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Math Help - Hard probability

  1. #1
    Senior Member DivideBy0's Avatar
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    Hard probability

    Edit: I think I've worked it out... winning in 5 is a more trivial case than winning in 4

    Jimmy and Robert play a best of 5 chess match. Jimmy has a probability of 0.4 of winning the first game against Robert. If he wins he becomes more confident and his probability of winning the next game rises to 0.55. However, if he loses the game his chance of winning the next is reduced to 0.3.

    Hence, if Pr(J_i) represents the probability of Jimmy winning the ith game and Pr(R_i) represents the probability of Robert winning the ith game, we have

    Pr(J_1)=0.4, Pr(J_{i+1}|J_i)=0.55, Pr(J_{i+1}|R_i)=0.3

    Pr(R_1)=0.6, Pr(R_{i+1}|J_i)=0.45, Pr(R_{i+1}|R_i)=0.7

    Find the probability that
    a) Jimmy wins in 5 games
    b) Robert wins in 5 games


    Is there any short&sweet way to do this? Or do I really have to draw out a diagram? It just seems that there are so many restrictions so using combinatorics looks hard. Thanks
    Last edited by DivideBy0; October 15th 2007 at 01:37 AM.
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