Unformatted text preview: Solutions to quiz # 3 (January 26) 1. We are in the situation of the asymmetric random walk with the absorbing barriers at x = 0 and x = 3. The particle starts at k = 1, moves one step to the right with probability p = 3 / 5 and one step to the left with probability q = 2 / 5. Hence the probability that it reaches x = 3 before it reaches x = 0 is 1 ( q/p ) 1 1 ( q/p ) 3 = 1 (2 / 3) 1 (2 / 3) 3 = 9 19 . If you forgot the formula, here is a quick way to find the answer. Let x 1 be the probability that A eventually wins if he initially has $1 and let x 2 be the probability that A eventually wins if he initially has $2. Then x 1 = (3 / 5) x 2 , since for A to win starting with $1, he has to win $1 in the first play and then to win eventually. Similarly, x 2 = (3 / 5) + (2 / 5) x 1 , since for A to win starting with $2, he can either win in the first play or lose in the first play and then win eventually. From this system of equations we obtain x 2 = (5 / 3) x 1 and (5 / 3) x 1 = (3 / 5) + (2...
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 Winter '12
 AlexanderBarvinok
 Probability, Trajectory, Harshad number, Elementary algebra, Start

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