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Math Help - League tables

  1. #1
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    League tables

    3 teams (A,B,C) play in a league of 20. Using historical data for the mean and SD of points totals one can compute P(A finishes season higher than B) "P(A>B)" etc.

    Q1. Given just P(A>B) and P(B>C) can one calculate P(A>C)?

    Q2 Given the above how do you then calc. P(A finishes higher than BOTH B & C)?
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  2. #2
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    Re: League tables

    Didn't we already do this? If you have 20 years of data, why not just take it from the data?

    There remain only six outcomes:

    ABC
    ACB
    BAC
    BCA
    CAB
    CBA

    Event(A>B) = U(ABC,ACB,CAB)
    Event(B>C) = U(ABC,BAC,BCA)
    Event(A>C) = U(ABC,ACB,BAC)
    Event(A>Both) = U(ABC,ACB)

    Off the top of this third rumination, I'd still say you need more information.
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  3. #3
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    Re: League tables

    I re-posted because I was still stuck without a formula relating P(A>B), P(B>C),
    P(A>C) and P(A>B&C). And still am! I understand your expressions but don't know how to calculate.

    I don't have 20 years of data - it's a league of 20 teams. But regardless of the data, I'm just interest in the theory and annoyed with myself that I cannot find a formula.

    Lets put some numbers in and you'll see my dilemma.

    1. If P(A>B) = P (B>C) = 0.5 then (intuitively) P(A>C) = 0.5 and P(A>B&C) = 0.3333

    In this scenario the three teams have equal strength.

    2. If P(A>B) = 1 and P(B>C) = 0.5 then answers are also intuitive, being 1 and 1.

    3. But say P(A>B) = 0.6 and P(B>C) = 0.55 then what formulae do I plug 0.6 and 0.55 into to get results for P(A>C) and P(A>both)?

    That's where I'm stuck.

    This formula must work for any scenario of course
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  4. #4
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    Re: League tables

    Quote Originally Posted by RichardJBentley View Post
    I don't have 20 years of data - it's a league of 20 teams.
    My mistake. If you have historical data, why not just take it from the data?

    annoyed with myself that I cannot find a formula.
    Perhaps it would be wiser to be annoyed becaue you do not recognize there is insufficient information.

    In this scenario the three teams have equal strength.
    If they have equal strength, each of the six is eqully likely.

    ABC
    ACB
    BAC
    BCA
    CAB
    CBA

    P(A>B) = 3/6
    P(B>C) = 3/6
    P(A>C) = 3/6
    P(A>B&C) = 2/6

    Where's the tricky part?
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  5. #5
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    Re: League tables

    OK, let's consider a subset of all possible team strengths. In this subset teams B & C have equal strength [P(B>C) = P(C>B) = 0.5] and A may vary. Since B & C are equal then P(A>B) = P(A>C) = P1 for present purposes. Some trial & error work gives the following definite results:

    If P1 = 0 then P(A>both B&C) = 0
    If P1 = 1/3 then P(A>both B&C) = 1/6
    If P1 = 1/2 then P(A>both B&C) = 1/3
    If P1 = 2/3 then P(A>both B&C) = 1/2
    If P1 = 1 then P(A>both B&C) = 1

    What I want is an equation which relates P1 to P(A > both B&C), so that I can plug in any value for P1 and get the result. That equation must deliver the above results and give the result for any value of P1 in between the above easy cases.

    Then I want to be able generalise that equation to the situation where B & C are not equal strength teams as in the subset above.

    There cannot be any information missing. There are no other relevant variables as far as I can see. I have first class degree in maths from a good english unversity but I'm just rusty after 25 years!!! I get the feeling there are equations for surfaces needed here somewhere and perhaps the equation of the line that solves the above subset is a line on the general surface?
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