Inscrutable probabilities and the replay argument

Suppose that we have two hypotheses about a long sequence X of zeroes and/or ones:

• S: The sequence has the underlying structure of a sequence of independent and identically distributed (iid) probabilistic phenomena with possible outcomes zero or one.

• U: The sequence is completely probabilistically unstructured, being completely probabilistically inscrutable.

What can we say about the evidential import of X on S and U?

Here is one thought. We have a picture of how a repeated sequence of independent and identically distributed binary events should look. Much as van Inwagen says in his discussion of the replay argument, we expect to have the proportion of ones to oscillate but then converge to some value. So if X looks like that, then we have evidential support for S, and if X doesn’t look like that—maybe there is wilder oscillation and no convergence—then we do not have evidential support for S.

But does the fact that the sequence X has a “nice gradual convergence” in frequencies support the structure hypothesis S? Van Inwagen in his replay argument against indeterministic free will seems to assume so.

Actually, convergence of frequencies does not support S (or U, for that matter). Here is why not. Let’s model hypothesis S as follows. There is some unknown real dispositional probability p between 0 and 1, and then our sequence X is taken by observing a sequence of iid results with the probability of the result being 1 being equal to p. Given p, any sequence of n events whose proportion of ones is m/n has probability p^m(1−p)^n − m, regardless of whether the sequence shows any “nice gradual convergence” of the frequency of ones to m/n or is just a seqence of n − m zeroes followed by m ones.

Of course, hypothesis S doesn’t say what the probability p of getting a one on a given experiment is. So, we need to suppose some probability distribution Q over the possible values of p in the interval [0,1], and then our probability of the sequence X of m ones and n − m zeroes will be ∫p^m(1−p)^n − mdQ(p). But it’s still true that this probability does not depend on the order between the ones and zeroes. The sequence could “look random”, or it could look as fishy as you like, and the probability of it on S would be exactly the same if the number of ones is the same.

On the other hand, on the hypothesis U, all possible sequences X of length n are probabilistically inscrutable, and hence we cannot say anything about any sequence being more likely than another—we might as well represent the probability of any sequence X of observations as just an interval-valued probability of [0,1].

So, no facts about the order of events in X make any difference between the hypotheses S and U. In particular, van Inwagen’s idea that an appearance of convergence is evidence for S is false.

Now, let’s say a little more about the Bayesian import of the observation X. Let’s suppose that S and I are our only two hypotheses, and that both have serious prior probabilities, say somewhere between 0.1 and 0.9. Further, let’s suppose that that the sequence X is long and has a decent amount of variation in it—for instance, let’s suppose that it has at least 50 zeroes and at least 50 ones. Because of this, P(X|S) will be some astronomically small positive number α. Indeed, we can prove that α is at most 1/2¹⁰⁰ ≈ 10⁻³⁰, for any distribution of the unknown probability p of getting a one.

On the other hand, P(X|U) will be completely probabilistically inscrutable, and hence can be represented as the full interval [0,1].

Here’s what follows from Bayes’s theorem combined with a natural way of handling inscrutable probabilities as a range of probability assignments: The posterior probability of S will be a range of probabilities that starts with something within an order of magnitude of α and ends with 1. Hence, our observation of X does not support S: upon observing X, we move from S having some moderate probability between 0.1 and 0.9, to a nearly completely inscrutable probability in a range starting with something astronomically small and ending at one. The upper end of the range is higher than what S started with but the lower end of the range is lower.

Thus, if we were to actually do the experiment that van Inwagen describes, and get the results that he thinks we would get (namely, a sequence converging to some probability), that would not support the hypothesis S that van Inwagen thinks it would support.