# 18 Allais Paradox

Suppose that there is an urn with 100 balls. There are 89 white balls, 10 blue balls and 1 red ball. You are asked to compare two sets of lotteries:

Question 1: Which of the following two lotteries do you prefer?

Lottery 1: A ball is drawn from the urn and you win $1 million no matter what color is drawn.

Lottery 2: A ball is drawn from the urn and you win $0 if the ball is red, $1 million if the ball is white, and $5 million if the ball is blue.

Question 2: Which of the following two lotteries do you prefer?

Lottery 3: A ball is drawn from the urn and you win $1 million if the ball is red, $0 if the ball is white, and $1 million if the ball is blue.

Lottery 4: A ball is drawn from the urn and you win 0M if the ball is red, $0 if the ball is white, and $5 million if the ball is blue.

You should answer the above questions before reading further.

The Allais paradox asks decision makers to form preferences over two sets of lotteries. The first two lotteries are the following, where \(1M\) means “1 million dollars” and \(5M\) means “5 million dollars”, \(0M\) means “0 dollars”: \[L_1=[1M:0.01, 1M:0.89, 1M:0.10]\quad \mbox{vs.}\quad L_2=[0M:0.01, 1M:0.89, 5M:0.10].\] Many decision makers report that they strictly prefer \(L_1\) to \(L_2\) (i.e., \(L_1\mathrel{P} L_2\)). After reporting their preference between \(L_1\) and \(L_2\), decision makers are asked to compare the following two lotteries: \[L_3=[1M:0.01, 0M:0.89, 1M:0.10]\quad\mbox{vs.}\quad L_4=[0M:0.01, 0M:0.89, 5M:0.10].\] Many decision makers report that they strictly prefer \(L_4\) to \(L_3\) (i.e., \(L_4\mathrel{P} L_3\)).

The observation of the Allais paradox is the following: While there is nothing irrational about each opinion by itself, reporting both that \(L_1\mathrel{P} L_2\) *and* \(L_4\mathrel{P} L_3\) is inconsistent with expected utility theory. That is, if a decision maker ranks lotteries by their expected utility with respect to some utility function, then: \[L_1\mathrel{P} L_2\mbox{ if, and only if, }L_3\mathrel{P}L_4.\]

This means that for any rational decision maker we have the following:

\(L_1\mathrel{P} L_2\) and \(L_3\mathrel{P} L_4\) is consistent with expected utility theory.

\(L_1\mathrel{P} L_2\) and \(L_4\mathrel{P} L_3\) is not consistent with expected utility theory.

\(L_2\mathrel{P} L_1\) and \(L_3\mathrel{P} L_4\) is not consistent with expected utility theory.

\(L_2\mathrel{P} L_1\) and \(L_4\mathrel{P} L_3\) is consistent with expected utility theory.

The problem is that many people report the preference \(L_1\mathrel{P} L_2\) and \(L_4\mathrel{P} L_3\) (these are called the *Allais preferences*), and so have preferences that are inconsistent with expected utility theory. We explain why the Allais preferences are inconsistent with expected utility theory in Section 18.1. Since, Allais preferences are inconsistent with expected utility theory, by the Von Neumann Morgenstern Theorem (Theorem 17.1), the Allais preferences must violate at least one of the axioms Compound Lottery (Chapter 14), Independence (Chapter 15), or Continuity (Chapter 16). We show that, assuming the Compound Lottery axiom and that \((P, I)\) are a rational preference (Definition 8.3) on the set of lotteries, that the Allais preferences violate the Independence Axiom in Section 18.2.

## 18.1 The Allais Preferences are Inconsistent with Expect Utility Theory

**Lemma 18.1 **If \(L_1, L_2, L_3\), and \(L_4\) are defined as in the Allais 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 \(u:\{0M, 1M, 5M\}\rightarrow\mathbb{R}\) 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:

\[\begin{align*} EU(L_1, u) &= EU([1M:0.01, 1M:0.89, 1M:0.10], u) \\ &= 0.01*u(1M) + 0.89 * u(1M) + 0.10 * u(1M) \\ &= u(1M) \\ \end{align*}\]

\[\begin{align*} EU(L_2, u) &= EU([0M:0.01, 1M:0.89, 5M:0.10], u) \\ &= 0.01*u(0M) + 0.89 * u(1M) + 0.10 * u(5M) \\ \end{align*}\]

Since \(EU(L_1, u) > EU(L_2, u)\), we have that: \[u(1M) > 0.01*u(0M) + 0.89 * u(1M) + 0.10 * u(5M).\]

Subtracting \(0.89 * u(1M)\) from both sides of the inequality gives the following: \[0.11 * u(1M) > 0.01 * u(0M) + 0.10 * u(5M).\]

Now, the expected utility calculations for \(L_3\) and \(L_4\) are:

\[\begin{align*} EU(L_3, u) &= EU([1M:0.01, 0M:0.89, 1M:0.10], u) \\ &= 0.01*u(1M) + 0.89 * u(0M) + 0.10 * u(1M) \\ &= 0.11 * u(1M) + 0.89 u(0M) \\ \end{align*}\]

\[\begin{align*} EU(L_4, u) &= EU([0M:0.01, 0M:0.89, 5M:0.10], u) \\ &= 0.01*u(0M) + 0.89 * u(0M) + 0.10 * u(5M) \\ &= 0.90 * u(0M) + 0.10 * u(5M) \\ \end{align*}\]

Since \(EU(L_4, u) > EU(L_3, u)\), we have that: \[0.90 * u(0M) + 0.10 * u(5M) > 0.11 * u(1M) + 0.89 u(0M).\]

If we subtract \(0.89 * u(0M)\) from both sides of the inequality, then we have that: \[0.01 * u(0M) + 0.10 * u(5M) > 0.11 * u(1M).\]

But, this is impossible since we cannot have that:

- \(0.11 * u(1M) > 0.01 * u(0M) + 0.10 * u(5M)\); and
- \(0.01 * u(0M) + 0.10 * u(5M) > 0.11 * u(1M).\)

## 18.2 The Allais Preferences are Inconsistent with the Independence Axiom

**Lemma 18.2 **Suppose that \(L_1, L_2, L_3\), and \(L_4\) are defined as in the Allais paradox, and that the decision maker satisfies the Compound Lottery axiom and that \((P, I)\) is a rational preference (Definition 8.3) on the set of lotteries. Then, \(L_1\mathrel{P} L_2\) and \(L_4\mathrel{P} L_3\) violates the Independence Axiom.

*Proof*. Suppose that \(L_1, L_2, L_3\), and \(L_4\) are defined as in the Allais paradox, the decision maker satisfies the Compound Lottery axiom, \((P, I)\) is a rational preference (Definition 8.3) on the set of lotteries, and that

- \(\left[1M:\frac{1}{100}, 1M:\frac{89}{100}, 1M:\frac{10}{100}\right]\mathrel{P}\left[0M:\frac{1}{100}, 1M:\frac{89}{100}, 5M:\frac{10}{100}\right]\), and
- \(\left[0M: \frac{1}{100}, 0M:\frac{89}{100}, 5M:\frac{10}{100}\right] \mathrel{P}\left[1M: \frac{1}{100}, 0M:\frac{89}{100}, 1M:\frac{10}{100}\right ]\).

Assume that the decision maker satisfies the Independence Axiom. We will show that this leads to a contradiction.

We first note the following two consequences of the Compound Lottery axiom:

Since \(s([1M: \frac{1}{11}, 1M:\frac{10}{11}]:\frac{11}{100}, [1M:1]:\frac{89}{100}]) = [1M: \frac{1}{100}, 1M:\frac{89}{100}, 1M:\frac{10}{100}]\), by the Compound Lottery axiom, \[\left [\left[1M: \frac{1}{11}, 1M:\frac{10}{11}\right]:\frac{11}{100}, \left[1M:1\right]:\frac{89}{100}\right ]\mathrel{I}\left[1M: \frac{1}{100}, 1M:\frac{89}{100}, 1M:\frac{10}{100}\right ].\]

Since \(s([[0M: \frac{1}{11}, 5M:\frac{10}{11}]: \frac{11}{100}, [1M:1]:\frac{89}{100}]) = [0M: \frac{1}{100}, 1M:\frac{89}{100}, 5M:\frac{10}{100}]\), by the Compound Lottery axiom, \[\left[0M: \frac{1}{100}, 1M:\frac{89}{100}, 5M:\frac{10}{100}\right]\mathrel{I}\left[\left[0M: \frac{1}{11}, 5M:\frac{10}{11}\right]: \frac{11}{100}, \left[1M:1\right]:\frac{89}{100}\right]\]

By first assumption of the Allais preferences, \(\left[1M:\frac{1}{100}, 1M:\frac{89}{100}, 1M:\frac{10}{100}\right]\mathrel{P}\left[0M:\frac{1}{100}, 1M:\frac{89}{100}, 5M:\frac{10}{100}\right]\), and so by transitivity of strict preference and indifference, we have that \[{\small \left [\left[1M: \frac{1}{11}, 1M:\frac{10}{11}\right]:\frac{11}{100}, \left[1M:1\right]:\frac{89}{100}\right ]\mathrel{P}\left[\left[0M: \frac{1}{11}, 5M:\frac{10}{11}\right]: \frac{11}{100}, \left[1M:1\right]:\frac{89}{100}\right]}.\]

By the Independence Axiom, where \(L\) is \(\left[1M: \frac{1}{11}, 1M:\frac{10}{11}\right]\), \(L'\) is \(\left[0M: \frac{1}{11}, 5M:\frac{10}{11}\right]\), \(L''\) is \(\left[1M:1\right]\), and \(r\) is \(\frac{11}{100}\), this implies that \[\left [1M: \frac{1}{11}, 1M:\frac{10}{11}\right]\mathrel{P}\left[0M: \frac{1}{11}, 5M:\frac{10}{11}\right].\]

Applying the Independence Axiom a second time where \(L\), \(L'\), and \(r\) are as above, and \(L''\) is \([0M:1]\), we have that \[{\small \left [\left[1M: \frac{1}{11}, 1M:\frac{10}{11}\right]:\frac{11}{100}, \left[0M:1\right]:\frac{89}{100}\right ]\mathrel{P}\left[\left[0M: \frac{1}{11}, 5M:\frac{10}{11}\right]: \frac{11}{100}, \left[0M:1\right]:\frac{89}{100}\right]}.\]

To see the contradiction, note the following two consequences of the Compound Lottery axiom:

Since \(s([1M: \frac{1}{11}, 1M:\frac{10}{11}]:\frac{11}{100}, [0M:1]:\frac{89}{100}]) = [1M: \frac{1}{100}, 0M:\frac{89}{100}, 1M:\frac{10}{100}]\), by the Compound Lottery axiom, \[\left[1M: \frac{1}{100}, 0M:\frac{89}{100}, 1M:\frac{10}{100}\right ]\mathrel{I}\left [\left[1M: \frac{1}{11}, 1M:\frac{10}{11}\right]:\frac{11}{100}, \left[0M:1\right]:\frac{89}{100}\right ].\]

Since \(s([[0M: \frac{1}{11}, 5M:\frac{10}{11}]: \frac{11}{100}, [0M:1]:\frac{89}{100}]) = [0M: \frac{1}{100}, 0M:\frac{89}{100}, 5M:\frac{10}{100}]\), by the Compound Lottery axiom, \[\left[\left[0M: \frac{1}{11}, 5M:\frac{10}{11}\right]: \frac{11}{100}, \left[0M:1\right]:\frac{89}{100}\right]\mathrel{I}\left[0M: \frac{1}{100}, 0M:\frac{89}{100}, 5M:\frac{10}{100}\right]\]

Since \(\left [\left[1M: \frac{1}{11}, 1M:\frac{10}{11}\right]:\frac{11}{100}, \left[0M:1\right]:\frac{89}{100}\right ]\mathrel{P}\left[\left[0M: \frac{1}{11}, 5M:\frac{10}{11}\right]: \frac{11}{100}, \left[0M:1\right]:\frac{89}{100}\right]\), by transitivity of strict preference and indifference, we have that \[\left[1M: \frac{1}{100}, 0M:\frac{89}{100}, 1M:\frac{10}{100}\right] \mathrel{P}\left[0M: \frac{1}{100}, 0M:\frac{89}{100}, 5M:\frac{10}{100}\right ].\]

But, this contradictions the second assumption about the Allais preferences that \(\left[0M: \frac{1}{100}, 0M:\frac{89}{100}, 5M:\frac{10}{100}\right] \mathrel{P}\left[1M: \frac{1}{100}, 0M:\frac{89}{100}, 1M:\frac{10}{100}\right ]\).