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Math Help - what is the appropriate approach?

  1. #1
    alv
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    Exclamation what is the appropriate approach?

    In a game, two players A and B shoot a target (both to the same target). Suppose that shooters are independent and the probability of (successful) shooting to target is relatively, pA and pB. If A shoots the target (successfully) before B, how much is it probable that A is the starter of the game?

    PS. would you please first give your opinion about the appropriate approach and why you choose that?


    probability models(distributions)

    Markov chains
    non-frequentist approaches like Bayesian solutions, or other things!
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  2. #2
    Flow Master
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    Quote Originally Posted by alv View Post
    In a game, two players A and B shoot a target (both to the same target). Suppose that shooters are independent and the probability of (successful) shooting to target is relatively, pA and pB. If A shoots the target (successfully) before B, how much is it probable that A is the starter of the game?






    PS. would you please first give your opinion about the appropriate approach and why you choose that?

    probability models(distributions)
    Markov chains
    non-frequentist approaches like Bayesian solutions, or other things!
    You want Pr(A starts | A wins).

    This suggests Bayes Theorem: Pr(A starts | A wins) = Pr(A starts and A Wins)/Pr(A wins).

    Pr(A wins) = Pr(A starts and A wins) + Pr(B starts and A wins).

    The calculation of two probabilities is suggested. Let q = (1 - p_A)(1 - p_B). Then:

    Pr(A starts and A wins) = p_A + q p_A + q^2 p_A + ..... = p_A (1 + q + q^2 + ....) = p_A \left( \frac{1}{1-q}\right) = \frac{p_A}{1-q}.

    Pr(B starts and A wins) = (1-p_B)p_A + q (1-p_B) p_A + q^2 (1-p_B) p_A + ..... = (1-p_B) p_A (1 + q + q^2 + ....) = \frac{(1-p_B)p_A}{1-q}.

    Therefore Pr(A starts | A wins) = \frac{1}{2 - p_B}.


    Note: If p_B = 0 then Pr(A starts | A wins) = 1/2, as you'd expect. If p_B = 1 then Pr(A starts | A wins) = 1, as you'd expect. If p_B = \frac{1}{2} then Pr(A starts | A wins) = \frac{2}{3}, as you'd expect.
    Last edited by mr fantastic; July 11th 2008 at 04:53 PM. Reason: Deleted my p.s.
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