15 Ellsberg Paradox
Suppose there is an urn containing 90 balls. You know that 30 of these balls are blue, while the remaining 60 balls are either yellow or green, but their exact distribution is unknown. You are asked to make a choice between two different sets of lotteries:
Question 1: Which of the following two lotteries do you prefer?
- Lottery 1: A ball is drawn from the urn. You win $1 million if the ball is blue, and $0 if it is yellow or green.
- Lottery 2: A ball is drawn from the urn. You win $1 million if the ball is yellow, and $0 if it is blue or green.
Question 2: Which of the following two lotteries do you prefer?
- Lottery 3: A ball is drawn from the urn. You win $1 million if the ball is either blue or green, and $0 if it is yellow.
- Lottery 4: A ball is drawn from the urn. You win $1 million if the ball is either yellow or green, and $0 if it is blue.
To make the comparison easier, here is a table summarizing the payouts of each lottery based on the color of the ball drawn:
| blue | yellow | green | |
|---|---|---|---|
| Lottery 1 | \(\$1,000,000\) | \(\$0\) | \(\$0\) |
| Lottery 2 | \(\$0\) | \(\$1,000,000\) | \(\$0\) |
| Lottery 3 | \(\$1,000,000\) | \(\$0\) | \(\$1,000,000\) |
| Lottery 4 | \(\$0\) | \(\$1,000,000\) | \(\$1,000,000\) |
You should answer the above questions before reading further. This is knowns as the Ellsberg paradox (Ellsberg 1961).
The Ellsberg paradox presents decision makers with the task of choosing between two sets of lotteries, where the challenge lies in the fact that some probabilities are unknown. Suppose there is an integer \(b\) such that \(30 \leq b \leq 60\) representing the number of blue balls, and \(y = 90 - 30 - b\) representing the number of yellow balls. Here, \(0M\) stands for “0 dollars” and \(1M\) stands for “1 million dollars”. The first question asks decision makers to compare the following two lotteries:
- \(L_1= \frac{30}{90}\cdot 1M + \frac{b}{90}\cdot 0M + \frac{y}{90}\cdot 0M\); and
- \(L_2=\frac{30}{90}\cdot 0M + \frac{b}{90}\cdot 1M + \frac{y}{90}\cdot 0M.\)
Many decision makers express a strict preference for \(L_1\) over \(L_2\) (i.e., \(L_1 \mathrel{P} L_2\)). After indicating their preference between \(L_1\) and \(L_2\), decision makers are then asked to compare the following pair of lotteries:
- \(L_3= \frac{30}{90}\cdot 1M + \frac{b}{90}\cdot 0M + \frac{y}{90}\cdot 1M\); and
- \(L_4=\frac{30}{90}\cdot 0M + \frac{b}{90}\cdot 1M + \frac{y}{90}\cdot 1M.\)
Many decision makers report a strict preference for \(L_4\) over \(L_3\) (i.e., \(L_4 \mathrel{P} L_3\)).
The key insight of the Ellsberg paradox is that, while each individual preference (either \(L_1 \mathrel{P} L_2\) or \(L_4 \mathrel{P} L_3\)) might seem rational on its own, expressing both preferences simultaneously is inconsistent with expected utility theory. Specifically, if a decision maker ranks lotteries according to their expected utility based on some utility function, then: \[L_1\mathrel{P} L_2\mbox{ if, and only if, }L_3\mathrel{P}L_4.\]
This implies the following for any rational decision maker:
- Preferring \(L_1 \mathrel{P} L_2\) and \(L_3 \mathrel{P} L_4\) is consistent with expected utility theory.
- Preferring \(L_1 \mathrel{P} L_2\) but not \(L_3 \mathrel{P} L_4\) is inconsistent with expected utility theory.
- Not preferring \(L_1 \mathrel{P} L_2\) but preferring \(L_3 \mathrel{P} L_4\) is inconsistent with expected utility theory.
- Not preferring \(L_1 \mathrel{P} L_2\) and not preferring \(L_3 \mathrel{P} L_4\) is consistent with expected utility theory.
The challenge lies in the fact that many decision makers prefer \(L_1 \mathrel{P} L_2\) because they want to avoid the ambiguity associated with lottery \(L_2\), where the probability of drawing a yellow or green ball is unknown. The complication arises when, for the same reason—avoiding ambiguous lotteries—these decision makers also prefer \(L_4\) over \(L_3\). However, this combination of preferences implies that they are not ranking lotteries based on their expected utilities. In other words, the desire to avoid ambiguity leads to preferences that are inconsistent with expected utility theory. Similar to the Allais Paradox, these preferences also violate the Independence Axiom.
15.1 The Ellsberg Preferences are Inconsistent with Expect Utility Theory
Lemma 15.1 If \(L_1, L_2, L_3\), and \(L_4\) are defined as in the Ellsberg paradox, then \(L_1\mathrel{P} L_2\) and \(L_4\mathrel{P} L_3\) is inconsistent with expect utility theory.
Proof. To see why \(L_1\mathrel{P} L_2\) and \(L_3\mathrel{P}L_4\) is inconsistent with expected utility theory, we will show that for any utility function \(u:\{0M, 1M, 5M\}\rightarrow\mathbb{R}\), it is impossible that \[EU(L_1, u) > EU(L_2, u)\quad\mbox{ and }\quad EU(L_4, u) > EU(L_3, u).\]
Suppose that \(u:\{0M, 1M\}\rightarrow\mathbb{R}\) is a utility function and that \(EU(L_1, u) > EU(L_2, u)\) and \(EU(L_4, u) > EU(L_3, u)\). We show that this leads to a contradiction. The expected utility calculations for \(L_1\) and \(L_2\) are:
\[\begin{align*} EU(L_1, u) &= EU(\frac{30}{90}\cdot 1M + \frac{b}{90}\cdot 0M + \frac{y}{90}\cdot 0M$, u) \\ &= \frac{30}{90} \times u(1M) + \frac{b}{90} \times u(0M) + \frac{y}{90} \times u(0M) \\ &= \frac{30}{90} \times u(1M) + \frac{b + y}{90} \times u(0M) \\ \end{align*}\]
\[\begin{align*} EU(L_2, u) &= EU(\frac{30}{90}\cdot 0M + \frac{b}{90}\cdot 1M + \frac{y}{90}\cdot 0M, u) \\ &= \frac{30}{90} \times u(0M) + \frac{b}{90} \times u(1M) + \frac{y}{90} \times u(0M) \\ &= \frac{30 + y}{90} \times u(0M) + \frac{b}{90} \times u(1M) \\ \end{align*}\]
Since \(EU(L_1, u) > EU(L_2, u)\), we have that: \[\frac{30}{90} \times u(1M) + \frac{b + y}{90} \times u(0M) > \frac{30 + y}{90} \times u(0M) + \frac{b}{90} \times u(1M)\]
Subtracting \(\frac{30}{90} \times u(1M)\) and \(\frac{30 + y}{90} \times u(0M)\) from both sides of the inequality gives the following: \[\frac{b + y - 30 - y}{90} \times u(0M) > \frac{b-30}{90} \times u(1M)\] Simplifying the probabilities, we have that: \[\frac{b - 30}{90} \times u(0M) > \frac{b-30}{90} \times u(1M)\]
Now, the expected utility calculations for \(L_3\) and \(L_4\) are:
\[\begin{align*} EU(L_3, u) &= EU(\frac{30}{90}\cdot 1M + \frac{b}{90}\cdot 0M + \frac{y}{90}\cdot 1M, u) \\ &= \frac{30}{90} \times u(1M) + \frac{b}{90} \times u(0M) + \frac{y}{90} \times u(1M) \\ &= \frac{30 + y}{90} \times u(1M) + \frac{b}{90} \times u(0M) \\ \end{align*}\]
\[\begin{align*} EU(L_4, u) &= EU(\frac{30}{90}\cdot 0M + \frac{b}{90}\cdot 1M + \frac{y}{90}\cdot 1M, u) \\ &= \frac{30}{90} \times u(0M) + \frac{b}{90} \times u(1M) + \frac{y}{90} \times u(1M) \\ &= \frac{30}{90} \times u(0M) + \frac{b + y}{90} \times u(1M) \\ \end{align*}\]
Since \(EU(L_4, u) > EU(L_3, u)\), we have that: \[\frac{30}{90} \times u(0M) + \frac{b + y}{90} \times u(1M) > \frac{30 + y}{90} \times u(1M) + \frac{b}{90} \times u(0M)\]
Subtracting \(\frac{30 + y}{90} \times u(1M)\) and \(\frac{30}{90} \times u(0M)\) from both sides of the inequality gives the following: \[\frac{b + y - 30 - y}{90} \times u(1M) > \frac{b-30}{90} \times u(0M)\] Simplifying the probabilities, we have that: \[\frac{b - 30}{90} \times u(1M) > \frac{b-30}{90} \times u(0M)\]
But, this is impossible since we cannot have that:
- \(\frac{b-30}{90} \times u(0M) > \frac{b - 30 }{90} \times u(1M)\), and
- \(\frac{b - 30}{90} \times u(1M) > \frac{b-30}{90} \times u(0M).\)