Friday, October 24, 2014

Yet yet another probability paradox

Start with a set M of countably infinitely many people, and a set D of countably infinitely many fair dice. Suppose that there are no natural orderings on the set D, and that each person in M has exactly one of the dice in D assigned to her. (Or if you prefer, these are sets of unique names of people and coins respectively.) You are a person in M, and you know what all the members of D are but have no information whatsoever on which member of D is yours. Now all the dice are simultaneously and independently tossed. Obviously, your probability that your die showed sixes is 1/6.

Then the set of all the dice that landed sixes is revealed to you. Call the revealed set D6.

Suppose—this will be no surprise, as it had probability one—that the set of six-landing dice is infinite and the set of non-six-landing dice is infinite as well. Before it was revealed to you which dice landed sixes, your probability that your die yielded a six was 1/6. Did that probability change after you learned which set was the set of dice that landed sixes?

There are three options:

  1. No, it didn't change at all—it stayed at 1/6.
  2. Yes, it changed to an undefined value.
  3. Yes, it changed to some other defined value.
To choose between the options, observe first that your current probability that your die landed six must now be exactly the same as the probability that your die is a member of D6. But the fact that D6 is in fact the set of the six-showing dice carries no information as whether your die is in D6. Since all the dice are independent and fair, learning which dice landed sixes is completely irrelevant to finding your die. So whatever probability you assign to your die being among the members of D6 after the revelation must be the same as the probability you assigned to it before the revelation.

So, if we choose option (1), then already before you found out that the double-six rollers were the members of D6, you would have already assigned probability 1/6 to your die being in D6. But there was no natural ordering on the set D of dice, so the set D6 will be epistemically on par with its complement WW6. Both are simply countably infinite sets with countably infinite complements, and we can easily define an isomorphism of D onto itself that swaps the two sets. So if prior to learning the dice results you assigned 1/6 to your die being in D, you should have equally assigned 1/6 to your die being in DD6. But that's incoherent, since it's a given that the die is in D or DD6 but 1/6+1/6=1/3<1. So it seems that (1) is not an option.

That leaves (2) and (3). But those options are very strange. They imply that in such infinite die rolling scenarios, more data can always destroy your reasonable initial probability assignments.

Now, you might think that the above scenario only works when you don't know which die is yours, and that's kind of a strange scenario. But one can modify the scenario to work even when you do know which die is yours, but there is some other unique feature you don't know about your die, say, which of infinitely many (metaphysically) possible exotic particles is hidden inside the die, which of infinitely many angels has your die as a personal favorite, or what an independent sequence of rolls of the die yielded. Then the set D will be set of these unique features, and D6 will be the set of these features among the dice that landed six.


IanS said...

The dice may have no natural order, but they must have some order, viz. the one used to show countability. Using this order, for any subset of D you can define a function R(n) = (number of elements of the subset in the first n elements of D) / n. Apply this to D6. With probability 1 you will find R(n) converges to 1/6. Do the same for the complement, and (again with probability 1), R(n) will converge to 5/6. So you need not treat the sets as equivalent.

IanS said...

The above comment may have missed the point a bit. You may have been (reliably) told that the dice are countable, but not seen the enumeration. You are shown D6. What is the probability that your die is in it? My answer: you have no idea. You are now told that D6 is dice that show 6. What is the probability now? My answer: 1/6. Why the change? How does the extra information help? My answer: It tells you that (with probability 1) average density of D6 in D (i.e. the limit of R(n) in the previous comment) is 1/6. You don’t need the explicit enumeration to know this - you only need to know it exists.

Rob K said...

Does it make a difference if the number of six-landing dice is finite? I suppose that in that case I should revise my probability for my dice being 6 to 0. Given that that is so, is it more surprising that I should revise my probability on learning that the set of six-landing dice is some particular infinite set?