A frog starts at point A on the 3 × 3 grid of points shown below, where every pair of points that are 1 unit apart is connected by an edge. Every second, the frog randomly chooses an edge adjacent to its current point, and moves 1 unit along that edge. The frog may revisit an edge or point. What is the probability that after four seconds, the frog is at point B? Express your answer as a common fraction.

Guest Jan 5, 2019

#1**+2 **

Look at the grid below

7 8 9

4 5 6

1 2 3

First.....note that only the corner points and the middlle point can be reached in 4 moves from 1....so points 2, 4 , 6 and 8 cannot be reached in any set of 4 moves from 1

To reach 1 from 1 we have the following sets of moves

(1, 2, 1, 2, 1) ( 1, 2, 5, 4, 1) (1, 4, 5, 2, 1) (1, 2, 3, 2, 1) (1, 4, 1, 4, 1) (1, 4, 7 , 4, 1)

To reach 3 from 1, we have these sets of moves

(1, 2, 3, 2, 3) ( 1, 2, 1, 2, 3 ) (1, 2, 3, 6, 3) (1, 2, 5, 6, 3) (1, 2, 3, 2, 1) ( 1, 4, 5, 6, 3)

To reach 7 from 1 we have the same number of sets of moves as that of reaching 3 from 1

To reach the middle point 5 from 1 we have these sets of moves

(1, 2, 3, 6 , 5) (1, 2, 1, 2 , 5) (1, 2, 3, 2, 5) ( 1, 2, 5, 4 , 5) (1, 2, 5, 8, 5) (1, 2, 5, 2, 5) (1, 2, 5, 6, 5)

(1, 4, 7, 4, 5) ( 1, 4, 7, 8, 5) (1, 4, 5, 6, 5) (1, 4, 5, 8, 5) ( 1, 4, 1, 4 , 5)

To reach 9 from 1 we have

(1, 2, 3, 6, 9) (1, 2, 5, 6, 9) ( 1, 2, 5, 8, 9) (1, 4, 5, 6 , 9) (1, 4, 5, 8, 9) (1, 4, 7 , 8 , 9)

So.....we have a total of 4(6) + 12 = 36 possible moves

And 6 of these reach 9 from 1 - B from A -.....so....the probability is 6 / 36 = 1 / 6

CPhill Jan 6, 2019