Goodman’s ‘New Riddle’ - fitelson.orgfitelson.org/grue/torun talk/grue_torun_no_pauses.pdf · Cognitive Sciences Core Faculty ... The problem with comparative theories that

Hempel Goodman Bayesian Analyses of “Grue”: Old and New References

Goodman’s ‘New Riddle’

Branden Fitelson

Department of PhilosophyGroup in Logic and the Methodology of Science

&Cognitive Sciences Core Faculty

University of California–Berkeley

[email protected]://fitelson.org/

Branden Fitelson Goodman’s ‘New Riddle’ fitelson.org


The Plan

This is mainly a historical talk (part of an ongoing bookproject on confirmation theory) with the following two aims:

I. To reconstruct what Goodman was actually up to — givingtwo “grue” arguments against formal confirmation theory:

An Indistinguishability/Underdetermination ArgumentA Triviality Argument

Two Notes: − Goodman initially aimed these arguments at Hempel’stheory of confirmation. But, he seemed to think that anyformal theory of confirmation would fall prey to them.

− Owing to time constraints, I will only discuss the first ofthese two arguments, which is more widely discussed(especially by Bayesians). See my [3] for the full story.

II. To explain (1) why Goodman’s first argument does gothrough w.r.t. Hempel’s theory, (2) why Bayesians havethought it also goes through w.r.t. their theory, and (3) whyBayesians have thereby conceded too much to Goodman.



Hempel [8], taking Nicod [12] as his point of departure,formulates a logical theory of instantial confirmation.

We won’t need all the gory technical details of Hempel’stheory. In fact, we’ll only need the following two properties:

(NC) [φa &ψa\ confirms [(∀y)(φy ⊃ ψy)\ and[φa & ∼ψa\ disconfirms [(∀y)(φy ⊃ ψy)\.

(M) If [φa\ confirms H, then [φa &ψa\ confirms H.

(NC) is called Nicod’s Criterion, because he endorsed it [12].(M) is a monotonicity property of the confirmation relation.

Both (NC) and (M) are quite controversial, especially forBayesian confirmation theorists. I’ll return to this later.

As we’ll soon see, (NC) and (M) are the key properties thatundergird Goodman’s first (underdetermination) argumentagainst formal confirmation theory. They jointly entail:

(†) [φx &ωx &ψx\ confirms both of the following “laws”:[(∀y)(φy ⊃ ψy)\ and [(∀y)[φy ⊃ (ωy ≡ ψy)]\.



Here’s the passage containing Goodman’s first argument [6]:

Now let me introduce another predicate less familiar than “green”. It is the

predicate “grue” and it applies to all things examined before t just in case

they are green but to other things just in case they are blue. Then at time twe have, for each evidence statement asserting that a given emerald is green,

a parallel evidence statement asserting that that emerald is grue. And the

statements that emerald a is grue, that emerald b is grue . . . will each confirm

the general hypothesis that all emeralds are grue. Thus according to our

definition, the prediction that all emeralds subsequently examined will be

green and the prediction that all will be grue are alike confirmed by evidence

statements describing the same observations. . . . Thus although we are well

aware which of the two incompatible predictions is genuinely confirmed,

they are equally well confirmed according to our present definition. . . .

Next, I will reconstruct this first part of the “New Riddle”,and evaluate it from Hempelian and Bayesian perspectives.

In the end, I’ll argue that Goodman is (at best) “half right”.



Let Ox Ö x is examined before t, Gx Ö x is green, Ex Ö xis an emerald, and Gx Ö Ox ≡ Gx (i.e., x is “grue”).

Now, we can state the two salient hypotheses, as follows:

(H1) All emeralds are green. Formally, H1 is: (∀x)(Ex ⊃ Gx).(H2) All emeralds are grue. Formally, H2 is: (∀x)(Ex ⊃ Gx).

Even more precisely, H2 is: (∀x)[Ex ⊃ (Ox ≡ Gx)].And, here are three salient (instantial) evidential statements:

(E1) Ea &Ga (E2) Ea & (Oa ≡ Ga) (E) Ea &Oa &GaGoodman’s conclusion is that there is some statement (!)that — on Hempel’s theory — confirms both H1 and H2.From (†), we see it’s the statement E that plays this role.

∴ Goodman’s first argument has the following conclusions:

(G) E confirms both H1 and H2.(G?) E confirms both H1 and H2 equally.

That is, H1 and H2 are confirmationally indistinguishable byE (for Hempel). Goodman suggests this is (intuitively) false.



For Bayesians, confirmation is a three-place relation,defined as follows. E confirms H, relative to backgroundcorpus K iff Pr(H | E &K) > Pr(H |K), for some suitable Pr.From a Bayesian point of view, (NC) is not generally true[4, 5, 11]. However, a closely related principle is correct:

(NC⊃) For all objects x and all (compatible) properties φ and ψ:[φx ⊃ ψx\ confirms [(∀y)(φy ⊃ ψy)\, and[φx & ∼ψx\ disconfirms [(∀y)(φy ⊃ ψy)\.

(NC⊃) makes sense because (∀y)(φy ⊃ ψy) ` φa ⊃ ψa,and φa & ∼ψa ` ∼(∀y)(φy ⊃ ψy). On the other hand,(NC) is less clear because (∀y)(φy ⊃ ψy) ø φa &ψa.Of course, if one assumes (M), then (NC⊃) entails (NC)! But,(M) is rather obviously false from a Bayesian point of view.Let Bx Ö x is a black card, let Ax Ö x is the ace of spades,and let Jx Ö x is the jack of clubs. Assuming (K) that wesample a card a at random from a standard deck, we have:

(7) Ba confirms Aa, relative to K, but

(8) Ba & Ja disconfirms (indeed, refutes!) Aa, relative to K.



In Hempel’s theory, (G) trades on (M). But, (M) is clearly falsefrom a Bayesian point of view. So, we’ve seen no goodreason for a Bayesian to accept (G). [In fact, (M) is a reasonto reject Hempel’s theory [1] — no “grue” needed for that!]Moreover, here’s a Bayesian counterexample to (G) & (G?) (ala Good [4, 5]) such that E confirms H1 but disconfirms H2.

(K) Either: (H1) there are 1000 green emeralds 900 of which have

been examined before t, no non-green emeralds, and 1 million

other things in the universe, or (H2) there are 100 green

emeralds that have been examined before t, no green emeralds

that have not been examined before t, 900 non-green emeralds

that have not been examined before t, and 1 million other things.

Now, imagine an urn containing true descriptions of eachobject in the universe, and let E Ö “Ea &Oa &Ga” bedrawn. E confirms H1 but E disconfirms H2, relative to K:

Pr(E |H1 &K) = 9001001000

>100

1001000= Pr(E |H2 &K)



But, all this shows is that neither (G) nor (G?) is a logicalconsequence of Bayesian confirmation theory. Of course, (G)and (G?) are consistent with Bayesian confirmation theory.As in the literature on the raven paradox [5, 9, 1], we turn tothe “appropriateness” of our probabilistic assumptions formodeling the confirmation relations of the “paradox”.Hempel [8, 9] explains that the raven paradox is about K’swhich do not entail that a has (or lacks) any of theproperties involved in the H’s (i.e., ±ravenhood/±blackness).Presumably, we should say the same thing about “grue”.But, it is often assumed (e.g., [13]) that K entails Oa. Thisdirectly contradicts the Hempelian line on “paradoxes”.Following Hempel, I have not assumed the object inquestion (a) has or lacks any of the properties E, G, or O.The key question is whether Pr(E |H2 &K) < Pr(E |H1 &K)can be motivated as a reasonable (“appropriate for ‘grue”’)probabilistic confirmation-theoretic modeling assumption.



In a recent manuscript, Roger White asks us to think aboutwhat a “grue” (viz., “anti-inductive”) world might be like:

To achieve our goal of making standard inductive methods unreliable in a

world, we are going to have to take a more aggressive . . . approach. Rather

than just assigning properties to objects and leaving it to chance how our

inductivist victims will obtain their data, we must somehow give them a

tendency to obtain misleading data. . . . We put only green emeralds in the

vicinity of the surface of the earth, burying the blue ones deep below, or

tucking them away in hard to find places. Then once our victims have

sampled sufficiently many green emeralds that they can inductively predict

that the next one will be green, we cause a blue one to rise to the surface

nearby. . . . Coordinating the . . . show will require either a very smart demon

overseeing it all, or the design of some intricate and peculiar physical laws.

Following this line, it may be reasonable to use models inwhich Pr(Oa |H2 &K) < Pr(Oa |H1 &K) to evaluate thesalient confirmation relations. If so, this opens the door toPr(E |H2 &K) < Pr(E |H1 &K). �Think inductive Pr [10, 12]!�



Some would-be Bayesians modelers [13] assume somethingeven stronger than Pr(Oa |K) = 1: that K entails Ea &Oa.

Relative to K’s that entail both Ea and Oa, both H1 and H2

entail E. Therefore, Pr(E |H1 &K) = Pr(E |H2 &K) = 1.

It follows immediately that E confirms both H1 and H2,relative to such K’s. This ensures Goodman’s claim (G).

Moreover, most Bayesians (e.g., [13]) seem to think that thisalso suffices for Goodman’s comparative claim (G?).But, this last step presupposes a “Law of Likelihood”:(LL) Pr(E |H1 &K) = Pr(E |H2 &K) =⇒ c(H1, E |K) = c(H2, E |K).

[where c(H, E |K) is some Bayesian relevance measure ofthe degree to which E confirms H, relative to corpus K.]

As I have explained elsewhere [2], (LL) is only true relativeto some relevance measures c. And, intuitively, (LL) is false.

E.g., if E ` H1 and E ø H2, then (intuitively) E confirms H1

more strongly than E confirms H2. But, this contradicts (LL).



The problem with comparative theories that ential (LL) [2] isthat they are sensitive only to the likelihoods, but not thecatch-all likelihoods: Pr(E | ∼H1 &K) and Pr(E | ∼H2 &K).All Bayesian theories of comparative confirmation agree on:

(WLL) Pr(E|H1&K) = Pr(E|H2&K)&&& Pr(E|∼H1&K) = Pr(E|∼H2&K)⇓

c(H1, E |K) = c(H2, E |K)But, in the case of Goodman’s “grue”, the second (catch-all)conjunct of the antecedent of (WLL) is (intuitively) false!

That is, intuitively, Pr(E | ∼H1 &K) < Pr(E | ∼H2 &K). Why?

As a result, any adequate Bayesian theory of comparativeconfirmation [2] will imply that c(H1,E |K) > c(H2,E |K).And, this is just what Goodman seems to think ought to bethe case! So, a proper Bayesian analysis of “grue” revealsthat Goodman’s first argument is, at best, only “half right”.

I.e., from a proper Bayesian point of view, even if (G) is true,(G?) is false. See [3] for more on Goodman’s two arguments.



Assume that Pr(·) has been conditionalized on the standardBayesian “grue” background corpus, which implies Oa & Ea.

Now, in order for a Bayesian to establish (G?), she needs:

Pr(Ga | Some OE are G ∨ Some OE are G)= Pr(Ga | Some OE are G ∨ Some OE are G)

To my knowledge, nobody has ever argued for this.

There are various ways this could be true, e.g., if facts aboutthe color of unexamined emeralds were (appropriately)irrelevant to facts about the color of examined emeralds.

But, what reason do we have to believe that? That alreadysounds like a kind of anti-inductivism (in probabilistic form).

I think there are reasons to reject this = in favor of a <.

∼H1 entails that there are non-green emeralds (which, forall you know, might be examined by you), but ∼H2 does not.

Even if that’s not a compelling reason, it remains true thatmore (Bayesian) work needs to be done to secure (G?).



[1] B. Fitelson, The paradox of confirmation, Philosophy Compass (B. Weathersonand C. Callender, eds.), Blacwkell (online publication), Oxford, November 2005,URL: http://dx.doi.org/10.1111/j.1747-9991.2005.00011.x.

[2] Likelihoodism, Bayesianism & Relational Confirmation, Synthese, to appear.

[3] Goodman’s ‘New Riddle’, manuscript, http://fitelson.org/grue.pdf.

[4] I.J. Good, The white shoe is a red herring, BJPS 17 (1967), 322.

[5] , The white shoe qua red herring is pink, BJPS 19 (1968), 156–157.

[6] N. Goodman, Fact, Fiction, and Forecast, Harvard University Press, 1955.

[7] C. Hempel, A purely syntactical definition of confirmation, JSL 8 (1943), 122–143.

[8] , Studies in the logic of confirmation, Mind 54 (1945), 1–26, 97–121.

[9] , The white shoe: no red herring, BJPS 18 (1967), 239–240.

[10] J. Keynes, A Treatise on Probability, Macmillan, 1921.

[11] P. Maher, Probability captures the logic of scientific confirmation, ContemporaryDebates in the Philosophy of Science (C. Hitchcock, ed.), Blackwell, 2004.

[12] J. Nicod, The logical problem of induction, (1923) in Geometry and Induction,University of California Press, 1970.

[13] E. Sober, No model, no inference: A Bayesian primer on the grue problem, inGrue! The New Riddle of Induction (D. Stalker ed.), Open Court, Chicago, 1994.

[14] R. White, A New Look at the Old Problem of Induction, unpublished manuscript.URL: http://philosophy.fas.nyu.edu/docs/IO/1180/induction.pdf.


http://dx.doi.org/10.1111/j.1747-9991.2005.00011.x

http://fitelson.org/grue.pdf

http://philosophy.fas.nyu.edu/docs/IO/1180/induction.pdf

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Goodman’s ‘New Riddle’ - fitelson.orgfitelson.org/grue/torun talk/grue_torun_no_pauses.pdf · Cognitive Sciences Core Faculty ... The problem with comparative theories that