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2020 AMC 12A Problem 11

Problem 11 of 25IntermediateAlgebraProbability & StatisticsProblem-Solving Techniques

A frog sitting at the point (1,2)(1, 2) begins a sequence of jumps, where each jump is parallel to one of the coordinate axes and has length 1,1, and the direction of each jump (up, down, right, or left) is chosen independently at random. The sequence ends when the frog reaches a side of the square with vertices (0,0),(0, 0), (0,4),(0, 4), (4,4),(4, 4), and (4,0).(4, 0). What is the probability that the sequence of jumps ends on a vertical side of the square?

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Solution

Let P(x,y)P(x, y) be the probability of ending on a vertical side. On a vertical side P=1,P = 1, on a horizontal side P=0,P = 0, and at an interior point PP is the average of its four neighbors. By left-right symmetry P(2,2)=12.P(2, 2) = \tfrac12. Let a=P(1,2),a = P(1, 2), b=P(1,1)=P(1,3),b = P(1, 1) = P(1, 3), and c=P(2,1)=P(2,3).c = P(2, 1) = P(2, 3). Then a=14(1+12+2b),a = \tfrac14\left(1 + \tfrac12 + 2b\right),   b=14(1+c+a),\;b = \tfrac14(1 + c + a), and c=14(2b+12).c = \tfrac14\left(2b + \tfrac12\right). Substituting gives b=12,b = \tfrac12, hence a=38+12b=58.a = \tfrac38 + \tfrac12 b = \tfrac58. Thus, B is the correct answer.
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Tagged: random walk · system of equations · symmetry

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