30초 요약
3Blue1Brown solves the ladybug clock puzzle by waiting for a first visit to a neighbor of the target. The resulting random-walk question is identical for all eleven possible last numbers, giving each probability 1/11.
- 1Equal-looking outcomesSimulations suggest the eleven possible last numbers are equally likely.
- 2Wait for a neighborThe ladybug eventually reaches a neighbor of the target with probability one.
- 3Reduce to a random walkFrom there the relevant walk must reach +10 before −1.
- 4Every number has probability 1/11The same argument applies to every number other than the starting point.
핵심 정리
- Simulations suggest the eleven possible last numbers are equally likely.
- The ladybug eventually reaches a neighbor of the target with probability one.
- From there the relevant walk must reach +10 before −1.
- The same argument applies to every number other than the starting point.
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The clock puzzle
A ladybug’s random walk colors numbers on a clock.
A ladybug starts at 12 on a clock and takes random steps clockwise or counterclockwise, each with equal probability. Every number she touches turns red. What is the probability that 6 is the last number to be colored?
When I simulate the walk many times and record the final number, all numbers from 1 through 11 appear about equally likely. That is surprising: one might initially expect 6, opposite the start, to be less likely than 1 or 11.
Wait for a neighbor
Reframe the question at the first visit to either neighbor of the target.
The key is to wait until the ladybug first reaches either neighbor of 6, namely 5 or 7. This eventually happens with probability one, so waiting for that moment does not change the probability of the final outcome.
Suppose she reaches 7 first. For 6 to be last, she must reach the still-untouched 5 by going around the other way before ever touching 6. That is a net ten steps clockwise before one step counterclockwise. Once she reaches 5 this way, every other number has been colored.
The same probability for all
The same random-walk problem applies to each possible last number.
Straighten the circle into a random walk: each step is +1 or −1 with equal probability, and the question is whether the walk reaches +10 before −1. We do not need to calculate this directly to finish the argument.
Nothing about the argument is special to 6. For 3, wait until the ladybug first reaches 2 or 4, then the same ten-steps-versus-one-step question applies. The probability is identical for every possible final number from 1 to 11, so each must have probability 1/11.
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