---
title: 'Solution to the ladybug clock puzzle'
source: 'https://youtube.com/watch?v=7gG91SZwBoE'
video_id: '7gG91SZwBoE'
date: 2026-08-08
duration_sec: 153
---

# Solution to the ladybug clock puzzle

> Source: [Solution to the ladybug clock puzzle](https://youtube.com/watch?v=7gG91SZwBoE)

## Summary

This video explains the solution to a probability puzzle involving a ladybug on a clock. The ladybug starts at 12 and takes random steps, coloring numbers red. The puzzle asks for the probability that the last number colored is the 6. The solution reveals a surprising symmetry: all numbers between 1 and 11 are equally likely to be last.

### Key Points

- **Puzzle Setup** [00:02] — A ladybug starts at 12 o'clock and takes random steps, coloring each number red. The puzzle asks for the probability that the last number colored is the 6.
- **Surprising Simulation Result** [00:27] — Simulation shows that all numbers between 1 and 11 seem equally likely to be the last colored, which is surprising because the 6 is farther from the start than 1 or 11.
- **Key Insight: Conditioning** [00:54] — The key is to wait until the ladybug hits a neighbor of the target number (5 or 7 for 6). This doesn't change the final outcome because it's bound to happen with probability 1.
- **Reduction to Random Walk** [01:47] — From the neighbor, the problem becomes a random walk: the probability of reaching the far neighbor (10 steps away) before the near one (1 step away). This is the same for all numbers.
- **Final Answer** [02:26] — Since the structure is identical for all numbers between 1 and 11, the probability of ending on any specific number is 1 in 11.

## Transcript

starts on the 12 of o'clock. She takes a series of random steps either clockwise or counterclockwise, each with random chance. Each number she touches gets colored red. And the puzzle is, what is the probability that the last number to
get colored is a six? If I simulate it many times in a row, recording which number happened to be the last one colored, here's what it looks like. Running [music] this many hundreds of times, a somewhat surprising
fact starts to make itself clear. All the numbers between one and 11 seem to one touched. I say this is surprising because initially it feels like ending in a six should be less likely than ending on the
numbers closer to the initial position, like the one or the 11. And yet, empirically, that just doesn't seem true. And there is a key insight to understand why. Rather than calculating the probability of ending on a six from
the initial position, start by letting the ladybug wander until the first moment that she hits one of the two neighbors of that six, the five or the seven. And then from there, ask the probability of ending in a six. That is
a different question, but the reason it doesn't change the answer is because for any run of the simulation, eventually she's going to hit the five or the seven at some point. That's bound to happen with probability one. Waiting for this
middle condition does nothing to change the probability of that final outcome. So, from this state, where she's on, say, the seven, what's required to end on the six? Well, she has to reach that untouched five before ever touching the
six, meaning she has to eventually end up 10 steps away clockwise on net before ever ending up one step away counterclockwise. If that happens, if all the other numbers, it's a done deal that six is going to be the last.
straighten things out and you're taking a random walk where each step is plus one or minus one with a 50/50 chance. And you want to know the probability of ending up at plus 10 before ever touching the negative numbers.
>> Again, this is a hard question to answer directly, but the point is that you don't actually have to answer it directly. There's nothing special about probability of ending on the three? Well, let the simulation run until the
first moment she hits a two or a four, which again is bound to happen eventually, and then from there it's the probability of taking on net 10 steps around one way before ending one step around the other way. Because
identical for all the numbers between one and 11, the answer must be one in one and 11, the answer must be one in 11.