BoundedRationality_TheAdaptiveToolbox.pdf

Comparing Fast and Frugal Heuristics and Optimal Models 149
has the higher value on a criterion. This inference is to be made on the basis of n
binary cues, valued either 1 or 0. From a formal point of view this task is a categorization
task. Here a pair (A, B) is to be categorized as XA>XB ovXB> XA
(where X denotes the criterion), based on cue information.
TWO HEURISTICS FOR CHOICE FROM
THE ADAPTIVE TOOLBOX
When searching for a word in a dictionary, we know which word comes first by
comparing letters in lexicographic fashion. The lexicographic ordering of digits
in numbers also makes their choice easy. For example, to establish that 110 is
larger than 101, we check the value of the first digits on the left and, since they
coincide, move to the second digits, which differ. At this moment we make the
decision that the number containing 1 as a second digit is larger than the number
containing a 0.
This is exactly how Take The Best and Minimalist operate. In their most general
form, these heuristics operate even if some of the objects are unknown to us.
Being unfamiliar with a city, for example, may be a hint that the city is small.
Thus, when comparing city populations, having heard of one city and not the
other is in itself useful information. When recognition is correlated with the criterion,
it is the very first building block of both Take The Best and Minimalist
(Goldstein and Gigerenzer 1999). For the purpose of this chapter I will analyze
these two heuristics as operating on sets of known objects to be compared on a
criterion based on binary cues, all of whose values are known and encoded by
0's and 1 's. If objects has a 1 on a cue and object B has a 0, we expect to have
X A >X B> at least most of the time. Take The Best can be described as follows:
Learning phase. Compute the cue validities defined by:
' Ri + Wt' (9.1)
where R. is the number of right (correct) inferences and Wt the number of
wrong (incorrect) inferences based on cue i alone, when one object has the
value 1 and the other has the value 0. For convenience, I define 0.5 < v- < 1,
which can always be obtained by inverting a cue that shows validity of less
than 0.5, i.e., change each 1 into a 0 and each 0 into a 1. Rank the cues according
to their validity.
Step 1. Ordered search: Pick the cue with the highest validity and look up the
cue values of the two objects.
Step 2. Stopping rule: If one object has a cue value of one ("1") and the other
has a value of zero ("0"), then stop search. Otherwise go back to Step
1 and pick the cue with the highest validity among the remaining
ones. If no cues are left, guess.