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#1 |
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![]() Say you had a horse race. You decided to choose 4 of 8 horses. Okay, well thats all well and good as 8c4=(I cant find my sci calc), and then I know the multiple to be 4!.Anyway, what I'm trying to get at is how shoould I figure a multiple into the strict combinatoric rules of odds that would account for handicapping. Specifically, if I knew that the favorite finishes in the money (thats 1st , 2nd or 3rd) x% of thetime and that my ability to choose the other three finishers correctly were some indeterminate variable yet to be developed that somehow could be used as a multiple.Please help! Thanks
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#2 |
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![]() you don't make sense. combinatorics give the number of possible events C(8,4) ways to select the top four finishers in order, but has nothing to do with their probability. vSo that information about one horse (the favorite) will shed no light on the relative order of the rest of the field. more data is needed to make a start at this.
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