Suppose an infinite collection of fair coins is going to be tossed. An outcome is then a specification of how each particular coin (maybe they are numbered) has landed. For any set of outcomes A, we can define this game:
- Game GA: You toss the coins. If the outcome is in A, you get a dollar. Otherwise, you get nothing.
Here’s a way of putting the paradox of nonmeasurable sets. Assume the Axiom of Choice. Suppose that for every set A of outcomes you assign a (countably additive) probability P(A) between 0 and 1 (not inclusive) of winning the game.
Suppose you are perfectly rational. Then there is a set A of outcomes where you will behave in the following way. First, you agree to pay an initial amount of money to play GA. After we have made the agreement, I offer you a special deal. For a certain extra increment of money and a certain particular coin C, after all the coins are tossed, regardless of how coin C has landed, I will give it an extra turn-over before we check whether you won (i.e., whether the outcome is in A). And you agree.
The paradoxical part is the “And you agree”. If the coins are fair, it should make no difference whether I give a coin an extra turn-over before you learn the outcomes (of course, with an unfair coin, it will in general make a difference). But for any probability measure on the collection of all outcomes of an infinite collection of coins, there will be a game GA and a coin C where you will think it is more likely that you will win if you give C an extra turn-over.
The above remarks are just a restatement of the fact that given the Axiom of Choice, there is no (countably additive) probability measure on the set of all outcomes of an infinite collection of coin flips that is invariant under all individual coin reversals.
This trick need not work if the probability measure is only finitely additive: given the Axiom of Choice, there is a finitely measure invariant under all individual coin reversals (because the group generated by the coin reversals is Abelian and hence amenable).
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