# Finding desirabilities and probabilities in Jeffrey's Logic of Decision

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:

A,B,G

T

¬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.

Solution:

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.

• 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 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 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 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 at 15:51
• 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 at 19:25