   flipping coins Solution using Markov chains: Pr(H|H) = 0.0    Pr(T|H) = 1.0 Pr(H|T) = 0.5    Pr(T|T) = 0.5 So, the transition probability matrix is: T = |0.0 1.0|       |0.5 0.5| Obtaining the stationary probabilities of the Markov chain gets us the solution to the problem. Briefly: (I - T')X = 0,      (n eqns in n variables)  (1) sum of prob = 1  (1 eqn in n variables)          (2) where I is the identity matrix and T' is the transpose of T. Drop ANY 1 equation from (1) and solve with (2). |1.0 0.0|    |0.0 0.5|    | 1.0  -0.5| |0.0 1.0| -  |1.0 0.5| = |-1.0  0.5| So, the equations are:     P(H) - 0.5P(T) = 0    (1)     - P(H) + 0.5P(T) = 0  (drop this)     P(H) +    P(T) = 1    (2) => P(H) = 1/3, P(T) = 2/3 This solution allows us to easily solve a class of these problems where the transitions may be different or dice may be used instead of coins! Velant Guy Friday, July 18, 2003 I have to say I'm impressed. Not that I followed the explanation, but I'm still impressed. David Clayworth Friday, August 1, 2003 Recent Topics Fog Creek Home