Wednesday, April 06, 2005

Logic of motivation

One interesting thing about desires (and motivational states in general, I think) is that they don't enter into the truth-functional relations that beliefs do.

Suppose I've been kidnapped, and I'm watching hungrily as my kidnapper sits down to dinner. Being hungry, I desire to eat the food that's on the table in front of the kidnapper. And hoping for an opportunity to escape, I desire that the food is poisoned. But I don't desire the conjunction of these states -- I don't desire that the food is poisoned and that I eat it.

Intending works this way too. Suppose I have a dart in each hand, and I'm throwing them simultaneously at a pair of small targets. If just one dart hits, I'll make $50. If zero or two darts hit, I don't get any money. Given how unlikely it is that I'll hit both targets, and how hard it is to hit even one, I may try to aim with each hand so as to maximize the probability of succeeding with that dart. I intend to hit the target on the left. I also intend to hit the target on the right. But I don't intend the conjunction -- I don't intend to hit both targets.

10 comments:

wirrdo said...

good thing to know when formulating an algebra of desire. it may be that we can and often desire very specific outcomes that makes generalizing motivations so difficult. ... like in that first example: suppose i'm kidnapped and i'm hungry. i may desire that the food be poisoned, the kidnapper eat it, that kidnapper to fall over dead, the keys to my handcuffs fall out of his/her pocket, the kidnapper's pet monkey to pick them up and bring them to me, and that i escape to freedom, after raiding the fridge for some non-poisioned food, and a beer. hooray beer. of course, all of that was assuming i was in handcuffs, and the kidnapper had a trained nice-temered pet chimp in a fez who may have shared in the desire of my escaping, which may or may not have been true... see? notice all the mays? that's why i like physics.

eww. i just remembered that desires have rank too(i.e. you'd probably desire the kidnapper to share his/her food before you'd desire the poisoning), which makes ethics resemble computer science more and more each day. so, like, examle two: the primary intent is to satisfy the desire of winning fifty dollars, cause god knows i need it. i, being right handed, will aim the dart in my right hand, and in the throw simply drop the dart in my left hand to the ground, guaranteeing that that dart will not hit. since i do not intend to hit both targets, i simplify the problem and intend to hit only one target, with the desired outcome of winning fifty dollars. there. that makes sense, doesn't it?

Chris said...

Neil,
Interesting. I recently read a paper using the fact that desires are not verifiable by external states (i.e., that they aren't intentional states, as intentional states are usually defined) to argue that conscious desire is logically impossible. The gist of the argument was simply that all conscious states are intentional states, desires are not intentional states, therefore desires are not conscious states. Unfortunately, I don't remember who the paper was by, or where I read it, but it was in a recent issue of some philosophy journal. I have it lying around somewhere, though. So, if you're interested, I can go find it (I don't know if you're now doing work in this area, or if you just thought this was interesting, and decided to post on it).

Bill said...

In partial response to the end of wirrdo's comment. I think the scenario being presented was this:

P_L = Prob. of hitting left target
P_R = Prob. of hitting right target
P_B = Prob. of hitting both targets

and assume that P_L is independendent of whether or not you through the right dart (resp. for P_R), and that P_L and P_R are independent events (ie. P_L*P_R = P_B).

P(extactly 1) = P_L(1-P_R) + P_R(1-P_L) = P_L + P_R -2*P_L*P_R

assume furthermore, that P_L = P_R = P. So we have the following 2 cases.

A) I throw one dart, my prob. winning is P.
B) I throw both, my prob of wining is 2P- P^2.

if P < 1/2, then B>A and you should throw both.

more sophisticated analysis can be done for the cases where my assumptions don't hold, but it doesn't change the point of the argument.

Bill said...

Correction: Case B) should have probablity 2P-2*P^2

Neil Sinhababu said...

Thanks for the math, Bill. Let me just say (in response to wirrdo) that the example can always be modified to make the P_L = P_R = P assumption true -- perhaps by moving the left target closer.

By the way, I presented the darts example to Michael Bratman this weekend after his talk at a colloquium. He wasn't that hot on it, but some of the faculty in the audience liked it a lot.

Justin said...

I think the suggestion that desires don't enter into truth functional relations like beliefs do is interesting, but I think the dart throwing example is flawed and fails to show what it's supposed to. Compare...

I'm at the bar watching Neil throw darts. I give him a 70% chance of hitting the dart board on the left. (This is a subjective probability of course.) It then seems fair to say that I *believe* that he'll hit the dart board on the left. (For the sake of my argument, it won't really matter how high we say a subjective probability assigned to P has to be before we count it as a belief that P, so long as the probability is less than 100% my point will follow. And it seems plausible it *should* be less than %100 -- I seem to believe lots of things I don't assign a %100 probability to.) I also assign a 70% probability to Neil hitting the dart board on the right. So, I believe that too. However, I *don't* believe that Neil will hit both dart boards. The math...

Pr(A & B) = Pr(A) x Pr(B).
In this case, Pr(L) = .7, Pr(R) = .7, so Pr (L & R) = .49

That is, I assign only a %49 chance that Neil will hit both dart boards, so I don't believe that he will both boards. I actually think it's (slightly) more likely that he won't.

This example doesn't show any interesting sense in which belief (or rational belief) fails to be closed under conjunction. Belief seems to behave just like intending is supposed to according to Neil's example.

Neil Sinhababu said...

I don't think Justin's comment picks up on what makes this example work. I can intend to do something even if I think the probability of success is below 50%. Even if my probability of hitting with the left dart is 5%, I can intend to hit with it. So the squaring of a number over .5 to get a number below .5 isn't what's going on here.

Blar said...

I did the math on my own when Neil originally posted this but I wasn't tracking the comments so I didn't realize that the discussion had turned to the numbers.

It's fairly easy to generalize Bill's results to the case where P_L does not equal P_R , as long as we continue to assume that they're independent so that P_B = P_L * P_R . The result is: If you can make more than half of your throws when you're throwing one dart with your good hand, then you should throw one dart. Otherwise, you should throw both darts at the targets.

Here's the math behind that. Let G be the probability of hitting the target with your good hand and W be the probability of hitting the target with your weak hand (for most people, G = P_R and W = P_L ). Then you should throw with both hands iff:
P(hit 1 out of 2) > P(hit 1 with good hand)
G(1-W)+W(1-G) > G
G(1-W)+W(1-G) > GW + G(1-W)
W(1-G) > GW
1-G > G
.5 > G

Another way to generalize the problem is to say that you're choosing two probabilities, p from the interval [0,G] and q from the interval [0,W], that represent how carefully you're aiming the darts. For instance, if p=G then you're trying to hit the target with your good hand, if p=0 then you're intentionally missing, and if p has some intermediate value then you're making a half-assed attempt to hit that target. Your goal is to maximize the probability of hitting exactly one target, which (assuming independence) equals p(1-q)+q(1-p). A little bit of calculus (setting the derivative of this equal to zero) shows that you only need to worry about the cases where p is 0, .5, or G and q is 0, .5, or W. If you check those nine cases, what you find is that the optimal solution is p=G, q=0 if G > .5 and p=G, q=W if G < .5 (there are other multiple equally good solutions in the special cases where G = .5 or G=W, but no better ones). In English, the result is: If you can make over half your shots with your good hand you should try to hit with only your good hand, and if not then you should try to hit with both hands.

(You can sort of see that the math should work out this way if you draw a graph of, with p on the x-axis and q on the y-axis, and look at the areas of the four rectangles in the unit square formed by a line at some value of p and a line at some value of q. Think about the areas that you'd add or subtract by moving p and q around.)

Anonymous said...

I THINK YOU ARE STUPID.

Anonymous said...

ALL OF YOU ARE STUPID.