Monday, April 28, 2025

Probabilities of regresses of chickens

Suppose we have a backwards-infinite sequence of asexually reproducing chickens, ..., c−3, c−2, c−1, c0 with cn having a chance pn of producing a new chicken cn + 1 (chicken c0 may or may not have succeeded; the earlier ones have succeeded). Suppose that the pn are all strictly between 0 and 1, and that the infinite product p−1p−2p−3... equals some number p strictly between 0 and 1.

Intuitively, we should be surprised that chicken c0 exists if p is low and not surprised if p is high. If we have observed c0 and are considering theories as to what the chances pn are, other things being equal, we should prefer the theories on which the product p is high to ones on which it’s low.

But what exactly does p measure? It seems to be some kind of a chance of us getting c0. But it doesn’t measure the unconditional probability of getting an infinite sequence of chickens leading up to c0. For that is very tiny indeed, since it is extremely unlikely that the world would contain chickens at all. It seems to be a kind of conditional probability. Let qn be the proposition that chicken cn exists. Then P(q0qn) = p0p−1p−2...pn, and so p is the limit of the conditional probabilities P(q0qn). It is plausible thus to think of p as a conditional probability of q0 on q−∞, which is the infinite disjunction of all the qn.

But q−∞ is a rather odd proposition. It is grounded in qn for every finite n, assuming that a disjunction, even an infinite one, is grounded in its true disjuncts. Thus every one of the qn is explanatorily prior to q−∞. But this means that P(q0q−∞) is actually a conditional probability of q0 on something that isn’t explanatorily prior to q0—indeed, that is explanatorily posterior to q0. This challenges the interpretation of p as a chance of getting chicken c0.

I am not quite sure what conclusion to draw from the above argument. Maybe it offers some support for causal finitism, by suggesting that things are weird when you have a backwards infinite causal sequence?

2 comments:

William said...

Since limit as n -> Inf of x^n where 0 < x < 1 is 0, isn't the probability of the next chicken (Inf + 1) actually p(x | 0), which is by Baye's rule 0/0, that is NaN?

Alexander R Pruss said...

That's assuming each chicken has equal reproduction probability. Not so in my story.