Hard probability

**DivideBy0** · October 15th 2007, 12:32 AM

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 $i$ th game and $Pr(R_i)$ represents the probability of Robert winning the $i$ th 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

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