A Problem About Levers

Probability Level 2

Suppose there were 10 levers and each configuration of these 10 levers would give you a different painting. You coded a robot that flips a lever, to give you a different painting each day at 7:00. On the $x$ th day, you see a same painting. What is $x$ ?

Note: Each lever can only be flipped a maximum of 10 times

The answer is 1025.

1 solution

Anonymous1 Assassin
May 29, 2021

We use the Pigeonhole Principle that:

if n items are put into m containers, with n>m>0, then at least one container must contain more than one item.

We're going to use that principle here with a bit of probability:

There are $2^{10}$ configurations of 10 levers, as each lever has two states, On and Off . Because of the pigeonhole principle, we need to find the smallest integer larger than $2^{10}$ which is $2^{10}+1$ .

$\therefore$ You have to wait $\color{#3D99F6} 2^{10}+1$ $=$ $\color{#D61F06} 1025$ days before a same painting appears again.

This is wrong. What if the robot flips one lever on the first day, then it keeps flipping another lever forever. You would NEVER reach the original painting.

Andrei Zonga - 2 weeks ago

Noted, I have changed the question.

Anonymous1 Assassin - 1 week, 5 days ago

A Problem About Levers

The answer is 1025.

1 solution

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