Say that a function P : F → [0,1] where F is a σ-algebra of subsets of Ω is chanceable provided that it is metaphysically possible to have a concrete (physical or not) stochastic process with a state space of the same cardinality as Ω and such that P coincides with the chances of that process under some isomorphism between Ω and the state space.
Here are some hypotheses ones might consider:
If P is chanceable, P is a finitely additive probability.
If P is chanceable, P is a countably additive probability.
If P is a finitely additive probability, P is chanceable.
If P is a countably additive probability, P is chanceable.
A product of chanceable countably additive probabilities is chanceable.
It would be nice if (2) and (4) were both true; or if (1) and (3) were.
I am inclined to think (5) is true, since if the Pi are chanceable, they could be implemented as chances of stochastic processes of causally isolated universes in a multiverse, and the result would have chances isomorphic to the product of the Pi.
I think (3) is true in the special case where Ω is finite.
I am skeptical of (4) (and hence of (3)). My skepticism comes from the following line of thought. Let Ω = ℵ1. Let F be the σ-algebra of countable and co-countable subsets (A is co-countable provided that Ω − A is countable). Define P(A) = 1 for the co-countable subsets and P(A) = 0 for the countable ones. This is a countably additive probability. Now let < be the ordinal ordering on ℵ1. Then if P is chanceable, it can be used to yield paradoxes very similar to those of a countably infinite fair lottery.
For instance, consider a two-person game (this will require the product of P with itself to be chanceable, not just P; but I think (5) is true) where each player independently gets an ordinal according to a chancy isomorph of P, and the one who gets the larger ordinal wins a dollar. Then each player will think the probability that the other player has the bigger ordinal is 1, and will pay an arbitrarily high fee to swap ordinals with them!
3 comments:
In the example, unless I’m missing a point, each player will pay up to a dollar to swap. But that doesn’t spoil the paradox.
My mistake indeed. In my excuse, my yesterday started with a 5:00 am flight from El Paso to Waco, and so I didn't get much sleep.
Perhaps if one needs to make the paradox nastier, one can say: "I'll do the following. I'll pay up to a dollar to swap, and I promise that if you still win, I'll give you a million."
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