Well, in terms of decisions, P(green) = 1⁄3 +- 1⁄9 means that I’d buy a bet on green for the price of a true randomised bet with probability 2⁄9, and sell for 4⁄9, with the caveats mentioned.
We might say that the price of a left boot is $15 +- $5 and the price of a right boot is $15 -+ $5.
Showing that it can’t be pumped just means that it’s consistent. It doesn’t mean it’s correct. Consistently wrong choices cost utility, and are not rational.
To be clear: you mean that my choices somehow cost utility, even if they’re consistent?
I would greatly love an example that compares a plain Bayesian analysis with an interval analysis.
It’s a good idea. But at the moment I think more basic questions are in dispute.
(This argument seems to suggest a “common-sense human” position between high ambiguity aversion and no ambiguity aversion, but most of us would find that untenable.)
Well then, P(green) = 1⁄3 +- 1⁄3 would be extreme ambiguity aversion (such as would match the adversary I think you are proposing), and P(green) = 1⁄3 exactly would be no ambiguity aversion , so something like P(green) = 1⁄3 +- 1⁄9 would be such a compromise, no? And why is that untenable?
To clarify: the aversary you have in mind, what powers does it have, exactly?
Generally speaking, an adversary would affect my behaviour, unless the loss of ambiguity aversion from the fact that all probabilities are known were exactly balanced by the gain in ambiguity aversion from the fact that said probabilities are under control of a (limited) aversary.
(Which is similar to saying that finding out the true distribution from which the urn was drawn would indeed affect your behaviour, unless you happened to find that the distribution was the prior you had in mind anyway.)
Once they start paying for equivalent options, then they get money-pumped.
Okay. Suppose there is an urn with 31 red balls, and 60 balls that are either green or blue. I choose to bet on red over green, and green-or-blue over red-or-blue. These are no longer equivalent options, and this is definitely not consistent with the laws of probability. Agreed?
(My prior probability interval is P(red) = 31⁄91 exactly, P(green) = (1/2 +- 1⁄6)(60/91), P(blue) = (1/2 -+ 1⁄6)(60/91).)
It sounds like you expected (and continue to expect!) to be able to money-pump me.
you would think it’s excessive to trade (20U,0U) for just 1U.
What bet did you have in mind that was worth (20U,0) ? One of the simplest examples, if P(green) = 1⁄3 +- 1⁄9, would be 70U if green, −20U if not green. Does it still seem excessive to be neutral to that bet, and to trade it for a certain 1U (with the caveats mentioned)
What if they were in the care of her future self who already flipped the coin? Why is this different?
This I don’t understand. She is her future self isn’t she?
Bonus scenario:
Oh boy!
There are two standard Elisberg-paradox urns, each paired with a coin. You are asked to pick one to get a reward for iff ((green and heads) or (blue and tails)). At first you are indifferent, as both are identical. However, before you make your selection, one of the coins is flipped. Are you still indifferent?
So there are two urns, one coin is going to be flipped. No matter what I’m offered a randomised bet on the second urn. If the coin comes up heads I’ll be offered a bet on green on the first urn, if the coin comes up tails I’ll be offered a bet on blue on the first urn. So looks like my options are:
A) choose urn 1 either way
B) choose urn 1 (i.e. green) if the coin comes up heads, choose urn 2 if the coin comes up tails
C) choose urn 2 if the coin comes up heads, choose urn 1 (i.e. blue) if the coin comes up tails
D) choose urn 2 either way
And to be pedantic: E) flip my own coin to randomise between options B and C.
I am indifferent between A, D, and E, which I prefer to B or C.
Generally, we seem to be really overanalysing the phrase “ought to flip a coin”.
I’m not sure what you mean. If it’s because the situation was too symetrical, I think I adressed that.
For example, you could add or remove a couple of red balls. I still choose red over green, and green-or-blue over red-or-blue. I think the fact that it still can’t lead to being dutch booked is going to be a surprise to many LW readers.
Well, it would push me away from ambiguity aversion, I would become indifferent between a bet on red and a bet on green, etc.
Put it another way: a frequentist could say to you: “Your Bayesian behaviour is a perfect frequentist model of a situation where:
You choose a bet
An urn is selected uniformly at random from the fictional population
An outcome occurs.
It seems totally unreasonable to apply it in the Ellsberg situation or similar ones. For instance, you would then not react if you were in fact told the distribution.”
And actually, as it happens, this isn’t too far from the sort of things you do hear in frequentist complaints about Bayesianism. You presumably reject this frequentist argument against you.
And I reject your Bayesian argument against me.
If money has logarithmic value to you, you are not risk neutral, the way I understand the term. How are you using the term?
For example, you would choose 1U with certainty over something like 10U ± 10U. You said that you would be still make the ambiguity-adverse choice if a few red balls were taken out, but what if almost all of them were removed?
If I had set P(green) = 1⁄3 +- 1⁄3, then yes. But in this case I’m not ambiguity averse to the extreme, like I mentioned. P(green) = 1⁄3 +- 1⁄9 was what I had, i.e. (1/2 +- 1⁄6)(2/3). The tie point would be 20 red balls, i.e. 1⁄4 exactly versus (1/2 +- 1⁄6)(3/4).
On a more abstract note, your stated reasons for your decision seem to be that you actually care about what might have happened for reasons other than the possibility of it actually happening (does this make sense and accurately describe your position?).
It makes sense, but I don’t feel this really describes me. I’m not sure how to clarify. Maybe an analogy:
What Irina and Joey’s mother wants is to not intend to favour either of her children.
Maybe. Though I put it to you that the mother wants nothing more than what is “best for her children”. Even if we did agree with her about what his best for each child separately, we might still disagree with her about what is “best for her children”.
Perhaps I just want the “best chance of winning”.
(ADDED:) If it helps, I don’t think the fact that it is she making the decision is the issue—she would wish the same thing to happen if her children were in someone else’s care.
I see. My cunning reply is thus:
Suppose you were told that, rather than being from an unknown sources, the urn was in fact selected uniformly at random from 61 urns. In the first urn, there are 30 red balls, and 60 green balls. In the second urn, there are 30 red balls, 1 blue ball, and 59 green balls, etc, and in the sixty-first urn, there are 30 red balls and 60 blue balls.
This seems like pretty significant information. The kind of information that should change your behavior.
Would it change your behavior?
It seems totally unreasonable to apply it in that situation or similar ones.
You mean:
My behaviour could be explained if I were actually Bayesian, and I believed X
But I have agreed that X is false
Therefore my behaviour is unreasonable.
(Where X is the existence of an opponent with certain properties.)
Am I fairly representing what you are saying?
For instance, you would then not react to the presence of an actual adversary.
Why’s that then? If there was an adversary, I could apply game theory just like anyone else, no?
Also, I think that to formalize potential connections between different events better, you should replace intervals of probabilities with functions from a parameter space to probabilities.
You may be interested in the general case of variational preferences then. But you could also just go to a finite tuple, rather than a pair, like I briefly mentioned. That would cover a lot of cases.
But for the purposes of this post I think 2 is sufficient.
Yes, replacing the new one. I.e. given a choice between trading the bet on green for a new randomised bet, we prefer to keep the bet on green. And no, the virtual interval is not part of any bet, it is persistent.
Agreed, the structural component is not normative. But to me, it is the structural part that seems benign.
If we assume the agent lives forever, and there’s always some uncertainty, then surely the world is thus and so. If the agent doesn’t live forever, then we’re into bounded rationality questions, and even transitivity is up in the air.
P6 is really both. Structurally, it forces there to be something like a coin that we can flip as many times as we want. But normatively, we can say that if the agent has blah blah blah preference, it shall be able to name a partition such that blah blah blah. See e.g. [rule 4]. This of course doesn’t address why we think such a thing is normative, but that’s another issue.
Your definition of total pre-order:
A total preorder % satisfies the following properties: For all x, y, and z, if x % y and y % z then x % z (transitivity). For all x and y, x % y or y % x (totality). (I substituted “%” for their symbol, since markdown doesn’t translate their symbol.) Let “A %B” represent “I am indifferent between, or prefer, A to B”.
Looks to me like it’s equivalent to what I wrote for rule 1. In particular, you say:
To wit, I am indifferent between A and B, and between B and C, but I prefer A to C. This satisfies the total preorder, but violates Rule 1.
No, this violates total pre-order, as you’ve written it.
Since you are indifferent between A and B, and between B and C: A%B, B%A, B%C, C%B. By transitivity, A%C and C%A. Therefore, you are indifferent between A and C.
The “other” type of indifference, you have neither A%B nor B%A (I called this incomparability). But it violates totality.
I don’t think Rule 1 is a requirement of rationality
Hope you’ll forgive me if I set this aside. I want to grant absolutely every hypothesis to the Bayesian, except the specific thing I intend to challenge.
Hmm! I don’t know if that’s been tried. Speaking for myself, 31 red balls wouldn’t reverse my preferences.
But you could also have said, “On the other hand, if people were willing to pay a premium to choose red over green and green-or-blue over red-or-blue...” I’m quite sure things along those lines have been tried.
Heh :-) I’m okay with people being more interested in the Ellsberg paradox than the Savage theorem. Sections headers are there for skipping ahead. There’s even colour :-)
I think it would be unfair to ask me to make the Savage theorem as readable as the Ellsberg paradox. For starters, the Ellsberg paradox can be described really quickly. The Savage theorem, even redux, can’t. Second, just about everyone here agrees with the conclusion of the Savage theorem, and disagrees with the Ellsberg-paradoxical behaviour.
My goal was just to make it clearer than the previous post -- and this is not an insult against the previous author, he presented the full theorem and I presented a redux version covering only the relevant part, as I explained in the boring rationality section before the boring representation theorem. I’d be happy if some people who did not understand the previous set of axioms understood the four rules here.
As for the rest, yes, consensus here so far (only a few hours in, of course, but still impressively unanimous) seems to be that it’s a bias. Of course, in that case, it’s a very famous bias, and it hasn’t been covered on LW before. I can still claim to have accomplished something I think, no? And if it turns out it’s not so irrational after all, well!
Meh. It should not really affect what I’ve said or what I intend to say later if you substitute “Violation of the rules of probability” or “of utility” for “paradox” (Ellsberg and Allais resp.) However paradox is what they’re generally called. And it’s shorter.
I replied to Manfred with the Ellsberg example having 31 instead of 30 red balls. Does that count as different? If so, do I lose utility?