Counting down from infinity

In one version of the Kalaam argument, Bill Craig argues against forming an infinite past by successive addition by asking something like this: Why would someone who had been counting down from infinity have been finished today rather than, say, yesterday? This argument puzzles me. After all, there is a perfectly good reason why she finished today: because today she reached zero and yesterday she was still on the number one. And yesterday she was on one because the day before she was on two. And so on.

Of course, one can object that such a regress generates no explanation. But then the Kalaam argument needs a Principle of Sufficient Reason that says that there must be explanations of such regressive facts and an account of explanation according to which the explanations cannot be found in the regresses themselves. And with these two assumptions in place, one doesn’t need the Kalaam argument to rule out an infinite past: one can just run a “Leibnizian style” cosmological argument directly.

Probability on infinite sets and the Kalaam argument

Suppose there is an infinite line of paving stones, labeled 1, 2, 3, ..., on each of which there is a blindfolded person. You are one of these persons. That's all you know. How likely is it you're on a number not divisible by ten? The obvious answer is: 9/10. But now I give you a bit more information. Yesterday, all the same people were standing on the paving stones, but differently arranged. At midnight, all the people were teleported, in such a way that the people who yesterday were standing on numbers divisible by ten are now standing on numbers not divisible by ten, and vice versa. Should you change your estimate of the likelihood you're on a number divisible by ten?

Suppose you stick to your current estimate. Then we can ask: How likely is it that you were yesterday on a number not divisible by ten? Exactly the same reasoning that led to your 9/10 answer now should give you a 9/10 answer to the back-dated question. But the two probabilities are inconsistent: you've assigned probability 9/10 to the proposition p₁ that yesterday you were on a number not divisible by ten and 9/10 to the proposition p₂ that today you are on a number not divisible by ten, even though p₁ holds if and only if p₂ does not (this violates finite additivity).

So you better not stick to your current estimate. You have two natural choices left. Switch to 1/2 or switch to 1/10. Switching to 1/2 is not reasonable. Let's imagine that today is the earlier day, and you have a button you can choose to press. If you press it, the big switch will happen—the folks on numbers divisible by ten will be swapped with the folks on numbers not divisible by ten. If you had switched to 1/2 in my earlier story, then if you press the button, you should also switch your probabilities to 1/2, while if you don't press the button, you should clearly stick with 9/10. But it's absurd that your decision whether to press the button or not should affect your probabilities (assume that there is no correlation between what decision you make and what number you're on).

Alright, so the right answer seems to be: switch to 1/10. But this means that the governing probabilities in infinite cases are those derived from the initial arrangements. Why should that be so?

Here is a suggestion. We assume that the initial arrangement came from some sort of a regular process, perhaps stochastic (where "regular" is understood in the same sense as "regularity" in discussions of laws). For instance, maybe God or a natural process brought about which squares the people go on by taking the people one by one, and assigning them to squares using some natural probability distribution, like probability 1/2 to 1, 1/4 to 2, 1/8 to 3, and so on, with the assignment being iterated until a vacant square is found (equivalently do it in one step: use this distribution but condition on vacancy). And, maybe, for most of the "regular" distributions, once enough people are laid down, we get about a 9/10 chance that the process will land you on a square not divisible by ten.

This assumes, however, that there is a process that puts people on squares. Suppose this assumption is false. Then there seems to be no reason to privilege the probability distribution from the first time the folks are put on squares. And our intuitions now lead to inconsistency: assigning 9/10 to p₁ and 9/10 to p₂.

Where has all this got us? I think there is an argument here that absurdity results from an actual, simultaneous infinity of uncaused objects. But if an actual infinity of objects is possible, and it is possible to have a contingent uncaused object, then it is very plausible (this is an ampliative inference) that it is possible to have an actual infinity of simultaneous contingent uncaused objects.

Therefore: either it is impossible to have an uncaused object or it is impossible to have an actual infinity of simultaneous contingent objects. But it is possible to have an actual infinity of simultaneous contingent objects if it is possible to have an infinite past. This follows by al-Ghazali's argument: just imagine at each past day a new immortal soul coming into existence, and observe that by now we'll have a simultaneous infinity of objects. So, it is either impossible to have an uncaused contingent object or it is impossible to have an infinite past. We thus have an argument for the disjunction of the premises of the Kalaam argument, which is kind of cool, since both of the premises of the argument have been disputed. Of course, it would be nicer to have an argument for their conjunction. But this is some progress. And it may be the further thought along these lines will yield more fruit.

A puzzle about uniform probabilities on infinite sets

This puzzle is inspired by a reflection on (a) a talk [PDF] by John Norton, and (b) the problem of finding probability measures on multiverses. It is very, very similar—quite likely equivalent—to an example [PDF] discussed by John Norton. Suppose you are one of infinitely many blindfolded people. Suppose that the natural numbers are written on the hats of the people, a different number for each person, with every natural number being on some person's hat. How likely is it that the number on your hat is divisible by three?

The obvious answer is: 1/3. But Norton's discussion of neutral evidence suggests that this obvious answer is mistaken. And here is one way to motivate the idea that the answer is mistaken. Suppose I further tell you this. Each person also have a number on her scarf, a different number for each person, with every natural number being on some person's scarf. Moreover, the following is true: the number on x's scarf is divisible by three if and only if the number on x's hat is not divisible by three. (Thus, you can have 3 on your scarf and 17 on your hat, but not 16 on your scarf and 22 on your hat.) This can be done, since the cardinality of numbers divisible by three equals the cardinality of numbers not divisible by three.

If you apply the earlier hat reasoning to the scarf numbers, it seems you conclude that the likelihood that the number on your scarf is divisible by three is 1/3. But this is incompatible with the conclusion from the hat reasoning, since if the likelihood that the scarf number is divisible by three is 1/3, the likelihood that the hat number is divisible by three must be 2/3.

If there are numbers on hats and scarves as above, symmetry, it seems, dictates that the probability of your hat number being divisible by three is the same as the probability of your scarf number being divisible by three, and hence is equal to 1/2. But this conclusion seems wrong. For the numbers on scarves, even if anti-correlated with those on the hats, should not affect the probability of the hat number being divisible by three. Nor should it matter in what order the hat and scarf numbers were written—hats first, and then scarves done so as to ensure the right anti-correlation between divisibilities, or scarves first, and then hats. But if the hat numbers are written first, then surely the probability of divisibility by three is 1/3, and this should not change from the mere fact that scarf numbers are then written.

One of several conclusions might be drawn:

1. Actual infinities are impossible.
2. Uniform priors on infinite discrete sets make no sense.
3. Probabilities on infinite sets are very subtle, and do not follow the standard probability calculus, but there is a very intricate account of dependence such that whether the hat numbers are assigned first or the scarf numbers are assigned first actually affects the probabilities. I don't know if this can be done—but when I think about it, it seems to me that it might be possible. I seem to be seeing glimpses of this, though the fact that as of writing this (a couple of hours after my return from Oxford) I've been up for 21 hours may be affecting the reliability of my intuitions.

Deep Thoughts XI

If you never get into trouble, you'll never have to get out of it. - Constable Keating, "Misunderstood" by P. G. Wodehouse.

I prefer tautologies for Deep Thoughts, but this almost one. The non-tautologous bit is that, after all, one might be in trouble but never have got into it, and yet still need to get out of it. One way is if one is in trouble from the beginning of one's existence. Maybe, then, one's coming into existence counts as a getting into trouble, so this is not a counterexample to Constable Keating's claim. The other way one might be in trouble but never have got into it is if one has infinite age and has always been in trouble. Of course, it might be that Kalaam considerations will show that that is in fact impossible. If so, then Constable Keating's claim will be a necessary truth. But it probably still won't be a tautology.