A gambling puzzle about nonmeasurable events

I have two sealed envelopes, labeled A and B. One contains $3 and the other nothing. You don’t know which is which. I am willing to sell either or both envelopes for $1 each. You have a fixed period of time to inform me whether you are buying neither, both, A, or B, after which time you pay any get to open any envelopes you bought.

Obviously, it makes sense for you to hand me $2 and buy both envelopes and profit by a dollar.

But suppose now that I tell you that I chose which envelope to put the $3 using a saturated nonmeasurable method. For instance, perhaps I chose a subset N of the points on the circumference of a spinner that has the properties that:

1. N is nonmeasurable,

2. the only measurable subsets of N have measure zero, and

3. the only measurable subsets of the complement of N have measure zero,

then I spun the spinner, and if the spinner landed in N, I put the $3 in envelope A, and otherwise in B.

Your purchase options are: Neither, Both, A and B. The probability that the $3 is in A is completely undefined (we should represent the probability as the full interval from 0 to 1) and the probability that the $3 is in B is completely undefined.

It seems then:

1. It’s clearly rationally permissible for you to go for Both.

2. Going for A is neither rationally better nor rationally worse than going for Both. For by going for A, you miss out on B. But the expected utility of purchasing B iscompletely undefined: it is a choice to pay $1 for a gamble that has a completely undefined probability of paying out $3. So, it is completely undefined whether Both is better than A or worse. If Both is permissible, so is A, then.

3. But by similar reasoning it is completely undefined whether going for Neither is better than or worse than going for A. For the expected payoff of A is completely undefined. So, if A is rationally permissible, so is Neither, then.

4. Swapping A and B in the reasoning in (2) shows that B is rationally permissible as well.

So now it seems that all four options are equally permissible. But something has gone wrong here: Clearly, Both beats Neither, and it’s irrational to go for Neither.

I think to get out of the above puzzle, we have to deny the initially plausible principle:

5. If an option is rationally permissible, and another option is neither better nor worse than it, then the latter is also permissible.

Here is another case where this principle needs to be denied. You have a choice between playing Pac Man, or eating one scoop of ice cream, or eating two. Playing Pac Man is neither better nor worse than either one or two scoops of ice cream. Two scoops of ice cream is better than one. It is clearly rationally permissible to play Pac Man. By (5), it’s permissible to eat one scoop of ice cream, then. But that’s not true, since two scoops beats one.

So, let’s deny (5). Now I think the reasonable thing to say is that Neither is irrational, but each of Both, A and B is rationally permissible. But there is still a puzzle in the vicinity. Suppose you are asked about your purchases envelope-by-envelope. First you’re offered the chance to buy A, and then a chance to buy B, and once a deal is declined, it’s gone. You have no rational obligation to buy A. After all, going for B alone is permissible. So, let’s say you decline A. Next you’re asked about B. At this point, A is out of the picture, and the question is whether to pay $1 for a completely undefined probability of getting $3. It’s permissible to decline that. So, you can permissibly decline B as well. So, let’s say you do so. Now by a pair of perfectly rational choices you ended up “doing something stupid”. This is a bit like Satan’s Apple. but with a finite number of choices.

The puzzle above seems familiar. I may have read it somewhere and it stuck in my subconscious.