Open futurism and many-worlds quantum mechanics

I’ve been thinking about some odd parallels between the many-worlds interpretation of quantum mechanics and open future views.

On both sets of views, in the case of genuinely chancy future events there is strictly no fact of the matter about what will turn out. On many-worlds, the wavefunction provides a big superposition of the options, but for no one option is it true that it will eventuate. The same is true for open future views, except that what we have instead of a superposition depends on the particular temporal logic chosen.

Yet, despite no fact about outcomes, on both sets of views one would like to be able to make probabilistic predictions about “the outcome”. For instance, one wants to say that if one tosses an indeterministic coin, it is moderately likely that the coin will land on heads and extremely unlikely that it will land on heads. In both cases, this is highly problematic, because on both views it is certain that it is not true that the coin will land on heads. So how can something that is certainly not going to happen be more likely than another event? In both cases, there is a literature trying to answer this problem (and I am not convinced by it).

Anyway, I wonder how far we can take the parallel. The wavefunction in the many-worlds interpretation is a superposition of many options about what the present is like, and is interpreted as a plurality of worlds in which different options are true. Why not do the same in the open-future case? Why not just say that there are now many worlds, including some where the coin will land on heads, some where the coin will land on tails, and some where it will land on edge? After all, if it is reasonable to interpret the superposition this way, why is it not reasonable to interpret the temporal logic this way?

There is, however, one crucial difference. The open futurist insists that reality will collapse: that once the coin lands, there will be a fact about which way it landed. On many-worlds, there is no collapse: there is never a fact about how the coin landed. Nonetheless, this could be accommodated in a many-worlds interpretation of an open-future view: we just suppose that once the coin lands, a lot of the worlds disappear.

So what if there is a parallel? Why does it matter?

Well, here are some things that we might say.

First, in both cases, there is an underlying metaphysics (a non-classical truth assignment to future facts, or a giant superposition), and then we need to interpret that underlying metaphysics. I wonder if it might not be true:

1. A many-worlds interpretation of the underlying metaphysics is reasonable in the quantum case if and only if it is reasonable in the open-future case.

Suppose (1) is true. Most people think a many-worlds interpretation of open-future is absurd. But then why isn’t the many-worlds interpretation of quantum mechanics (or, more precisely, a quantum mechanics with exceptionlessly unitary evolution and all the facts supervening on the wavefunction) also absurd?

Second, it may well be that the open-futurist finds plausible the standard criticism of many-worlds interpretations that it does not make sense of probabilistic predictions. If so, then they should probably find equally problematic probabilistic predictions on open-future views.

Classical probability theory is not enough

Here’s a quick argument that classical probability cannot capture all probabilistic phenomena even if we restrict our attention to phenomena where numbers should be assigned. Consider a nonmeasurable event E, maybe a dart hitting a nonmeasurable subset of the target, and consider a fair coin flip that is causally isolated from E. Let H and T be the heads and tails results of the flip. Then let A be this disjunctive event:

• (E and H) or (not-E and T).

Intuitively, event A clearly has probability 1. If E happens, the probability of A is 1/2 (heads) and if E doesn’t happen, it’s also 1/2 (tails). (The argument uses finite conglomerability, but it is also highly intuitive.)

So a precise number should be assigned to A, namely 1/2. And ditto to H. But we cannot have these assignments in classical probability theory. For if we did that, then we would also have to assign a probability to the conjunction of H and A, which is equivalent to the conjunction of E and H. But we cannot assign a probability to the conjunction of E and H, because E and H are independent, and so we would have a precise probability for E, namely P(E)P(H)/P(H)=P(E&H)/P(H), contrary to the nonmeasurability of E.

The Probabilistic Counterexampler

Every so often someone asks me if some piece of probabilistic reasoning works. For instance, today I got a query from a grad student whether

1. P(A|C)>P(A|B) implies P(A|B ∨ C)>P(A|B).

Of course, I could think about it each time somebody asks me something. But why think when a computer can solve a problem by brute force?

So, last spring I wrote a quick and dirty python program that looks for counterexamples to questions like that simply by considering situations with three dice, and iterating over all the possible combinations of subsets A, B and C of the state space (with some reduction due to symmetries).

The program is still quick and dirty, but at least I made the premises and conclusions not be hardcoded. You can get it here.

For instance, for the query above, you can run:

python probab-reasoning.py "P(a,c)>P(a,b)" "P(a,b|c)>P(a,b)" 

(The vertical bars are disjunction, not conditional probability. Conditional probability uses commas.) The result is:

a={1}, b={1, 2}, c={1}
a={1}, b={1, 2, 3}, c={1}
a={1}, b={1, 2, 3}, c={1, 2}
a={1}, b={1, 2, 3}, c={1, 3}
a={1}, b={1, 2, 3}, c={1, 4}
...

So, lots of counterexamples. On the other hand, you can do this:

python probab-reasoning.py "P(a)*P(b)==P(a&b)" "P(b)>0" "P(a,b)==P(a)" 

and it will tell you no counterexamples were found. Of course, that doesn’t prove that the result is true, but in this case it is.

The general operation is that you install python (either 2.7 or 3.x) and use a commandline to run:

python probab-reason.py premise1 premise2 ... conclusions

You can use any single letter variables for events, other than P, and the operations & (conjunction), | (disjunction) and ~ (negation) between the events. You can use conditional probability P(a,b) and unconditional probability P(a). You can use standard arithmetical and comparison operators on probabilities. Make sure that you use python’s operators. For instance, equality is ==, not =. You should also use python’s boolean operations when you are not working with events: e.g., “P(a)==1 and P(b)==0.5”.

Any premise or conclusion that requires conditionalization on a probability zero event to evaluate automatically counts as false.

You can use up to five single-letter variables and you can also specify the number of sides the die has prior to listing the premises. E.g.:

python probab-reasoning.py 8 "P(a)*P(b)==P(a&b)" "P(b)>0" "P(a,b)==P(a)" 

Two kinds of non-measurable events

Non-measurable events are ones to which the probability function in the situation assigns no probability. Philosophically speaking, non-measurable events come in two varieties:

1. Non-measurable events that should not have any probability assignment.

2. Non-measurable events that should have a probability assignment.

Type (1) non-measurable events are the kinds of weird events that can be constructed from the Hausdorff and Banach-Tarski paradoxes, as well as perhaps (this is less clear) the Vitali non-measurable sets.

But I think there are also type (2) non-measurable events relative to standard choices of probability functions. For instance, suppose that in each universe of an infinite multiverse a fair coin is tossed countably infinitely often.

How likely is it that in at least one universe all the coin tosses are heads? If the universes form a countable infinity, classical probability theory gives an answer: zero. But if the universes form an uncountable infinity, classical probability theory gives no answer at all—the standard completed product measure makes the event be non-measurable. However, intuitively, there should be an answer in at least some cases. If the number of universes is much larger than the number of possible countable sequences of coin tosses (i.e., is much larger than 2^ω), we would expect the probability to be 1 or close to it. We can coherently extend the standard probability function to give that answer. But we can also coherently extend it to give a different answer, including the answer that the probability of an all-heads universe is zero, even if the number of universes is a gigantic infinite cardinality.

We don’t want to just make up an answer here. We want the answer to be derivable in some way resembling the proof of the theorem that if you toss a coin infinitely many times, you’ve got probability 1 of getting heads at least once.

I suppose we could take it to be a metaphysical axiom that if you have K disjoint collections each with M coin tosses, then if K and M are infinite and K > M, then with probability one at least one collection yields all heads. But it would be nice to have more than just intuition here, and in similar problems.

Regular probability comparisons imply the Banach-Tarski Paradox

Paper posted here (forthcoming in Synthese). Among goodies in the paper is a proof that the order extension principle (even in a weak form) implies the Banach-Tarski paradox, and a new argument against commensurability in decision theory. This is a very technical paper, so reader beware.

The Axiom of Choice in some claims about probabilities

I spent the last week trying to get clear on the logical interconnections between a number of results about probabilities that are relevant to formal epistemology and that use a version of the Axiom of Choice in proof, such as:

1. For every non-empty set Ω, there is an ordered field K and a K-valued probability function that assigns non-zero finitely additive probability to every non-empty subset of Ω.
2. For every non-empty set Ω, there is a full finitely additive conditional probability on Ω (i.e., a Popper function with all non-empty subsets normal).
3. The Banach-Tarski Paradox holds: one can decompose a three-dimensional ball into a finite number of pieces that can be moved around and made into two balls of the same size.
4. There are Lebesgue non-measurable sets in the unit interval [0,1].

All of these results require some version of the Axiom of Choice. It turns out that there is a very simple map of their logical interconnections in Zermelo-Fraenkel (ZF) set theory:

• BPI→(1)→(2)→(3)→(4),

where BPI is the Boolean Prime Ideal theorem, a weaker version of the Axiom of Choice.

The proof from BPI to (1) is standard--just let K be an ultrapower of the reals with an appropriate ultrafilter. That from (1) to (2) is almost immediate: just define the conditional probabilities via the ratio formula and take the standard part. Pawlikowski's proof of Banach-Tarski easily adapts to use (2) (officially, he uses Hahn-Banach). Finally, Foreman and Wehrung show in ZF that every subset of Rⁿ is Lebesgue measurable iff every subset of [0,1] is. But it follows from (2) that not every subset of R³ is Lebesgue measurable.

This has important consequences. Without the Axiom of Choice, one can prove that either (a) there are sets that have no regular probabilities no matter what ordered field is chosen for the values and no full conditional probabilities, or (b) the Banach-Tarski Paradox holds and hence there are no rigid-motion-invariant probabilities on regions of three-dimensional space big enough to hold a ball. And in either case, Bayesianism has a problem.

The Axiom of Choice gives and takes away probabilities

Suppose you have assigned coherent (i.e., finitely additive) probabilities to a collection of options, but then you come upon a refinement of this collection of options, including more fine-grained ones. For instance, previously you had assigned probabilities to propositions about which individuals in a population had brown or non-brown eyes. But now you realize you should refine the non-brown-eye group into the blue, green and none of the above groups, as well as considering people's hair-color. It is intuitively plausible that if your initial probabilities about brown versus non-brown eyes were coherent, you should be able to come up with a coherent assignment of probabilities to the refined cases, e.g., by equally dividing up the probabilities of the non-brown eye category between the three newly recognized suboptions. One way to state the above in full generality is this:

1. Whenever P is a finitely additive probability assignment on an algebra F of subsets of a space Ω, and F is a subalgebra of a finer algebra G of subsets of Ω, then P can be extended to G.

Is this true? Well, the Axiom of Choice implies it is and I think this was first proved by Tarski. The Axiom of Choice gives probabilities (or at least implies that they exist).

But the Axiom of Choice also takes away probabilities. One famous case is that of nonmeasurable subsets of an interval, but that's about countably additive stuff, while I am right now talking of finitely additive stuff. One way it does this is by implying the Banach-Tarski paradox:

2. A solid three-dimensional ball can be decomposed into a finite number of subsets which can be moved rigidly to produce two balls of the same size as the original. It follows that no region in three dimensions that has the room to contain a ball has a (finitely additive) probability assignment on all its subsets that is invariant under rigid motions.

Now, for those who, like Bayesians, think epistemology is basically probability theory, (1) is going to be attractive but (2) is going to be paradoxical and repugnant. These thinkers may be tempted to give up the Axiom of Choice in order to deny (2). But I think they are likely to still want (1). And since (1) is known not to actually be equivalent to the Axiom but weaker, there might seem to be some hope.

Question: Can we coherently deny (2) and accept (1) in Zermelo-Fraenkel Set Theory (ZF) without Choice?

It turns out that the hope is vain. For:

Theorem. Claim (1) implies claim (2) in ZF.

For Luxemburg (1969) proved that (1) is equivalent to the Hahn-Banach Theorem, while Pawlikowski (1991) proved that the Hahn-Banach Theorem implies the Banach-Tarski Paradox.

So we cannot get away from this (at least not without even more radical revision to set theory). If we want probability existence results like (1), we must accept probability nonexistence results like (2).

YouTube talks on Principle of Sufficient Reason and on the measure problem in cosmology

For most of the week I was at the Santa Cruz Philosophy of Cosmology institute. Thursday I gave a talk on the Principle of Sufficient Reason and on the measure problem in cosmology for multiverses. These talks and many talks by smart people are available here. Click on "Program", then scroll down close to the bottom to Thursday 7/11 if you want mine.

Or just click right here for part 1 and part 2 of my talk on Youtube.

Here are the slides to part 1 (cosmological argument) and part 2 (measure problem).

Visits to a nonmeasurable set and a new sceptical worry

Let X₁,X₂,... be a sequence of independent, identically distributed random variables. Let H be a set of values, and let R_n(H) be the proportion of X₁,...,X_n that are in H. Thus R_n(H)=V_n(H)/n, where V_n(H) is the number of times that the sequence X₁,...,X_n has visited H. We can call R_n(H) the rate of visits to H.

The strong Law of Large Numbers then shows that if H is a measurable set, then, almost surely (i.e., with probability one), R_n(H) converges to P(X₁ in H). We can use X₁ (or any of the other variables, since they are identically distributed) to induce a probability measure P₀ on the set of possible values via the formula P₀(H)=P(X₁ in H). Thus, for measurable H, almost surely, lim R_n(H)=P₀(H). I.e., the asymptotic rate of visits to a measurable set H is equal to the probability of that set.

But what if H is nonmeasurable? We could consider the general case, but let's simplify and make things more interesting. What if H is maximally nonmeasurable? A set H is maximally nonmeasurable with respect to a probability measure P₀ if and only if:

• All the measurable subsets of H have measure zero.
• All the measurable supersets of H have measure one.

On a reasonable assignment of interval-valued probabilities, a maximally nonmeasurable set is one that gets the full interval [0,1].

Such sets are intuitively a mess. So what should we expect the rate of visits to a maximally nonmeasurable set to behave like. It was my intuition that we can expect the rate of visits to be a mess—to not converge to any particular value.[note 1] Here's a precise way to formulate the question. Let B be any non-empty proper subset of the interval [0,1]. Form the following subsets of our original P-probability space:

• L(H): the set of points of the probability space such that lim R_n(H) exists.
• L_B(H): the set of points of the probability space such that lim R_n(H) exists and falls in B.
• I_B(H): the set of points of the probability space such that liminf R_n(H) falls in B.
• S_B(H): the set of points of the probability space such that limsup R_n(H) falls in B.

My intuition that we should expect the rate of visits to be a nonconvergent mess is an intuition that L(H) should have high probability, or, if it is itself nonmeasurable, it should contain a measurable subset of high probability. If some proofs that I haven't checked all the details of are correct, this intuition is wrong.

Conjecture (Theorem if my proofs are right): The sets L(H), I_B(H), S_B(H) and L_B(H) are all maximally nonmeasurable if H is maximally nonmeasurable.

If this is correct, then there is basically nothing probabilistic you can say about the asymptotic convergence of R_n(H) for a maximally nonmeasurable set H. You can't say that the rate of visits probably will converge (no surprise there) and you can't say that it probably won't.

So what? Well, consider now a new sceptical problem. We perform some experiment E a thousand times, and 405 times we get outcome H. We very reasonably want to conclude that the circumstances of the experiment E have approximately a 40% tendency of producing outcome H. And the greater the number of experiments we do, as long as the observed rate of H's is around 40%, the more confident we are of this judgment, with our confidence going to one in the limit.

But wait! What about the sceptical hypothesis that the objective chances are such that H is maximally nonmeasurable given E? It is tempting to say: "Well, that could be true, but the longer our sequence of experiments with a rate of around 40%, the more confident we should be that H is measurable and has measure around 40%. We just wouldn't expect to get such nice convergence if H were maximally nonmeasurable." However, the Conjecture, assuming it's correct, shows that as a piece of probabilistic reasoning, this is completely wrong. For it is neither likely nor unlikely on the maximal nonmeasurability hypothesis that we would observe an asymptotic rate of 40% (or of any other value). To see this, let B be the singleton {0.4}, and note that L_B(H) is maximally nonmeasurable. Thus, its interval-valued probability is all of [0,1], and we can have no probabilistic expectations about it.

If the maximal nonmeasurability hypothesis cannot be ruled out a posteriori, and yet must be ruled out, then it must be ruled out a priori. I think our best hope is a postulate that outcomes are always at least partly measurable, i.e., aren't maximally nonmeasurable. And that's a kind of Principle of Sufficient Reason.

I think my (unchecked) proofs of the Conjecture can generalize to give a more complicated result in the case of nonmaximally measurable sets.

Absolutely nonmeasurable sets

The ideal of a non-zero (point) probability assignment to all possibilities is incoherent for cardinality reasons. Moreover, as Alan Hajek has insisted, the existence of nonmeasurable sets provides further difficulties.

One might try to get around both issues by problem-specific Bayesianism, where one only insists on a probability assignment specific to a particular problem at hand. This gets around my no-go theorem, since that theorem shows that there is no single non-zero probability assignment to all the possibilities there are. But in any given probabilistic calculation, the collection of possibilities is restricted to some set, and then there could well be a generalized probability (e.g., satisfying the axioms here) for that problem.

One might even have some hope that problem-specific Bayesianism could handle the issue of nonmeasurable sets. For there are isometrically invariant extensions of Lebesgue measure (i.e., extensions invariant under translation, rotation and reflection) that make some Lebesgue nonmeasurable sets be measurable.

But no such luck. Start by noting that there are absolutely nonmeasurable sets. A bounded absolutely nonmeasurable set (I'm making up this technical term) is a subset A of n-dimensional Euclidean space Rⁿ such that there is no isometrically invariant probability measure that (a) makes A measurable, (b) assigns finite measure to every bounded measurable subset of Rⁿ, (c) assigns non-zero measure to some bounded subset of Rⁿ. The Hausdorff Paradox then shows that there is a bounded absolutely nonmeasurable set if n=3, assuming the Axiom of Choice.

In fact, from the Hausdorff Paradox we can prove that there is a bounded subset A of R³ such there is no isometrically invariant generalized finitely additive probability measure, e.g., in the sense of this post, on the cube [0,1]³ or on the three-dimensional ball of unit radius that makes A measurable.

So the problem-specific approach also runs into trouble, at least assuming the Axiom of Choice. And the Axiom of Choice (or, more weakly, the Boolean Prime Ideal Theorem--I don't know if this makes a difference, but in any case BPI has no intuitive support beyond the fact that AC implies it) is also assumed by hyperreal extensions of probability theory.

Of course, if one allows for interval-valued measures, that's different kettle of fish.

Hausdorff Paradox and conditional probabilities

The Hausdorff Paradox shows that given the Axiom of Choice, there is no finitely additive probability measure defined for all subsets of the surface of a ball that is invariant under rotations—that assigns the same measure to a subset of the surface of the sphere and to a rotation about any axis through the center of that subset. Because of results like that, the standard Lebesgue measure on the surface of a ball is only defined for some subsets, i.e., the measurable ones.

Now, in classical probability, we can only define conditional probability when we condition on an event A with non-zero probability. We then use the formula P(B|A)=P(B&A)/P(A). Some have tried to come up with axiomatizations that allow for conditioning on all non-empty measurable sets, including zero-probability ones.

We should not hold our breath for success. Here's why. Let C be a solid ball of unit volume. Let P be Lebesgue measure on C: the measure of a subset of C is just the volume of the subset. We expect the Lebesgue measure on C to be invariant under rotations about all axes through the center of the ball. We would also expect that the conditional probability to be thus invariant. I.e., if r is a rotation about an axis through the center, we would expect it to be that P(r(B)|r(A))=P(B|A). Unfortunately this cannot be done, at least not if one assumes the axioms that P(−|A) is a finitely additive measure on the P-measurable subsets of C and P(A|A)=1. For let A be the surface of a ball concentric with C but of smaller radius. Then the volume of A is zero: A is a two-dimensional surface, after all. Moreover, Lebesgue measure has the property that every subset of a set with zero measure is also measurable (and has zero measure). So P(−|A) will be a finitely additive probability measure on all subsets of A. If we have our rotation invariance condition, then P(B|A)=P(r(B)|r(A))=P(r(B)|A) since r(A)=A (the sphere A is invariant under rotation through its center). So, P(−|A) will be a finitely additive rotation-invariant probability measure on all subsets of A, which violates the Hausdorff Paradox (assuming the Axiom of Choice).

Put it differently: Any conditional probability assignment that allows conditioning on all non-empty subsets will exhibit an unacceptable rotational bias.

Amusing probability case

Write down a decimal point. Then choose a digit at random, with equal probability 1/10 of each possible digit. Repeat ad infinitum, with all the digits chosen independently. Let X be the number you've written down the infinite decimal expansion of.

Suppose you find out that X is going to be either 1/4 or 1/3. Which of the two is more likely? Answer: 1/4. For there are two ways of getting 1/4: 0.250000... and 0.249999.... But there is only one way of getting 1/3: 0.333333..., and each infinite sequence is equally likely. Thus, intuitively P(X=1/4 | X=1/3 or X=1/4)=2/3. Surprised?

Another interesting fact here. In the technical probability-theory sense, X is uniformly distributed on the interval [0,1]. But in the intuitive sense, it's not. So the technical probability-theory sense does not capture the notion of uniform distribution.

Similarly, the technical probability-theory sense of independence does not capture the intuitive notion of independence. Suppose that a random process uniformly picks out a number Y in the interval [0,1], and suppose you get a dollar if and only if the number is 1/2. Let A be the event that the number picked out is 1/2 and let B be the event that you get a dollar. Then P(A&B)=P(A)=0=P(A)P(B), and hence in the probability-theoretic sense A and B are independent. But intuitively they are far from independent: B is entirely determined by A.

Maybe a better definition of independence for philosophical (though maybe not mathematical) purposes is that both P(A|B)=P(A) and P(B|A)=P(B). And then conditional probabilities should not be defined by ratios of unconditional probabilities.

Attitudes to risk and the law of large numbers

People do things that seem to be irrational in respect of maximizing expected utilities. For instance, art collectors buy insurance, even though it seems that the expected payoff of buying insurance is negative—or else the insurance company wouldn't be selling it (some cases of insurance can be handled by distinguishing utilities from dollar amounts, as I do here, but I am inclined to think luxury items like art are not a case like that). Likewise, people buy lottery tickets, and choose the "wrong" option in the Allais Paradox.

Now, there are all sorts of clever decision-theoretic ways of modeling these phenomena and coming up with variations on utility-maximization that handle them. But rather than doing that I want to say something else about these cases.

Why is it good to maximize expected utilities in our choices (and let's bracket all deontic constraints here—let's suppose that none of the choices are deontically significant)? Well, a standard and plausible justification involves the Law of Large Numbers [LLN] (I actually wonder if we shouldn't be using the Central Limit Theorem instead—that might even strengthen the point I am going to make). Suppose you choose between option A and option B in a large number of independent trials. Then, on moderate assumptions on A and B, the LLN applies and says that if the number of trials N is large, probably the payoff for choosing A each time will be relatively close to NE[A] and the payoff for choosing B each time will be relatively close to NE[B], where E[A] and E[B] are the expected utilities of A and B, respectively. And so if E[A]>E[B], you will probably do better in the long run by choosing A rather than by choosing B, and you can (on moderate assumptions on A and B, again) make the probability that you will do better by choosing A as high as you like by making the number of trials large.

But here's the thing. My earthly life is finite (and I have no idea how decision theory is going to apply in the next life). I am not going to have an infinite number of trials. So how well this LLN-based argument works depends on how fast the convergence of observed average payoff to the statistically expected payoff in the LLN is. If the convergence is too slow relative to the expected number of A/B-type choices in my life, the argument is irrelevant. But now here's the kicker. The rate of convergence in the LLN depends on the shape of the distributions of A and B, and does so in such a way that the lop-sided distributions involved in the problems mentioned in the first paragraph of the paper are going to give particularly slow convergence. In other words, the standard LLN-based argument for expected utility maximization applies poorly precisely to the sorts of cases where people don't go for expected utility maximization.

That said, I don't actually think this cuts it as a justification of people's attitudes towards things like lotteries and insurance. Here is why. Take the case of lotteries. With a small number of repetitions, the observed average payoff of playing the lottery will likely be rather smaller than the expected value of the payoff, because the expected value of the payoff depends on winning, and probably you won't win with a small number of repetitions. So taking into account the deviation from the LLN actually disfavors playing the lottery. The same goes for insurance and Allais: taking into account the deviation from the LLN should, if anything, tell against insuring and choosing the "wrong" gamble in Allais.

Maybe there is a more complex explanation--but not justification--here. Maybe people sense (consciously or not—there might be some evolutionary mechanism here) that these cases don't play nice with the LLN, and so they don't do expected utility maximization, but do something heuristic, and the heuristic fails.

Objective probabilities

Suppose:

1. W is a set of possible worlds (or maybe situations?)
2. L is a first-order language suitable for talking about what is going on at a member of W, and with a finite symbol-set
3. S is the set of strings, of finite or countably infinite length, but with a starting point (i.e., "ababababab..." is acceptable, but "...ababababab" is not) in the symbol-set of L
4. e(s) is the proposition expressed by a sentence s of L
5. B_W(s) is the claim that s is a member of S such that e(s) is true at exactly one member of W
6. r is a random variable whose values range over the members of S and have the following property: P(r=s)=(n+1)^−(l(s)+1), where n is the number of symbols in the symbol-set of L and l(s) is the length of s; thus, r simply chooses a random string in S, letter by letter, with an equal likelihood of any particular letter or of ending the string there.

Then it seems we can define a probability of a first-order[note 1] proposition p relative to the worlds in W as follows: P_L,W(p)=P(e(r) entails p|B_W(r)).

If the language L is somehow natural for describing the members of W, then it makes sense to think of P_L,W as defining a natural probability measure for what goes on in members of W. If p is W-impossible, i.e., if p holds at no member of W, then P_L,W(p)=0.

What is particularly nice about P_L,W is that it favors worlds with simpler laws. Thus, it is a probability measure particularly well-suited to making scientific inferences.

A serious technical difficulty with the above definition is that P_L,W(p) will not be defined for all p, but only for those p for which the set of sentences r such that e(r) entails p is measurable. One can avoid this difficulty by restricting the ps for which P_L,W(p) is defined, or by replacing the Axiom of Choice with the axiom that all subsets of the reals are measurable.

A second technical difficulty is that P(B_W(r)) might be zero. This difficulty will be avoided if we have at least one finitely simple world, where a world w is finitely simple if and only if there is a finite sentence r such e(r) is true at w and only at w. I suspect (again, I haven't written out the proof) that in that case we get the following interesting theorem: With probability one, we are in a finitely simple world. This suggests that the measure P might be useful for inductive purposes—it seems to be a measure that prefers simpler worlds.

Dembski's definition of specified complexity

A central part of Dembski's definition of specified complexity is a way of measuring whether an event E is surprising. This is not just a probabilistic measure. If you roll eleven dice and get the "unsurprising" sequence 62354544555, this sequence has the same probability 1/6¹¹ as the intuitively more "surprising" sequences 12345654321 or 11111111111. It would be a mistake (a mistake actually made by some commenters on the design argument) to conclude from the probabilistic equality that there is no difference in surprisingness, since what one should conclude that is that surprisingness is not just a matter of the probabilities. Instead of talking about "surprisingness", however, Dembski talks about "specification". The idea is that you can "specify" the sequences 12345654321 or 11111111111 ahead of time in a neat way. The first you specify as the only sequence of eleven dice throws consisting of a strict monotonic ending precisely where a strict monotonic decline begins. The second is one of only six sequences of eleven dice throws that each yield the same result.

I will describe Dembski's account of specification, and that will be somewhat technical, and then I will criticize it, and consider a way of fixing it up which is not entirely satisfactory.

Dembski proposes a measure of specification.[note 1] Suppose we have a probability space S (e.g., the space of all sequence of eleven dice throws) with a probability measure P_H (defined by some chance hypothesis H). Let f be a "detached" real-valued function on S (a lot more on detachment later). Then an event E in the probability space S is just a measurable subset of the probability space. For any real-valued function f defined on S and real number y, let f_y be the set of all points x in S such that f(x)≥y. This is an event in S. Indeed, f_y is the event of being at a point x in our probability space where f(x) is at least y.

We now say that an event E in S is specified to significance a provided that there is a function f on S "detached" from E (a lot more on detachment later on) and a real number y such that f_y contains E and P_H(f_y)<a.

For instance, in our eleven dice throw case, if x is a sequence of eleven dice throw results, let f(x) be the greatest number n such that at least n of the throw results in x are the same. Then, f₁₁ is equivalent to the event that all eleven of the dice throws were the same. Let E be the event of the sequence 11111111111 occurring. Then E is contained in f₁₁, and P_H(f₁₁)=1/6¹¹<10^-8, and so E is specified to significance 10^-8, as long as we can say that f is detached from E. Similarly, we can let f be the length of the largest interval over which a sequence of dice throws is monotonic increasing plus the length of the largest interval over which a sequence of dice throws is monotonic decreasing, and then our sequence 12345654321 will be a member of f₁₂, and if f is detachable, we can thus compute a significance for this result.

The crucial part of the account is the notion of "detachability". Without such a condition, every improbable event E is significant. Given our intuitively unsurprising sequence 62354544555 (which was as a matter of fact generated by a pretty random process: I made it by using random.org[note 2]). Let f be the function assigning 1 to the sequence 62354544555 and 0 to every other sequence. Then our given sequence is the only member of f₁, and so without any detachability condition on f, we would conclude that we have specification to a high degree of significance. But of course this is cheating. The function f was jerryrigged by me to detect the event we were looking at, and one can always thus jerryrig a function. To check for specification, however, we need a function f that could in principle have been specified beforehand, i.e., before we found out what the result of the dice throwing experiment was. If we get significance with such a function, then we can have some confidence that our event E is specified.

Dembski, thus, owes us an account of detachability. In No Free Lunch, he offers the following:

a rejection function f is detachable from E if and only if a subject possesses background knowledge K that is conditionally independent of E (i.e., P(E|H& K) = P(E|H)) and such that K explicitly and univocally identifies the function f.

Or, to put it in our notation, f is detachable from E iff the epistemic agent has background knowledge K such that P_H(E|K)=P_H(E). It is hard to overstress how central this notion of detachability is to Dembski's account of specification, and therefore to his notion of specified complexity, and thus to his project.

But there is a serious problems with detachability: I am not sure that the independence condition P_H(E)=P_H(E|K) makes much sense. Ordinarily, the expression P(...|K) makes sense only if K is an event in the probability space or K is a random variable on the probability space (i.e., a measurable function on the probability space). In this case, K is "knowledge". This is ambiguous between the content of the knowledge and the state of knowing. Let's suppose first that K is the content of the knowledge—that's, after all, what we normally mean in probabilistic epistemology when we talk of conditioning on knowledge. So, K is some proposition which, presumably, expresses some event—probabilities are defined with respect to events, not propositions, strictly speaking.[note 3] What is this proposition and event? The knowledge is supposed to "identify" the function f. It seems, then, that K is a proposition of the form "There is a unique function f such that D(f)", where D is an explicit and univocal identification.

But on this reading of "knowledge", the definition threatens uselessness. Let K be the proposition that there is a unique function f such that f(x)=1 if and only if x equals 62354544555 and f(x)=0 otherwise. This function f was our paradigm of a non-detachable function. But what is P_H(E|K)? Well, K is a necessary truth: It is a fact of mathematics that there is a unique function as described. If P_H is an objective probability, then all necessary truths have probability 1, and so to condition on a necessary truth changes nothing: P_H(E|K)=P_H(E), and we get detachability for free for f, and indeed for every other function.

So on the reading where K is the content of the knowledge, if necessary truths get unit probability, Dembski's definition is pretty much useless—every function that has a finite mathematical description becomes detachable, since truths about whether a given finite mathematical description uniquely describes a function are necessary truths.

But perhaps P_H is an epistemic probability, so that necessary truths might have probability less than 1. One problem with this is that much of the nice probabilistic apparatus now breaks down. How on earth do we define a probability space in such a way that we can assign probabilities less than 1 to necessary truths? Do we partition the space of possibilities-and-impossibilities into regions where it is true that there is a unique function f such that f(x)=1 iff x=62354544555 and f(x)=0 otherwise and regions where this is false? I am not sure what we can make of probabilities in the regions where this is false. Presumably they are regions where mathematics breaks down. How do we avoid incoherence in applying probability theory—as Dembski wants to!—over the space of possibilities-and-impossibilities?

Moreover, it seems to me that on any reasonable notion of epistemic probabilities, those necessary truths that the epistemic agent would immediately see as necessary truths were they presented to her should get probability 1. Any epistemic agent who is sufficiently smart to follow Dembski's arguments and who knows set theory would immediately see as a necessary truth the claim that there is a unique function f on S such that f(x)=1 iff x=62354544555 and f(x)=0 otherwise. So even if we allow that some necessary truths, such as that horses are mammals, might get epistemic probabilities less than 1, the ones that matter for Dembski are not like that—they are self-evident necessary truths in the sense that once you understand them, you understand that they are true. The prospects for an account of epistemic probability that does not assign 1 to such necessary truths strike me as unpromising, though I think this is the route Dembski actually wants to go according to Remark 2.5.7 of No Free Lunch.

Besides, as a matter of fact, any agent who is sufficiently smart to understand Dembski's methods will be one who will assign 1 to the claim that there is a unique function f as above. So on the objective probability reading, Dembski's definition of detachability applies to all finitely specifiable functions. On the epistemic, it does so too, at least for agents who are sufficiently smart. This makes Dembski's definition just about useless for any legitimate purposes.

Let's now try the second interpretation of K, where K is not the content of the knowledge, but the event of the agent's actually knowing the identification of f. This is more promising, I think. Let p be the proposition that there is a unique function f on S such that f(x)=1 iff x=62354544555 and f(x)=0 otherwise. Let us suppose, then, that K is the event of the agent knowing that p. It is essential, we've seen, to judging f to be non-detachable that P_H(K) be not equal to 1. This requires a theory of knowledge where for an agent to know p is more than just for an agent to be in a position to know p, as when the agent knows things that self-evidently entail p. An actual explicit belief is required for knowledge on this view. Seen this way, P_H(K)<1, since the agent might never have thought about K. Since K is a bona fide event on this view, we can apply probability theory without any worries about dealing with incoherence. So far so good.

But new problems show up. It is essential to Dembski's application of his theory to Intelligent Design that it apply in cases where people have only thought of f after seeing the event E—cases of "old evidence". Take, for instance, Dembski's example of the guy whose allegedly random choices of ballot orderings heavily favored one party. Dembski proposes a function f that counts the number of times that one party is on the top of the ballot. But I bet that Dembski did not actually think of this function before he heard of the event E of skewed ballot orderings. Moreover, hearing of the event surely made him at least slightly more likely to think of this function. If he never heard of this event, he might never have thought about the issue of ballot orderings, and hence about functions counting them. There is surely some probabilistic dependence between Dembski's knowing that there is such a function and the event E. Similarly, seeing the sequence 11111111111 does make one more likely to think of the function counting the number of repetitions. One might have thought of that function anyway, but the chance of thinking of it is higher when one does see the result. Hence, there is no independence, and, thus, no detachability.

This problem is particularly egregious in some of the biological cases that ultimately one might want to apply Dembski's theory to. Let's consider the event E that there is intelligent life. Let K be any state of knowledge identifying a function. Surely, there is probabilistic dependence between E and K. After all, P_H(K|~E)=0, since were there no intelligent life, nobody would know anything, as there would be nobody to do the knowing. Thus, P_H(E|K)=1, which entails that E and K are not probabilistically independent unless P(E)=1.

So the problem is that in just about no interesting case where we already knew about E will f be detachable from E, and yet the paradigmatic applications of Dembski's theory to Intelligent Design are precisely such cases. Here is a suggestion for how to fix this up (inspired by some ideas in Dembski's The Design Inference). We allow a little bit of dependence between E and K, but require that the amount of dependence not be too big. My intuition is that smaller the significance a of the specification (note that the smaller the significance a, the more significant the specification—that's how it goes in statistics), the more dependence we can permit. To do that right, we'd have to choose an appropriate measure of dependence, but since I'm just sketching this, I will leave out the details.

However, there is a difficulty. The difficulty is that in "flagship cases" of Intelligent Design, such as the arising of intelligence or of reproducing life-forms, there is a lot of dependence between E and K, since our language is in large part designed (consciously or not) for discussing these kinds of events. It is in large part because reproducing life-forms are abundant on earth that our language makes it easy to describe reproduction, and that our language makes it easy to describe reproduction significantly increases the probability that we will think of functions f that involve reproductive concepts. In these cases, the amount of dependence between E and K will be quite large.

There may still be cases where there is little dependence, at least relative to some background data. These will be cases where our language did not develop to describe the particular cases observed but developed to describe other cases, perhaps similar to the ones observed but largely probabilistically independent of them. Thus, our language about mechanics and propulsion plainly did not develop to describe bacterial flagella, and it may be that the existence of bacterial flagella is probabilistically independent of the things for which our language developed. So maybe the above account works if K is a state of knowing a specification that includes bacterial flagella. Or not! There are hard questions here. One of the hard questions is with regard to how particular K is. Is K the event of one particular knower, say William Dembski, having the identification of f? If so, then there is a lot of probabilistic dependence between the existence of bacterial flagella and K, since the probability of Dembski's existing in a world where there are no bacterial flagella is very low, since history would have gone very differently without bacterial flagella, and probably Dembski would never have come into existence.

Or is K the event of some knower or other having the identification of f? Then, to evaluate the dependence between K and the existence of bacterial flagella we would have to examine the almost intractable question of what a world without bacterial flagella would have been like.

An argument against an infinite past

This is a version of an argument by Bill Craig, with probability in place of the Principle of Sufficient Reason. I don't actually think this argument is sound, but the premises might well be plausible to a number of people. Suppose, then, for a reductio, that it is possible for a world to have an infinite past. Let H be the following hypothesis: The world has an infinite past and future (nobody who allows an infinite past will balk at an infinite future, surely), and contains Jones, who counts up from minus infinity (not inclusive) to zero (inclusive), uttering one number a day. Thus, on some day he uttered "-4848", and on the next he uttered "-4847" and so on. Then on some day he finished by uttering "0".

For any time t, let E_t be the hypothesis that Jones has finished counting at a time t* such that t - 1day < t* ≤ t, i.e., that Jones has finished within in the 24 hours preceding t. Let p(t)=P(E_t|H). Since H does not mention any specific times, by the principle of indifference, p(t) has to have the same value for every value of t. Thus, for all t, p(t)=p(0).

But now consider the following infinite sequence of events: ...,E_{-3 days},E_{-2 days},E_{-1 day},E₀,E_{1 day},E_{2 days},E_{3 days},.... Given H, it is certain that exactly one of them happens. Thus, P(... or E_{-3 days} or E_{-2 days} or E_{-1 day} or E₀ or E_{1 day} or E_{2 days}, or E_{3 days} or ...|H) = 1. Moreover, these events are mutually exclusive, so the left hand side of this equation is equal to: ...+p(-3 days)+p(-2 days)+p(-1 day)+p(0)+p(1 day)+p(2 days)+p(3 days)+.... But each of the summands here is the same, namely p(0). If p(0) is positive, then this sum is infinite, and hence not equal to 1. If p(0) is zero, then this sum is zero, and hence not equal to 1. And p(0) can't be negative since it's a probability. Hence, impossibility ensues no matter what value p(0) has. (And, no, infinitesimals won't help. That was shown by Tim McGrew--see this paper of mine.) If all of this works, then we need to reject as absurd the assumption that an infinite past is possible. And once we reject this assumption, the Kalaam argument becomes available.

There are two weak points in the argument. The first is that there is an actual difference between the hypotheses E_t for different values of t. If one accepts an A-theory of time, according to which what time it is now is an objective feature of the universe, then one has to agree there is a difference between these hypotheses--it is an objectively different thing for Jones to finish counting today than to have finished counting yesterday. Likewise, if one takes a substantival theory of time, one will see a difference. But the Leibnizian like me, who takes time to be purely relational, will not see a difference between the hypotheses: if one shifts over the history of the world by a day, one changes nothing. The second weak point is the assumption that one can apply classical probability theory to events like E_t conditioned on H, which, again, I am suspicious of. (But I accept the Principle of Sufficient Reason, and that can be used in place of the probabilistic reasoning.)