Friday, August 31, 2012

Could you please correct your miscalculation, Dr. Dembski?

William A. Dembski wrote "a long article […] on conservation of information. " at Evolution News and Views (ENV), an outlet of the Discovery Institute. Others have commented on more sophisticated problems, either at Uncommon Descent or at The Skeptical Zone. Here I want just to correct some simple math which occurs in a toy example used in the article:
To see how this works, let's consider a toy problem. Imagine that your search space consists of only six items, labeled 1 through 6. Let's say your target is item 6 and that you're going to search this space by rolling a fair die once. If it lands on 6, your search is successful; otherwise, it's unsuccessful. So your probability of success is 1/6. Now let's say you want to increase the probability of success to 1/2. You therefore find a machine that flips a fair coin and delivers item 6 to you if it lands heads and delivers some other item in the search space if it land tails. What a great machine, you think. It significantly boosts the probability of obtaining item 6 (from 1/6 to 1/2).
But then a troubling question crosses your mind: Where did this machine that raises your probability of success come from? A machine that tosses a fair coin and that delivers item 6 if the coin lands heads and some other item in the search space if it lands tails is easily reconfigured. It can just as easily deliver item 5 if it lands heads and some other item if it lands tails. Likewise for all the remaining items in the search space: a machine such as the one described can privilege any one of the six items in the search space, delivering it with probability 1/2 at the expense of the others. So how did you get the machine that privileges item 6? Well, you had to search among all those machines that flip coins and with probability 1/2 deliver a given item, selecting the one that delivers item 6 when it lands heads. And what's the probability of finding such a machine? To keep things simple, let's imagine that our machine delivers item 6 with probability 1/2 and each of items 1 through 5 with equal probability, that is, with probability 1/10. Accordingly, this machine is one of six possible machines configured in essentially the same way. There's another machine that flips a coin, delivers item 1 from the original search space if it lands heads, and delivers any one of 2 through 6 with probability 1/10 each if the coin lands tails. And so on. Thus, of these six machines, one delivers item 6 with probability 1/2 and the remaining five machines deliver item 6 with probability 1/10. Since there are six machines, only one of which delivers item 6 (our target) with high probability, and since only labels and no intrinsic property distinguishes one machine from any other in this setup (the machines are, as mathematicians would say, isomorphic), the principle of indifference applies to these machines and prescribes that the probability of getting the machine that delivers item 6 with probability 1/2 is the same as that of getting any other machine, and is therefore 1/6. But a probability of 1/6 to find a machine that delivers item 6 with probability 1/2 is no better than our original probability of 1/6 of finding the target simply by tossing a die. In fact, once we have this machine, we still have only a 50-50 chance of locating item 6. Finding this machine incurs a probability cost of 1/6, and once this cost is incurred we still have a probability cost of 1/2 of finding item 6. Since probability costs increase as probabilities decrease, we're actually worse off than we were at the start, where we simply had to roll a die that, with probability 1/6, locates item 6. 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. Conservation of information says that this is always a danger when we try to increase the probability of success of a search -- that the search, instead of becoming easier, remains as difficult as before or may even, as in this example, become more difficult once additional underlying information costs, associated with improving the search and often hidden, as in this case by finding a suitable machine, are factored in.
Can you spot the error in his calculation? The probability to find the correct machine and then the target is indeed 1/12, but the probability to find the target via chosing a machine at random at first is 1/6, thanks to the symmetry of the problem: The probability for a success is 1/6 * 1/2 + 5/6 * 1/10 = 1/6. So the problem didn't become more difficult.

80 comments:

  1. Could you please explain why your "+" should not be a "*".

    Thank you

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  2. Elementary, my dear Watson - my answer to you at UD is still in moderation, so I'll copy it for you:


    It’s called the Law of total probability. Consider the events:

    T: the target 6 is identified and
    S: the better machine is chosen

    Then P(S) = 1/6, P(T|S) = 1/2, P(T|not(S)) = 1/10 and by said law we get:

    P(T) = P(T|S)*P(S) + P(T|not(S))*P(not(S)) = 1/2 * 1/6 + 1/10 * 5/6 = 1/6

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  3. Apparently the First Law of ID is "Length is Strength."

    The strength of an argument is directly proportional to the quantity of verbiage expressing it, multiplied by the mean length of the words employed.

    This weeds out all but the most dedicated critics.

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  4. Umm DiEB that is just restating your post as opposed to explaining it.

    So perhaps you could provide a demonstration that proves what you say is true.

    To Petrushka, apparently the first law of your position is throw out as many bald assertions and misrepresentations as possible

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  5. OMG. wMad fails probability 101.

    Joe - Dr. Dembski calculates success only if you get the right coin. He assumes that if you get the wrong coin, the probability is zero.

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  6. http://theskepticalzone.com/wp/?p=1120&cpage=1#comment-15418

    Joe. It takes longer to answer a question than to ask it, but here it is.

    ReplyDelete
  7. Don't know if this will wpublish, but here's the BASIC program and results that I think models Dembski's six machines.

    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

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  8. I believe you intended to say "the right machine" rather than "the right coin", Bob. I don't know if that's a probability fail or a more general translation from English to math fail, but fail it is.

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  9. Well I've modeled six machines. In five of them you have a one in ten chance, and in one of them you have a one in two chance.

    What is the combined probability of getting a "y"?

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  10. Who are you going to believe, Petrushka? Dembski or your lying eyes?

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  11. I haven't had any math courses since 1963. I'm waiting for others to see if I did the model correctly.

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  12. Thanks, Petrushka, for modeling De. Dembski's toy problem - it works!

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  13. Given the two-tiered search- level 1 for the machine and level 2 for the 6- each of the six pathways (pathway being via one of thesix machines) to 6 starts with a probability of 1/12. And that is half as good as the original.

    If you get the correct machine at level one then your odds jump to 1/2, which is 3 times better than the original. And if you get one of the other machines your odds step to 1/10, which is 2/3 worse than the original.

    And the odds of getting a path that is worse than the original is 5/6.


    And Bob Oh- you are correct and that is why DiEB is incorrect

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  14. Here's a version that follows Dembski's description.

    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%)






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  15. Here's a version that follows Dembski's description.

    How can that be seeing he is only talking about ONE possible path, ie the path to get the 50/50 chance at hitting a 6?

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  16. Joe, there is no correct machine. All the machines have some probability of producing a six. My code clearly picks a machine first and then a "position" from the array representing that machine.

    So it is a two tier search.

    You can't just ignore the possibility that the "wrong" machine will still produce a six.

    Anyway, I'm sure Dr. Dembski will hear about this and respond.

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  17. Petrushka-

    There is a 1/6 chance of picking any one machine. So you are already at 16.7% after the first step. So how can you get, as a result, 17% after TWO rounds? You have to include the odds of picking that particular machine with your calculations. Then you have the odds at finding the 6 on top of that.

    And YOU don't get to change what Dembski is talking about and no one is ignoring the possibilty that the "wrong" machine will produce a 6. So stop the strawman already.

    You people just can't leave it at what he is talking about. Someone trying to increase their odds 1/6 wants the machine that will give him a 50/50 chance. He is trying to make it easier.

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  18. Joe,

    arent you excluding the probability of picking one of the other 5 machines that each has the probability of choosing 6 ( 5/6 * 1/10 = 1/12)? Does you consider this irrelevant to Dembski's point?

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  19. I'm betting Dembsmi would rather admit to a momentary lapse of reason than remain on record with this. The "target" is a 6. You have to find the probability of finding a 6. There is no wrong machine. There are only machines having different probabilities of producing a 6. The calculation would be the same if some had no 6s.

    Now I am certainly not claiming to be smarter than Dembski. I didn't find the error. I didn't even understand it at first. It's been about 45 years since I took statistics. I really only understood it as I programmed the model.

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  20. paragwinn,

    I am not excluding anything. EACH machine has a 1/6 chance of being picked. And you have to factor that in.

    It is like rolling a dice to make the first pick- ie picking a machine.

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  21. Petrushka,

    Your program is wrong as it does not account for the odds of the first choice- the first choice is choosing a machine.

    Yes the final target is a 6 but the FIRST target is a machine.

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  22. The machine is chosen in the line following Randomize, "j=Int(6*Rnd+1)." That's choosing the machine, Joe. It couldn't be more explicit.

    Think about this. Having choosen the machine, what next?

    The error is not in the arithmetic. It's in the logic. There is no difference between having multiple machines and having just one big one.

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  23. There's one other thing about the so-called search for a search. Suppose the machines were stores, and the goal is not digits but lowering one's costs.

    It doesn't take much imagination to realize that over time one would learn which store minimizes cost. Learning and evolution are less like climbing hills than they are like following the terraine into valleys. Or like water conforming to the bottom of a pond.

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  24. “is it easier to roll a die to get your 6, a 1/6 probability than it is to also have to search for a machine that will allow you to find your 6 by say flipping a coin? And if there are 6 machines out there that will help you find your 6, but only one is the 50/50, is that less or more difficult than just rolling a die?”

    And if the machines were in stores you would have to search for those stores.

    So again which is the easiest route? Rolling a die vs finding a store then finding a machine that then has to do something that has either a 1/2 or a 1/10 chance of getting you to your target?

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  25. Having choosen the machine, what next?

    You have to use the machine to see if you get what you are searching for.

    But think this, if you roll a die to pick your machine you have already used the 1/6 chance. yes the odds are 1/1 that you will pick a machine, but you don't know which.

    And before all that you have to search for the machines.

    OR you could just roll the die in the forst place.

    Which is easier? And if you factor in the manufacture of the machines vs the manufacture of the die....

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  26. Joe. Dembski's search for a search is just a two diminsional array of elements. I accidentally modeled as three dimensions and got the same answer. Guess what? It could be a search for a search for a search for a search and it would still be an array of elements. The target is the element. The only thing that affects probability is the ratio of targets to elements.

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  27. Petrushka,

    Way to avoid responding to everything I said. Nothing says that you are a coward more than your avoidance of your opponent's posts.

    Good luck with that...

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  28. You answered your own question. The probility of choosing a machine is one. One times the remaining probability is the same as the remaining probility.

    Since you have no 6 detector and no information about the contents after choosing the machine, the probability remains the total number of 6s divided by the number of digits in all the machines.

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  29. You still fail to follow along. You need to search for the machines in the first place.

    And your target for the first search is the machine that gives you a 50/50 chance of success. Only an evo would leave behind 1/6 odds in search for 1/10 odds and think he would be better off.

    “is it easier to roll a die to get your 6, a 1/6 probability than it is to also have to search for a machine that will allow you to find your 6 by say flipping a coin? And if there are 6 machines out there that will help you find your 6, but only one is the 50/50, is that less or more difficult than just rolling a die?”

    Everyone reading knows why you are avoiding answering those questions. Thank you

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  30. You are just highlighting Dembski's muddled concept. He says the machines are unlabeled and appear identical. So having chosen one, how do you know what kind it is? Do you need a ten sided die or would a coin suffice? How do you query the machine?

    Dembski avoids discussion of the mechanics of querrying. But his description is of an array such as onlooker posted at UD. And which I programmed.

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  31. You are just highlighting Dembski's muddled concept.

    Just because you are muddled doesn't mean Dembski's concept is. And thank you for continuing to avoid those questions. The questions that get to the very heart of Dembski's concept. But then again there is your muddle problem....

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  32. Looking at Debbski's description of choosing the machine and querrying it, onlooker's array and my program are an exact model. If you disagree, find a teacher of statistics to correct us.

    Or email Dr Dembski and ask him to comment. It's his thread at UD.

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  33. “is it easier to roll a die to get your 6, a 1/6 probability than it is to also have to search for a machine that will allow you to find your 6 by say flipping a coin? And if there are 6 machines out there that will help you find your 6, but only one is the 50/50, is that less or more difficult than just rolling a die?”

    Everyone reading knows why you are avoiding answering those questions. Thank you

    Ignore your teeth and they will go away. Ignoring those questions makes your "argument" go away. Thank you.

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  34. This can be settled very easily. Write your own program that models Dembski's machines and his description of choosing and querying them. Your choice of language.

    Or ask Gil to do it.

    Then publish it here with the results.

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  35. I've answered. You don't like the answer. He starts with the problem as you quoted, but when he describes the machines and the method of querying them it is as I have programmed.

    I await his response.

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  36. No, you didn't answer me. Now you are lying.

    You do realize that people who read this aren't as dishonest as you are being.

    He starts with the problem as you quoted, but when he describes the machines and the method of querying them it is as I have programmed.

    Nope, that doesn't answer my questions. Try again.

    And please hold your breath while you wait for his response.

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  37. Is it easier to write a program or to roll a die?

    Well first to write a program you have to search for a computer, the compiler, etc.- you just refuse to get it. No wonder you always retreat to equivocation...

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  38. We'll add basics probability theory to the long list of things (biology, genetics, physics, chemistry, genetic algorithms, paleontology, archaeology, the Bible, etc.) that Joe G is completely clueless on.

    It is pretty funny to see him try and defend Dembski's boner though.

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  39. And we can add Thorton to the list of cowards who refuse to answer my questions. And your cowardly false accusations are duly noted.

    But nice to see that you are still upset because I have exposed your ignorance on more than one occasion.

    That pretty much settles it that rolling a die is easier and Dembski is once again correct.

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  40. Joe can't code. Its unfair to ask him to use computers, he's not as capable as you with them.

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  41. There are lots of programmers at UD. They could add one more refinement. Instead of onlooker's matrix they could use Demski's coin toss.

    It won't change the result.

    Joe: After choosing the machine, how do you know which machine you chose? After querying the machine, how do you know which machine you chose?

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  42. Joe, W. Dembski wrote the article - therefore it is fair to use his methods and definitions. And in the toy example, he isn't interested in the physical realization of the experiment, but in its mathematics. That's why a computer simulation is spot on. And when he talks about the "cost" of the operation, he doesn't factor in the manufacture of the machines vs the manufacture of the die...., he has it defined as "probability cost" or "information cost". That's why no one answers your questions for the cost - as you doesn't use cost in the sense of W. Dembski. Isn't that what's called equivocation?

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  43. DiEB- DEMBSKI writes:

    But then a troubling question crosses your mind: Where did this machine that raises your probability of success come from?

    Then he finishes with:

    Conservation of information says that this is always a danger when we try to increase the probability of success of a search -- that the search, instead of becoming easier, remains as difficult as before or may even, as in this example, become more difficult once additional underlying information costs, associated with improving the search and often hidden, as in this case by finding a suitable machine, are factored in.

    IOW DiEB you are selectively reading Dembski which is a sure sign of a coward.

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  44. Nice to see RichTARD chime in with its usual false accusations.

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  45. Petruska,

    How do you know which machine your program picks?

    And is it easier to write a program or roll a die?

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  46. Joe: " Finding this machine incurs a probability cost of 1/6," At least in the toy example, W. Dembski talks about costs only in the sense of probability or information costs. Please reread the example carefully - which is neither a sign of cowardice nor of special braveness IMO.

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  47. So you are just going to ignore what I posted. Got it.

    That's fine just don't think that you are actually making a case.

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  48. Is searching for, finding and using a suitable machine easier or more difficult than just rolling a die?

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  49. Joe, I don't ignore what you are posting (look here). In the real world, building a machine which flips a coin - which you have to provide - etc. may be often more difficult than throwing a die. But that's not how difficulty or cost is defined in W. Dembski's pure mathematical context.

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  50. DiEB- reread what I posted. It is all part of how it is defined.

    The point being is Dembski is talking about the real world. The more layers of search you add the more complex the search becomes.

    To me that is as simple as 1 + 1 = 2

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  51. "Imagine that your search space consists of only six items, labeled 1 through 6."

    "The point being is Dembski is talking about the real world."

    Joe: 0
    Dembski: 1

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  52. Joe, this is a serious question. Following Dembski's example as he describes it, having picked one of six machines and tossed the coin and gotten a result, how do you know which machine you picked? How do you know whether or not you picked the good one?

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  53. Petrushka- I am STILL waiting for YOU to answer my questions.

    Until then...

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  54. Hey petrushka- my car is a physical system- can it learn and adapt?

    Also which has the better chance of success- a purely blind search, ie one that doesn't even know the target or a targeted search?

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  55. Petrushka, what part of your program factors in the probabilities of finding the machines in the first place?

    Once you factor that in will your final %s go up or down?

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  56. Joe,
    Why don't you just write a model of your own, the way you see it as working? Then the behaviour of the models can be compared directly?

    It's only a few lines of code.

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  57. Joe:n Dembski’s scenario we can get the final target of a 6 without securing the “6″ machine. And we would never know if the 6 we got was from the “6″ machine or any of the other 5 machines.

    And if you can’t tell the machines apart- isomorphic/ the principle of indifference- then the ONLY thing you will know is if your search was successful or not. You will never know why.



    Thanks for answering my question. You finally understand that Sembski's scenario makes no provision for knowing which configuration you chose. Therefore his calculation makes no sense in his context.

    It would make no sense in real life either, because if one of a hundred slot machines in a casino started paying off at a rate higher than the coin input, it would be overwhelmed by customers rather quickly.

    I'm left wondering what his point could possibly be.

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  58. Anonymous, First YOU tell me how to factor in the probabilities for finding the machines. Or shut up because you obvioulsy don't even understand what Dembski is saying.

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  59. Petrushka,

    I TOLD you what Dembski's point is. If you would have the balls to answer my questions, as opposed to being the coward you are, you would see the point.

    Of course Dembski's calculation makes no snse in your context. Your context is meaningless.

    So you can shut up too...

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  60. Joe G

    Anonymous, First YOU tell me how to factor in the probabilities for finding the machines. Or shut up because you obvioulsy don't even understand what Dembski is saying.


    You're so predictable Joe. Every time you get caught in an idiotic blunder like now you try to muddy the water by bringing up all sorts of meaningless rabbit trails.

    Why aren't you whining about the probability of finding a fair die to roll? Or the probability you'll have the lights fail and you won't be able to read the die? They have just as much relevance to Dembski's example as the other stupid smoke trails you're spewing.

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  61. This comment has been removed by a blog administrator.

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  62. Joe G

    Dembski's whole point was how difficult it is when you add layers.


    And his whole point was rendered void because he made a freshman math error, the same error you are too stupid to grasp. In his example the probability of ending up with a 6 didn't get more difficult, it stayed the same.

    In real life, evolution doesn't have "layers" or "searches for searches". In real life species search the immediate vicinity of the working solution they already have. They don't have to blindly search the space of all possible solutions as Dembski naively assumes. Dembski's whole article is meaningless blather designed to prop up his idiotic 'evolution can't do it' claims.

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  63. No Thorton. The only error he made was thinking that people who read it wouldn't be infantile.

    In real life evolutionism doesn't do anything. natural selection doesn't do anything. Your entire position is evidence-free nonsense. THAT is why you have to flalil away at news ideas- they frighten you.

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  64. Joe, if you start posting off topic rants, strings of vile obscenities and links to pornography like you did at TSZ, I bet you can get yourself banned here too.

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  65. This comment has been removed by a blog administrator.

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  66. Joe,
    "Anonymous, First YOU tell me how to factor in the probabilities for finding the machines."

    If you don't understand what Dembski is describing that's fine. Why don't you write your model as best you can anyway?

    Note: If your "e" key is missing on the computer you use to create the model I'm afraid you'll have to calculate how the probability of finding the machine changes yourself. Likewise a power cut etc.

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  67. Anonymous,

    I understand what Dembski is describing. Obviously you do not. Do not blame me because you are an imbecile.

    But anyway DiEB is being a typical evoTARD faggot so I have nothing else to say here.

    ReplyDelete
    Replies
    1. Joe, I won't delete this comment, but I will delete any comment in which such language is used against any other commenter.

      Delete
    2. DiEB-

      I'm finished here- do whatever you want.

      That you can't see that searching for and locating six such machines is quite a bit more difficult than rolling a die tells me all I need to know about you.

      THAT is what I would be saying to the class as I walked out. And I bet most, if not all, of the rest would follow me.

      Delete
    3. 1) Sometimes the complexity of a algorithm which uses a random generator is defined as the number of random numbers which has to be generated - as this is the most costly operation. By this definition, the cost for looking for the machine and then rolling a die would have been twice as much the cost of just rolling a die. But that's not the definition of cost W. Dembski uses - and it would be an equivocation to insert the above definition out of the blue into an article where W. Dembski is using his own definition of costs, the probability costs.

      2) At UncommonDescent you write to W. Dembski : You should have had the other 5 machines give a zero probability of hitting a 6- no output at all. That way there is definitely a suitable machine that you are searching for and ignoring the other 5 is valid in a real world scenario.
      The six machines W. Dembski uses in his toy example stand in for the infinite number of possible guessing strategies (they can be represented perhaps as points on a 5-simplex). A fundamental property of this set of measures is the symmetry - and this is reflected by W. Dembski's choice of measures.

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    4. DiEB- The probability costs are 1) machines that do the job and 2) searching for and locating them.

      That you refuse to grasp that simple concept, even though it has been explained to you, tells me I have stayed here much too long

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    5. W. Dembski will be surprised - he writes Finding this machine incurs a probability cost of 1/6. Your concept may be simple, but just not the concept W. Dembski is using.

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  68. Joe: Suppose all the machines were identical but you still had to choose one. Would that double your cost?

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  69. Petrushka-

    You STILL have to search and locate the machines- without even knowing if they exist.

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  70. Joe G

    You STILL have to search and locate the machines- without even knowing if they exist.


    You STILL have to search and locate the die to roll - without even knowing if it exists too. Where are those costs in your calculations?

    Can't have it both ways.

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    Replies
    1. Umm you already had the die. As I said you are too simple to even follow along

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    2. Then you already knew where the machines were too.

      Maybe if you hadn't walked out of so many school classes you wouldn't be so ignorant. Maybe you'd understand that probability cost given in the example by Dembski doesn't mean monetary/logistic/mechanical "costs".

      But you did, and you are, so you don't.

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    3. Nope- again I see that your inability to follow along has you spewing nonsense as if it means something.

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  71. DiEB,

    I see that English is not your first language. The 1/6 cost was CHOOSING the correct machine out of the six. IOW it does not include searching for the machines.

    As I said you are a waste of time.

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