What is the surprising outcome of simulating the ladybug clock puzzle?

Empirical simulations show that all numbers from 1 to 11 are roughly equally likely to be the last number colored. This is surprising because intuition might suggest numbers closer to the starting point (12) would be less likely.

What is the key insight to solving the ladybug clock puzzle?

The key insight is to not calculate the probability from the initial position directly. Instead, consider the moment the ladybug first hits one of the neighbors of the target number (e.g., 5 or 7 for the number 6).

Why does focusing on the neighbors simplify the problem?

Hitting one of the neighbors of the target number is guaranteed to happen eventually with probability one. This condition simplifies the problem without altering the final probability outcome.

How is the puzzle related to a random walk?

After setting the condition of hitting a neighbor, the problem can be reframed as a simple random walk on a line. It becomes about the probability of reaching a certain point (+10) before reaching another point (-1) with equal step probabilities.

What is the probability that any number (1-11) is the last colored?

The probability for any specific number between 1 and 11 to be the last colored is 1/11. This is because the solution structure and probability calculation are identical for all these numbers.

Solution to the ladybug clock puzzle

3Blue1Brown

Education3 min read3 min video

Feb 16, 2026|861,423 views|24,224|509

Save to Pod

Key Moments

TL;DR

Ladybug on a clock puzzle: Probability of any number being last is 1/11.

Key Insights

The probability puzzle involves a ladybug coloring a clock face by taking random steps.

The surprising empirical result is that all numbers are equally likely to be the last one colored.

A key insight involves reframing the problem by waiting for the ladybug to reach a neighbor of the target number.

This reframing simplifies the problem to a one-dimensional random walk.

The symmetry of the rephrased problem implies that the probability for any number being the last is the same.

The final probability for any specific number being the last to be colored is 1/11.

THE PUZZLE SETUP

The puzzle begins with a ladybug at the 12 on a clock face. It then takes a series of random steps, moving either clockwise or counterclockwise with equal probability. As the ladybug lands on each number, that number is colored red. The objective is to determine the probability that the number six is the very last number to be colored on the clock.

EMPIRICAL OBSERVATIONS AND SURPRISE

Simulations of this puzzle reveal a counterintuitive outcome. When the simulation is run many times, and the last number colored is recorded, it appears that all numbers from 1 to 11 have approximately an equal chance of being the last one colored. This finding is surprising because one might intuitively expect numbers closer to the starting position (12), such as 1 or 11, to be colored earlier and therefore less likely to be the last.

A STRATEGIC REFRAMING OF THE PROBLEM

To understand this surprising result, a key strategic insight is to change the perspective of when we start considering the probability. Instead of focusing on the initial start at 12, we can wait for the ladybug to touch one of the two neighbors of the target number (in this case, 5 or 7 for the number 6). This moment is chosen because hitting 5 or 7 is guaranteed to happen eventually with probability one.

TRANSFORMING INTO A ONE-DIMENSIONAL WALK

Once the ladybug hits a neighbor (say, the 7), the problem is simplified. To determine if 6 is the last number colored, we now need to calculate the probability that the ladybug reaches the untouched 5 before it reaches the original target of 6. This is equivalent to a one-dimensional random walk problem: imagine straightening the clock into a line, and the ladybug needs to reach a point 10 steps away in one direction (coloring all numbers) before it reaches a point 1 step away in the opposite direction.

SYMMETRY AND EQUAL PROBABILITY

The critical realization is that there is nothing unique about the number six. The same logic can be applied to find the probability of any other number (e.g., 3) being the last colored. For the number 3, we would wait for the ladybug to hit its neighbors, 2 or 4. From that point, the problem becomes: what is the probability of reaching 10 steps away (to color the remaining numbers) before reaching 1 step away in the other direction?

THE FINAL PROBABILITY CALCULATION

Since the reframed problem of reaching a boundary of 10 steps away before a boundary of 1 step away has identical probabilities regardless of the starting neighbor and the target number, the probability must be the same for all numbers between 1 and 11. Given there are 11 such numbers, and they are all equally likely to be the last colored, the probability for any specific number, such as six, is simply 1 divided by 11.

Mentioned in This Episode

●Tools

●Concepts

Ladybug Clock Puzzle: Solving Strategy

Practical takeaways from this episode

Do This

Look at the neighbors of the target number (e.g., 5 or 7 for target 6) as a starting point.

Understand that hitting one of these neighbors is a certainty.

Consider the problem as a net number of steps needed to reach a boundary before another.

Recognize that the logic applies equally to numbers 1 through 11.

Avoid This

Don't assume numbers closer to the start (12) are less likely to be the last.

Don't try to calculate the probability directly from the initial position.

Don't think there's anything unique about the number 6 compared to 3 or any other non-12 number.

Common Questions

It's a probability puzzle where a ladybug starts at 12 on a clock and moves randomly, coloring each number it lands on. The question is what is the probability that a specific number, like 6, is the last one to be colored.