Circular Motion 2
A frog is situated on a circular arrangement of 100 lily pads, each uniquely numbered from 0 to 99 in increasing order counter‐clockwise, starting with 0. The frog begins its journey on lily pad 0 and can hop to either the closest lily pad on its left or right with equal probability for each hop. What is the probability that the frog lands on lily pad 50 for the first time only after it has visited every other lily pad (excluding lily pad 50)?
Answer
To solve this problem, we note that for the frog to land on lily pad 50 last (i.e., only after visiting all other pads), it must initially hop to either lily pad 49 or lily pad 51. These two pads are equidistant and therefore equally likely due to symmetry.
Without loss of generality, assume the first hop is to lily pad 49. At that point, we can ``relabel'' lily pad 49 as our new starting point (call it 0), then consider a hop toward lily pad 50 as \(-1\) and a hop away from 50 as \(+1\). In this new labeling, the problem transforms into one of a ``random walker'' on the integers, needing to reach \(98\) (corresponding to lily pad 51; not 50, think about it) before reaching \(-1\) (corresponding to lilypad 50). Since this is a symmetric random walk with two absorbing barriers, the probability of reaching \(98\) before \(-1\) is the same as the probability that lily pad 50 remains unvisited until every other pad has been visited. This probability is:
Hence, the desired probability that the frog lands on lily pad 50 for the first time only after it has visited every other lily pad is