I'm trying to solve problem 16 (Ch. 5) of Jeffrey's Logic of Decision. The problem says:

Suppose that A and B are pairwise incompatible propositions, and suppose that the preference ranking is as follows:



¬G, ¬(A∨B)

¬A, ¬B

Suppose that des G = 1 and des T = 0

b) Assuming that des ¬G = -2, find des ¬A, prob A, prob B, prob G.

My attempt to solve it:

By (5-5) prob G = 1 / (1 - (des G / des ¬G))

So prob G = 1 / (1 - (1 / -2)) = 2/3

By problem 11, if A and B are ranked together but not with T, then prob A = prob B iff ¬A and ¬B are ranked together, which is the case.

By (5-6) des ¬(A∨B) = -(prob A∨B / prob ¬(A∨B)) * des A∨B = -2, since it's ranked together with ¬G.

By (5-2) des A∨B = (prob A * des A + prob B * des B) / (prob A + prob B) = 1, since prob A = prob B, and both des A and des B are equal to 1.

So, by substitution:

-2 = -(prob A∨B / prob ¬(A∨B)) * (prob A * des A + prob B * des B) / (prob A + prob B)

= -(prob A∨B / prob ¬(A∨B)) * 1.

That's as far as I get.

I've also tried to solve for ¬(A∨B), since it is logically equivalent to ¬A¬B, with equation (5-1)(e) page 81 but it doesn't get me anywhere.

What am I missing?

Would appreciate any help.

Thanks for your time and consideration.

NOTE: It's not homework. It's just me trying to understand the theory.



As we sorted out in the comments below, everything until your conclusion prob(A)=prob(B) is correct. So I will start there.

You have shown:

By (5-2) des A∨B = (prob A * des A + prob B * des B) / (prob A + prob B) = 1, since prob A = prob B.

But to apply (5-2), one has to show that prob(A∨B)≠0: if this were the case, then 0=prob(A∨B)des(A∨B)+prob¬(A∨B)des¬(A∨B)=des¬(A∨B), which contradicts the ranking of ¬(A∨B).

des ¬(A∨B)=-2, since it is ranked together with ¬G. This allows us to compute prob(A∨B) via prob¬(A∨B)=1/(1-(des¬(A∨B)/des(A∨B)))=1/3. Since A and B are mutually exclusive, and we already know that prob(A)=prob(B) we have prob(A∨B)=2⋅prob(A) and via 1-1/3=1-prob¬(A∨B)=prob(A∨B)=2⋅prob(A) we conclude prob(A)=1/3.

Now to compute des(¬A), use prob(A)=1/(1-(des(A)/des(¬A))). Solving this for des(¬A) yields des(¬A)=-1/2.

That does not fit the assumed order:

That seems to be an error with the exercise: If des(¬A)=-prob(A)/prob(¬A)*1 = -prob(A)/(1-prob(A))≤-2, one can conclude prob(A)≥ 2/3. With the same argument prob(B)≥ 2/3. But A and B are supposed to be mutually incompatible, and the sum of their probabilities would exceed 1, which is impossible.

Thanks for pointing out all the errors.

  • 1
    Your answer was very helpful. Thank you. But, I've been studying it and there's something that doesn't fit. The problem states that ¬G is ranked above ¬A, therefore des ¬G > des ¬A. However, in the answer des ¬A > des ¬G (-1/2 > -2). – martin Jan 13 '19 at 12:47
  • That is a good point. Assuming that problem 11 is really applicable (and it should be, if it says exactly what you have written there), the only cause of trouble I can see is (5-6). So what exactly does (5-6) say? More input would be helpful. Just to be sure, prob means probability, right? – Jishin Noben Jan 13 '19 at 13:09
  • Yes, prob is probability. (5-6) says this exactly page 86: "if des T=0 and des X ≠ 1, we have (5-6) des ¬X = -(prob X / prob ¬X) (des X)." – martin Jan 13 '19 at 14:50
  • Then (5-6) is simply not applicable here. By the way, you used (5-6) nevertheless in your reasoning without showing that this assumption is satisfied. There might be an error as well. I will look into it when I get home and update my answer accordingly. – Jishin Noben Jan 13 '19 at 15:51
  • 1
    In relation to page 86, I found a mistake in the book. The assumption of (5-6) is not that des X ≠ 1 but that prob X ≠ 1, which makes more sense. This is clear in the 1965 edition of the book. – martin Jan 17 '19 at 19:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.