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!
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.
ReplyDeleteMy 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.
ReplyDeletePerhaps 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."
ReplyDeleteAren't there subsets A in your example, which are countabel and simultaniously co-countable?
ReplyDeleteIf so, then what is exactly P(countable and co-countable A) in your example?
What are you even talking about?!?
Is the following infinite lottery "chanceable" by your definition?
Ω=ℕ with P: F→[0;1], such that P({n})=λ·(1-λ)^n (a geometric distribution) for any (n∈ℕ)∈(σ-algebra of ℕ) F with some λ∈(0;1).
Is this geometric distribution a countably additive probability?
If so, then there is at least one countably additive "chanceable" probability.
Marriages are "chanceable" as bachelors are "chanceable".
"Married bahcelors" are not "chanceable".
But that, "Married bahcelors" being not "chanceable", doesn't imply, that therefore marriages and or bachelors are also not "chanceable".