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The von Neumann-Morgenstern theorem states that, assuming a person's preferences under risk satisfy certain rationality axioms, then there exists a utility function u, the von Neumann utility function, such that the person will tend to maximize the expected value of u. For this reason, the hypothesis that people satisfy the von Neumann-Morgenstern rationality axioms is known as expected utility theory. Now one of the major challenges for expected utility theory is the Ellsberg Paradox. It goes as follows.

Suppose you have an urn which has a total of 90 balls, 30 of which are red and each of the other 60 balls is either black or yellow. And suppose a ball is drawn at random from the urn. Then would you rather have lottery A, where you get 100 dollars if a red ball is drawn, or lottery B, where you get 100 dollars if a black ball is drawn? Most people would prefer lottery A. And would you rather have lottery C, where you get 100 dollars if you draw a red or yellow ball, or lottery D, where you get 100 dollars if you draw a black or yellow ball? Most people would prefer lottery D. But the thing is, preferring both lottery A to lottery B and lottery D to lottery C is inconsistent with expected utility theory; see this Wikipedia article for the proof.

I'd like to understand this logic a little better. Consider a new urn that just has 60 balls, and each ball is either black or yellow. Then would you rather have lottery A', where you get 30 dollars guaranteed, or lottery B', where you get a dollar for every black ball in the urn? I think most people would prefer lottery A'. And would you rather have lottery C', where you get 30 dollars plus a dollar for every yellow ball in the urn, or lottery D', where you get 60 dollars guaranteed? I think most people would prefer lottery D'.

So my question is, does it violate expected utility theory to prefer both A' to B' and D' to C'? I think it's consistent with expected utility theory. Assuming I'm right, couldn't you just change the currency from "dollars" to "lottery tickets" everywhere in my example, where a lottery ticket entitles you to a 1/90 chance of getting a 100 dollars? That would transform my example into the example in Ellsburg's paradox. So where's the flaw in my reasoning?

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The short answer seems to be yes your example violates expected utility... It mostly seems to me like a simple transformation of the first example you gave (but you got rid of the red balls).

As mentioned in other answers expected utility is not equipped to handle uncertainty because it deals with taking expectations and expectations cannot be computed when you don't know probabilities. For that reason, I think it is relevant to my answer to specify the distinction between what risk and uncertainty are.

  • Risk: This deals with facing a set of probabilities over outcomes in which the agent knows/understands the possible outcomes that he is facing
  • Uncertainty: This deals with an agent facing an unknown set of probabilities over possible outcomes

In a previous version of my answer, I wasn't careful in how I thought about and explained these ideas and I have clarified a few ideas in my head since then which hopefully translates to a clarified answer. Now to answer your question:

We have an urn with 60 balls. We are uncertain about the number of black balls and yellow balls in this urn. Imagine we present this urn with the above specified lotteries to an individual (with a monotonic utility function $u$) and ask them to choose between $L_{A'}$ and $L_{B'}$. Without loss of generality, imagine they choose to take $L_{A'}$.

This tells us by revealed preference $L_{A'} \succ L_{B'}$ which then implies

$$u(30) > u(B)$$

where $B$ is the number of black balls that are believed to be in the urn (no probabilities here, this is some number the decision maker arrived at and at this point we don't know how). This implies

$$B < 30$$

Now imagine that you give the agent the choice between $L_{C'}$ and $L_{D'}$ and the agent chooses to face lottery $D'$. Then

$$u(60) > u(30 + (60-B))$$

where $(60-B)$ is the number of yellow balls that the agent must think are in the urn. This implies

$$60 > 30 + (60-B) \Rightarrow B > 30$$

This means that an agent cannot prefer $L_{A'}$ to $L_{B'}$ and prefer $L_{C'}$ to $L_{D'}$ because that would imply $B > 30$ and $B < 30$. It is difficult to talk about where $B$ comes from because it is a hard concept to think about when we are used to working in expected value terms, this agent cannot compute any type of expectation over the sets of values that he thinks $B$ lies in. As an example of this, let me use a max-min approach.

Imagine that when he is presented the urn and lotteries that he is told that the number of black balls is either 15 or 45, but the agent doesn't know with what probability it is 15 and what probability it is 45.

  • Lottery $A'$ will give him $u(30)$ utility units for sure
  • Lottery $B'$ will give him either $u(15)$ utility units or $u(45)$ utility units.
  • Lottery $C'$ will give him either $u(75)$ utility units or $u(45)$ utility units.
  • Lottery $D'$ will give him $u(60)$

The agent is worried about uncertainty (an interpretation of the max-min is that he thinks it is a magic urn that tries its best to pay the least amount as possible, so it will choose the value that leaves him worse off if given a choice). Then using this in his decision process when comparing $L_{A'}$ and $L_{B'}$ he thinks these lotteries will pay the following amounts

  • $L_{A'}$ will pay $u(30)$
  • $L_{B'}$ will pay $\min(u(15), u(45)) = u(15)$

thus he chooses lottery $A'$. Now consider $L_{C'}$ and $L_{D'}$. He thinks these lotteries will pay

  • $L_{C'}$ will pay $\min(u(45), u(75)) = u(45)$
  • $L_{D'}$ will pay $u(60)$

thus he would choose lottery $D'$. This is one example of how uncertainty can be treated in a problem like this, but by no means is the only.

Note: Earlier I thought had a piece of my answer that talked about priors over the value of $B$, but this wasn't carefully thought through. You can have multiple values that you might imagine $B$ takes, but as soon as you assign a probability distribution to any of these values then you have left the realm of uncertainty and moved to the realm of risk.

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  • $\begingroup$ What if you don't have a belief regarding the exact numbers of balls, but you believe that there is, say a 50% chance that B < 30? Is that sort of belief not allowed in expected utility theory? A belief like that isn't really Knightian uncertainty, because it's completely quantifiable. So what sort of utility theory would accommodate a probability distribution over the different possible numbers of black balls? $\endgroup$
    – Keshav Srinivasan
    May 2, 2015 at 20:15
  • $\begingroup$ Having a probability distribution over multiple possibilities of black balls is exactly the same. It doesn't matter whether you have expectations of expectations of expectations of the number of black balls. It will eventually spit out a number of balls $B$ that you think are in the urn. If you don't have that then you can't deal with expected utility. $\endgroup$
    – cc7768
    May 2, 2015 at 20:24
  • $\begingroup$ Well, let's consider my example, and suppose you believe that there's a 50% chance that there are 15 black balls and a 50% chance that there are 45 black balls. Then are you telling me that it violates expected utility theory to both prefer a guaranteed 30 dollars to a 50% chance of 15 dollars and a 50% chance of 45 dollars, and to prefer a guaranteed 60 dollars to a 50% chance of 45 dollars and a 50% chance of 75 dollars? I think it's completely consistent with expected-utility theory; it's standard risk-aversion in a vNM utility function. $\endgroup$
    – Keshav Srinivasan
    May 2, 2015 at 22:31
  • $\begingroup$ Firstly, I have no sense of risk-aversion in my example. My example only treats a risk-neutral version of this problem. Secondly, there isn't very much channel for risk the way that you phrased the problem because each of the lotteries are only a single outcome instead of mixing 2 outcomes, so for my entire argument, I could just add utility functions around both sides of each inequality and it goes through. Thirdly, an important distinction to make is the difference between example and counter-example. $\endgroup$
    – cc7768
    May 3, 2015 at 0:12
  • $\begingroup$ An example is not sufficient to prove something is true, but a counter-example is sufficient to prove that it is not true. What you give in your comment is precisely the one example in which my inequalities turn into equalities and thus it is ok. If you move it outside of that knife-edge case (where the random variables both have expectation 30) then I think you will see that it falls apart. At the very least consider the counter-example of your expectation on the balls is $B = 0$ and $Y=60$ $\endgroup$
    – cc7768
    May 3, 2015 at 0:15
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"...one of the major challenges for expected-value theory is the Ellsberg Paradox."

Hmm, but in what way? Elsberg paradox is not like other related paradoxes surrounding Expected Utility theory, (like Allais' or Machina's). In the Elsberg paradox something is missing from the framework into which Expected Utility theory confines itself: there are unknown probabilities, or "Knightian Uncertainty" -and Expected Utility theory is not constructed to operate in such an environment.

But then, Expected Utility theory is not refuted by the Ellsberg Paradox. What the paradox does, is bringing to our attention that there exist real-world situations where Expected Utility is inapplicable. I.e. it cannot be a universal theory. But that should not be surprising: what theory in Social Sciences holds universally through space and time?

The value of the Ellsberg paradox is that it presses us to modify Expected Utility theory, or come up with some complementary theory to handle cases where there exists "Knightian Uncertainty", and decision makers neither know the actual probabilities nor do they form subjective probabilities -but they decide in ways we have yet to fully understand how to model (although "ambiguity aversion", which is the predominant way to rationalize the Ellsberg Paradox has by now been formalized to a large degree).

As regards the decision-making framework you create in the second part of the answer, I note that the "Ellsberg's Paradox" is an observed empirical phenomenon. You create another framework and you just say "I think people would prefer this or that" -and you neither have actual observations to back that up, nor even a theoretical and/or mathematical argument as to why you think "people would prefer" what you think they will prefer. So if the framework you created is equivalent to the one characterizing "Ellsberg's Paradox" (and you should write down the mathematics to show that), then I am afraid recorded experience refutes your impression about "what people would prefer" in the framework you created.

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  • $\begingroup$ Your answer doesn't answer my question at all. I was just guessing what I think most people would do in my scenario, but that's not the question I was asking. My actual question was just, is preferring A' to B' and D' to C' inconsistent with expected utility theory or not? And if it is consistent with expected utility theory, why does it not imply that the results of Ellsberg's paradox are also consistent with expected utility theory? $\endgroup$
    – Keshav Srinivasan
    Apr 26, 2015 at 0:50
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    $\begingroup$ This is the second time that you are changing a question of yours by posting a "clarification comment" on an answer. $\endgroup$
    – Alecos Papadopoulos
    Apr 26, 2015 at 0:55
  • $\begingroup$ I'm not changing my question. I'm saying the same thing I said above: "So my question is, does it violate expected utility theory to prefer both A' to B' and D' to C'? I think it's consistent with expected utility theory. Assuming I'm right, couldn't you just change the currency from "dollars" to "lottery tickets" everywhere in my example, where a lottery ticket entitles you to a 1/90 chance of getting a 100 dollars? That would transform my example into the example in Ellsburg's paradox. So where's the flaw in my reasoning?" $\endgroup$
    – Keshav Srinivasan
    Apr 26, 2015 at 0:58
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    $\begingroup$ Alecos Papadopoulos did answer your question: "there exist real-world situations where Expected Utility is inapplicable" Without beliefs about the distribution you cannot have N-M utility functions. Given any beliefs a 'rational' consumer could not prefer both A' to B' and D' to C'. Real-life people are not very good at dealing with probability (and use heuristic algorithms), so it is quite possible that real-life A' to B', D' to C' preferences exist. $\endgroup$
    – Giskard
    Apr 26, 2015 at 4:26
  • $\begingroup$ @denesp My question isn't asking how applicable expected-utility theory is to the real world, it's about A', B', C', and D'. Now if you can prove that someone who is vNM-rational cannot prefer both A' to B' and D' to C', I suggest you post an answer of your own. $\endgroup$
    – Keshav Srinivasan
    Apr 26, 2015 at 5:18
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Formally Ellsberg paradox is defined in a different choice environment that expected utility. Ellsberg paradox is a problem for Savage subjective utility where there is ambiguity. Superficially they look the same because the have the same affine form, but in Savage the probabilities over states are not objective (are not a primitive) rather they are part of the representation. Also Expected Utility is defined over lotteries, while Savage Subjective Utility is defined over "acts". A nice explanation of Ellsberg paradox and Savage Expected Subjective Utility is given in http://lesswrong.com/lw/9e4/the_savage_theorem_and_the_ellsberg_paradox/

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  • $\begingroup$ OK, but this still doesn't answer my question. Even if we use Savage's axioms rather than the vNM axioms, the question remains, is preferring both A' to B' and D' to C' inconsistent with Savage's axioms? $\endgroup$
    – Keshav Srinivasan
    Apr 29, 2015 at 10:44

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