Some Annotations to the Previous Post

1. Joe, at this point I'd advice students to draw a decision tree. Some would draw one with six nodes in the first layer, representing the machines $M_1, M_2, ... , M_6$ and then 36 nodes in the second layer, representing each of the possible outcomes from $1,2,...,6$ for each of the machines. At each of the branches, they put the possibility to get from one node to the next, and at the end of the the diagram they write down the 36 probabilities for the outcomes which they get by multiplying the probabilities on the branches which lead to the outcome. However, others would opt for a much easier design, summarizing the machines $M_2, M_3, ..., M_6$ as $\overline{M_1}$, and the non-desirable outcomes $\{1,2, ...,5\}$ as $\overline{6}$, which leads to the following graph:

It is easy to see that in this tree there are two paths to success, one of which was ignored by W. Dembski.

2. Let's assume that W. Dembski is correct, and that " it in the end [it is] even more improbable that we'll find item 6", i.e., that the probability to find a six with this arrangement is 1/12. Then we should find with a probability of 11/12 another number - each with the probability of 11/60. This means that we can use the six perfectly symmetric machines in an experiment where e.g. the 1 is asked for, and we would get a success in 18.6% of all searches - an improvement to the 16.7% for a blind search...

3. A similar problem happens at W. Dembski's Horizontal No Free Lunch theorem in his paper The Search for a Search: the situation is absolutely symmetrical, each of the searches which are feasible in his scenario will have the same probability of success if you take the average over all possible targets. So one would expect the average active information to be the same. But that is not the case: W. Dembski's definition of average active information leads to an outcome (random search isn't equally good, but often even better) which doesn't stand a test.

4. At the moment I've got two comments in moderation at “Conservation of Information Made Simple” at ENV - for nearly two days.

DiEb September 1, 2012 at 12:22 am Your comment is awaiting moderation. W. Dembski: The probability of finding item 6 using this machine, once we factor in the probabilistic cost of securing the machine, therefore ends up being 1/6 x 1/2 = 1/12. So our attempt to increase the probability of finding item 6 by locating a more effective search for that item has actually backfired, making it in the end even more improbable that we’ll find item 6. Yet often, as in this example, we may actually do worse by trying to improve the probability of a successful search. The part in bold fond isn’t true – it isn’t more improbable to find item 6, it is exactly as improbable as before, we don’t do actually worse. @Chance Ratcliff: IMO the final cost of B after securing the correct machine would be P(B|A) and not P(B∩A). “I found the target” – “Sure, but you didn’t use the right machine, so we can’t give you any points for the answer” – that’s just not how it works….

DiEb September 1, 2012 at 11:28 am Your comment is awaiting moderation. @Chance Ratcliff 1) Do you agree that the probability to find the target “6″ using the two-layered system of at first choosing a machine at random and then let the machine choose the target is 1/6 ? 2) W. Dembski says: “So our attempt to increase the probability of finding item 6 by locating a more effective search for that item has actually backfired, making it in the end even more improbable that we’ll find item 6.” How do you square this with your answer to 1) ?

5. Though some of the comments have been uncivil, I'm glad that no one has been baited into a shouting-match. In this case I'd follow Lizzy's lead at The Sceptical Zone and move the offensive comments to a special thread (swamp is a good name).

6. Imagine the following scenario: Instead of six machines you have ten copies of each machine - so sixty machines at all. Each machine is used to find the target, machines which fail to perform the task are destroyed while a successful machine is copied once. After one go the number of machines would be reduced to perhaps ten, five able to find the target with a probability of 1/2, the other five with a probability of 1/10. If you repeat the trials you'll find that only copies of the more successful machine are still in the contest even after a few trials...