Solving the Ladybug Random Walk Puzzle: Counterintuitive Probability Explained

The Ladybug Random Walk Puzzle

• 💡 The puzzle involves a ladybug taking a random walk on a 12-hour clock, starting at 12.
• 🎯 At each second, the ladybug moves one step to an adjacent number (clockwise or counterclockwise) with equal 50/50 probability.
• 🔑 The goal is to determine the probability that the number opposite the starting point (6) will be the very last number visited.
• ⚠️ Initial intuition suggests the furthest point (6) would be most likely to be last, but this intuition is incorrect.

Solving a Simpler Case (N=4 Clock)

• 🔬 To understand the problem, a simpler version with a 4-number clock (0, 1, 2, 3) is analyzed, starting at 0, with 2 being the opposite.
• 🧩 For 2 to be the last number visited, the ladybug must visit both 1 and 3 before landing on 2.
• 🧠 This scenario transforms into a one-dimensional random walk problem, specifically a "Gambler's Ruin" problem.

The Gambler's Ruin Principle

• 📊 In a one-dimensional random walk, the probability of hitting one end before the other is calculated as the distance to the "loss wall" divided by the total distance between both walls.
• ✅ Applying this to the N=4 clock, if the ladybug goes to 1 first, the probability of reaching 3 before 2 is 1/3.
• ✨ Considering both initial paths (to 1 or 3), the total probability for 2 to be last on the 4-number clock is 1/3.

Surprising Symmetry and General Theorem

• 🤯 Interestingly, the probability that a neighboring point (like 1) is the last visited on the N=4 clock is also 1/3.
• 🚀 This reveals a general theorem: for any symmetric random walk on a cycle of size n, every non-starting node has an equal probability of 1 / (n - 1) of being the last visited.
• 📌 The distance from the starting point does not matter due to the perfect symmetry of the circle.

Solution to the 12-Hour Clock Puzzle

• 📈 For the original 12-hour clock, with 12 numbers total and starting at 12, there are 11 other numbers that could be last.
• ✅ Using the theorem, the probability that 6 (or any other non-starting number) is the last visited is 1 / (12 - 1) = 1/11.
• 💡 This means every number, regardless of its distance from the starting point, has an 1/11th chance of being the last one visited, defying common intuition.