The Law(?) of Conservation of Information

(Preamble: I apologize in advance for cluttering TSZ with these three posts. There are very few people on either side of the debate that actually care about the details of this “conservation of information” stuff, but these posts make good on some claims I made at UD.)

For the past three years Dembski has been promoting his Law of Conservation of Information (LCI), most recently here. The paper he most often promotes is this one, which begins as follows:

Laws of nature are universal in scope, hold with unfailing regularity, and receive support from a wide array of facts and observations. The Law of Conservation of Information (LCI) is such a law.

Dembski hasn’t proven that the LCI is universal, and in fact he claims that it can’t be proven, but he also claims that to date it has always been confirmed. He doesn’t say whether he as actually tried to find counterexamples, but the reality is that they are trivial to come up with. This post demonstrates one very simple counterexample.

Definitions

First we need to clarify Dembski’s terminology. In his LCI math, a search is described by a probability distribution over a sample space Ω. In other words, a search is nothing more than an Ω-valued random variable. Execution of the search consists of a single query, which is simply a realization of the random variable. The search is deemed successful if the realized outcome resides in target T ⊆ Ω. (We must be careful to not read teleology into the terms search, query, and target, despite the terms’ connotations. Obviously, Dembski’s framework must not presuppose teleology if it is to be used to detect design.)

If a search’s parameters depend on the outcome of a preceding search, then the preceding search is a search for a search. It’s this hierarchy of two searches that is the subject of the LCI, which we can state as follows.

Given a search S, we define:

  • q as the probability of S succeeding
  • p2 as the probability that S would succeed if it were a uniform distribution
  • p1 as the probability that a uniformly distributed search-for-a-search would yield a search at least as good as S

The LCI says that p1 ≤ p2/q.

Counterexample

In thinking of a counterexample to the LCI, we should remember that this two-level search hierarchy is nothing more than a chain of two random variables. (Dembski’s search hierarchy is like a Markov chain, except that each transition is from one state space to another, rather than within the same state space.) One of the simplest examples of a chain of random variables is a one-dimensional random walk. Think of a system that periodically changes state, with each state transition represented by a shift to the left or to the right on an state diagram. If we know at a certain point in time that it is in one of, say, three states, namely n-1 or n or n+1, then after the next transition it will be in n-2, n-1, n, n+1, or n+2, as in the following diagram:

Assume that the system is always equally likely to shift left as to shift right, and let the “target” be defined as the center node n. If the state at time t is, say, n-1, then the probability of success q is 1/2. Of the three original states, two (namely n-1 and n+1) yield this probability of success, so p1 is 2/3. Finally, p2 is 1/5 since the target consists of only one of the final five states. The LCI says that p1 ≤ p2/q. Plugging in our numbers for this example, we get 2/3 ≤ (1/5)/(1/2), which is clearly false.

Of course, the LCI does hold under certain conditions. To show that the LCI to biological evolution, Dembski needs to show that his mathematical model of evolution meets those conditions. This model would necessarily include the higher-level search that gave rise to the evolutionary process. As will be shown in the next post, the good news for Dembski is that any process can be modeled such that it obeys the LCI. The bad news is that any process can also be modeled such that it violates the LCI.

7 thoughts on “The Law(?) of Conservation of Information

  1. DiEB claims that Dembski made a rookie mistake in arithmetic, and Joe has challenged that claim.

    I’m not a very sophisticated programmer, but it looked like something I cold swing at. So heres my set of six virtual machines loaded with Ys and Ns to what I think are Dembski’s specifications. The results of a few tens of thousands of searches are  included.

    Option Explicit
    Option Base 1
    Private m(6, 6, 10) As String

    Private Sub LoadIt()
        Dim i As Integer, j As Integer, k As Integer
        ‘ load 5 bad machines
        For i = 1 To 5
            For j = 1 To 6
                For k = 1 To 10
                    Select Case k
                        Case 1 To 9
                            m(i, j, k) = “n”
                        Case Else
                            m(i, j, k) = “y”
                    End Select
                Next k
            Next j
        Next i
        ‘ load 1 good machine
        For j = 1 To 6
            For k = 1 To 10
                Select Case k
                    Case 1 To 5
                        m(6, j, k) = “n”
                    Case Else
                        m(6, j, k) = “y”
                End Select
            Next k
        Next j
       
    End Sub

    Private Sub searchIt()
       
        Dim f As Integer, i As Integer, j As Integer, k As Integer, result As String
        Dim Ys As Integer, Ns As Integer, outp As Integer
        LoadIt
        Ys = 0
        Ns = 0
        For f = 1 To 10000
            Randomize
            i = Int(6 * Rnd + 1)
            j = Int(6 * Rnd + 1)
            k = Int(10 * Rnd + 1)
            result = m(i, j, k)
            Select Case result
                Case “n”
                    Ns = Ns + 1
                Case “y”
                    Ys = Ys + 1
            End Select
           
        Next f
        outp = 0
        outp = (Ys / 10000) * 100
        Debug.Print “Ys = ” & Ys & ” (” & outp & “%) — Ns = ” & Ns
    End Sub

    Results:

    Ys = 1720 (17%) — Ns = 8280
    Ys = 1876 (19%) — Ns = 8124
    Ys = 1563 (16%) — Ns = 8437
    Ys = 1685 (17%) — Ns = 8315
    Ys = 1800 (18%) — Ns = 8200
    Ys = 1691 (17%) — Ns = 8309
    Ys = 1742 (17%) — Ns = 8258
    Ys = 1758 (18%) — Ns = 8242
    Ys = 1405 (14%) — Ns = 8595

    I seem to have stumbled on the expected answer, but I defer to others whether this is Dembski’s array of six machines.

  2. I see that R0b has posted a simpler and more accurate model. I could change my arrays to match his, but the results would be the same. None of this is relevant to evolution because the question of probabilities is empirical.

  3. Here’s a version that conforms to R0b’s post. I think it is exactly what Dembski described.

    Option Explicit
    Option Base 1
    Private m(6, 10) As Integer

    Private Sub LoadIt()
        Dim j As Integer, k As Integer
        ‘ load 6 machines
        For j = 1 To 6
            For k = 1 To 10
                Select Case k
                    Case 1 To 5
                        m(j, k) = j
                    Case Else
                        m(j, k) = IIf(k – 5 >= j, k – 4, k – 5)
                End Select
                Debug.Print m(j, k);
            Next k
            Debug.Print “”
        Next j

    End Sub

    Private Sub searchIt()
       
        Dim f As Integer, i As Integer, j As Integer, Result() As Integer
        Dim k As Integer, OutP As Integer
        LoadIt
        For i = 1 To 10
            ReDim Result(6) As Integer
            For f = 1 To 10000
                Randomize
                j = Int(6 * Rnd + 1)
                k = Int(10 * Rnd + 1)
                Result(m(j, k)) = Result(m(j, k)) + 1
            Next f
            OutP = 0
            OutP = (Result(6) / 10000) * 100
            Debug.Print “Sixes = ” & Result(6) & ” (” & OutP & “%)”
        Next i
    End Sub

    Results:

     1  1  1  1  1  2  3  4  5  6
     2  2  2  2  2  1  3  4  5  6
     3  3  3  3  3  1  2  4  5  6
     4  4  4  4  4  1  2  3  5  6
     5  5  5  5  5  1  2  3  4  6
     6  6  6  6  6  1  2  3  4  5
    Sixes = 1459 (15%)
    Sixes = 1827 (18%)
    Sixes = 1685 (17%)
    Sixes = 1678 (17%)
    Sixes = 1719 (17%)
    Sixes = 1562 (16%)
    Sixes = 1753 (18%)
    Sixes = 1485 (15%)
    Sixes = 1530 (15%)
    Sixes = 1686 (17%)
     1  1  1  1  1  2  3  4  5  6
     2  2  2  2  2  1  3  4  5  6
     3  3  3  3  3  1  2  4  5  6
     4  4  4  4  4  1  2  3  5  6
     5  5  5  5  5  1  2  3  4  6
     6  6  6  6  6  1  2  3  4  5
    Sixes = 1654 (17%)
    Sixes = 1679 (17%)
    Sixes = 1540 (15%)
    Sixes = 1822 (18%)
    Sixes = 1592 (16%)
    Sixes = 1827 (18%)
    Sixes = 2029 (20%)
    Sixes = 1952 (20%)
    Sixes = 1768 (18%)
    Sixes = 1574 (16%)
     1  1  1  1  1  2  3  4  5  6
     2  2  2  2  2  1  3  4  5  6
     3  3  3  3  3  1  2  4  5  6
     4  4  4  4  4  1  2  3  5  6
     5  5  5  5  5  1  2  3  4  6
     6  6  6  6  6  1  2  3  4  5
    Sixes = 1842 (18%)
    Sixes = 1407 (14%)
    Sixes = 1736 (17%)
    Sixes = 1510 (15%)
    Sixes = 1725 (17%)
    Sixes = 1845 (18%)
    Sixes = 1526 (15%)
    Sixes = 1982 (20%)
    Sixes = 1723 (17%)
    Sixes = 1680 (17%)

     

     

     

  4. Dieb is right (it hardly needs a computer programme to prove it). But I don’t think it is big problem for Dembski. I think his point would be that having got a six via the machine which increases your chances of getting a six you might think you had improved your overall odds. In fact the chances of getting six via that route are lower than by simple throwing the dice.

    Actually this is just another way of looking at R0b’s balls (if you will pardon the expression) otherwise known as Betrand’s Box which I think is a brilliant way of thinking clearly aboout this.

    Suppose there was a large number of six different colours of balls all in the same proportion. You want a gold one. You have two options. You can just choose a ball at random from the master bag. Or you can choose a box which has 5 of one colour and one of each other colour – but you don’t know which colour – and then choose a ball. It is rather obvious that your chances of getting a gold ball remain the same.

    That is essentially the NFL theory. The LCI adds a corollary that the probability of selecting a box which increases your chances of getting a gold ball is always so low that your overall chances of getting a gold ball by that route are not increased. It is a slightly different statement and to my mind not very interesting except outside the bizarre interpretation he has put on it.

    The point being that the LCI does not apply at all to the evolution of life or indeed most of the real world. In evolution of life a box already exists which greatly increases the chances of gold balls (it is called the environment it works by eliminating the other colours). The proportion of gold balls before entering this box is absolutely unrelated to the probability of this box existing.

    I just thought of an even simper way of expressing this using Dembski’s own example.  His example assumes that we know we want a six in advance.  The way evolution works we don’t know what number we want.  There is a machine out there (and in his model one such machine must exist) which makes one number more likely. We discover which number it is because that’s the one the machine creates. The machine (the environment) defines the target (whatever is needed to survive) – not the other way round.

  5. The error itself isn’t devastating, but it punctures the aura of infallibility that Dembski has in the ID community, not the least, because it slipped through ID’s peer review. It might mean very little in absolute terms, but it exposes the inability of the ID community to think clearly about probability.

    Dembski has devoted quite a bit of time and effort to crying “squirrel”! Look over there! It’s a shell game where the object is obfuscation and misdirection.

  6. Perhaps the most revealing thing about this episode is not what it says about Dembski but what it says about his followers.

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