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Thread: Book's formula for run probability-- correct or not?

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    Book's formula for run probability-- correct or not?




    Hi, group,

    Semi-retiree here with newfound interest in probability, slowly working my way through the problems in Wm Feller's excellent book. Problem 10 in section II.10 is: At a parking lot there are twelve places arranged in a row. A man observed that there were eight cars parked, and that the four empty spaces were adjacent to each other (formed one run). Given that there are four empty places, is this arrangement surprising (indicative of nonrandomness)?

    I've attached two files: (1) Feller excerpt.pdf = a scan of the relevant passage from Feller's book, and (2) Feller.xls = Microsoft Excel file of my computations.

    There are C( 12,8 ) = 495 possible random distributions of 8 cars in 12 spaces, so each distribution has equal probability of 1/495. The Excel spreadsheet is a breakdown of these 495 distributions tabulated according to the number of runs of cars and spaces, using the formula given in file Feller Excerpt.pdf.

    The formula seems to give the correct result, EXCEPT for the case where the number of runs of cars is the same as the number of runs of spaces. In this case the formula seems to be off by a factor of two.

    This is easy to see. Consider the case where there is one run of cars and one run of spaces. The formula says one distribution. However there are clearly two, one being 8 cars followed by 4 spaces, and the other its mirror image, 4 spaces followed by 8 cars. The same argument applies to the other cases where you have equal runs. You can list them out, they're not that many.

    Finally, you can check the results by summing all the distributions. Feller's formula sums to 375, which is off by 240 from the previously computed number of 495, while my count agrees (see the spreadsheet).

    Am I wrong? If so, where?
    Attached Files

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    Re: Book's formula for run probability-- correct or not?

    Bump! Hello... anybody?

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    Re: Book's formula for run probability-- correct or not?

    Your calculations look correct to me and add up to the expected {12 \choose 8} = 495

    The formula in the book(or at least in the part you excerpted) is specifically for the case of n alpha runs and n + 1 beta runs.

    When you have 2 runs of cars and 1 run of spaces, that can only happen one way: cars|spaces|cars.

    When you have 2 runs of cars and runs of spaces that can also only happen one way: spaces|cars|spaces|cars|spaces.

    When you have 2 runs of cars and 2 runs of spaces that can happen 2 ways cars|spaces|cars|spaces or spaces|cars|spaces|cars

  4. The Following User Says Thank You to asterisk For This Useful Post:

    X-man (11-20-2013)

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    Re: Book's formula for run probability-- correct or not?


    Thanks, asterisk, you're probably right, with probability approaching 1.00...

    You're correct that the formula in the excerpt is for n_1 alphas and n_1+1 betas, but I don't just apply this formula, since I did compute the cases where I have n_1, n_1 -1, and n_1 +1 runs. I basically use the logic behind the formula's derivation, reasoning as follows: If there are n runs of spaces, there can only be n-1, n, and n+1 runs of cars, so that there are respectively C(8-1,n-2), C(8-1,n-1), and C(8-1,n) ways to arrange them, by Feller's lemma. Similarly there are C(4-1,n-1) ways to arrange the 4 spaces in n runs; then the total number of arrangements SHOULD BE the product C( 4-1,n-1 )*[ C( 8-1,n-2 ) + C( 8-1,n-1 ) + C( 8-1,n ) ].

    Except the reasoning breaks down when the number of runs of cars and spaces are the same; in this case we appeal to a visual inspection to see that that's the case.

    I guess what threw me is the ad-hoc feel of the exception. To me it somehow smacks of a nonmathematical way of thinking. But I've seen proofs before consisting of the single line, "by inspection," so I guess that's just me.

    But thanks again, asterisk. Just wanted to make sure I wasn't missing anything.

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