Saturday, February 10, 2018

Counting goods

Suppose I am choosing between receiving two goods, A and B, or one, namely C, where all the goods are equal. Obviously, I should go for the two. But why?

Maybe what we should say is this. Since A is at least as good as C, and B is non-negative, I have at least as good reason to go for the two goods as to go for the one. This uses the plausible assumption that if one adds a good to a good, one gets something at least as good. (It would be plausible to say that one gets something better, but infinitary cases provide a counterexample.) But there is no parallel argument that it is at least as good to go for the one good as to go for the two. Hence, it is false that I have at least as good reason to go for the one as to go for the two. Thus, I have better reason to go for the two.

This line of thought might actually solve the puzzles in these two posts: headaches and future sufferings. And it's very simple and obvious. But I missed it. Or am I missing something now?


Martin Cooke said...

Mod 2 arithmetic is a counter-example?
A = B = C = 1
A + B = 0

Martin Cooke said...

Goods might be like that?
E.g. suppose you are on fire:
a tub of water would put it out,
but if another tub would drown you then
you would be no better off with 2 than none.

Martin Cooke said...

Or the eating of cakes:
Eating any one cake is good,
but eating any two makes you feel sick.

Martin Cooke said...

Or what if you need only one of the goods
and others use goods that you do not use?

Martin Cooke said...

I think that last 'thought' of mine is probably the general case with goods, either because of God or else because evolution favors such 'altruism' (a species can, if numerous enough, keep other species off its food).

I wonder about the infinite case too:
Prime numbers are fewer than odd numbers, in some highly intuitive sense (e.g. if there were sheep S1, S2, S3 and so forth), and yet you might be led to choose them by your rule; e.g.
A = the sum of every other prime numbered good
B = the sum of all the other prime numbered goods
C = the sum of all the odd numbered goods

Martin Cooke said...

I am assuming there that A, B and C are equal goods,
perhaps because they have the same cardinality
and the numbered goods (e.g. sheep) are roughly equal.

A + B = the sum of the prime numbered goods
(which in my example is roughly half of C).

(Note that if I add a sheep, say S, to the numbered sheep
then I do have, in a very intuitive sense, more sheep.)

Alexander R Pruss said...

The cake situation is one where B isn't a good (to you) given A. So I'm not worried about that, or about the case where B is useless to you given A.

But the infinite case is one I should have noticed as a counterexample to my thesis, as it is precisely the sort of case I had in mind, and I meant the thesis not to be counterexampled by it. Oops!

Maybe I need to explicitly assume that A, B and C are finite.

Anyway, I think my bigger point goes through. In the infinitary case, what goes wrong with the argument is that my sentence "But there is no parallel argument that it is at least as good to go for the one good as to go for the two" is false. In the finite case, maybe apart from some other special cases, that sentence is true, and its truth helps explain why the two goods are better to go for.

Martin Cooke said...

Upon reflection, I'm not sure that I do have a counterexample: I equivocated between A, B and C being equal, and A + B being a tiny fraction of C (which I miscalled "half of C").

Would your bigger point have gone through, though? In the finite case, you can simply say that what is preferable about the 2 goods is that extra good.

(Btw I note that a good-for-you could involve other people's goods-for-them, e.g. your looking after the welfare of sheep; and then the relevant arithmetic would be my intuitive counting, where A + B was indeed a tiny fraction of C, not equal to C, and so was different to standard arithmetic, even if it did not yield a counter-example.)

Brent said...
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IanS said...

Maybe you can make comparisons between infinite collections work if you add an extra principle: to compare collections with common elements, ignore the common elements and compare the remainder collections. This deals with cases like prime numbered items vs odd-numbered ones - the comparison reduces to {#2} vs {#9, #15, ….}. The principle seems pretty intuitive: the goods that you must choose in any case seem irrelevant to the choice between alternatives.

Philip Rand said...
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