# Causal Decision Theory violates the Independence of Irrelevant Alternatives

9 July 2019

As I’ll use the name here, the independence of irrelevant alternatives (IIA) says that adding an additional option to the menu can’t transform an impermissible choice into a permissible one. An old story from Sidney Morgenbesser illustrates the seeming irrationality of violating this principle: asked to decide between steak and chicken, a man says “I’d rather have the steak”. The waiter tells him that they also have fish, to which he responds: “Oh, in that case, I’ll have the chicken”. This behavior looks irrational, and a principle like IIA explains why. I recently realized that causal decision theory (CDT) doesn’t abide by the IIA. Not just orthodox CDT, but basically any theory worth calling ‘causalist’ will end up violating the principle. The point of today’s post is to explain why.

### 1. Stage Setting

Orthodox CDT measures the choiceworthiness of acts with their utility, $\mathcal{U}$, where $$\mathcal{U}(A) \stackrel{\text{df}}{=} \sum_K \Pr(K) \cdot \mathcal{D}(KA)$$ (In the above, the $K$s are a partition of states of nature, and $\mathcal{D}(KA)$ says how strongly you desire that you perform $A$ in the state of nature $K$.) Orthodox CDT says that an act is permissible iff it maximizes utility.

Let me use ‘$\mathbf{M}$’ for a menu of options, and I’ll use “$\mathcal{P}(\mathbf{M})$” for the permissible options on the menu $\mathbf{M}$. Then, orthodox CDT says that an act is permissible iff its utility is no less than any alternative act, $$\mathcal{P}(\mathbf{M}) = \{ A \in \mathbf{M} \mid (\forall B \in \mathbf{M}) \mathcal{U}(A) \geqslant \mathcal{U}(B) \}$$

One noteworthy feature of the measure $\mathcal{U}$ is that its values can depend upon how likely you think you are to select each act. Let’s write “$\mathcal{U}_A(B)$” for the utility you would assign to the act $B$, were you to learn only that you had performed the act $A$: $$\mathcal{U}_A(B) \stackrel{\text{df}}{=} \sum_K \Pr(K \mid A) \cdot \mathcal{D}(KA)$$ Then, in a choice between two options, $A$ and $B$, both of the following situations are possible:

Self-Undermining Choice Once chosen, every act is worse than the alternative. \begin{aligned} \mathcal{U}_A(B) > \mathcal{U}_A(A) \qquad \text{ and } \qquad \mathcal{U}_B(A) > \mathcal{U}_B(B) \end{aligned} Self-Reinforcing Choice Once chosen, every act is better than the alternative. \begin{aligned} \mathcal{U}_A(A) > \mathcal{U}_A(B) \qquad\text{ and } \qquad \mathcal{U}_B(B) > \mathcal{U}_B(A) \end{aligned}

This can lead CDT’s verdicts to change as you make up your mind about what to. In a self-undermining choice, once you follow CDT’s advice and intend to do the act it called rational, it will change its mind and begin to call you irrational. In a self-reinforcing choice, if you disregard its advice and do what it said was irrational, CDT will change its mind and call you rational for doing so.

I’ve come to think that this is a reason to doubt orthodox CDT. But this feature won’t be relevant to anything I’m saying today. All I will need to appeal to here is the following, minimal committment of CDT:

Minimal CDT. In a choice between two options, $A$ and $B$, if the utility of $A$ exceeds the utility of $B$, and it would continue to do so whether you choose $A$ or $B$, then $B$ is impermissible. $$\left( \mathcal{U}_A(A) > \mathcal{U}_A(B) \text{ and } \mathcal{U}_B(A) > \mathcal{U}_B(B) \right) \Rightarrow B \notin \mathcal{P}(\{ A, B \})$$

For instance, in Newcomb’s Problem, whether you one box or two box, the utility of two-boxing will exceed the utility of one-boxing. So Minimal CDT says that one-boxing is impermissible.

### 2. CDT violates IIA

What I’ll show here is that the following three principles are jointly inconsistent.

Independence of Irrelevant Alternatives (IIA)

Given any two menus of options where the first is a subset of the second, and $A$ appears on both, if it is not permissible to choose $A$ from the smaller menu, then it is not permissible to choose $A$ from the larger menu. $$A \in \mathbf{M} \subseteq \mathbf{M}^+ \Rightarrow \left[ A \notin \mathcal{P}(\mathbf{M}) \Rightarrow A \notin \mathcal{P}(\mathbf{M}^+) \right]$$

Minimal CDT

In a choice between two options, $A$ and $B$, if the utility of $A$ exceeds the utility of $B$, and it would continue to do so whether you choose $A$ or $B$, then $B$ is impermissible. $$\left( \mathcal{U}_A(A) > \mathcal{U}_A(B) \text{ and } \mathcal{U}_B(A) > \mathcal{U}_B(B) \right) \Rightarrow B \notin \mathcal{P}({ A, B })$$

No Dilemmas

Given any menu of options, some option is permissible to choose from that menu. $$\forall \mathbf{M} \quad \mathcal{P}(\mathbf{M}) \neq \varnothing$$

Since CDT is clearly committed to Minimal CDT and No Dilemmas, it must reject the IIA.

To see why these three principles are jointly inconsistent, consider the following decision:

You must choose between three boxes, labeled ‘$A$’, ‘$B$’, and ‘$C$’. You can take one, and only one, of the boxes. Yesterday, a reliable predictor made a prediction about how you would choose. Their predictions are 80% reliable—so, conditional on you taking box $X$, you’re 80% sure that this is what they predicted you’d do. If they predicted that you would take $A$, then they left nothing in $A$, 10 dollars in $C$, and bill for 10 dollars in $B$. If they predicted that you would take $B$, then they left nothing in $B$, 10 dollars in $A$, and a bill for 10 dollars in $C$. If they predicted that you would take $C$, then they left nothing in $C$, 10 dollars in $B$, and a bill for 10 dollars in $A$.

If we use “$K_X$” for the state of nature in which it was predicted that you would take box $X$, then the desirabilities and probabilities for this decision are shown in the matrices below.

$$\begin{array}{r | c c c} \mathcal{D}(\text{Row Col}) & K_A & K_B & K_C \\\hline A & 0 & 10 & -10 \\\ B & -10 & 0 & 10 \\\ C & 10 & -10 & 0 \end{array} \qquad \begin{array}{r | c c c} \Pr(\text{ Col} \mid \text{ Row}) & A & B & C \\\hline K_A & 0.8 & 0.1 & 0.1 \\\ K_B & 0.1 & 0.8 & 0.1 \\\ K_C & 0.1 & 0.1 & 0.8 \end{array}$$

Multiplying the left-hand-side matrix by the right-hand-side matrix gives us the following matrix of the utility you would assign to the row act, were you to learn only that you had performed the column act:

\begin{array}{r | c c c} \mathcal{U}_{\text{Col}}(\text{Row}) & A & B & C \\\hline A & 0 & 7 & -7 \\\
B & -7 & 0 & 7 \\\
C & 7 & -7 & 0
\end{array}

Now, suppose that, instead of being given a choice between $A$, $B$, and $C$, you were instead offered a choice between just $A$ and $B$—$C$ is taken off of the menu (though there’s still a 10% chance that the predictor falsely predicted that you would take $C$). In that case, notice that both $\mathcal{U}_A(A) > \mathcal{U}_A(B)$ and $\mathcal{U}_B(A) > \mathcal{U}_B(B)$. So Minimal CDT says that, in a decision between $A$ and $B$, $B$ is an impermissible choice, $B \notin \mathcal{P}(\{ A, B \})$. Also notice that a choice between $B$ and $C$ is exactly like a choice between $A$ and $B$. As is a choice between $C$ and $A$.

$$\begin{array}{r | c c} \mathcal{U}_{\text{Col}}(\text{Row}) & A & B \\\hline A & 0 & 7 \\\ B & -7 & 0 \end{array}$$

$$\begin{array}{r | c c} \mathcal{U}_{\text{Col}}(\text{Row}) & B & C \\\hline B & 0 & 7 \\\ C & -7 & 0 \end{array}$$

$$\begin{array}{r | c c} \mathcal{U}_{\text{Col}}(\text{Row}) & C & A \\\hline C & 0 & 7 \\\ A & -7 & 0 \end{array}$$

That is: both $\mathcal{U}_B(B) > \mathcal{U}_B( C )$ and $\mathcal{U}_C(B) > \mathcal{U}_C( C )$. So Minimal CDT says that, in a decision between $B$ and $C$, $C$ is an impermissible choice, $C \notin \mathcal{P}(\{ B, C \})$. And both $\mathcal{U}_C( C ) > \mathcal{U}_C(A)$ and $\mathcal{U}_A( C ) > \mathcal{U}_A(A)$. So Minimal CDT says that, in a decision between $C$ and $A$, $A$ is an impermissible choice, $A \notin \mathcal{P}(\{ C, A \})$.

Now, consider what it is permissible to choose from the full menu of options, $\{ A, B, C \}$. By No Dilemmas, some option on this menu is permissible. Suppose it is permissible to choose $A$, $A \in \mathcal{P}(\{ A, B, C \})$. Then, $A \notin \mathcal{P}(\{ C, A \})$ and $A \in \mathcal{P}(\{ C, A, B \})$. This violates IIA. Suppose, on the other hand, that it is permissible to choose $B$, $B \in \mathcal{P}(\{ A, B, C \})$. Then, $B \notin \mathcal{P}(\{ A, B \})$ and $B \in \mathcal{P}(\{ A, B, C \})$. And this violates IIA. Suppose, finally, that it is permissible to choose $C$, $C \in \mathcal{P}(\{ A, B, C \})$. Then, $C \notin \mathcal{P}(\{ B, C \})$ and $B \in \mathcal{P}(\{ B, C, A \})$. Again, this violates IIA. So: whichever of $A, B,$ and $C$ is permissible, there will be a violation of IIA.

So: if we assume Minimal CDT and No Dilemmas, then we will have violations of IIA.