You are waiting for one of three things: a bus that runs every 20 minutes on a timetable you don't know, so the wait is 10 minutes on average; a bus that comes at any moment, every 10 minutes on average; a reply to a job application, which takes 10 days on average. You've waited 10 minutes (10 days for the job), and nothing yet. How long is left now: less than 10, the same 10, or more? Lock in an answer for each case, then check it.
At the start, all three waits are 10 on average. You've waited 10. The timetable bus now comes in 5 minutes on average. The random bus, just as at the start, in 10. And the job reply is still 15 days away on average, and the longer the silence, the more is left: after 20 days of silence, another 20. Time already spent doesn't come back, but it tells you what kind of thing you are waiting for.
This puzzle is about what the time you have already waited says about the time that is left. Reliability theory calls it the mean residual life.
This puzzle is about what the time you have already waited says about the time that is left. Reliability theory calls it the mean residual life, and depending on the distribution it can fall, stay put or grow.
The bus runs strictly every 20 minutes, but you don't know the timetable and turn up at an arbitrary moment. Between your arrival and the bus anything from 0 to 20 minutes passes, 10 on average.
Buses come independently of each other, at random moments, every 10 minutes on average. As the bus paradox shows, if you turn up at an arbitrary moment, the wait is also 10 minutes on average.
You've applied for a job. Companies reply in 10 days on average, but very unevenly: about half reply within 5 days, one in eight stays silent for more than 20 days, and some reply months later.
In each case you've already waited 10 (minutes or days), and nothing yet. How long is left on average: less than 10, the same 10, or more?
"I've waited so long, it must be soon now"
That is true only for the timetable bus: its remainder drops to 5 minutes. The random bus doesn't care how long you've waited; it is still 10 away. And after 10 days of silence the job reply is still 15 days away on average: a long silence is a bad sign.
Where the waits end: the blue timetable stays strictly within 0–20, the green random buses mostly end by 20–30, and the job replies (yellow) have a long tail.
For each wait, pick how much is left on average compared with the 10 at the start. Lock in your answer before reading on.
A breakdown of your answer will appear here.
Hard to decide? Open the hints one at a time: the easy one first, then stronger ones.
Ask whether each wait has a limit. Is there a moment after which the timetable bus must already have come?
Buses with no timetable come independently: the chance is the same every minute. Does the bus know how long you've been standing there?
Some companies are quick and some are slow. Which ones are most likely to be still silent 10 days after you apply?
The next rounds show how long is left in each case: a waiting room, a simulation and the explanation. Answer round 1 first, or go on?
You chose to always show all rounds.
Each dot is a person who started waiting at time zero. Start the clock or move the slider. When the bus or the reply arrives, the dot turns hollow. On the right you see how many people are still waiting and how long they have left on average. This is not a formula, it is the actual fate of these 300 people.
In the job-reply queue, the shade of a dot shows how quick the company is: light ones reply within days, dark ones take weeks. Move the slider to 10, then to 20, and watch which dots are left.
The computer plays out many waits of each kind. For every moment t we keep only those still waiting and work out how long they have left on average. Lines are the exact formulas, dots are the simulation.
Left: three straight lines start at the same point, 10 at the outset, and fan out. The timetable falls to zero at minute 20, the random bus stays at 10, and the job reply grows by half of every day already waited. Right: the shares still waiting. After 20 nobody is left in the timetable queue, 13.5% are left waiting for the random bus and 12.5% for a reply. Those two shares are almost equal, but after that the replies thin out far more slowly: their tail is long. On the left, the job-reply dots beyond t = 30 scatter more because few people are left there.
Once you answer, the simulation code appears here so you can run it yourself.
The same simulation, short and with no libraries. Run the JavaScript in your browser console (F12) or in Node; the Python needs only the standard library.
// Three waits, each 10 on average. How long is left for those who have already waited t?
const N = 400000;
const waits = {
'timetable': () => 20 * Math.random(), // a bus every 20 min: 0 to 20
'random bus': () => -10 * Math.log(1 - Math.random()), // exponential, mean 10
'job reply': () => 20 * (Math.pow(1 - Math.random(), -1 / 3) - 1) // Lomax α = 3, λ = 20, mean 10
};
for (const [name, draw] of Object.entries(waits)) {
const xs = Array.from({ length: N }, draw);
const out = [0, 10, 20].map(t => {
const left = xs.filter(x => x > t); // only those still waiting
if (!left.length) return `t=${t}: –`;
const avg = left.reduce((s, x) => s + x - t, 0) / left.length;
return `t=${t}: ${avg.toFixed(2)}`;
});
console.log(name.padEnd(14), out.join(' '));
}
# Three waits, each 10 on average. How long is left for those who have already waited t?
import random
N = 400_000
waits = {
"timetable": lambda: 20 * random.random(), # a bus every 20 min: 0 to 20
"random bus": lambda: random.expovariate(1 / 10), # exponential, mean 10
"job reply": lambda: 20 * ((1 - random.random()) ** (-1 / 3) - 1), # Lomax α = 3, λ = 20, mean 10
}
for name, draw in waits.items():
xs = [draw() for _ in range(N)]
out = []
for t in (0, 10, 20):
left = [x - t for x in xs if x > t] # only those still waiting
out.append(f"t={t}: {sum(left) / len(left):.2f}" if left else f"t={t}: –")
print(f"{name:14}", " ".join(out))
The minutes already spent can't be refunded, so the decision to stay or leave should only look forward. But what you see ahead depends on what the time already spent has told you about your wait.
“In plain words” is hidden on all puzzles.
Simplify the model to two kinds of companies. You send 100 applications: 75 to quick companies that reply in 5 days on average, and 25 to slow ones that take 25 days on average. Together that is the same 10 days on average.
Ten days pass. Only about 10 of the quick companies are still silent (light dots), against about 17 of the slow ones (dark). So if nobody has replied yet, you are most likely waiting on a slow company, and there is a long way to go: about 17 days on average in this example.
This card shows the mechanism, not the exact number: in the page's model the companies' speeds are spread smoothly, and there 15 days are left after 10.
The bus runs every 20 minutes, so it comes any time from 0 to 20 minutes after you arrive. You've waited 10 and it isn't here. So it comes in the second half, between minute 10 and minute 20, on average in the middle: 5 more minutes.
Here the time spent really does work for you: the wait has a limit, and every minute brings you closer to it. After 15 minutes 2.5 are left, after 19 only half a minute.
Buses with no timetable come independently of one another, so in every minute the chance of a bus is the same: about 1 in 10. It doesn't depend on how long you've been standing there. Whether you've waited 10 minutes or 30, the same 10 minutes lie ahead on average.
In the same way, a coin is never "due" for tails after five heads in a row. Here the time spent says nothing either way: it is plain bad luck that doesn't change the future. More in the bus paradox.
The sunk cost fallacy goes: "I've already put in so much, it would be a waste to leave now." But what you've put in doesn't come back whether you stay or go. The sensible question is just one: how much longer, and is it worth it.
This puzzle adds an important detail: the time already waited doesn't enter the decision as a cost, but it does enter as information. If the wait has a limit (a timetable, a kitchen timer), staying gets more worthwhile with every minute. If it is a mix of quick and very slow cases (replies, repairs, projects), a long silence means you have most likely drawn a slow one, and "it would be a waste to leave" is the worst argument exactly when it feels strongest.
“Rigorous” is hidden on all puzzles.
| Waited | Timetable | Random | Job reply | Still waiting for a reply |
|---|---|---|---|---|
| 0 | 10 | 10 | 10 | 100% |
| 5 | 7.5 | 10 | 12.5 | 51.2% |
| 10 | 5 | 10 | 15 | 29.6% |
| 20 | – | 10 | 20 | 12.5% (1 in 8) |
| 40 | – | 10 | 30 | 3.7% (1 in 27) |
| 60 | – | 10 | 40 | 1.6% (1 in 64) |
After 20 minutes the timetable bus has surely come, hence the dash. One in eight is still waiting for a reply after 20 days, and those people have 20 more days to go on average, as long as they have already waited. After 60 days one in 64 is still waiting, with 40 more to go.
X is the full wait, S(x) is the share of people who wait longer than x. The mean remainder after t is taken only over those still waiting.
Timetable: uniform from 0 to 20.
No timetable: the exponential distribution with mean 10, the only continuous one with no memory.
Job reply: the Lomax distribution (Pareto type II) with α = 3 and λ = 20. Mean λ/(α − 1) = 10, median 20(21/3 − 1) ≈ 5.2.
Where the long tail comes from: each company replies "without memory" but at its own rate r, and the rates differ (a gamma distribution). The mixture of exponentials is exactly the Lomax distribution.
A handy way to describe the difference between the three cases is the completion rate (hazard rate) h(t) = f(t)/S(t): the chance that the wait ends in the next minute, given that it hasn't ended yet. For the timetable it rises: 1/(20 − t). For the random bus it is constant: 1/10. For the job reply it falls: 3/(20 + t). When h(t) rises the remainder shrinks; when it falls the remainder grows.
“Deeper” is hidden on all puzzles.
In "The Fractal Geometry of Nature" (1982) Benoit Mandelbrot described the rule using careers: however long a person's body of work has already run, on average it will run as long again. That is exactly the line m(t) ≈ t, like the job reply for large t. In "Antifragile" (2012) Nassim Taleb used the name for books, ideas and technologies: a book that has been in print for 40 years will most likely last decades more.
The Lindy effect only holds for things that don't age. People and cars wear out, and for them the mean remainder falls with age, as with the timetable.
Engineers draw the failure rate as a "bathtub curve". At first it falls: some units have factory defects and fail right away, and one that has run for a week most likely has none. That is our job reply. Then it stays flat for a long time: failures are random, like buses with no timetable. At the end it rises: the unit wears out, like the timetable bus that inevitably draws closer.
That is why new hardware is often run for a few days before shipping (burn-in): the units that survive the start are more reliable afterwards.
The job reply is just one point on a scale. You can slide smoothly from the random bus to tails where the wait drags on without end, and see when it is worth leaving.
The mean is 10 everywhere. What changes is the Lomax tail parameter α, with λ = 10(α − 1) chosen to keep the mean fixed. Then m(t) = 10 + t/(α − 1): every day already waited adds 1/(α − 1) of a day to the forecast.
As α grows the tail shortens, and in the limit α → ∞ you get the random bus: a flat line at 10. As α approaches 2, every day waited adds almost a whole day, as in the Lindy effect. At α ≤ 2 the variance becomes infinite: the simulation jumps from run to run, because a few rare, very long waits decide everything. At α ≤ 1 even the mean becomes infinite.
And if some companies never reply, the effect is stronger still. Say one company in five stays silent for good. Then after 10 days of silence the share of these among the unanswered rises from 20% to 46%, and after 20 days to 67%. Silence more and more often means "never".
Before saying "it must be soon", ask which of the three cases your wait resembles: does it have a limit, is it memoryless, or is it a mix of quick and very slow ones.
| Situation | Type | What the time waited says |
|---|---|---|
| A kitchen timer, a timetable bus, a lecture that ends on schedule | Has a limit | The longer you've waited, the less is left. Here "soon now" is true. |
| A radioactive atom decaying, buses with no timetable, calls to a call centre in a quiet hour | Memoryless | Nothing. An atom that has lasted a thousand years has the same chance to decay as a new one. |
| A job reply, a repair "for a couple of days", a project that is already late | A mix, heavy tail | The longer the silence, the more is left. It pays to set yourself a deadline in advance and act once it passes: send other applications, find another repairer. |
| A human life | Ageing | The older a person, the less time is left on average, even though the age they can expect to reach goes up: those who made it to 80 have already survived the risks of youth. |
The name "Lindy's Law" first appeared in an essay by Albert Goldman in The New Republic. Lindy's was a New York deli where comedians met to talk show business. In Goldman's version the law said a comedian has a limited stock of material, so the more often he appears, the sooner he runs out.
Benoit Mandelbrot took the name for "The Fractal Geometry of Nature" but made it mean the opposite: the longer a body of work has already run, the longer it will run, on average as long again. He added that when it does stop, it breaks off on average at half of its promise. Toby Ord later noted that Goldman and Mandelbrot may have been describing quite different laws.
The astrophysicist Richard Gott saw the Berlin Wall in 1969, when it was 8 years old. He reasoned that he had come at a random moment in its existence and gave a 50% chance that it would stand another 2⅔ to 24 years. The Wall fell 20 years later. In 1993 Gott described this "Copernican principle" for forecasts in Nature.
In a paper in Nature, Richard Dawkins and Tamsin Carlisle named the sunk cost fallacy after the supersonic Concorde, which the British and French governments kept funding long after any payoff was in sight. They showed that the same fallacy tempts theorists of animal behaviour when they explain parental care.
Hal Arkes and Catherine Blumer (1985) sold Ohio University Theater season tickets for 1982–83 at three prices: full price, or with a $2 or $7 discount. In the first half of the season the full-price buyers went to 4.11 shows on average, the discount buyers to about 3.3. In the second half there was hardly any difference. Later researchers think the effect is weaker than it looked.
Radioactive decay is the purest example of a memoryless wait. That is why an atom has a half-life but no age: a uranium atom that has sat in rock for a billion years has the same chance of decaying in the next second as one formed just now.
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