This problem requires the von Neumann-Morgenstern theorem to solve.

Let L = {A, B, C, D} be a set of lotteries. You are indifferent between receiving A for sure and a lottery that gives you B with probability 0.9 and C with probability 0.1. You are also indifferent between receiving A for sure and a lottery that gives you B with probability 0.6 and D with probability 0.4. You preferences satisfy the vonn Neumann-Morgenstern axioms.

(i) What do you prefer most, C or D?

(ii) Calculate the (relative) difference in utility between B and C, and between B and D.

(iii) If we stipulate that your utility of B is 1 and your utility of C is 0, what are your utilities of A and D?

I did the problem like this:

For (i), it was obvious that D is preferred to D, because in the lottery with D I was willing to accept a lower chance of receiving B than I was in the lottery with C.

u(A) = 0.9u(B) + 0.1u(C)

u(A) = 0.6u(B) + 0.4u(D)

Therefore,

0.9u(B) + 0.1u(C) = 0.6u(B) + 0.4u(D)

Subtract both sides by 0.6u(B) and get

0.3u(B) + 0.1u(C) = 0.4u(D)

Let u(B) = 1, and u(C) = 0. Therefore, by the above equation:

0.3 = 0.4u(D)

Divide both sides by 0.4 and get

0.75 = u(D)

Now for the answer to (ii), if u(B) = 1, u(C) = 0, and u(D) = 0.75, then

u(B) - u(c) = 1

and

u(B) - u(d) = 0.25

Therefore, the difference between B and C is 4 times the difference between B and D.

Finally, for (iii),

Since u(A) = 0.9u(B) + 0.1u(C)

u(A) = 0.9

We can confirm this by looking at the other equation too:

u(A) = 0.6u(B) + 0.4u(D) = 0.6 + 0.4*0.75 = 0.6 + 0.3 = 0.9

So, u(A) = 0.9, and u(D) = 0.75.

I looked at the answer in the book, and this is exactly what the author put. But then I looked at the errata list for the book on his website (http://www.martinpeterson.org/errata.pdf)

and it says the following:

p. 115,... in the solution to 5.7 all that can be concluded is that 0.9B + 0.1C = 0.6B + 0.4D.

Did I make the same mistake as the author? If so, what is the mistake?