100 prisoners are imprisoned in solitary cells. Each cell is windowless and soundproof (i.e., there’s absolutely no way to exchange information among them once they are imprisoned). There’s a central room with one light bulb; the bulb is initially off. No prisoner can see the light bulb from his or her own cell. Each day, the warden picks a prisoner equally at random, and that prisoner visits the central room; at the end of the day the prisoner is returned to his cell. While in the room, the prisoner can toggle the bulb if he or she wishes. Also, the prisoner has the option of asserting the claim that all 100 prisoners have been to the room. If this assertion is false (that is, some prisoners still haven’t been to the room), they loose. If it is indeed true, all prisoners are set free. Thus, the assertion should only be made if the prisoner is 100% certain of its validity.
The solution to this puzzle is well known. Before being imprisoned, they agree on a protocol. Here, a counter is elected: a special prisoner that behaves differently from everyone else. Whenever someone (who’s not the counter) goes into the room, and find the switch in the off position, it turns it on. But it only does this once. And whenever the counter goes into the room and finds the switch in the on position, it turns it off, and counts plus one. When the counter reaches 99, then he knows for certain that everyone was called before. Now here’s the real question we want to tackle:
How long would it take on average, using the aforementioned strategy, for the prisoners to be released?
First approach
Let $C$ be a random variable that denotes the counter being called, and $NC$ a noncounter. Then:
\[\begin{eqnarray*} C &\sim& \text{NB}(1; \frac{1}{100}) \\ NC(x) &\sim& \text{NB}(1; \frac{x}{100}) \end{eqnarray*}\]… where $x$ is the number of prisoners that are yet to be accounted for; this step is important, since every time a prisoner turns on a switch, the probability of picking one that is yet to perform such action decreases.
Finally, we can express the expected value as:
\[\sum_{i=1}^{99} { \mathbb{E}[C] + \mathbb{E}[NC(100i)]}\]Let’s remember some useful stuff
Recall the linearity of expectation; a negative binomial can be expanded as something like this:
case  calls  probability 

p  1  $p$ 
qp  2  $pq$ 
qqp  3  $pq^2$ 
…  …  … 
…  n  $pq^{n1}$ 
The mean (or expected value) is, by definition, the sum of all possible values multiplied by their respective probabilities:
\[\begin{equation*} \mu_X = \mathbb{E}[X] = \sum_{x\in X}x\cdot p(x) \end{equation*}\]Hence, considering $q = (1p)$:
\[1p + 2pq + 3pq^2 + \cdots + npq^{n1} = \sum_{i=1}^{\infty} np(1p)^{n1} = \frac{1}{p}\]It is easy to see that the infinite sum converges. And since $p=\frac{1}{100}$, then $\mathbb{E}[C] = \class{bghighlight}{100}$. The conclusion is actually very intuitive: on average, one has to call 100 prisoners before our counter is selected.
I heard something about “distributions”…
Alternatively, we also know that the expected value of a random variable following a negative binomial distribution, up to and including the success event happening ($r = 1$, where $r$ is the number of successes that need to happen), is:
\[\mathbb{E}[C] = r\frac{1p}{p} +1= 1\frac{1\frac{1}{100}}{\frac{1}{100}} + 1 = \class{bghighlight}{100}\]Similarly, the expected value for a prisioner that has never turned on the switch, and considering that the probability depends on how many prisoners still have that property^{1}:
\[\mathbb{E}[NC(x)] = 1\frac{1\frac{x}{100}}{\frac{x}{100}} + 1 = \frac{100}{x}\]…where \(x = \{99, 98, \ldots, 1\}\). We now have all the elements to calculate the expected value, so:
$\phantom{\equiv}~\sum_{i=1}^{99} \mathbb{E}[C] + \mathbb{E}[NC(100i)]$
$\equiv$ { since $\Sigma (a+b) = \Sigma a+\Sigma b$ }
$\phantom{\equiv}~\sum_{i=1}^{99} \mathbb{E}[C] + \sum_{i=1}^{99} \mathbb{E}[NC(100i)]$
$\equiv$ { solving left side }
$\phantom{\equiv}~9900 + \sum_{i=1}^{99} \mathbb{E}[NC(100i)]$
$\equiv$ { by definition }
$\phantom{\equiv}~9900 + \sum_{i=1}^{99} \frac{100}{100i}$
$\equiv$ { solving right side }
$\phantom{\equiv}~9900 + 517.74 = \class{bghighlight}{10417.74}$
Simulation
I like to solve the same problem using more than one technique; so, as always, we shall simulate it:
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const avg = (r) => r.reduce((a, b) => a+b) / r.length
const simIter = simulation()
const random = (min, max) => Math.floor(Math.random() *
(Math.floor(max)  Math.ceil(min) + 1)) +
Math.ceil(min)
function *simulation() {
while(true) {
const mem = new Set()
const counterGuy = random(1, 100)
let counter = 0
let sw = false
let steps = 0
while(counter < 99) {
steps++
const p = random(1, 100)
if (p == counterGuy) {
if (sw == true) {
counter++
sw = false
}
} else {
if (sw == false && !mem.has(p)) {
sw = true
mem.add(p)
}
}
}
yield steps;
}
}
const samples = Array.from(Array(500), simIter.next, simIter).map(o => o.value)
console.log(`[${samples.length} simulations] Expected value = ${avg(samples)}`);
Fancy chart
Of course, the Expected Value is exactly what the name says: expected. On average, it is 10417.74, but in reality it is a random variable that follows a normal distribution, centered at that value. We can run the above simulation, say, $n = 500$ times, and then plot an histogram to get the feeling of it, like so:
You can add 500 more simulations to the run, 1000, or 10000, or start fresh.

Expressing the same using linearity of expectation is left as an exercise to the reader. ↩︎