Bijection Probability AMC10/12 Intermediate
2020


Problem - 4736

A frog sitting at the point $(1, 2)$ begins a sequence of jumps, where each jump is parallel to one of the coordinate axes and has length $1$, 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 $(0,0), (0,4), (4,4),$ and $(4,0)$. What is the probability that the sequence of jumps ends on a vertical side of the square?


There are four cases for the frog's first move:

  • Move left. It will immediately hit a vertical line. Hence, the desired probability in this case is $\frac{1}{4}$.
  • Move right. It will be the center of the square. Thus, by the principle of symmetry, the frog will have a $50\%$ chance of ending up with a vertical side. This means the overall probability in this case is $\frac{1}{4}\times\frac{1}{2}=\frac{1}{8}$.
  • Move up. It will be on the main diagonal of the square which is a line of symmetry too. Hence, the frog will have a $50\%$ chance of ending up with a vertical side. The overall probability is $\frac{1}{4}\times\frac{1}{2}=\frac{1}{8}$.
  • Move down. This is the same as the previous case by the principle of symmetry.

Therefore, the final answer is $$\frac{1}{4}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\boxed{\frac{5}{8}}$$

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