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June 30th, 2007, 09:03 PM  #1 
Member Joined: Mar 2007 Posts: 57 Thanks: 0  coin tossing problem/ markov chain problem
The following two problems, if I'm right, are equivalent. I got an answer of 11/25 but I'm not sure. I drew a tree diagram (didn't have to but I found that simpler) and then used first step analysis. a markov chain problem: Let { X_t }_t>=0 and { Y_t }_t>=0 be two markov chains each with state space S = { 0,1 } and transition probabilities P(0,0) = P(0,1) = P(1,0) = P(1,1) = 1/2 Both chains started in state 0 at time t=0. Denote C_t = I_{ t : X_t = X_t1 = 1 } ( t ) D_t = I_{ t : Y_t = Y_t1 = 1 } ( t ) where I is the indicator function and denote A = inf { t : C_t = 1 } B = inf { t : D_t = 1 } What is the probability that B < A ? a coin tossing problem: Two fair coins, coin X and coin Y, are tossed together, repeatedly. Let A = the event that (the first time that) 2 heads come up in a row from coin X B = the event that (the first time that) 2 heads come up in a row from coin Y What is the probability that event B occurs before event A? my working: using the context from the second problem, let: P = probability of B before A Q = P given that a head came up in the previous roll from coin X, while a tail came up in the previous roll from coin Y then Q = P/4 + (1/2 + P/4 + Q/4)/4 Q = (5P+2)/15 P = P/4 + Q/4 + (1/2 + P/4 + Q/4)/4 + (1/4 + P/4)/4 P = 11/25 Please correct me if I'm wrong. Thanks! 

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