The waiting time paradox: why is my bus always late? (2018)

https://jakevdp.github.io/blog/2018/09/13/waiting-time-paradox/

by skadamat • 2 years ago

219 190 2 years ago

jakevdp.github.io

gpvos• 2 years ago

It's not just that the bus is always late, it's also that when you are late yourself, the bus is always on time and just leaving.

dhosek2• 2 years ago

When I first started going into the office regularly back in February, I would stop in a 7-Eleven a block away from my “L” stop on my way in.¹ Every day for the first couple of weeks, I would watch the train leaving the stop right when I walked out of 7-Eleven, regardless of when I left my apartment.

My solution has been to stop looking at the station when I leave 7-Eleven.

⸻

1. I still do.

thesuitonym• 2 years ago

Same energy as ``Doctor, it hurts when I raise my arm.''

``Then stop raising your arm.''

Obscurity4340• 2 years ago

I feel this is my bones

chongli• 2 years ago

Yes, or my favourite: the bus shows late and then later and later and later on the “next bus” feature on my phone, then eventually it jumps 20 minutes indicating the bus has been cancelled for that cycle, so I start walking home, only for the bus to show up when I’m just too far from the stop to reach it!

edgarvaldes2• 2 years ago

A variation: I used to take the bus instead of a 15 minute walk (it was late in the day, I was tired, etc). Sometimes the bus would take more than 15 minutes to arrive, so I questioned my decision to continue waiting. When I finally walked home, the bus would catch up with me a couple of blocks before I got home.

JeremyNT• 2 years ago

Yes. This is the key piece to understand this "paradox":

you arrive at a random time

So, um, if you intend to take public transit, it's best to not arrive at a random time. Looking at the time tables and planning around them is public transit user 101.

bluGill8• 2 years ago

Only for infrequent service. Many riders have service so frequent they don't check or plan as they know the bus will be there not long after they arrive, no matter when they arrive. This might even be a majority as such service is so convenient that people who live there don't bother with alternatives like a owting a car.

YZF2• 2 years ago

or worse. leaves two minutes early when you're right on time.

chii• 2 years ago

what you didnt realize is that the "early" bus is actually the late bus from the previous timeslot.

magicalhippo• 2 years ago

Ah, but it's on time those times you are late but really need to get that one departure for some important meeting or similar.

edit: interesting post with a different ending than I imagined.

Been thinking about using some statistical methods to give me some better estimates of the busses I take to work. Like "given that it's $today, which is a Tuesday, it's 17:30, and the display says the bus is 7 minutes delayed, how long til it will actually come?

jeffbee2• 2 years ago

some statistical methods

Anything would be better than what we have. When I studied NextBus predictions for the SF Muni 1-California line I found that the predicted arrival time was not even correlated with the actual arrival time. If they had taken all the radios and computers out of the system and just showed a random number as the minutes to arrival that would have been every bit as good as NextBus.

dontlikeyoueith• 2 years ago

If they had taken all the radios and computers out of the system and just showed a random number as the minutes to arrival that would have been every bit as good as NextBus.

Bold of you to assume they hadn't already done that.

Remember, lowest bidder wins the contract.

SllX2• 2 years ago

Dunno what year you were doing this, but all the buses that travel California Street at some point are bad about this and you are correct that it might as well be a d20.

The N Judah is usually on time or a little early, except going outbound between 6:30pm and 7:30pm. If you’re waiting for the F, you should mentally add 2 minutes to whatever time you see. The J Church going downtown becomes increasingly late the closer to downtown it actually gets (partly because it often loses the signal lottery with the N Judah since it gets stalled out at the Market Street intersection even if it was scheduled to go in ahead of time). When the L trains were running, it would always seem to depart from the Zoo-side terminus on time if you were waiting there, but if you were waiting up the street on Taraval, you could reliably add 5 to 10 minutes to your wait time.

Also at the Embarcadero, there can never be two working upwards escalators going from the MUNI platform to the eastern exits at the same time. It simply cannot happen. If one gets fixed, the other will break, even if it was just fixed a few days ago.

cbsks• 2 years ago

(2018), but this article is timeless…

Past discussion: https://news.ycombinator.com/item?id=18321062

krackers• 2 years ago

incognito124• 2 years ago

But, somehow, lighting a cigarette at the station makes the bus spawn instantly. 100% reproducible.

onlyrealcuzzo• 2 years ago

As a non smoker, I'll buy some cigarettes and see if I can reproduce.

Thanks for the tip!

Something1234• 2 years ago

Waste of time because the universe knows you don't actually want that pleasure of the cigarette and will be further punished by the cigarette for your hubris.

bregma• 2 years ago

Filter tip only, I hope.

Terr_• 2 years ago

I've been told by the manufacturers that cigarettes increase the odds of reproducing.

https://xkcd.com/583/

prmoustache2• 2 years ago

I don't understand all those smokers that are lighting a cigarettes seconds or 1 minute before the bus reaches the halt. Most of the time you can see the bus from a distance yet they still light them up.

laweijfmvo2• 2 years ago

maybe they know they can't smoke on the bus, so they "need" to smoke as late as possible before getting on to survive the ride? idk, not a smoker.

humanfromearth9• 2 years ago

For me it is shorter : I don't understand all those smokers.

akshaybhalotia• 2 years ago

I wish there were an r/AngryUpvote thing here

prmoustache2• 2 years ago

I would say just use the app that says at which bus stop the next bus is and when it estimates its arrival.

OK sometimes the app / bus location system fucks up but most of the time or their are unexpected road road work or traffic accident that suddently forces it to be slower but most of the time it is pretty much accurate.

edgarvaldes2• 2 years ago

The waiting time paradox is not only about the bus.

xnorswap10• 2 years ago

My favourite corollary of this is that even if you win the lottery jackpot, then you win less than the average lottery winner.

Average Jackpot prize is JackpotPool/Average winners.

Average Jackpot prize given you win is JackpotPool/(1+Average winners).

The number of expected other winners on the date you win is the same as the average number of winners. Your winning ticket doesn't affect the average number of winners.

This is similar to the classroom paradox where there are more winners when the prize is poorly split, so the average observed jackpot prize is less than the average jackpot prize averaged over events.

pif• 2 years ago

The number of expected other winners on the date you win is the same as the average number of winners.

Sorry, but no! The total number of expected winners (including you) is the same as the average number of winners.

xnorswap10• 2 years ago

No, it's 1+average number of winners.

If the odds of winning are 1 in 14 million, and 28 million tickets are sold, then you expect there to be 2 winners.

If look at your ticket and see you've won the lottery, then the odds of winners are still 1 in 14 million, and out of the 27,999,999 other tickets sold, you expect 2 other winners, and now expect 3 winners total, given you have won.

cortesoft4• 2 years ago

To understand why your totally understandable conclusion is wrong, it helps me to think about what it means to determine the average number of other winners when I win.

The reason is similar to the Monty Hall Problem (https://en.wikipedia.org/wiki/Monty_Hall_problem)

To understand, lets think about the simplest representation of this problem... 2 people playing the lottery, and a 50/50 chance to win.

So, we can map out all the possible combinations:

A wins (50%) and B wins (50%) - 25% of the time

A wins (50%) and B loses (50%) - 25% of the time

A loses (50%) and B wins (50%) - 25% of the time

A loses (50%) and B loses (50%) - 25% of the time

So we have 4 even outcomes, so to figure out the average number of winners, we just add up the total number of winners in all the situations and divide by 4... so two winners in the first scenario, plus one winner in scenario 2, plus one winner in scenario 3, and zero winners in scenario 4, for 4 total winners in all situations... divide that by 4, and we see we have an average of 1 winner per scenario.

This makes sense... with 50/50 chance of winning with 2 people leads to an average of 1 winner per draw.

Now lets see what happens if we check for situations where player A wins; in our example, that is the first two scenarios. We throw out scenario 3 and 4, since player A loses in those two scenarios.

So scenario one has 2 winners (A + B) while scenario two has 1 winner (just A)... so in two (even probability) outcomes where A is a winner, we have a total of 3 winners... divide that 3 by the two scenarios, and we get an average of 1.5 winners per scenario where A is a winner.

Why does this happen? In this simple example it is easy to see why... we removed the 1/4 chance where we have ZERO winners, which was bringing down the average.

This same thing happens no matter how many players and what the odds are... by selecting only the scenarios where a specific player wins, we are removing all the possible outcomes where zero people win.

mitthrowaway25• 2 years ago

You're both right! This is where the subjective Bayesian framework helps clarify things. The passive-voice term "expected winners" leaves ambiguous a key idea: Expected by whom?

The number of winners you expect depends on what information you have, namely, whether or not you know that you are holding a winning lottery ticket or not!

mecsred2• 2 years ago

Your winning ticket doesn't affect the average number of winners.

I think this is a good hint that the conclusion isn't true. Just think about what it would mean if this were true for a sample of lotto winners. For a winner, if they win, their average number of winners is higher than the global average. Repeat this logic for each individual winner... And every winner wins with a higher number of winners than the average. Which is clearly impossible.

It would be true if you were guaranteed to win, since that's the assumption you have conditioned the probability on, but that's not a lottery then. If you want to get the actual expected value across all samples you need to take a weighted sum including the expected value when you don't win.

xnorswap10• 2 years ago

Which is clearly impossible.

It's just like the average pupil being in a larger than average class size, it's not impossible!

Take a situation where you have 499 lotteries with zero winners, and 1 lottery with 1000 winners.

There are on average 2 winners per lottery.

From the perspective of all the winners, there was an average of 1000 winners.

That's the very basis of the paradox in the article.

Now, in that case, the lottery would be investigated for fraud. But the paradox plays out in a much gentler sense.

kgwgk11• 2 years ago

Average Jackpot prize is JackpotPool/Average winners.
Average Jackpot prize given you win is JackpotPool/(1+Average winners).

That doesn't make a lot of sense.

Maybe you mean that most winners get less than the average prize.

Let's say that there is $1m jackpot and there could be one, two, three or four winners (with equal probability).

To simplify the calculation, let's say that each outcome happens once.

The average prize is $400k (4 x $1m / (1+2+3+4)).

A winner has 40% probability of getting just $250k and 30% probability of getting $333k.

----

Edit: Or maybe you tried to say something like the following but didn't get it right because "average winners" means different things when you win and when you don't.

Average Jackpot prize is JackpotPool/Average winners when there are one or more winners
Average Jackpot prize given you win is JackpotPool/(1+Average winners when there are zero or more winners).

xnorswap10• 2 years ago

The key here is that you don't care what happens when you don't win, you don't care how much other people win.

What you care about, is the expected amount you win, given that you have a winning ticket.

Let's say there are N players, and let's say anyone has a 1 in X independent chance to win.

If you don't buy a ticket, there are N/X expected winners.

If you do buy a ticket, it doesn't affect whether other people win or not.

There are still N/X expected other winners.

Your participation doesn't reduce the expected number of people, who are not yourself, that will win.

This isn't a Monty hall problem, because Monty Hall introduced new information.

Buying a ticket doesn't introduce new information.

With Prob of (X-1)/X, you lose, and go home unhappy.

With Prob of 1/X, you win. And now there are 1 + N winners.

Your buying a ticket therefore increased the overall expected number of winners by 1/X. That is correct.

Conditioned on you winning, there are 1+N expected winners.

Conditioned on you losing, there are N expected winners.

GuB-42• 2 years ago

That's true if you are cheating, for example by knowing the numbers in advance, guaranteeing a win. The cheater is the "+1" in your argument, an extra player with a 100% win rate.

But if you are not, and pick a random time where you win, on average, you will win as much as the average lottery winner.

For the classroom paradox to work, you have to take the average prize per draw after splitting, not the average prize per winner.

For example, if there are 9 winners in the first draw and 1 in the second, then there are 5 winners on average, so the average prize is 1/5. If you are one of the winners, there is 9/10 chance you are among the 9 and only win 1/9, which is less than average, but there is also 1/10 change of winning full prize, which is much better than average. If you take a weighed average of these (9/10*1/9+1/10*1) you get 1/5, back to the average prize. The average individual prize per draw is (1/9+1)/2=5/9, but it is kind of a meaningless number.

Another way to see it is that most of the times, you will win less than average, but the few times you win more, then you will win big. But isn't it what lotteries are all about?

nonameiguess• 2 years ago

This is (technically) wrong, but not for the reasons I've seen others give so far. Your reasoning is basically fine, but your definition of an average jackpot prize is not. If we have k lottery winners and we denote each individual prize as n_i, then the average prize is sum(n_1 ... n_k) / k. It's pretty easy to see that number cannot possibly be larger than all individual n_i and thus it cannot be the case that "you" won less than the average prize for all possible yous. Some winners win less than average and some win more, or they all win exactly the same amount.

On the other hand, your analytically computed expected winning is indeed less than an analytically computed expected average prize, when conditioned on the fact that you won, because you are more likely than not to be in a lottery that has more winners than the average lottery. This is mathematically the same phenomenon as the thing where the perceived average class size if you sample random students is greater than the actual average class size, because more students will be in the larger classes. This doesn't mean every class is larger than the average class, which is not possible. It just means that if you randomly select a student, you have a better than 50/50 chance of selecting someone in a larger than average class.

ruuda• 2 years ago

Another one is that your friends on average have more friends than you. (Because you are more likely to be friends with people who have many friends than with people who have few friends.)

FabHK4• 2 years ago

Related: Suppose Bitcoin's difficulty is tuned correctly to the target block time of 10 mins/block. Then, if you pick a block uniformly from the list of blocks, its expected length is 10 minutes. However, if you pick a point in time uniformly, the expected length of the block it's in is 20 minutes.

laweijfmvo2• 2 years ago

my take away is that if you're lucky enough to live in a place that has such a bus schedule, you can just ignore the schedule and show up whenever you want and only wait 10 minutes. sounds lovely!

i80and• 2 years ago

Tell me about it. Reliable 45 minute bus headways would be a dream where I live

bluGill8• 2 years ago

30 minutes is the worst I will accept. I have better things to do than sit outside a locked door for an hour waiting for the person with the key to arrive. I want every 5 mintes but I will be able to deal with up to half hour. More than that and I guess I must drive.

was_a_dev2• 2 years ago

Being able to track a bus via GPS works well like this. It doens't matter if the bus is 20 minutes late, if you can check that the bus is 4 minutes away.

A bit like usnig the metro. You don't know the schedule, you just know the realative time to the next train

dekhn• 2 years ago

I got asked a variation on this in an interview several decades ago. "What is the expected waiting time for a bus that arrives on average every ten minutes and you show up at a random time". I was sure it was 5 but they actually wanted me to do the math from the article, in my head, in 30 minutes. I did not pass that interview and did not get the job (which was a good thing long term).

I really enjoy having 10-20 lines that produces the expectation value directly by simulation; that's a fast way for me to understand the underlying values.

memming• 2 years ago

Renewal theory! https://en.wikipedia.org/wiki/Renewal_theory

kqr• 2 years ago

This is one of my favourite queueing theory-adjacent consequences.

As the article notes, it's the same reason the average coin in a sequence of tosses will be in a longer run than the average run length.

ChrisArchitect• 2 years ago

(2018)

Some discussion then: https://news.ycombinator.com/item?id=18321062

c_moscardi• 2 years ago

Related reading; explains the same concept quite well IMO with NYC subway data. This is where I learned about this concept.

[1] https://erikbern.com/2016/04/04/nyc-subway-math

[2] https://erikbern.com/2016/07/09/waiting-time-math.html

stonemetal122• 2 years ago

when the average span between arrivals is N minutes, the average span experienced by riders is 2N minutes.

Who is arriving in the first part of the sentence? At first I thought he meant the bus arrival, thus N = 10, and 2N would be 20. But then he says

The average wait time is also close to 10 minutes, just as the waiting time paradox predicted.

10 isn't 20 so ???

outop2• 2 years ago

The average time between two buses (based on the Poisson model used in TFA) is N minutes. But you are more likely to arrive in a long interval than a short one. So if you turn up to the station at a random time, the average time between the last bus that departed before you got there and the next departure, is 2N minutes.

stonemetal122• 2 years ago

Doesn't change the fact that N is 10 and evidently 5 in the same sentence. Does author think only 'N' can be used for variables and readers are left to figure out which 'N's are the same and which are different?

taeric6• 2 years ago

This seems more to say that the average time of all passengers waiting will be close to the interval, but that the average time for any individual in a given stop will be closer to half? (Similarly, if you are discussing the longest time you will wait throughout the day and you have multiple stops you have to wait at, it will drift up to the the interval time.)

That is, they sound like similar questions, but they are not. How long can one random person expect to wait at a stop is different from how long a population will wait at a given spot. In large because a person can only arrive at a single time in the waiting interval, but more passengers become less likely the closer to departure time.

(I realize I didn't word all of this as a question, but I am not asserting I'm correct here. Genuinely curious if I understand correctly.)

maeil3• 2 years ago

It's not about population vs. individual. The underlying principle is the assumption that we (the people taking the bus) arrive at a random time. And we're more likely to arrive in a shit interval (because they're longer) than in a lucky interval (because they're shorter).

Here's a trivial example.

Buses are supposed to arrive at a 10 minute interval: 12:00, 12:10 and 12:20. But today the second bus arrives a bit early, at 12:07. So they arrive at 12:00, 12:07 and 12:20. We arrive at the bus stop at a random moment >12:00 and <= 12:20.

If we arrive in the interval 12:00-12:07, our average waiting time will be 3.5 minutes. What's the chance that we do arrive in this interval? 7 mins/20 mins.

If we arrive in the interval 12:07-12:20, our average waiting time will be 6.5 minutes. What's the chance that we do arrive in this interval? 13 mins/20 mins.

So our expected waiting time is not 5 minutes but 3.5 * 7/20 + 6.5 * 13/20 = 5.45.

Basically "the shit intervals are longer so we're more likely to arrive in them. the lucky intervals are shorter so we're less likely to arrive in them.". If we arrive at a random time, which is the core assumption here.

Now you might say "but 5.45 doesn't feel close to 2N". And that's where the other assumption that probably does not reflect reality comes in - the bus arrival times are simulated as uniform random numbers. I mean, it depends on where you live, haha. But it's pretty much a worst case scenario, so in reality it's not as bad. Which the writer shows using the real-world data.

Nevertheless, unless it's the Japanese subway which always arrives exactly on time, it's always going to be bigger than 2N.

And what if we don't arrive at a random time, but arrive according to some pattern guided by the bus schedule? That change everything.

Still, it's actually pretty common to arrive at a random time, and buses (and some subways, or other things in life) do tend to arrive not exactly on time, in which case it holds. To some extent.

taeric6• 2 years ago

This seems to be attacking from a different perspective, though? You are requiring the change from a bus that is not to schedule. This article pointed out that that was not necessary. And, indeed, you can presume perfectly scheduled busses and still see a distribution quirk where the average wait time of the population is at the interval level. Right?

Projectiboga• 2 years ago

In combinatorics we calculated that the typical wait time is very close to the actual planned interval.

rjmunro• 2 years ago

There's another thing that happens with busses that makes it worse.

The further behind the previous bus a bus is, the more people will arrive at the bus stop. The more people there are at the stop, the longer the bus has to spend picking them all up and selling them tickets etc. Therefore the delayed bus will tend to experience more delay. The bus behind them will have less people to pick up, so it will spend a shorter time at stops and tend to catch up with the first bus, so the two busses are dragged towards each other.

jjbinx007• 2 years ago

Also buses are more likely to let other buses out in traffic so that's another reason why you get clumps of buses arriving rather than regularly spaced ones

ajuc10• 2 years ago

It's a law here in Poland that everybody has to let the buses leaving a bus stop to enter the lane before them.

I think most people complaining about buses in this thread just live in a city where public transport isn't a priority so it barely works :/

The city buses I've seen in USA have 1 or 2 doors. It's already wrong - it makes the boarding time unnecessarily long. Then there's the tickets - drivers shouldn't be selling or checking the tickets. You should buy tickets in a ticket machine or on your smartphone. And they shouldn't be checked every time - it takes too long. Have a group of people who board random buses and check the tickets there while the bus is driving so as not to waste anybody's time.

Bus schedules and routes should be designed with randomness in mind. There should be a small buffer (1 minute is enough if boarding is quick) to zero the randomness on each bus stop. Most bus stops should be mandatory so that 3 consecutive bus stops without passangers don't wreck the whole schedule (and then it spreads to other buses because you have to wait for 5 minutes at a bus stop for your departure time and you block entrance for other buses' passangers which makes boarding longer).

If you just put an intercity bus (that can work with 1 door and driver selling the tickets) and use it as a city bus that stops every 1-5 minutes - it won't work.

City buses should be optimized for latency not throughput.

Ylpertnodi• 2 years ago

the longer the bus has to spend picking them all up and selling them tickets etc.

In my country, apart from an app/ online, you can buy a ticket pretty much anywhere. I guess someone worked out that bus drivers with money are a potential theft risk, and also that selling tickets on the bus takes time and makes busses late(r than they would be).

matrix2003• 2 years ago

As a rider, I also just find it more convenient to buy tickets in an app.

I can link my payment method, and purchase tickets in seconds whenever I’m ready.

bluGill8• 2 years ago

If you rarely ride though cash is easier. I won't use the app again so I don't want it. Fortunately in the us multiples of $1 are good price points so exact change put it in the safe works well.

jerlam• 2 years ago

The slickest process I've seen is to just swipe your credit card, without any setup whatsoever.

ajuc10• 2 years ago

In my city the buses have ticket machines inside them, there's also ticket machines at the bus stops, and you can buy tickets on your smartphone. Or at small street shops but it's last resort.

Most people that drive often just have monthly tickets so they don't have to do anything - just get in/out of the bus.

Drivers are banned from selling tickets - they only do the driving. And nobody checks if you bought a ticket on every ride - there's a random check every now and then and if you're caught you pay a high fine. But you have maybe 1% chance of being checked at any given ride.

mitthrowaway25• 2 years ago

That bus with more riders on board also has a higher probability of needing to stop to let people off at each location as well, slowing it down even further!

ajuc10• 2 years ago

This is part of a good route design - most bus stops should be "mandatory" - which means the bus stops there no matter what. Some bus stops are "optional" - driver only stops there if there's somebody waiting or if somebody in the bus presses the "STOP" button near the doors. It's marked on the timetable which bus stop is optional.

It's not worth it to make every stop optional because then the routes become too unpredictable and scheduling is hard. Usually there's like 5-10% of optional bus stops on each route - only in the places where very few people get in/out.

Gravityloss2• 2 years ago

Robotic buses could be made smaller than driver buses since the cost of driver doesn't need to be amortized as many passengers as possible. Then you could implement optional stop skipping. At the end of the spectrum you have Uber X ie taxi with ride sharing.

emmelaich2• 2 years ago

https://www.keoride.com.au/what-is-keoride

the Keoride app matches customers who are travelling in the same direction and calculates an optimised flexible route to pick up and drop off customers close to their destination, you can even track the vehicle and get updated on your ETA in real time.

bluGill8• 2 years ago

This is why good back office daspatch is needed. If the bus is late slow the following but and/or add another.

ajuc10• 2 years ago

That's why city buses have 3 or 4 double doors and there's ticket machines inside (and drivers don't sell tickets). The time to board rarely goes over 15 seconds.

Compare:

https://www.lubus.info/images/stories/taborbus/5122-57.jpg vs https://www.chicagobus.org/system/photos/250/large/DSC00925....

That's double the boarding time at every stop right there.

The schedule is also designed in such a way that the bus is usually ~1 minute ahead of time and can wait for the proper time to depart from each bus stop - zeroing the randomness on each stop. If it gets too delayed on one part of the route it can catch up on next few bus stops.

On intercity routes there's fewer bus stops so usually there's just 1 door and the driver sells the tickets.

bhuber• 2 years ago

This phenomenon consistently happened to my college bus system, but on an even worse scale. The main bus line did a loop around campus, which took ~20 min to complete and buses scheduled every 5 minutes. In reality, you got a caravan of 4 busses arriving every 20 minutes, with the first one totally full and the last practically empty.

theluketaylor• 2 years ago

When I was a teen in Calgary the transit agency was really good at dealing with issues like this during peak periods. They would pair or triple busses together and alternate stops. If someone requested the stop the drivers would radio to coordinate. Sometimes both buses would have a requested stop, but they would work together so only one bus allowed new riders on. The non-loading bus would quickly drop off passengers and leave while the other stayed behind to handle new riders. Nearly all the stops had dedicated out of traffic space for the bus, so the leap-frog maneuver was really simple. A small amount of low cost infrastructure and some operational cooperation enabled much better service.

rvb2• 2 years ago

This phenomenon is called "bus bunching". My friends, two profs from Georgia Tech and UChicago, came up with one solution for it. They wrote a paper about their solution[1], and then built a startup that has successfully implemented it at a bunch of places[2].

[1]: [A self-coördinating bus route to resist bus bunching](https://doi.org/10.1016%2Fj.trb.2011.11.001)

[2]: [NAU’s new bus system makes for shorter wait times for riders](https://news.nau.edu/nau-bus-schedules/)

eichin• 2 years ago

That's why "dispatcher" is an actual job.

slater• 2 years ago

Isn't that when the second bus just sits idling at one stop for 5-10 mins? That's what they do here in SF ¯\_(ツ)_/¯

amiga386• 2 years ago

This is https://en.wikipedia.org/wiki/Bus_bunching

soperj• 2 years ago

If you track the busses, this should be as easy as changing one bus to "bus full" and have the emptier bus behind it picking up the passengers for a while. That will speed up the fuller bus and slow down the bus behind it.

blahedo• 2 years ago

In Chicago (I assume also other places) the leading bus will sometimes announce "after this stop this bus will run express to Western" (or wherever) "if you need a stop before then, transfer to the bus behind this one". I wish they'd do it more often, it really helps to de-bunch the bus routes.

tunesmith• 2 years ago

Some bus systems handle this (partially) by only allowing passengers to disembark from the lead bus. Stop, open the back door, don't open the front door, take off. I don't know either way, but the belief is that it helps smooth it out over time.

whiterock• 2 years ago

There are still buses that sell tickets :O May I ask where? This has been shut down years ago where I live for the time it takes as you say.

fragmede• 2 years ago

this is called bus bunching and is a well studied phenomenon.

remram2• 2 years ago

This seems easily fixable for buses, since they can overtake each other. Once the empty bus is in front, it is the one arriving first to pick passengers, while the full bus empties.

In practice I see buses slowing down or stopping to keep their ordering, and I don't understand why.

(Of course that doesn't work for trains which can't overtake)

fasa99• 2 years ago

What do they call it with packets, "quality of service"? FIFO.

It's a good plan but you have to make sure fairness is ensured. It's the same feeling as, if I go get a burger & fries, the guy behind me orders identically, burger and fries, and somehow gets served first. The instinct is to say, "what the heck, why does HE get fast service than me in a straightforwardly unfair manner?!?"

So you'd need the empty bus and full bus to hit the same spot at the same time, fill the empty bus as the full one unloads. But which bus leaves first? If the former empty bus leaves first, you're being unfair QoS to the people who are seated on the full bus (which, to their perspective, at their prior stop, was the first-in-order-on bus so therefore should be the first-in-order-off bus. On the other hand, if the full bus leaves first, why did you make incoming passengers board the empty one? They should have boarded the full one to get to their destination faster.

The solution, which we see in practice, is to have the empty bus wait 5 or 10 minutes and let the full one get ahead. If we are seeing rounded fairness to all parties as the priority.

maeil3• 2 years ago

This was easily the most memorable thing I learned during my statistics degree! Nothing else has stuck with me this well.

mass_and_energy• 2 years ago

Does this relate in any way to the phenomenon of "lighting a smoke to make the bus come?" you see, you're waiting for the bus and after a few minutes you realize "man, I could have had a smoke by now" so you light a smoke, but sure enough the bus will come before you can finish your cigarette. This seems to happen every time you light the cigarette waiting for the bus. So this time you get to the stop and light your cigarette right away so that the bus comes, to no avail. What gives?

drexlspivey• 2 years ago

Same thing is true for Bitcoin block times (also a Poisson process), a block is expected to arrive every 10 minutes on average but if 10 minutes (or 15 or 20) have passed since the last block the expected time for the next block is still 10 minutes.

kwhitefoot2• 2 years ago

I haven't read the article but just to answer the question in the title: Buses must always be late because a bus that leaves early is even more useless.

asdff2• 2 years ago

Well, they leave early all the time too.

kwhitefoot2• 2 years ago

Really? Where? That happened about twenty years ago where I live in Norway. The result was that bus drivers got severe tellings off, mostly from elderly women who had been used to arriving at the bus stop just in time. After a couple of weeks the practice was stopped.

fiforpg• 2 years ago

There is a mildly related math puzzle I learned at some point in high school (iirc):

someone comes to a subway station at (uniform) random times between 6p and 8p; he notices that the first train he observes arriving at the same station is 3 times more often inbound than outbound. He also knows that time intervals between the trains going in the opposite directions are fixed and equal — say, always 6 minutes, so the only random event here is when this person arrives at the platform. Explain how this is possible.

CRConrad• 2 years ago

The outbound ones are scheduled a minute and a half after the inbound. (For neat integers, should have been twice as often, or 8 minutes between them.)

But the bigger question: What's the difference between "an inbound train" and "an outbound train"? Aren't almost all trains usually both inbound and outbound? First they arrive at a station, then they go on to the next. Or even at a "terminus"-type station; first they arrive at the terminus, then they go back in the other direction. How does he decide whether a train is an inbound or an outbound one, while it is standing still?

wodenokoto• 2 years ago

Why does he times tau by N when calculating the bus arrival times in the beginning?

rhymer• 2 years ago

If the bus arrives on time, the arrival time would be [tau, 2 * tau, ..., N * tau].

One way to simulate "random" arrival time is to draw uniform points in the interval [0, N * tau].

It turns out the inter-arrival time generated this way is approximately exponential:

1. the difference of consecutive ordered uniformly distribution random variables follows a Beta(1, N) distribution [1].

2. As N goes to infinity, N * Beta(1, N) converges to Exponential(1) [2].

3. Since we scale the rand() by N * tau, the inter arrival time will follow an Exponential(1 / tau) distribution (as N goes to infinity), which has an expected value of tau [3].

Edit: I just realized that the author did mention this simulation is only an approximation in the side note.

[1] https://en.wikipedia.org/wiki/Order_statistic#The_joint_dist...

[2] https://en.wikipedia.org/wiki/Beta_distribution#Special_and_...

[3] https://en.wikipedia.org/wiki/Exponential_distribution#Relat...

The waiting time paradox: why is my bus always late? (2018)

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