
flipping coins
Solution using Markov chains:
Pr(HH) = 0.0 Pr(TH) = 1.0
Pr(HT) = 0.5 Pr(TT) = 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 01, 2003
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