# 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.

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 $2^{N-1}/{2^N-1}$.