Question about the Ellsberg Paradox in Expected Utility Theory
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.