Question about the Ellsberg Paradox in Expected Utility Theory 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?
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

$$u(30) &gt; u(B)$$

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

$$B &lt; 30$$

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

$$u(60) &gt; u(30 + (60-B))$$

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

$$60 &gt; 30 + (60-B) \Rightarrow B &gt; 30$$

This means that an agent cannot prefer $L_{A'}$ to $L_{B'}$ and prefer $L_{C'}$ to $L_{D'}$ because that would imply $B &gt; 30$ and $B &lt; 30$. 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.