Bayesian Blues: Possibilities with Probability 0

If my read of the landscape is correct (and it may not be), Bayesian views (relatively broadly construed) enjoy a certain dominance in formal epistemology. No doubt this is because the various explorations into Bayesian formalisms have provided philosophers with rich theoretical resources and aided in the development of many significant advances in formal epistemology. However, despite whatever virtues they have, these approaches wind up saddled with some results that are, prima facie, highly counter-intuitive. The specific counter-intuitive result that I’ve been thinking about recently has to do with possibilities that get assigned a probability of 0.

Preliminaries: A probability measure is said to be regular if all and only contradictions are assigned probability 0 (or, in possible worlds talk, for any set of possible worlds A, the probability of A is greater than 0 if A is not the empty set). Given the standard axioms of probability, this guarantees that all and only tautologies are assigned probability 1.

Regular probability measures are not very popular these days, at least in part because of issues arising when assigning probabilities to infinitely many exclusive propositions. For a quick illustration, imagine that some deity is going to pick one member of the set of integers, and you are trying to assign probabilities to the options. There is no single positive real value v that one can assign to each and every possible selection, without violating the axioms of probability. So, one could either assign different values to different integers (for instance, .5 to 1, .25 to 2, .125 to 3, and so forth), which requires favoring some as more likely than others, or, alternately, one could assign 0 to each individual integer, which would avoid having to favor some options as more likely than others. This option would require the rejection of a regular probability measure.

The Set-Up: Consider a case where some all-powerful deity is going to throw a dart so that it hits a point on the surface of the earth at random. Now, take the point that is at the center of the surface of Kansas (call it K). On the standard Bayesian picture, the probability that the dart hits K is 0. We can assign positive probabilities to various regions of the surface of the Earth (for instance, the probability that the dart lands in the northern hemisphere is somewhere near .5), but not to the individual points.

The Worry: Take the point on the moon’s surface that is closest to Earth (call it M). The probability that the dart hits M is also 0. Consequently, one who endorses this approach is committed to the view that it is just as probable that the dart will hit M as it is that the dart will hit K. K is no more likely a destination for the dart than M, even though it is not possible for the dart to hit M, but entirely possible for the dart to hit K. These non-probable possibilities would then seem to be on a par with outright impossibilities. But this is highly counterintuitive. While I might think Jones ridiculously irrational for betting any money at all on the proposition that the dart will hit K, I would take a dimmer view of Smith for betting on M. But there is no way to account for this.

The Response: The worry is overstated. It is not true that there is no way to account for it. The hard-and-fast result is that the explanation of why Jones is doing better than Smith cannot be cashed out in terms of K being more probable a destination than M, but that isn’t the only way to account for why Jones is better off than Smith. For one thing, it is possible that Jones wins, but not possible that Smith wins. But there are other things to be said as well. I won’t go into them here, in part because I am not familiar enough with the relevant details to present them cogently, but also because even assuming a perfect explanation of the relative evaluations of Jones and Smith, and the like, there is a lingering worry.

The Real Worry: Forget Jones and Smith, forget evaluations of the rationality of agents making bets, etc., absent the invocation of a non-standard system of numbers (such as infinitesimals), the Bayesian simply cannot affirm that K is a more likely or more probable destination for the dart than M. They have equal probabilities, and thus, are equally probable. And, at the end of the day, it really does seem as though it is more likely or more probable that the dart hits K than that it hits M.

Now, I’m not presenting this as an incontrovertible claim that needs to be captured by any viable theory. If the best view there is cannot capture the claim, it may well be that the claim needs to go, and not the theory. But it is still a legitimate worry for the view that it is incompatible with claiming that every possibility is more probable than any impossibility.

One Response to Bayesian Blues: Possibilities with Probability 0

  1. Ken says:

    Hi,Interesting post! From a practical point of view, the solution is to use probability densities instead of absolute probabilities. The probability density at any particular point on earth will be non-zero, whereas the probability density at any point on the moon will be zero.

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