Re: Dembski and Caesar cyphers

From: Iain Strachan (iain.strachan@eudoramail.com)
Date: Fri Nov 17 2000 - 14:39:45 EST

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    A small correction to my last post.

    I stated that Dembski's probability bound of 10^-150 was equivalent
    to the same person winning the lottery 13 times in succession. I
    should not have done the calculation in my head. The actual figure
    is 22 times in succession. (Or a specified person winning it 21
    times in succession). (chance of winning lottery is around 1 in 13
    million; 150/log10(13,000,000) = approximately 21.

    It makes little difference to the argument. It would be all over the
    tabloids if the same person won it twice in a row, and three times in
    a row would prompt an enquiry into the fair running of the lottery
    (i.e. based on a "design" inference) for certain.

    Iain.

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