L(a, x, y) I L(a, B, y)  5, via lemma just proved 
7  L(a, B, y) P L(a, B, B)  1 and 6, via ordering condition, part O6 
8  y P B  7, via betterprizes condition 
9 
Line 8 is impossible. 
B is defined as a basic prize that is at least as good as any other basic prize. 
10 
Line 5 leads to an impossibility 
9 
11 
Line 5 is not true. 
10 
12 
It is not true that x I B. 
11 
13 
B P x 
4, 2, and 12 
14 
y P B or y I B or B P y 
ordering condition, part O4 
15 
y I B 
considering middle possibility 
16 
L(a, x, y) I L(a, x, B) 
15, via lemma just proved 
17 
L(a, x, B) P L(a, B, B) 
1 and 16, via ordering condition, part O6 
18 
x P B 
17, via betterprizes condition 
19 
Line 18 is impossible. 
B is defined as a basic prize that is at least as good as any other basic prize. 
20 
Line 15 leads to an impossibility. 
19 
21 
Line 15 is not true. 
20 
22 
It is not true that y I B. 
21 
23 
B P y 
14, 3, and 22 
24 
L(a, B, B) P L(a, B, y) 
23, via betterprizes condition 
25 
L(a, B, y) P L(a, x, y) 
13, via betterprizes condition 
26 
L(a, B, B) P L(a, x, y) 
24 and 25, via ordering condition, part O5 
27 
It is not the case that L(a, x, y) P L(a, B, B) 
26, via ordering condition, part O1 
28 
Line 1 leads to a contradiction. 
1 and 27 
29 
Line 1 is false. 
28 
30 
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). 
29 
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, B, L) P L(a, B, B).  assumption for indirect proof 
3  L P B  2, via betterprizes condition 
4 
3 is impossible. 
1 
5 
There is no number a and lottery L of degree less than n for which L(a, B, L) P L(a, B, B). 
2–4 
 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 betterprizes 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
 proof that if no lottery of degree less than n is preferred to B, then there is no number a and no lotteries L_{1} and L_{2} of degree less than n for which L(a, L_{1}, L_{2}) P L(a, B, B):
This can be proved using a proof just like that used in problem 4b, with three changes:
 every reference to basic prize x must be replaced with a reference to lottery L_{1}
 every reference to basic prize y must be replaced with a reference to lottery L_{2}
 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 L_{1} or L_{2}) is preferred to B.”
no.  claim  justification 
1  No lottery of degree greater than 0 is preferred to L(a, B, B) for any number a.  stated as conclusion of exercise 5 
2 
L(a, L(a, B, B), L(a, B, B) is not preferred to L(a, B, B). 
1, since L(a, L(a, B, B), L(a, B, B) is a lottery of degree greater than 0 
3  Suppose, for some number a, L(a, B, B) P B.  assumption for indirect proof 
4  L(a, L(a, B, B), B) P L(a, B, B)  3, via betterprizes condition 
5 
L(a, L(a, B, B), L(a, B, B)) P L(a, L(a, B, B), B) 
3, via betterprizes condition 
6 
L(a, L(a, B, B), L(a, B, B)) P L(a, B, B) 
4 and 5, via ordering condition, part O5 
7 
Line 3 leads to a contradiction. 
2 and 6 
8 
Line 3 is false. 
7 
9 
For no number a, L(a, B, B) P B. 
8 
In class, Joe Morgan suggested a different strategy, which is used in the following proof:
no.  claim  justification 
1  No lottery of degree greater than 0 is preferred to L(a, B, B) for any number a.  stated as conclusion of exercise 5 
2  Suppose, for some number a, L(a, B, B) P B.  assumption for indirect proof 
3  L(a, L(a, B, B), B) P L(a, B, B)  2, via betterprizes condition 
4 
Line 3 is impossible. 
1 
5 
Line 2 leads to an impossibility. 
4 
6 
Line 2 is false. 
5 
7 
For no number a, L(a, B, B) P B. 
6 
After class, Courtney Gustafson suggested the following, also different, approach:
no.  claim  justification 
1  Suppose, for some number a, L(a, B, B) P B.  assumption for indirect proof 
2  L(a, B, L(a, B, B)) P L(a, B, B)  1, via betterprizes condition 
3  L(a, L(a, B, B), L(a, B, B)) I L(a, B, B)  reductionofcompoundlotteries condition 
4 
L(a, B, L(a, B, B)) P L(a, L(a, B, B), L(a, B, B)) 
2 and 3, via ordering condition, part O7 
5 
B P L(a, B, B) 
4, via betterprizes condition 
6 
Line 1 leads to a contradiction. 
1 and 5 
7 
Line 1 is false. 
6 
8 
For no number a, L(a, B, B) P B. 
7 
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, L(a, B, B), B)  1, via betterprizes condition 
3  L(a, L(a, B, B), B) P L(a, L(a, B, B), L(a, B, B))  1, via betterprizes condition 
4  L(a, B, B) P L(a, L(a, B, B), L(a, B, B))  2 and 3, via ordering condition, part O5 
5 
L(a, L(a, B, B), L(a, B, B)) I L(a, B, B) 
reductionofcompoundlotteries condition 
6 
It is not the case that L(a, B, B) P L(a, L(a, B, B), L(a, B, B)). 
5, via ordering condition, part O3 
7 
Line 1 leads to a contradiction. 
4 and 6 
8 
Line 1 is false. 
7 
9 
For no number a, B P L(a, B, B). 
8 
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 betterprizes condition 3 L(a, L(a, B, B), L(a, B, B)) I L(a, B, B) reductionofcompoundlotteries 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 betterprizes 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
no.  claim  justification 
1  B P x and x P W  assumption for proving conditional 
2  B P x  1, conjunction elimination 
3  x P W  1, conjunction elimination 
4  Suppose L(a, x, x) P x.  considering one possibility 
5 
L(a, L(a, x, x), x) P L(a, x, x) 
4, via betterprizes condition 
6 
L(a, L(a, x, x), L(a, x, x)) P L(a, L(a, x, x), x) 
4, via betterprizes condition 
7 
L(a, L(a, x, x), L(a, x, x)) P L(a, x, x) 
5 and 6, via ordering condition, part O5 
8 
L(a, L(a, x, x), L(a, x, x)) I L(a, x, x) 
reductionofcompoundlotteries condition 
9 
It is not the case that L(a, L(a, x, x), L(a, x, x)) P L(a, x, x) 
8, via ordering condition, part O3 
10 
Line 4 leads to a contradiction. 
7 and 9 
11 
Line 4 is false. 
10 
12 
It is not the case that L(a, x, x) P x. 
11 
13 
Suppose x P L(a, x, x). 
considering a second possibility 
14 
L(a, x, x) P L(a, L(a, x, x), x) 
13, via betterprizes condition 
15 
L(a, L(a, x, x), x) P L(a, L(a, x, x), L(a, x, x)) 
13, via betterprizes condition 
16 
L(a, x, x) P L(a, L(a, x, x), L(a, x, x)) 
14 and 15, via ordering condition, part O5 
17 
L(a, L(a, x, x), L(a, x, x)) I L(a, x, x) 
reductionofcompoundlotteries condition 
18 
It is not the case that L(a, x, x) P L(a, L(a, x, x), L(a, x, x)) 
17, via ordering condition, part O3 
19 
Line 13 leads to a contradiction. 
18 and 18 
20 
Line 13 is false. 
19 
21 
It is not the case that x P L(a, x, x). 
20 
22 
L(a, x, x) P x, or x P L(a, x, x), or L(a, x, x) I x. 
ordering condition, part O4 
23 
L(a, x, x) I x 
22, 12, and 21 
(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 43 (p. 98)
no.  claim  justification 
1  c = I(1) – I(0)  statement k 
2  I(1) = u’(u^{–1}(1)) and I(0) = u’(u^{–1}(0))  statement a 
3  1 > 0  the number line in your secondgrade classroom 
4  u^{–1}(1) P u^{–1}(0)  3, along with the fact that it is assumed that u satisfies condition 1 on p. 90 
5  u’(x) > u’(y) if and only if xPy  It is given that u’ satisfies condition 1 on p. 97. 
6  u’(u^{–1}(1)) > u’(u^{–1}(0))  4 and 5 
7 
I(1) > I(0) 
6, with substitution from 2 
8 
I(1) – I(0) > 0 
7, subtract I(0) from each side 
9 
c > 0 
8, with substitution from 1 
no.  claim  justification 
1  k is a number on the uscale.  given 
2  Let r and s be numbers on the uscale such that k is between them or is one or both of them, with r ≤ s.  entitled to this by virtue of the fact that k is on the uscale, and a uscale exists 
3  Let x be the lottery L((s – k)/(s – r), u^{–1}(r), u^{–1}(s))  defining lottery x 
4  u(x) = ((s – k)/(s – r))*u(u^{–1}(r)) + (1 – ((s – k)/(s – r))*u(u^{–1}(s))  3, along with the fact that it is assumed that u satisfies condition 1 on p. 90 
5  u(x) = ((s – k)/(s – r))*r + ((s – r)/(s – r) – ((s – k)/(s – r))*s  4, simplified 
6  u(x) = ((s – k)/(s – r))*r + ((s – r – s + k)/(s – r))*s  5, simplified 
7 
u(x) = ((s – k)/(s – r))*r + ((k – r)/(s – r))*s 
6, simplified 
8 
u(x) = ((s – k)*r + (k – r)*s)/(s – r) 
7, simplified 
9 
u(x) = (sr – kr + ks – rs)/(s – r) 
8, simplified 
10 
u(x) = (ks – kr)/(s – r) 
9, simplified 
11 
u(x) = k(s – r)/(s – r) 
10, factored 
12 
u(x) = k 
11, simplified 
To transform a 0to1 scale into a 1to100 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 –5to5 scale into a 0to1 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.
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 NeumannMorgenstern 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).
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 milesperhour scale for measuring speed, the feetandinches scale for representing length, etc.
no.  claim  justification 
1 
Define lotteries a, b, and c as follows: a = bet heads on fair coin b = bet heads on biased coin c = bet tails on biased coin Also, let X be the utility of winning $1, and let y be the utility of winning $0. Finally, let p be probability that the agent (subjectively) assigns to the winning side of a coin coming up. 
just defining some variables 
2  The choices most people make are aPb and bIc.  given in statement of problem 
3  u(a) > u(b)  2, via property 1 on p. 90 
4 
EU(a) > EU(b) 
3, via property 3 on p. 90 
5  (1/2)X + (1/2)Y > pX + (1 – p)Y  4, with substitution from 1 
6  (1/2)X + (1/2)Y > pX + Y – pY  5, simplified 
7  pY – pX > Y – (1/2)Y – (1/2)X  6, rearranged 
8  pY – pX > (1/2)Y – (1/2)X  7, simplified 
9  p(Y – X) > (1/2)(Y – X)  8, factored 
10  p > 1/2  9, divide each side by Y – X 
11  u(b) = u(c)  2, via property 2 on p. 90 
12  EU(b) = EU(c)  11, via property 3 on p. 90 
13  pX + (1 – p)Y = (1 – p)X + pY  12, with substitution from 1 
14  pX + Y – pY = X – pX + pY  13, simplified 
15  pX – pY + pX – pY = X – Y  14, rearranged 
16  2pX – 2pY = X – Y  15, simplified 
17  2p(X – Y) = X – Y  16, factored 
18  2p = 1  17, divide each side by X – Y 
19  p = 1/2  18, divide each side by 2 
20 
Line 2 leads to a contradiction. 
10 and 19 
21 
The choices most people make are not consistent with the combined theories of utility and subjective probability. 
20 
not chosen  chosen  
be devout  payoff is bad, probability is low  payoff is quite good, probability is high 
don’t be devout  payoff is just above bad, probability is high  payoff is excellent, probability is low 
bad genes  good genes  
abstain from smoking  payoff is bad, probability is low  payoff is quite good, probability is high 
smoke freely  payoff is just above bad, probability is high  payoff is excellent, probability is low 
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
= 19/40Clearly 19/32 is greater than 19/40, so the first act has the higher expected utility, and the physician should use the penicillin.
Pen. 50% Effective (50%)  Pen. 75% Effective (50%)  
Ant. X 70% Effective (50%)  Ant. X 40% Effective (50%)  Ant. X 70% Effective (50%)  Ant. X 40% Effective (50%)  
Cured  Not Cured  Cured  Not Cured  Cured  Not Cured  Cured  Not Cured  
Give Penicillin  0.125  0.125  0.125  0.125  0.1875  0.0625  0.1875  0.0625 
Give Ant. X  0.175  0.075  0.1  0.15  0.175  0.075  0.1  0.15 