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Unformatted text preview: runner runs barefooted is 1 / ( k + 1). 1 26. Let the state be the ordering of the deck of n cards, so there are n ! states. The transition probabilities are P ( i 1 ,i 2 , ··· ,i n ) , ( i j ,i 1 ,i 2 , ··· ,i j1 ,i j +1 , ··· ,i n ) = 1 n This Markov Chain is doubly stochastic, so in the limit all n ! states are equally likely. 35. The transition probabilities are P = 1 1 1 / 2 1 / 2 1 / 3 1 / 3 1 / 3 1 / 4 1 / 4 1 / 4 1 / 4 0 The limiting probabilities are obtained from π = π 1 + 1 / 2 π 2 + 1 / 3 π 3 + 1 / 4 π 4 π 1 = 1 / 2 π 2 + 1 / 3 π 3 + 1 / 4 π 4 π 2 = 1 / 3 π 3 + 1 / 4 π 4 π 3 = 1 / 4 π 4 π 4 = π π + π 1 + π 2 + π 3 + π 4 = 1 The solution is, π = π 4 = 12 / 37 ,π 1 = 6 / 37 ,π 2 = 4 / 37 ,π 3 = 3 / 37 . 2...
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 Spring '08
 Lim
 Operations Research, Probability theory, Stochastic process, Markov chain, Andrey Markov, Random walk, transition probabilities

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