# The Ellsberg paradox and money pumps

10 28 January 2012 05:34PM

Followup to: The Savage theorem and the Ellsberg paradox

In the previous post, I presented a simple version of Savage's theorem, and I introduced the Ellsberg paradox. At the end of the post, I mentioned a strong Bayesian thesis, which can be summarised: "There is always a price to pay for leaving the Bayesian Way."1 But not always, it turns out. I claimed that there was a method that is Ellsberg-paradoxical, therefore non-Bayesian, but can't be money-pumped (or "Dutch booked"). I will present the method in this post.

I'm afraid this is another long post. There's a short summary of the method at the very end, if you want to skip the jibba jabba and get right to the specification. Before trying to money-pump it, I'd suggest reading at least the two highlighted dialogues.

## Ambiguity aversion

To recap the Ellsberg paradox: there's an urn with 30 red balls and 60 other balls that are either green or blue, in unknown proportions. Most people, when asked to choose between betting on red or on green, choose red, but, when asked to choose between betting on red-or-blue or on green-or-blue, choose green-or-blue. For some people this behaviour persists even after due calculation and reflection. This behaviour is non-Bayesian, and is the prototypical example of ambiguity aversion.

There were some major themes that came out in the comments on that post. One theme was that I Fail Technical Writing Forever. I'll try to redeem myself.

Another theme was that the setup given may be a bit too symmetrical. The Bayesian answer would be indifference, and really, you can break ties however you want. However the paradoxical preferences are typically strict, rather than just tie-breaking behaviour. (And when it's not strict, we shouldn't call it ambiguity aversion.) One suggestion was to add or remove a couple of red balls. Speaking for myself, I would still make the paradoxical choices.

A third theme was that ambiguity aversion might be a good heuristic if betting against someone who may know something you don't. Now, no such opponent was specified, and speaking for myself, I'm not inferring one when I make the paradoxical choices. Still, let me admit that it's not contrived to infer a mischievous experimenter from the Ellsberg setup. One commentator puts it better than me:

Betting generally includes an adversary who wants you to lose money so they win in. Possibly in psychology experiments [this might not apply] ... But generally, ignoring the possibility of someone wanting to win money off you when they offer you a bet is a bad idea.

Now betting is supposed to be a metaphor for options with possibly unknown results. In which case sometimes you still need to account for the possibility that the options were made available by an adversary who wants you to choose badly, but less often. And you should also account for the possibility that they were from other people who wanted you to choose well, or that the options were not determined by any intelligent being or process trying to predict your choices, so you don't need to account for an anticorrelation between your choice and the best choice. Except for your own biases.

We can take betting on the Ellsberg urn as a stand-in for various decisions under ambiguous circumstances. Ambiguity aversion can be Bayesian if we assume the right sort of correlation between the options offered and the state of the world, or the right sort of correlation between the choice made and the state of the world. In that case just about anything can Bayesian. But sometimes the opponent will not have extra information, nor extra power. There might not even be any opponent as such. If we assume there are no such correlations, then ambiguity aversion is non-Bayesian.

The final theme was: so what? Ambiguity aversion is just another cognitive bias. One commentator specifically complained that I spent too much time talking about various abstractions and not enough time talking about how ambiguity aversion could be money-pumped. I will fix that now: I claim that ambiguity aversion cannot be money-pumped, and the rest of this post is about my claim.

I'll start with a bit of name-dropping and some whig history, to make myself sound more credible than I really am2. In the last twenty years or so many models of ambiguity averse reasoning have been constructed. Choquet expected utility3 and maxmin expected utility4 were early proposed models of ambiguity aversion. Later multiplier preferences5 were the result of applying the ideas of robust control to macroeconomic models. This results in ambiguity aversion, though it was not explicitly motivated by the Ellsberg paradox. More recently, variational preferences6 generalises both multiplier preferences and maxmin expected utility. What I'm going to present is a finitary case of variational preferences, with some of my own amateur mathematical fiddling for rhetorical purposes.

## Probability intervals

The starting idea is simple enough, and may have already occurred to some LW readers. Instead of using a prior probability for events, can we not use an interval of probabilities? What should our betting behaviour be for an event with probability 50%, plus or minus 10%?

There are some different ways of filling in the details. So to be quite clear, I'm not proposing the following as the One True Probability Theory, and I am not claiming that the following is descriptive of many people's behaviour. What follows is just one way of making ambiguity aversion work, and perhaps the simplest way. This makes sense, given my aim: I should just describe a simple method that leaves the Bayesian Way, but does not pay.

Now, sometimes disjoint ambiguous events together make an event with known probability. Or even a certainty, as in an event and its negation. If we want probability intervals to be additive (and let's say that we do) then what we really want are oriented intervals. I'll use +- or -+ (pronounced: plus-or-minus, minus-or-plus) to indicate two opposite orientations. So, if P(X) = 1/2 +- 1/10, then P(not X) = 1/2 -+ 1/10, and these add up to 1 exactly.

Such oriented intervals are equivalent to ordered pairs of numbers. Sometimes it's more helpful to think of them as oriented intervals, but sometimes it's more helpful to think of them as pairs. So 1/2 +- 1/10 is the pair (3/5,2/5). And 1/2 -+ 1/10 is (2/5,3/5), the same numbers in the opposite order. The sum of these is (1,1), which is 1 exactly.

You may wonder, if we can use ordered pairs, can we use triples, or longer lists? Yes, this method can be made to work with those too. And we can still think in terms of centre, length, and orientation. The orientation can go off in all sorts of directions, instead of just two. But for my purposes, I'll just stick with two.

You might also ask, can we set P(X) = 1/2 +- 1/2? No, this method just won't handle it. A restriction of this method is that neither of the pair can be 0 or 1, except when they're both 0 or both 1. The way we will be using these intervals, 1/2 +- 1/2 would be the extreme case of ambiguity aversion. 1/2 +- 1/10 represents a lesser amount of ambiguity aversion, a sort of compromise between worst-case and average-case behaviour.

To decide among bets (having the same two outcomes), compute their probability intervals. Sometimes, the intervals will not overlap. Then it's unambiguous which is more likely, so it's clear what to pick. In general, whether they overlap or not, pick the one with the largest minimum -- though we will see there are three caveats when they do overlap. If P(X) = 1/2 +- 1/10, we would be indifferent between a bet on X and on not X: the minimum is 2/5 in either case. If P(Y) = 1/2 exactly, then we would strictly prefer a bet on Y to a bet on X.

Which leads to the first caveat: sometimes, given two options, it's strictly better to randomise. Let's suppose Y represents a fair coin. So P(Y) = 1/2 exactly, as we said. But also, Y is independent of X. P(X and Y) = 1/4 +- 1/20, and so on. This means that P((X and not Y) or (Y and not X)) = 1/2 exactly also. So we're indifferent between a bet on X and a bet on not X, but we strictly prefer the randomised bet.

In general, randomisation will be strictly better if you have two choices with overlapping intervals of opposite orientations. The best randomisation ratio will be the one that gives a bet with zero-length interval.

Now let us reconsider the Ellsberg urn. We did say the urn can be a metaphor for various situations. Generally these situations will not be symmetrical. But, even in symmetrical scenarios, we should still re-think how we apply the principle of indifference. I argue that the underlying idea is really this: if our information has a symmetry, then our decisions should have that same symmetry. If we switch green and blue, our information about the Ellsberg urn doesn't change. The situation is indistinguishable, so we should behave the same way. It follows that we should be indifferent between a bet on green and a bet on blue. Then, for the Bayesian, it follows that P(red) = P(green) = P(blue) = 1/3. Period.

But for us, there is a degree of freedom, even in this symmetrical situation. We know what the probability of red is, so of course P(red) = 1/3 exactly. But we can set, say7, P(green) = 1/3 +- 1/9, and P(blue) = 1/3 -+ 1/9. So we get P(red or green) = 2/3 +- 1/9, P(red or blue) = 2/3 -+ 1/9, P(green or blue) = 2/3 exactly, and of course P(red or green or blue) = 1 exactly.

So: red is 1/3 exactly, but the minimum of green is 2/9. (green or blue) is 2/3 exactly, but the minimum of (red or blue) is 5/9. So choose red over green, and (green or blue) over (red or blue). That's the paradoxical behaviour. Note that neither pair of choices offered in the Ellsberg paradox has the type of overlap that favours randomisation.

Once we have a decision procedure for the two-outcome case, then we can tack on any utility function, as I explained in the previous post. The result here is what you would expect: we get oriented expected utility intervals, obtained by multiplying the oriented probability intervals by the utility. When deciding, we pick the one whose interval has the largest minimum. So for example, a bet which pays 15U on red (using U for "utils", the abstract units of measurement of the utility function) has expected utility 5U exactly. A bet which pays 18U on green has expected utility 6U +- 2U, the minimum is 4U. So pick the bet on red over that.

Operationally, probability is associated with the "fair price" at which we are willing to bet. A probability interval indicates that there is no fair price. Instead we have a spread: we buy bets at their low price and sell at their high price. At least, we do that if we have no outstanding bets, or more generally, if the expected utility interval on our outstanding bets has zero-length. The second caveat is that if this interval has length, then it affects our price: we also sell bets of the same orientation at their low price, and buy bets of the opposite orientation at their high price, until the length of this interval is used up. The midpoint of the expected utility interval on our outstanding bets will be irrelevant.

This can be confusing, so it's time for an analogy.

## Bootsianism

If you are Bayesian and risk-neutral (and if bets pay in "utils" rather than cash, you are risk-neutral by definition) then outstanding bets have no effect on further betting behaviour. However, if you are risk-averse, as is the most common case, then this is no longer true. The more money you've already got on the line, the less willing you will be to bet.

But besides risk attitude, there could also be interference effects from non-monetary payouts. For example, if you are dealing in boots, then you wouldn't buy a single boot for half the price of a pair, and neither would you sell one of your boots for half the price of a pair. Unless you happened to already have unmatched boots, then you would sell those at a lower price, or buy boots of the opposite orientation at a higher price, until you had no more unmatched boots. If you were otherwise risk-neutral with respect to boots, then your behaviour would not depend on the number of pairs you have, just on the number and orientation of your unmatched boots.

This closely resembles the non-Bayesian behaviour above. In fact, for the Ellsberg urn, we could just say that a bet on red is worth a pair of boots, a bet on green is worth two left boots, and a bet on blue is worth two right boots. Without saying anything further, it's clear that we would strictly prefer red (a pair) over green (two lefts), but we would also strictly prefer green-or-blue (two pairs) over red-or-blue (one left and three rights). That's the paradoxical behaviour, but you know you can't money-pump boots.

 A: I'll buy that pair of boots for 30 zorkmids. B: Okay, here's your pair of boots. A: And here's your 30 zorkmids. Thank you. B: Thank you. Say, didn't you just buy an identical pair this morning? A: Yeah, I did. Then a dingo ate the right one. I've got the left one here. Never worn. B: How narratively convenient! How much would you sell it for? A: Hmm, how about 10 zorkmids? B: Really, 10 zorkmids? So, do you think right boots are more valuable than left boots? A: No, of course not. Why? B: Arbitrage! A: Gesundheit. B: Thanks. I'll buy a left boot from you for 10 zorkmids. A: Great! Here's your left boot. B: And here's your 10 zorkmids. Thank you. A: Thank you! B: And I'll buy a right boot from you for 10 zorkmids. A: Errrm... Sorry? Why would I agree to that? B: You just sold me a left boot for 10 zorkmids. Well, you yourself said rights aren't more valuable than lefts. So, logically, you should be willing to sell me a right boot for 10 zorkmids. A: What? No.

## Boots' rule

So much for the static case. But what do we do with new information? How do we handle conditional probabilities?

We still get P(A|B) by dividing P(A and B) by P(B). It will be easier to think in terms of pairs here. So for example P(red) = 1/3 exactly = (1/3,1/3) and P(red or green) = 2/3 +- 1/9 = (7/9,5/9), so P(red|red or green) = (3/7,3/5) = 18/35 -+ 3/35. And similarly P(green|red or green) = (1/3 +- 1/9)/(2/3 +- 1/9) = 17/35 +- 3/35.

This rule covers the dynamic passive case, where we update probabilities based on what we observe, before betting. The third and final caveat is in the active case, when information comes in between bets. Now, we saw that the length and orientation of the interval on expected utility of outstanding bets affects further betting behaviour. There is actually a separate update rule for this quantity. It is about as simple as it gets: do nothing. The interval can change when we make choices, and its midpoint can shift due to external events, but its length and orientation do not update.

You might expect the update rule for this quantity to follow from the way the expected utility updates, which follows from the way probability updates. But it has a mind of its own. So even if we are keeping track of our bets, we'd still need to keep track of this extra variable separately.

Sometimes it may be easier to think in terms of the total expected utility interval of our outstanding bets, but sometimes it may be easier to think of this in terms of having a "virtual" interval that cancels the change in the length and orientation of the "real" expected utility interval. The midpoint of this virtual interval is irrelevant and can be taken to always be zero. So, on update, compute the prior expected utility interval of outstanding bets, subtract the posterior expected utility interval from it, and add this difference to the virtual interval. Reset its midpoint to zero, keeping only the length and orientation.

That can also be confusing, so let's have another analogy.

## Yo' mama's so illogical...

I recently came across this example by Mark Machina:

 M: Children, I only have one treat, I can only give it to one of you. I: Me, mama! J: No, give it to me! M: No. Rather than give it to either of you, it's better if I toss a coin. Heads, it goes to Irina, tails, it goes to Joey. ... M: Heads. Irina gets it. J: But mama! M: Fair is fair. I: Yeah Joey! J: But mama, you yourself said it's better to toss a coin than to give it to either of us. So, logically, instead of giving it to Irina you should toss a coin again. M: Nice try, Joey.

Instead of giving the treat to either child, she strictly prefers to toss a coin and give the treat to the winner. But after the coin is tossed, she strictly prefers to give the treat to the winner rather than toss again.

This cannot be explained in terms of maximising expected utility, in the typical sense of "utility". And of course only known probabilities are involved here, so there's no question as to whether her beliefs are probabilistically sophisticated or not. But it could be said that she is still maximising the expected value of an extended objective function. This extended objective function does not just consider who gets a treat, but also considers who "had a fair chance". She is unfair if she gives the treat to either child outright, but fair if she tosses a coin. That fairness doesn't go away when the result of the coin toss is known.

Or something like that. There are surely other ways of dissecting the mother's behaviour. But no matter what, it's going to have to take the coin toss into account, even though the coin, in and of itself, has no relevance to the situation.

Let's go back to the urn. Green and blue have the type of overlap that favours randomisation: P((green and heads) or (blue and tails)) = 1/3 exactly. A bet paying 9U on this event has expected utility of 3U exactly. Let's say we took this bet. Now say the coin comes up heads. We can update the probabilities as per above. The answer is that P(green) = 1/3 +- 1/9 as it was before. That makes sense because it's an independent event: knowing the result of the coin toss gives no information about the urn. The difference is that we now have an outstanding bet that pays 9U if the ball is green. The expected utility would therefore be 3U +- 1U. Except, the expected utility interval was zero-length before the coin was tossed, so it remains zero-length. Equivalently, the virtual interval becomes -+ 1U, so that the effective total is 3U exactly. (In this example, the midpoint of the expected utility interval didn't change either. That's not generally the case.) A bet randomised on a new coin toss would have expected utility 3U, plus the virtual interval of -+ 1U, for an effective total of 3U -+ 1U. So we would strictly prefer to keep the bet on green rather than re-randomise.

Let's compare this with a trivial example: let's say we took a bet that pays 9U if the ball drawn from the urn is green. The expected utility of this bet is 3U +- 1U. For some unrelated reason, a coin is tossed, and it comes up heads. The coin has also nothing to do with the urn or my bet. I still have a bet of 9U on green, and its expected utility is still 3U +- 1U.

But the difference between these two examples is just in the counterfactual: if the coin had come up tails, in the first example I would have had a bet of 9U on blue, and in the second example I would have had a bet of 9U on green. But the coin came up heads, and in both examples I end up with a bet of 9U on green. The virtual interval has some spooky dependency on what could have happened, just like "had a fair chance". It is the ghost of a departed bet.

I expect many on LW are wondering what happened. There was supposed to be a proof that anything that isn't Bayesian can be punished. Actually, this threat comes with some hidden assumptions, which I hope these analogies have helped to illustrate. A boot is an example of something which has no fair price, even if a pair of boots has one. A mother with two children and one treat is an example where some counterfactuals are not forgotten. The hidden assumptions fail in our case, just as they can fail in these other contexts where Bayesianism is not at issue. This can be stated more rigorously8, but that is basically how it's possible. Now We Know. And Knowing is Half the Battle.

## Notes

1. Taken almost verbatim from Eliezer Yudkowsky's post on the Allais paradox.
2. And footnotes pointing to some tangentially relevant journal articles make me sound extra credible.
3. For Choquet expected utility see: D. Schmeidler, Subjective probability and expected utility without additivity, Econometrica 57 (1989) pp 571-587.
4. For maxmin expected utility see: I. Gilboa and D. Schmeidler, Maxmin expected utility with a non-unique prior, J. Math. Econ. 18 (1989) pp 141-153.
5. For multiplier preferences see: L.P. Hansen and T.J. Sargeant, Robust control and model uncertainty, Amer. Econ. Rev. 91 (2001) pp 60-66.
6. For variational preferences see: F. Maccheroni, M. Marinacci, and A. Rustichini, Dynamic variational preferences, J. Econ. Theory 128 (2006) pp 4-44.
7. Any length between 0 and 1/3 works. But here's where I pulled 1/9 from: a Bayesian might assign exactly 1/61 prior probability to the 61 possible urn compositions, and the result is roughly approximated by the Laplacian rule of succession, which prescribes a pseudocount of one green and one blue ball. A similar thing with probability intervals is roughly approximated by using a pseudocount of 3/2 +- 1/2 green and 3/2 -+ 1/2 blue balls.
8. To quickly relate this back to Savage's rules: rules 1 and 3 guarantee that there's no static money pump. Rule 2 then is supposed to guarantee that there is no dynamic money pump. But it is stronger than necessary for that purpose. I claim that this method obeys rules 1, 3, and a weaker version of rule 2, and that it is dynamically consistent. For dynamic consistency of variational preferences in general, see footnotes above. This method is a special case, for which I wrote up a simpler proof.

## Appendix A: method summary

 Events are assigned a pair of prior probabilities, which can also be thought of as an oriented probability interval. e.g. (3/5,2/5) can also be thought of as 1/2 +- 1/10. Neither side of the pair can be 0 or 1, except when they're both 0 or both 1. Each side of the pair is additive: if A and B are disjoint, and P(A) = (x,y), and P(B) = (u,v), then P(A or B) = (x+u,y+v). Each side of the pair updates by Bayes' rule: if P(A and B) = (x,y), and P(B) = (u,v), then P(A|B) = (x/u,y/v). Given a utility function, each bet will then have an expected utility interval: multiply the probability intervals by the utility for each possible outcome. There is also a virtual expected utility interval to keep track of. The midpoint of this interval is always zero. When updating the virtual expected utility interval, compute the prior expected utility interval of the outstanding bet(s), subtract the posterior expected utility interval from it, and add this difference to the virtual expected utility interval. Throw away the midpoint (reset the midpoint of the interval to zero, keeping just the length and orientation). To decide among bets: compute the expected utility intervals of each of them -- including already outstanding bets, and including the virtual expected utility interval. Rank them according to the minimum values of the intervals. Implicitly when presented with options we are also presented with the option to randomise among them, and sometimes this is strictly better than any of the pure options.

## Appendix B: obligatory image for LW posts on this topic

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Comment author: 29 January 2012 08:04:16PM 3 points [-]

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.

I couldn't tell what position you're taking. Is ambiguity aversion something more than just a common bias? Personally, I understand the intuitive appeal, but on reflection I am indifferent to a bet on red-or-blue or on green-or-blue. If I have additional information about the distribution of green and blue (including knowing anything about who's offering the wagers, or in some cases, that the wager is offered at all), then I'll update my prior of equal probability of blue or green.

Am I doing it wrong? should I prefer one or the other? How much should I prefer it (how much, as a fraction of the bet, should I pay to be allowed to make the bet I prefer)?

For your other examples, boots' rule and yo mama, both are trivial cases of decomposing "value" too far. There's nothing inconsistent about a mother preferring to have a visible fairness mechanism, and to follow that mechanism when it is present. There's nothing wrong with valuing left and right boots differently based on your expected needs. Neither of these have ANYTHING to do with probability calculation, they're just a way to point out that value is not simple. Neither of these present any difficulty to standard decision theory.

I can't tell as of yet whether intervals are helpful in calculating something (and if so, what) that always gets the same answer as the simpler (to me) Bayesian calculation, or if they're able to make different (and better) beliefs in some case.

I would greatly love an example that compares a plain Bayesian analysis with an interval analysis. Start with a prior and a prior-interval, update both based on some discovery, and then propose a wager where the two methods give different beliefs.

Comment author: 30 January 2012 10:12:47PM 0 points [-]

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.

Comment author: 06 February 2012 01:15:14PM 0 points [-]

Instead of using a prior probability for events, can we not use an interval of probabilities?

Intervals of probability seem to reduce to probability if you consider the origin of the interval. Suppose in the Ellsberg paradox that the proportion of blue and green balls was determined, initially, by a coin flip (or series of coin flips). In this view, there is no ambiguity at all, just classical probabilities - so you seem to posit some distinction based on how something was setup. Where do you draw the line; when does something become genuinely ambiguous?

The boots and the mother example can all be dealt with using standard Bayesian techniques (you take utility over worlds, and worlds with one boot are not very valuable, worlds with two are; and the memories of the kids are relevant to their happiness), and you can re-express what is intuitively an "interval of probability" as a Bayesian behaviour over multiple, non-independent bets.

To be clear: you mean that my choices somehow cost utility, even if they're consistent?

You would pay to remove ambiguity. And ambiguity removal doesn't increase expected utility, so Bayesian agents would outperform you in situations where some agents had ambiguity-reducing knowledge.

Comment author: 08 February 2012 02:06:15AM *  0 points [-]

Suppose in the Ellsberg paradox that the proportion of blue and green balls was determined, initially, by a coin flip (or series of coin flips). In this view, there is no ambiguity at all, just classical probabilities

Correct.

Where do you draw the line

1) I have no reason to think A is more likely than B and I have no reason to think B is more likely than A

2) I have good reason to think A is as likely as B.

These are different of course. I argue the difference matters.

The boots and the mother example can all be dealt with using standard Bayesian techniques

Correct. See last paragraph of the post.

You would pay to remove ambiguity. And ambiguity removal doesn't increase expected utility, so Bayesian agents would outperform you in situations where some agents had ambiguity-reducing knowledge.

If you mean something like: red has probability 1/3, and green has probability 1/3 "on average", then I dispute "on average" -- that is circular.

The advantage of a money pump or "Dutch book" argument is that you don't need such assumptions to show that the behaviour in question is suboptimal. (Un)fortunately there is a gap between Bayesian reasoning and what money pump arguments can justify.

(Incidentally, if you determine the contents of the urn by flipping a coin for each of the 60 balls to determine whether it is green or blue, then this matters to the Bayesian too -- this gives you the binomial prior, whereas I think most Bayesians would want to use the uniform prior by default. Doesn't affect the first draw, but it would affect multiple draws.)

Comment author: 09 February 2012 12:32:36PM 0 points [-]

Incidentally, if you determine the contents of the urn by flipping a coin for each of the 60 balls to determine whether it is green or blue, then this matters to the Bayesian too -- this gives you the binomial prior, whereas I think most Bayesians would want to use the uniform prior by default. Doesn't affect the first draw, but it would affect multiple draws

But it still remains that in a many circumstances (such as single draws in this setup), there exists information that a Bayesian will find useless and an ambiguity-averter will find valuable. If agents have the opportunity to sell this information, the Bayesian will get a free bonus.

From a more financial persepective, the ambiguity-averter gives up the opportunity to be a market-maker: a Bayesian can quote a price and be willing to either buy and sell at that price (plus a small fee), wherease the ambiguity-averter's required spread is pushed up by the ambiguity (so all other agents will shop with the Bayesian).

Also, the ambiguity-averter has to keep track of more connected trades than a Bayesian does. Yes, for shoes, whether other deals are offered becomes relevant; but trades that are truly independent of each other (in utility terms) can be treated so by a Bayesian but not by an ambiguity-averter.

Comment author: 10 February 2012 03:21:46AM *  1 point [-]

But it still remains that in a many circumstances (such as single draws in this setup), there exists information that a Bayesian will find useless and an ambiguity-averter will find valuable. If agents have the opportunity to sell this information, the Bayesian will get a free bonus.

How does this work, then? Can you justify that the bonus is free without circularity?

From a more financial persepective, the ambiguity-averter gives up the opportunity to be a market-maker: a Bayesian can quote a price and be willing to either buy and sell at that price (plus a small fee), wherease the ambiguity-averter's required spread is pushed up by the ambiguity (so all other agents will shop with the Bayesian).

Sure. There may be circularity concerns here as well though. Also, if one expects there to be a market for something, that should be accounted for. In the extreme case, I have no inherent use for cash, my utility consists entirely in the expected market.

Also, the ambiguity-averter has to keep track of more connected trades than a Bayesian does. Yes, for shoes, whether other deals are offered becomes relevant; but trades that are truly independent of each other (in utility terms) can be treated so by a Bayesian but not by an ambiguity-averter.

I also gave the example of risk-aversion though. If trades pay in cash, risk-averse Bayesians can't totally separate them either. But generally I won't dispute that the ideal use of this method is more complex than the ideal Bayesian reasoner.

Comment author: 11 February 2012 10:22:40AM 0 points [-]

I wonder if you can express your result in a simpler fashion... Model your agent as a combination of a buying agent and a selling agent. The buying agent will always pay less than a Bayesian, the selling agent will always sell for more. Hence (a bit of hand waving here) the combined agent will never lose money to a money pump. The problem is that it won't pick up 'free' money.

Comment author: 10 February 2012 02:17:01PM 0 points [-]

How does this work, then? Can you justify that the bonus is free without circularity?

For two agents, I can.

Imagine a setup with two agents, otherwise identical, except that one owns a 1/2+-1/4 bet and the other owns 1/2. A government agency wishes to promote trade, and so will offer 0.1 to any agents that do trade (a one-off gift).

If the two agents are Bayesian, they will trade; if they are ambiguity averse, they won't. So the final setup is strictly identical to the start one (two identical agents, one owning 1/2+- 1/4, one owning 1/2) except that the Bayesian are each 0.1 richer.

Comment author: 12 February 2012 05:29:16PM 1 point [-]

Right, except this doesn't seem to have anything to do with ambiguity aversion.

Imagine that one agent owns \$100 and the other owns a rock. A government agency wishes to promote trade, and so will offer \$10 to any agents that do trade (a one-off gift). If the two agents believe that a rock is worth more than \$90, they will trade; if they don't, they won't, etc etc

Comment author: 13 February 2012 11:58:24AM 1 point [-]

But it has everything to do with ambiguity aversion: the trade only fails because of it. If we reach into the system, and remove ambiguity aversion for this one situation, then we end up unarguably better (because of the symmetry).

Yes, sometimes the subsidy will be so high that even the ambiguity averse will trade, or sometimes so low that even Bayesians won't trade; but there will always be a middle ground where Bayesians win.

As I said elsewhere, ambiguity aversion seems like the combination of an agent who will always buy below the price a Bayesian would pay, and another who will always sell above the price a Bayesian would pay. Seen like that, your case that they cannot be arbitraged is plausible. But a rock cannot be arbitraged either, so that's not sufficient.

This example hits the ambiguity averter exactly where it hurts, exploiting the fact that there are deals they will not undertake either as buyer or seller.

Comment author: 23 February 2012 08:11:27PM 2 points [-]

No, (un)fortunately it is not so.

I say this has nothing to do with ambiguity aversion, because we can replace (1/2, 1/2+-1/4, 1/10) with all sorts of things which don't involve uncertainty. We can make anyone "leave money on the table". In my previous message, using (\$100, a rock, \$10), I "proved" that a rock ought to be worth at least \$90.

If this is still unclear, then I offer your example back to you with one minor change: the trading incentive is still 1/10, and one agent still has 1/2+-1/4, but instead the other agent has 1/4. The Bayesian agent holding 1/2+-1/4 thinks it's worth more than 1/4 plus 1/10, so it refuses to trade. Whereas the ambiguity averse agents are under no such illustion.

So, the boot's on the other foot: we trade, and you don't. If your example was correct, then mine would be too. But presumably you don't agree that you are "leaving money on the table".

Comment author: 30 January 2012 10:49:18PM 0 points [-]

Yes. I mean that, when your choice is different from what standard (or for some cases, timeless) decision theory calculates for the same prior beliefs and outcome->utility mapping, you're losing utility. I can't tell if you think that this theory does have different outcomes, or if you think that this is "just" a simplification that gives the same outcomes.

Comment author: 30 January 2012 10:56:34PM 0 points [-]

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?

Comment author: 31 January 2012 07:36:27PM 0 points [-]

From Manfred's comments (with which I agree), it looks like yes, you lose utility by failing to buy a bet that has positive EV. You lose half as much if you flip a coin, because sometimes the coin is right...

Comment author: 28 January 2012 11:01:34PM 2 points [-]

The Ellsberg paradox is boring because it can have no possible effect on utility or the log score of your beliefs (note that this means it should not be conflated with the mother with the candy). This was, in fact, the main criticism of the previous post. Given that, the fact that it can't lead to being dutch booked is trivial. I would rather this post was a lot shorter and had fewer generalizations.

Comment author: 29 January 2012 03:41:54PM 1 point [-]

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.

Comment author: 29 January 2012 06:55:34PM 1 point [-]

I agree, I don't think there is any way to dutch-book someone for being wrong but consistent with the laws of probability (that is, still assigning 1/3 probabilities to r,g,b even when that's wrong). They simply lose money on average. But this is an extra fact, unrelated to the triviality that is not being able to dutch-book someone based on an arbitrary choice between two equivalent options. Once they start paying for equivalent options, then they get money-pumped.

Comment author: 30 January 2012 09:49:19PM 0 points [-]

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.

Comment author: 30 January 2012 10:56:45PM *  0 points [-]

I'm confused what your notation means. Let's drop the asymmetry for now and just focus on the fact that you appear to be violating the laws of probability. Does your (1/2 +- 1/6) notation mean that if I would give you a dollar if you drew a green ball, you would be willing to pay 1/3 of a dollar for that bet (bet 1)? Ditto for red (bet 2)? But then if you paid me a dollar if the ball came up (green-or-red), you would be willing to accept 1/2 of a dollar for that bet (bet 3)?

In that case, the dutch book consists of bets like (bet 1) + (bet 2) + (bet 3): you pay me 1/3, you pay me 1/3, I pay you 1/2 (so you paid me 1/6th of a dollar total). Then if the ball's green I pay you a dollar, if it's red I pay you a dollar, and if it's (green-or-red) you pay me a dollar.

Comment author: 30 January 2012 11:04:50PM *  1 point [-]

If the bet pays \$273 if I drew a red ball, I'd buy or sell that bet for \$93. For green, I'd buy that bet for \$60 and sell it for \$120. For red-or-green, I would buy that for \$153 and sell it for \$213. Same for blue and red-or-blue. For green-or-blue, I'd buy or sell that for \$180.

(Appendix A has an exact specification, and you may wish to (re-)read the boot dialogue.)

[ADDED: sorry, I missed "let's drop the asymmetry" .. then, if the bet pays \$9 on red, buy or sell for \$3; green, buy \$2 sell \$4; red-or-green, buy \$5 sell \$7; blue, red-or-blue same, green-or-blue, buy or sell \$6. Assuming risk neutrality for \$, etc etc no purchase necessary must be over 18 void in Quebec.]

Comment author: 31 January 2012 02:11:31AM 0 points [-]

Ah, I see. But now you'll get type 2 dutch booked - you'll pass up on certain money if someone offers you a winning bet that requires you to buy.

Comment author: 31 January 2012 07:27:59PM *  0 points [-]

I guess you mean: you offer me a bet on green for \$2.50 and a bet on blue for \$2.50, and I'd refuse either. But I'd take both, which would be a bet on green-or-blue for \$5. So no, no dutch book here either.

Or do you have something else in mind?

Comment author: 31 January 2012 08:13:37PM *  0 points [-]

I mean that I could offer you \$9 on green for 2.50, \$9 on blue for 2.50, and \$9 on red for 3.01, and you wouldn't take any of those bets, despite, in total, having a certainty of making 99 cents. This "type 2" dutch book argument (not really a dutch book, but it's showing a similar thing for the same reasons) is based on the principle that if you're passing up free money, you're doing something wrong :P

Comment author: 31 January 2012 10:14:28PM 1 point [-]

I wouldn't take any of them individually, but I would take green and blue together. Why would you take the red bet in this case?

Comment author: 31 January 2012 10:10:59PM *  0 points [-]

I wouldn't take any of them individually (except red), but I'd take all of them together. Why is that not allowed?

Comment author: 29 January 2012 07:21:17PM 0 points [-]

I think the fact that it still can't lead to being dutch booked is going to be a surprise to many LW readers.

I would not make this prediction. I would think anyone who would understand that claim without having to look up "dutch book" should find that obvious.

Comment author: 28 January 2012 08:57:01PM *  1 point [-]

This is a perfect game theoretic model of a situation where:

1. You choose a bet.
2. Your opponent chooses parameters that influence the chance of different outcomes.
3. An outcome occurs.

It seems totally unreasonable to apply it in that situation or similar ones. It seems unreasonable to apply it in other situations. For instance, you would then not react to the presence of an actual adversary.

(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.)

Comment author: 28 January 2012 09:51:21PM *  0 points [-]

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.

Comment author: 28 January 2012 11:26:51PM 0 points [-]

My cunning argument is thus:

Suppose you were told that, rather than being unknown, the frequencies had not yet been decided, but would instead be chosen by a nefarious adversary after you made your bet.

This seems like pretty significant information. The kind of information that should change your behavior.

Comment author: 28 January 2012 11:59:45PM 0 points [-]

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.

Comment author: 29 January 2012 07:53:06AM 1 point [-]

In which direction should it change my behavior? What does it push me towards?

Comment author: 29 January 2012 03:38:54PM *  2 points [-]

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:

1. You choose a bet

2. An urn is selected uniformly at random from the fictional population

3. 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.

Comment author: 29 January 2012 08:10:00PM 0 points [-]

There seems to be an issue of magnitude here. There are 3 possible ways the urn can be filled:

1. It could be selected uniformly at random
2. It could be selected through some unknown process: uniformly at random, biased against me, biased towards blue, biased towards green, always exactly 30/30, etc.
3. It could be selected so as to exactly minimize my profits

2 seems a lot more like 1 than it does like 3. Even without using any Bayesian reasoning, a range is a lot more like the middle of the range than it is like one end of the range.

(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.)

An alternative way of talking about it:

The point I am making is that it is much more clear which direction my new information is supposed to influence you then your information is supposed to influence me. If a variable x is in the range [0,1], finding out that it is actually 0 is very strongly biasing information. For instance, almost every value x could have been before is strictly higher than the new known value. But finding out that it is 1/2 does not have a clear direction of bias. Maybe it should make you switch to more confidently betting x is high, maybe it should make you switch to more confidently betting x is low. I don't know, it depends on details of the case, and is not very robust to slight changes in the situation.

Comment author: 30 January 2012 09:57:57PM 1 point [-]

(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.)

Comment author: 30 January 2012 10:06:21PM 0 points [-]

I don't get what this range signifies. There should be a data point about how ambiguous it is, which you could use or not use to influence actions. (For instance, if someone says they looekd in the urn and it seemed about even, that reduced ambiguity.) But then you want to convert that into a range, which does not refer to the actual range of frequencies (which could be 1/3 +- 1/3) and is dependent on your degree of aversion, but then you want to convert that into a decision?

Comment author: 30 January 2012 10:27:51PM 0 points [-]

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.

Comment author: 06 February 2012 12:56:59PM 0 points [-]

Instead we have a spread: we buy bets at their low price and sell at their high price.

How does the central limit theorem apply here? If you have a lot of independent deals of the form 1/2 +- 1/4, say, then when confronted with a million such bets, their value is very close to 1/2. I don't see how your model deals with these kinds of situations - randomising between +- 1/4 and -+ 1/4, maybe? Seems clumsy.

Comment author: 08 February 2012 02:02:04AM 0 points [-]

If you mean repeated draws from the same urn, then they'd all have the same orientation. If you mean draws from different unrelated urns, then you'd need to add dimensions. It wouldn't converge the way I think you're suggesting.

Comment author: 09 February 2012 11:49:06AM 0 points [-]

The ratio of risk to return goes down with many independent draws (variances add, but standard deviations don't). It's one of the reasons investors are keen to diversify over uncorrelated investments (though again, risk-avoidance is enough to explain that behaviour in Bayesian framework).

Comment author: 03 February 2012 02:10:15AM 0 points [-]

This oriented-probability-interval stuff does seem to perform as advertised. But I just want to point to another, in my opinion simpler, way to rationally refuse to play Savage-style expected utility games. The simple way contests the axioms dealing with preference, rather than those dealing with probability. If some options are incomparable, Savage's argument fails. (Of course if an agent is forced to choose between incomparable options, it will choose, but that doesn't mean it considers one of the options "better" in a straightforward way, nor that a classical utility function can be derived.)

What if Principle of Indifference-inspired probability theories can actually be made to work? Would you end your defiance of classical utility theory?

Upvoted, first and foremost for appendix B.

Comment author: 04 February 2012 07:04:07PM 1 point [-]

Here's an alternate interpretation of this method:

If two events have probability intervals that don't overlap, or they overlap but they have the same orientation and neither contains the other, then I'll say that one event is unambiguously more likely than the other. If two events have the exact same probability intervals (including orientation), then I'll say that are equally likely. Otherwise they are incomparable.

Under this interpretation, I claim that I do obey rule 2 (see prev post): if A is unambiguously more likely than B, then (A but not B) is unambiguously more likely than (B but not A), and conversely. I still obey rule 3: (A or B) is either unambiguously more likely than A, or they are equally likely. I also claim I still obey rule 4. Finally, I claim "unambiguously more likely" is transitive, but it is not total: there are incomparable events. So I break that part of rule 1.

Passing to utility, I'll also have "unambiguously better", "equally good", and "incomparable".

Of course if an agent is forced to choose between incomparable options, it will choose, but that doesn't mean it considers one of the options "better" in a straightforward way,

Exactly. But there's a major catch: unlike with equal choices, an agent cannot choose arbitrarily between incomparable choices. This is because incomparability is intransitive. If the agent doesn't resolve ambiguities coherently, it can get money pumped. For instance, an 18U bet on green and a 15U bet on red are incomparable. Say it picks red. A 15U bet on green and a 15U bet on red are also incomparable. Say it picks green. But then, an 18U bet on green is unambiguously better than a 15U bet on green.

The rest of the post is then about one method to resolve imcomparability coherently.

I personally think this interpretation is more natural. I also think it will be even less palatable to most LW readers.

Comment author: 29 January 2012 07:44:54AM 0 points [-]

Switching between utils and dollars is very confusing. You should be very clear when you do so, or you risk getting confused between cases of risk aversion with simple declining marginal utility.

If money has logarithmic value to you, then existing unresolved bets SHOULD have an effect on your preferences even if you're risk-neutral.

Comment author: 29 January 2012 03:29:20PM 1 point [-]

If money has logarithmic value to you, you are not risk neutral, the way I understand the term. How are you using the term?

Comment author: 29 January 2012 07:12:13PM 1 point [-]

Hmm. I think you're right: I've never connected the terms in that way, using "risk-neutral" in tems of utility rather than money. Looking at it more closely, it appears it's more commonly used for money, which would be risk-seeking in terms of utility, and probably non-optimal. (note: I also recognize that most people, including me, over-estimate the decline massively, and for small wagers it should be very close to linear).

Comment author: [deleted] 29 January 2012 10:49:03PM 4 points [-]

The standard meaning of “risk-{adverse, neutral, seeking} in terms of X”, AFAICT, is that your utility is a {concave downward, linear, concave upward} function of X, and hence you cannot be risk-adverse or risk-seeking in terms of utility (assuming the von Neumann-Morgenstern axioms).

Comment author: 28 January 2012 07:44:41PM 0 points [-]

A bet randomised on a new coin toss would have expected utility 3U, plus the virtual interval of -+ 1U, for an effective total of 3U -+ 1U. So we would strictly prefer to keep the bet on green rather than re-randomise.

I'm don't understand what is happening here. Why is the virtual interval being added as a term in the expected utility, but the first bet is not? Is the new bet replacing the old one? If so, shouldn't the virtual interval be considered part of the old bet and also get removed?

Comment author: 28 January 2012 08:16:21PM 1 point [-]

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.

Comment author: 28 January 2012 11:33:53PM *  0 points [-]

OK, now I understand why this is a necessary part of the framework.

I do think there is a problem with strictly choosing the lesser of the two utilities. 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?

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?). I don't think humans actually care about such things. Probability is in the mind; a difference in what might have happened is a difference in states of knowledge about states of knowledge. A sentence like "I know now that my irresponsible actions could have resulted in injuries or deaths" isn't actually true given determinism, it's about what you know believe you should have known in the past. [1] [2]

Getting back to the topic, people's desires about counterfactuals are desires about their own minds. What Irina and Joey's mother wants is to not intend to favour either of her children. [3] In reality, the coin is just as determininstic as her decision. Her preference for randomness is about her mind, not reality.

[1] True randomness like that postulated by some interpretations of QM is different and I'm not saying that people absolutely couldn't have preferences about truly random couterfactuals. Such a world would have to be pretty weird though. It would have to be timeful, for instance, since the randomness would have to be fundamentally indeterminite before it happens, rather than just not known yet, and timeful physics doesn't even make sense to me.

[2] This is itself a counterfactual, but that's irrelevent for this context.

[3] Well, my model of her prefers flipping a coin to drawing green or blue balls from an urn, but my model of her does not agree with me on a lot of things. If she were a Bayesian decision theorist, I would expect her to be indifferent between the coin and the urn, but prefer either to having to choose for herself.

Comment author: 29 January 2012 12:17:53AM *  1 point [-]

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.

Comment author: 29 January 2012 01:27:03AM *  1 point [-]

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).

Well utility is invariant under positive affine transformations, so you could have 30U +- 10U and shift the origin so you have 10U +- 10U. More intuitively, if you have 30U +- 10U, you can regard this as 20U + (20U,0U) and you would be willing to trade this for 21U, but you're guaranteed the first 20U and you would think it's excessive to trade (20U,0U) for just 1U.

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".

Interesting.

(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.

What if they were in the care of her future self who already flipped the coin? Why is this different?

Bonus scenario: 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?

Comment author: 29 January 2012 03:54:54PM *  0 points [-]

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".

Comment author: 06 February 2012 05:03:42AM 0 points [-]

Huh, my explanations in that last post were really bad. I may have used a level of detail calibrated for simpler points, or I may have just not given enough thought to my level of detail in the first place.

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 I told you that the balls were either all green or all blue? Would you regard that as (20U,0U) (that was basically the bet I was imagining but, on reflection, it is not obvious that you would assign it that expected utility)? Would you think it equivalent to the (20U,0U) bet you mentioned and not preferrable to 1U?

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 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.

So in the standard Ellisberg paradox, you wouldn't act nonbayesianally if you were told "The reason I'm asking you to choose between red and green rather than red and blue is because of a coin flip.", but you'd still prefer red if all three options were allowed? I guess that is at least consistent.

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?

This is getting at a similar idea as the last one. What seems like the same option, like green or Irina, becomes more valuable when there is an interval due to a random event, even though the random event has already occurred and the result is now known with certainty. This seems to be going against the whole idea of probability being about mental states; even though the uncertainty has been resolved, its status as 'random' still matters.

Comment author: 08 February 2012 02:53:23AM *  1 point [-]

What if I told you that the balls were either all green or all blue?

Hmm. Well, with the interval prior I had in mind (footnote 7), this would result in very high (but not complete) ambiguity. My guess is that's a limitation of two dimensions -- it'll handle updating on draws from the urn but not "internals" like that. But I'm guessing. (1/2 +- 1/6) seems like a reasonable prior interval for a structureless event.

So in the standard Ellisberg paradox, you wouldn't act nonbayesianally if you were told "The reason I'm asking you to choose between red and green rather than red and blue is because of a coin flip."

If I take the statement at face value, sure.

but you'd still prefer red if all three options were allowed?

Yes, but again I could flip a coin to decide between green and blue then.

This seems to be going against the whole idea of probability being about mental states;

Well, okay. I don't think this method has any metaphysical consequences, so I should be able to adopt your stance on probability. I'd say (for the sake of argument) that the probability intervals are still about the mental states that I think you mean. However these mental states still leave the correct course of action underdetermined, and the virtual interval represents one degree of freedom. There is no rule for selecting the prior virtual interval. 0 is the obvious value, but any initial value is still dynamically consistent.

Comment author: 08 February 2012 05:28:14AM 0 points [-]

My guess is that's a limitation of two dimensions -- it'll handle updating on draws from the urn but not "internals" like that. But I'm guessing. (1/2 +- 1/6) seems like a reasonable prior interval for a structureless event.

Would a single ball that is either green or blue work?

0 is the obvious value, but any initial value is still dynamically consistent.

I agree that your decision procedure is consistent, not susceptible to Dutch books, etc.

I don't think this method has any metaphysical consequences, so I should be able to adopt your stance on probability. I'd say (for the sake of argument) that the probability intervals are still about the mental states that I think you mean.

I don't think this is true. Whether or not you flip the coin, you have the same information about the number of green balls in the urn, so, while the total information is different, the part about the green balls is the same. In order to follow your decision algorithm while believing that probability is about incomplete information, you have to always use all your knowledge in decisions, even knowledge that, like the coin flip, is 'uncorrelated', if I can use that word for something that isn't being assigned a probability, with what you are betting on. This is consistent with the letter of what I wrote, but I think that a bet that is about whether a green ball will be drawn next should use your knowledge about the number of green balls in the urn, not your entire mental state.

Comment author: 10 February 2012 02:55:21AM 1 point [-]

Would a single ball that is either green or blue work?

That still seems like a structureless event. No abstract example comes to mind, but there must be concrete cases where Bayesians disagree wildly about the prior probability of an event (<5% vs >95%). Some of these cases should be candidates for very high (but not complete) ambiguity.

I think that a bet that is about whether a green ball will be drawn next should use your knowledge about the number of green balls in the urn, not your entire mental state.

I think you're really saying two things: the correct decision is a function of (present, relevant) probability, and probability is in the mind. I'd say the former is your key proposition. It would be sufficient to rule out that an agent's internal variables, like the virtual interval, could have any effect. I'd say the metaphysical status of probability is a red herring (but I would also accept a 50:50 green-blue herring).

Of course even for Bayesians there are equiprobable options, so decisions can't be entirely a function of probability. More precisely, your key proposition is that probabilistic indecision has to be transitive. Ambiguity would be an example of intransitive indecision.