# Chance of m-th of n light bulbs being on after switching on the light.

The following seems to be a pretty basic combinatorics / probability exercise. I thought of a solution, but I'm not sure if is correct. And I definitely can't shake the feeling that, even if it is correct, it's not a very good (or general) solution.

The problem:

Say there are $N$ (numbered, i.e. distinguishable) light bulbs. When we flick a switch, between 1 and $N$ of them will turn on (i.e. we know one light bulb turns on, at least). What is the chance that the light bulb with number $m$ will be on after flicking the switch? (I guess it needs to be added that they all have equal chances to turn on or off).

What I came up with:

Say there are $N$ light bulbs. So there are $2^N$ different configurations of these $N$ light bulbs to be on or off (including all of them being off, for now), i.e. $N$ ordered draws over $\{0,1\}$ with replacement.

Now say we add one additional light bulb, with number $n+1$. This means replacing each of the above configurations with two new configurations, identical to the original, except that in one of the new configurations, bulb $n+1$ is 'on', and in the the other $n+1$ is 'off'.

So we know that exactly half of the configurations of $N+1$ light bulbs contain bulb $n+1$ being off, and the other half contains bulb $n+1$ being on, so $2^N$ configurations of $2^{N+1}$ in total contain bulb $n+1$ 'on'.

We then exclude the one configuration where all bulbs are off, i.e. the total number of configurations is $2^{N+1} - 1$. Note that the configurations where $n+1$ is on, and the removed configuration where all bulbs are 'off' are disjoint.

The result is then:

For $N$ light bulbs, the chance that, after flicking the switch, the light bulb with given number $m$ is 'on' is $2^{N-1} / 2^{N}-1$.

Please correct me if the above is plain wrong (maybe), or if it's way too convoluted (almost certainly), or if there's a solution that better generalizes to similar problems.

## 2 Answers

Your line of thought is in the right direction but the proof can be simplified. Each state of the bulbs can be thought of as a $N$-tuple binary vector. There are $2^N$ such possibilities (including the possibility that all bulbs are off). Now if you want a particular bulb to be on, say the $m^{\text{th}}$ bulb. Then you want those vectors where the $m^{\text{th}}$ coordinate is $1$ and the remaining $N-1$ coordinates can be either $0$ or $1$. So the number of such vectors will be $2^{N-1}$. Hence the probability will be exactly what you obtained.

• Thanks. I guess what still mainly causes me conceptual trouble is the (formal) distinction between sample space and event space. In the solution I (or you) developed, the sample space are the states 'on' and 'off', correct? I guess I'm wondering then if there's a way to derive it s.t. the sample space is the $N$ lamps, and the drawing is either onto lamp states? lamp positions? ... not sure what that would look like, if it were possible – Bert Zangle Oct 12 '15 at 12:03

Your solution is totally correct but adding the $N+1$- th bulb was unnecessary. There are $2^N$ different configurations of all bulbs and $2^N - 1$ various states after flicking a switch (a state when all bulbs are 'off' is excluded). Let's say the $m$-th bulb is 'on'. There are $2^{N-1}$ such configurations because with the $m$-th bulb being 'on' we just need to count all configurations of other $N-1$ bulbs. Indeed, the answer is $2^{N-1}/{2^N-1}$.