‘Divide-and-choose’ in list-based decision problems

Abstract

When encountering a set of alternatives displayed in the form of a list, the decision maker usually determines a particular alternative, after which she stops checking the remaining ones, and chooses an alternative from those observed so far. We present a framework in which both decision problems are explicitly modeled, and axiomatically characterize a ‘divide-and-choose’ rule which unifies successive choice and satisficing choice.

Appendix 1

In order to show the independence of the above five axioms, let us consider the following examples of sc-rules each of which satisfies four of the axioms but violates the fifth one.

Example 3

Fix $x^{*}\in X$ and define $C^{1}$ as follows: for all $\ell \in \Lambda$, $C^{1}\left( \ell \right) =\left\langle \ell _{k},k\right\rangle$ if $\ell _{k}=x^{*}$ and $k\in \left\{ 1,le(\ell )\right\}$; $C^{1}\left( \ell \right) =\left\langle \ell _{k-1},k-1\right\rangle$ if $\ell _{k}=x^{*}$ and $k\in \left\{ 2,\ldots ,le(\ell )-1\right\}$; and $C^{1}\left( \ell \right) =\left\langle \ell _{1},le(\ell )\right\rangle$, otherwise. This sc-rule violates axiom 1 as for $a,b,c,x^{*}\in X$, $C(a,b,x^{*},c)=\left\langle b,2\right\rangle$ but $C(a,b,c)=\left\langle a,3\right\rangle$. On the other hand, $C^{1}$ satisfies axiom 2: the decision maker continues his search till the end of any list which either does not contain $x^{*}$ or $x^{*}$ is the last option in the list; otherwise, the premise in axiom 2 is not satisfied. Clearly, $C^{1}$ fulfills our third requirement since the decision maker always stops either before $x^{*}$ or immediately after $x^{*}$ and selects it (of course, when $x^{*}$ is present in the list). The proposed rule also satisfies axiom 4 since $C^{1}\left( \ell \right) =\left\langle x,le(\ell )\right\rangle$ with $x=\ell _{q}$ and $q<le(\ell )$ can happen only if $x^{*}\notin X(\ell )$. We have in this case that $C^{1}(x,y)=\left\langle x,2\right\rangle$ follows for any $y\in \left\{ \ell _{q+1},\ldots ,\ell _{le(\ell )}\right\}$ as required by axiom 4. Finally, in order to see that $C^{1}$ satisfies axiom 5 as well, notice that $C^{1}(x,y)=\left\langle x,2\right\rangle$ and $C^{1}(\ell )=\left\langle x,le(\ell )\right\rangle$ imply $x^{*}\notin X(\ell )\cup \left\{ y\right\}$ and thus, $x=\ell _{1}$. Thus, $C^{1}(\ell ,y)=\left\langle x,le(\ell )+1\right\rangle$ follows.

Example 4

Let $C^{2}$ be such that for all $\ell \in \Lambda$, $C^{2}\left( \ell \right) =\left\langle \ell _{le(\ell )},le(\ell )\right\rangle$ if $le(\ell )\le 2$ and $C^{2}\left( \ell \right) =\left\langle \ell _{2},2\right\rangle$, otherwise. Axiom 1 is satisfied by $C^{2}$ as $C^{2}\left( \ell \right) =\left\langle x,k\right\rangle$ with $k<le(\ell )$ implies $C^{2}\left( \ell \right) =\left\langle \ell _{2},2\right\rangle$ and thus, it holds for any $\ell \in \Lambda$ with $le(\ell )>2$. In order to see that $C^{2}$ violates axiom 2 let us consider the options $x,y,z\in X$ and note that we have the following outcomes: $C^{2}(x,y)=\left\langle y,2\right\rangle$, $C^{2}(y,z)=\left\langle z,2\right\rangle$, and $C^{2}(x,y,z)=\left\langle y,2\right\rangle$. From $C^{2}(x,y,z)=\left\langle y,2\right\rangle$ we have a clear violation of axiom 2 as this axiom requires the decision maker to stop after z in the list (x, y, z). The other three axioms are vacuously satisfied since they impose restrictions on a sc-rule C only if, for $x,y\in X$, $C(x,y)=\left\langle x,1\right\rangle$, $C(x,y)=\left\langle x,2\right\rangle$, or $C(\ell )=\left\langle x,le(\ell )\right\rangle$ with $x\ne \ell _{le(\ell )}$. Notice that the proposed sc-rule $C^{2}$ never induces such outcomes.

Example 5

Fix $x^{*}\in X$ and define $C^{3}$ as follows: for all $\ell \in \Lambda$, $C^{3}\left( \ell \right) =\left\langle \ell _{1},1\right\rangle$ if $\ell _{1}=x^{*}$ and $C^{3}(\ell )=\left\langle \ell _{le(\ell )},le(\ell )\right\rangle$, otherwise. This sc-rule violates axiom 3 as for instance for $y,z\in X$, $C^{3}(x^{*},y)=\left\langle x^{*},1\right\rangle$ but $C^{3}(y,x^{*},z)=\left\langle z,3\right\rangle$, i.e., the decision maker has continued her search after $x^{*}$ was displayed to her in the list $\left( y,x^{*},z\right)$. Clearly, $C^{3}$ satisfies axiom 1 as $C^{3}\left( \ell \right) =\left\langle x,k\right\rangle$ with $k<le(\ell )$ can happen only if $\ell _{1}=x^{*}$ and thus, $C^{3}\left( \ell \right) =\left\langle x^{*},1\right\rangle$ holds for any such list. This sc-rule also satisfies axiom 2: the decision maker continues her search till the end of any list not containing $x^{*}$ at first place, while for all other lists we note that $C^{3}(\ell )=\left\langle x,le(\ell )\right\rangle$ implies $x=x^{*}=\ell _{1}=\ell _{le(\ell )}$ and thus, $C^{3}(\ell _{le(\ell )},y)=C^{3}(x^{*},y)\ne \left\langle \ell _{le(\ell )},1\right\rangle =\left\langle x^{*},1\right\rangle$ can never happen due to the definition of $C^{3}$. Finally, it is easy to see that $C^{3}$ vacuously satisfies axioms 4 and 5, the reason being that a decision maker applying $C^{3}$ never selects an option different from the last one, provided that she has encountered all alternatives in a list.

Example 6

Take the sc-rule $C^{4}$ defined for all $\ell \in \Lambda$ by $C^{4}\left( \ell \right) =\left\langle \ell _{le(\ell )},le(\ell )\right\rangle$ if $le(\ell )\le 2$ and $C^{4}(\ell )=\left\langle \ell _{2},le(\ell )\right\rangle$, otherwise. Notice that for $x,y,z\in X$ we have $C^{4}(x,y,z)=\left\langle y,3\right\rangle$ and $C^{4}(y,z)=\left\langle z,2\right\rangle$ in violation of axiom 4. Clearly, $C^{4}$ satisfies axioms 1, 2 and 3, the reason being that the decision maker always stops at the end of a list (implying that axiom 1 is vacuously satisfied and as required by axiom 2) and that for $x,y\in X$ the situation $C^{4}(x,y)=\left\langle x,1\right\rangle$ never happens (and this is a premise in axiom 3). As $C^{4}(x,y)=\left\langle x,2\right\rangle$ never happens either, axiom 5 is satisfied as well.

Example 7

Define $C^{5}$ as follows: for all $\ell \in \Lambda$, $C^{5}(\ell )=\left\langle \ell _{1},le(\ell )\right\rangle$ if $le(\ell )<3$ and $C^{5}(\ell )=\left\langle \ell _{3},le(\ell )\right\rangle$, otherwise. This rule satisfies axioms 1 and 2 (the decision maker always stops at the end of a list), axiom 3 (as $C^{4}(x,y)=\left\langle x,1\right\rangle$ never happens), and axiom 4 (since each of $C^{5}(x,y)=\left\langle x,2\right\rangle$ for $x,y\in X$, $C^{5}(\ell _{1})=\left\langle \ell _{1},1\right\rangle$, and $C^{5}(\ell _{1},\ell _{2},\ell _{3})=\left\langle \ell _{3},3\right\rangle$ clearly holds). Notice however that, for $x,y,z\in X$, we have $C^{5}(x,y)=\left\langle x,2\right\rangle =C^{5}(x,z)$ and $C^{5}(x,y,z)=\left\langle z,3\right\rangle$ in violation of axiom 5.