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:
N is nonmeasurable,
the only measurable subsets of N have measure zero, and
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:
It’s clearly rationally permissible for you to go for Both.
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.
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.
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:
- 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.
4 comments:
Elga Subjective probabilities should be sharp uses a similar example to argue against imprecise credences. Responses include Chandler Subjective probabilities need not be sharp and Bradley & Steele Should subjective probabilities be sharp?.
Thanks! That paper may be why this seemed familiar.
I like de Finetti’s line (paraphrasing) that strictly, probabilistic decision theory applies to setups fully specified in advance. ‘Fully specified’ includes (at least in principle) the information you will be given and the options you will be offered.
If you know in advance that you will be offered the choice of none, A, B or both, you should choose both. If you know that your only options will be to accept or reject A, either is permissible.
The catch comes when you don’t know what options you will be offered. Then, as you describe, you can be led astray by a sequence of decisions, each of which ‘seemed like a good idea at the time’. Such is life, and such are the limits of decision theory. Or so I’m suggesting.
Standard expectation maximization (with sharp credences) seems to avoid this issue – maximizing expectation at each step maximizes total expectation. But there is a cost: it evaluates the distribution of payoffs by a single number i.e. the expectation. Note that many of the standard puzzles in decision theory (e.g. St. Petersburg, Pascal’s Wager) involve highly skewed distributions that are not well represented by their expectations.
Does this mean we should reject 2 - specifically the claim that Both isn't better than choosing either A or B? I could see this as an example of external circumstances giving bounds to the nonmeasurability here - while either A or B don't have any probability of containing the $3 by themselves, both of them together exhaust the options connected to the NMS and so give a certainty of 1 to you having $3.
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