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    Simple but annoying probability question




    A, B and C are in three way race - If the probability of A beating B is p, and the probability of A beating C is q, what is the probability of A beating B and C?

    It sound like it should be pq. But if if the three are equally good, then p = 0.5, q = 0.5 But the probability of A winning is 1/3, not 1/4.

    What is the general answer for p and q?

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    Re: Simple but annoying probability question

    P[A beating B and C] < min(p,q)
    All things are known because we want to believe in them.

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    Re: Simple but annoying probability question

    Thanks Mean Joe. But I was hoping for some way of actually calculating the answer. Or are you saying that there is no general answer? kat

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    Re: Simple but annoying probability question

    I don't think there is a general answer, as I'd also have to consider the probability of B beating C.
    Even if I knew that, these kinds of questions are tough.
    All things are known because we want to believe in them.

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    Re: Simple but annoying probability question

    Consider your original problem with the the racers that are equally good. I put a magic spell on b and c such that they always finish in a tie with each other. In this situation you can see that the probability that A bears both of them is just 0.5. We already saw a case where the probability is 1/3. So from this we can see that at the very least we don't have enough info to get a definitive conclusion.
    I don't have emotions and sometimes that makes me very sad.

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    Re: Simple but annoying probability question

    I like that concrete example, Dason, and having seen that it's not hard to arrange for any value between 1/2 and 1/3. Cheers, kat

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    Re: Simple but annoying probability question

    Mathematically, on average the probability is 1/3. Note that X1 and X2 are iid U~[0,1].

    U=\min \left ( X1,X2 \right )

    and

    \min \left ( X1,X2) \right )\leq Z.

    As such:

    \Pr \left \{ \min \left ( X1,X2 \right )> Z \right \}= \Pr \left \{ X1> Z \right \}\cap \Pr\left \{ X2> Z \right \}

    \Pr \left \{ U> Z \right \}=\Pr \left \{ X1> Z \right \}\times \Pr \left \{ X2> Z \right \}

    \Pr \left \{ U> Z \right \}=\left ( 1-F_{X1}\left ( Z \right ) \right )\left ( 1-F_{X2}\left ( Z \right ) \right )

    F_{U}\left ( Z \right )=1-\left ( 1-F_{X1} \right )\left ( 1-F_{X2} \right )

    F_{min\left ( X1,X2 \right )}^{Z}=F_{X1}\left ( Z \right )+F_{X2}\left ( Z \right )-F_{X1}\left ( Z \right )\times F_{X2}\left ( Z \right )

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    Re: Simple but annoying probability question

    Or visually, for the equal likelihood case, you can imagine a 1x1x1 cube for AxBxC with a random point inside. The volume where A beats both B and C is a tetrahedron of volume 1/3.

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    Re: Simple but annoying probability question


    Yes, in short, the right-hand side of the three expressions associated with the last equation I provided will yield probability density functions that provide expectations of: 1/2, 1/2, and 2/3, respectively (from left to right).

    Hence, given these expectations, we have: 1/2 + 1/2 - 2/3 = 1/3 as the expected value; which is what you should be looking for, katxt.

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