How to interpret ratios calculation in H. Simon's watchmaker parable?

  1. Hello! I posted this question also at Math Exchange. But since it is not getting many views there, let me try here too. I hope that's OK. There are perhaps no interesting mathematics per se here, but it is about their interpretation/usage.

    In 1962 Herbert Simon wrote a paper in which he presents his watchmaker parable about how piecemeal construction is more robust than one-shot construction. I won't repeat the full argument here - see the link - though the gist is below. My question is about the 'straightforward calculation'. While the math itself is simple as can be, I can't seem to fully pinpoint why it makes sense to use the ratios as he does.

    To summarize the givens: To complete a watch...
    • ... Tempus has to construct 1 assembly (of 1000 parts), has a .991000 chance to achieve that, and loses (the time of making) 100 parts on average when he fails;
    • ... Hora has to construct 111 assemblies (of 10 parts each), has a .9910 chance to achieve each assembly, and loses (the time of making) 5 parts on average when he fails.
    And the question is: How much longer does it take Tempus to complete a watch?

    (Note that Simon does quite a bit of rounding to get to his ~4000 figure -- from 3774.542... -- but that is not the issue here.)

    So, Hora has to build 111 assemblies, against 1 for Tempus:


    But building one assembly is easier for Hora than for Tempus, so it makes sense to want to throw in the achievement probabilities. Now it has to be easier for Hora to make an assembly (bringing down the 111 figure, relatively), so the multiplier needs to be small, relatively, which you get by taking the inverses of the probabilities:

    [itex]\frac{111}{1} × \frac{\frac{1}{.99^{10}}}{\frac{1}{.99^{1000}}} = \frac{111}{1} × \frac{.99^{1000}}{.99^{10}}[/itex]

    But how to understand - in words - this inverse-taking? I would be inclined to want to multiply e.g. 111 with .9910 (rather than .991000) from the idea that Hora is accumulating probabilities as he goes along building assemblies. Except that doesn't fit with the reasoning in the previous paragraph, and furthermore 111 times .9910 equals 100.386... which is not a probability of any kind... I'm lost: what are we actually calculating here? And then when the average losses are added in to obtain

    [itex]\frac{111}{1} × \frac{.99^{1000}}{.99^{10}} × \frac{5}{100}[/itex]

    I get still more lost as to its interpretation.

    Can anyone clarify? Much obliged.
  2. jcsd
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