So, let $\Omega = \{\omega_1, \omega_2, \dots, \omega_N\}$ be our finite search space with $N$ elements. We are looking for a single element $\omega_k$, so we try to maximize the fitness function $f = \chi_{\omega_k}$. To keep everything finite, we don't allow repetitions, i.e., in our search each place can only be visited once. This is - as Macready and Wolpert observed - always possible by keeping a look-up table and thus doesn't change the set-up. Therefore, our search is completed in at most $N$ steps."The search is defined to be a six-tuple consisting of the initiator, terminator, inspector, navigator, nominator, and discriminator. The paper studies the question of picking a search at random, and that would imply picking each of the six components at random. We did not consider it necessary to specifically state that each individual component was also selected at random. That would seem to be implied.

(BTW: The claim that

*"each of the six components [is picked] at random"*seems not to apply to the

*inspector*: this is a fixed function for a search - in our case, the

*inspector*returns the value of the fitness function. Of course, you can say that we pick the

*inspector*at random out of the set of the one possible

*inspector*.)

Let's take a look at all the searches which are ended by their

*terminator*only after the $N$-s step, i.e., the subset of all exhaustive searches. The price question:

**What is the probability to find the target in such an exhaustive search?**Until now, everyone looking at such problems would have thought that this probability is

**one**: we certainly visited $\omega_k$ and spotted that the function $f$ takes it maximum there. But in the world of Dembski, Ewert, and Marks it is not, as a random

*discriminator*takes its toll - and

*discriminators*aren't obliged to return the target if it was found and identified...

Counterintuitive? That is a flattering description: the

*discriminator*'s purpose seems to be to turn even a search which is successful by all human standards into a guess to fit the

*idée fixe*that each search can be "represented" by a measure on the search space.

**Addendum:**We can drop the condition of not having repetitions in our searches and just look at those searches which are terminated only after the whole search space was visited:

*terminators*with this property exist. Such searches may have length $N$, but can be much longer. The result is the same: the probability of finding the target during a complete enumeration of the search space is (much) less than one. I have to ask:

**What good is a model in which an exhaustive search doesn't fare much better than a single guess?**