The waiting time paradox: why is my bus always late? (2018) | Hacker News Reader

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

jakevdp.github.io

219 pointsskadamat190 comments

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.

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

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.

>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).

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.

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.

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

tap

Often the most expensive though. You're paying the highest individual fare rate, possibly plus card processing fees.

If you buy a transit pass or use their app you can get significant discounts.

in big cities in the uk (edinburgh, london, etc.) paying with a contactless tap of a debit card each journey will never result in paying more than the cheapest daily, weekly etc. travel pass. single tickets are gbp 2.00 and a daaily pass is gbp 5.00 so individual journeys are recorded and counted (when paid for with the same card, obviously) and at the end of the day, if more than three journeys have been made the total charge for the day is capped at gbp 5.00 - and the weekly total is also capped at the value of a weekly pass, etc.

> daily, weekly etc. travel pass

At least in London it stops there, i.e. the automatic capping doesn't do longer than weekly capping. Also the weekly capping only runs on a fixed Monday to Sunday period, whereas an individually bought weekly travelcard can start on any day of the week.

A lot of systems that work this way also cap your fare so that you never pay more than the day rate. Say the day cap is £5, and an individual fare costs £2, then if you take two journeys you'll pay £2 each, but from the fifth journey on, you're capped and only pay £1. In some places, there's day and week caps, so if you travel every day of the week, the price will be capped again once you've travelled enough that week.

The Oyster system in London is a classic example of this - originally you needed to load money onto a separate card, now you can do it with your bank card instead, but it still all works the same. I believe it's also the cheapest option - the daily and weekly caps are cheaper than their paper ticket equivalents. You can probably get better deals in specific cases, but generally the Oyster system is the way to go. I know they're slowly bringing that concept in in other cities in the UK as well, and there are schemes around Europe that use an app with GPS as the "tap on, tap off" mechanism.

In Vancouver you can tap your credit card but also tap a fare card that gives you a discount. I'd have thought buying a ticket on a bus would be a thing of the past- wondering where the parent is from.

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.

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!

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.

OTOH, it's extremely annoying to sit on a stopped bus when no one is boarding or leaving. That discourages use of mass transit.

It takes like 10 seconds. And you have to keep the schedule anyway - if you skip this bus stop you'll be waiting at the next one longer.

For express or intercity busses, that makes sense, but for high-frequency regular bus routes, I can't imagine that working. It means thar bus stops would have to be extremely sparse, or else the bus trips would need to be extremely slow.

Exactly the opposite. It sucks for intercity buses, cause there's no point. Bus stops are rare and buses don't "bunch up". It's essential for city buses.

This is how it works in every big city in Poland, it's working great. More cities started to use this system over time, because it improves the scheduling so much.

The point of city buses is that they drive in traffic anyway - they rarely drive over 50 km/h and they stop every 5 minutes. How regular they are is MUCH more important than how fast they drive.

If you skip 3 stops because nobody waited there - you get to the 4th bus stop 5 minutes too early and wait for 5 minutes there - potentially blocking the bus stop for others and wrecking havoc with the scheduling. Much better to split these 5 minutes between the bus stops where nobody is blocked.

It's like in gamedev - you don't want to optimize happy case cause you're making the situation WORSE. If your fastest frame takes 5 ms instead of 10 ms it changes nothing at best (and makes for more jerky movement at worst). If your longest frame takes 15 ms instead of 18 ms - it means you can keep consistent 60 FPS now - and that's a HUGE win.

City buses stop much more often than every 5 minutes in my experience. It's more like every couple of blocks, sometimes every block, at least in the densely populated areas.

That is way too often, it makes service too slow. People have things todo and sitting on the bus is not one oi them (even if they are doing something it is rarely what they want todo.

what is right is a compromise but everx block is too much

I agree, but a five minute drive is about a 30 minute walk (for an able-bodied person), which is way too far between bus stops for a non-express bus. Especially if you have to catch a connecting bus going along one of the major streets your bus skipped by.

Yeah, I checked the routes I take and it's 40 minutes for 25 bus stops so about 1.5 minutes between stops.

For intercity busses, keeping an accurate schedule is essential. If you miss your bus because it ran ahead of schedule, it's not a five-minute wait for the next one; you'll possibly even be booking a hotel for the night.

For express busses, stops are far enough in between and all major locations, so you may as well stop at all of them.

For milkrun busses, where the frequency is so high, scheduling errors are only really a problem if the busses bunch up and create excessive gaps.

If a bus trip takes 45 minutes when a car takes 15, more people drive and then traffic gets bad. But busses with dedicated lanes and coordinated light-timing can go much faster than traffic, when they aren't stopping for passengers!

I think you and I must live in cities with very differently-run transit companies!

Same in Sydney Australia,

> If a bus is indicating ( min 5 secs ) and is pulling out from a bus stop they have right of way. Failure to Give Way is worth a $362 fine and [demerit] 3 points.

>bus stops would have to be extremely sparse, or else the bus trips would need to be extremely slow.

Way too many transit operators choose extremely slow. Having bus stops every 100m is popular because it offers almost door to door service. But when every single person separately boards it results in the vehicle being stopped 1/3 of its running time! People generally prefer faster bus routes (average 20 MPH) even if it requires them to walk a block to the stop versus a service that stops every block but averages 6 MPH (bicycle speed).

https://humantransit.org/2011/04/basics-walking-distance-to-...

You can do a study of the actual number of people who get on/off at each stop and then determine which ones should be optional. And at off-peak hours, almost all the stops are optional at least from what I've seen in Chicago.

> And at off-peak hours, almost all the stops are optional at least from what I've seen in Chicago.

Do you not have schedules at bus stops? If you skip almost all the bus stops you'll be like 10 minute early at the first non-empty bus stop, so you'll have to wait for these 10 minutes there (or you depart early which makes people miss their bus).

Potentially you'll be blocking the bus stop for these 10 minutes for other buses.

Why not split these 10 minutes between the empty bus stops instead?

From my experience in NY, where most stops are optional unless someone needs to get on/off, the schedule means very little. If the schedule says the busses come at 15 minute intervals, all you can assume is that the bus might hopefully come sometime within the next 15 minutes. There tend to be stops roughly every block, so having all stops be mandatory would make walking competitive with taking the bus.

This is how it works around me in Detroit. I know the intervals for different routes and for 15min or less it's not a huge deal to wait. There is a major route that runs only once an hour so I will look up the exact schedule or the live bus tracker for that one. The only mandatory stops are the stations on the ends of the lines / bus turn around points / major transit hubs like the train station or the Greyhound/etc terminal.

I often walk from the place I live at one end of the city here in Lublin to the artificial lake at the other end of the city and I drive a bus back. It goes through the city center.

It's 12 km on foot one way and probably like 20 km in the bus the other way (it can't drive on the pedestrian/bike path along the river). I walk these 12 km in 2 hours and the bus takes 40 minutes to take me back. There's 25 bus stops on the route I take (and a few more later). There's 2 optional bus stops but they are past the point where I get off the bus.

So that's 1.5 minutes between bus stops (including the stops themselves which take around 5-15 seconds usually).

What's wrong with walking being competetive with buses BTW? The point is that it's better than driving in a car.

NYC is a big, dense city, it already takes forever to get around via public transport. If busses were about as fast as walking, people would just use cars.

There's nothing all that bad about driving a car, so public trasport has to be at least better than walking to be better than driving.

For example, looking at some of the commutes my family does (I work fully remotely), one has to spend 45 minutes to travel ~12km via subway, which would be 20 minutes via car, or 2 hours if walking. Another has to spend 2 hours to travel ~50km via subway and bus (all still within the city) or 30 minutes via car, with the equivalent walk apparently taking 7 hours.

So, in the first case, that's an extra hour and a half they're losing per day to just transport, and in the second case that's 3 hours per day being lost to transport. Now add in time spent taking kids places, and basically you end up spending most of your life outside of work just getting around instead of actually doing stuff.

But you have to search for a parking space and walk from the parking space to the actual destination anyway. Which is like another 10 minutes on top of everything. So buses win over cars in the city center (as long as you can plan to the exact minute so you don't waste time waiting at the bus stops).

My city has actually recently switched to making all stops optional, claiming that it improves efficiency. Let’s see how that will go.

It does if nobody rides the bus you can move faster but somehow you need to ensure the faster but doesn't attract so many riders that you are stopping too often. that generally means watching this and if you get close doingisomething different.

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.

Buses already do this.

If nobody is waiting and nobody asks to get off they don’t stop. If nobody asks to get off and there’s a second bus right behind, drivers skip the stop.

It amuses me how many people don't know how busses work on that front. I'm assuming most people have more of a train mentality when it comes to this?

I'd guess most people here don't actually take buses on a regular basis. (Or ever. Or any public transit at all.)

Yes of course. But it's not so useful since the buses are so big and they arrive so infrequently. With robobuses you could have smaller more frequent arrivals and then have higher probability of skipping. If the robobus understands the bus stop person's gestures that are they trying to stop it or not...

The cost of a human operator is not even remotely close to as important a factor as you think it is.

SF recently acquired 33 new buses for $1.7 million apiece. Throw in maintenance and fuel costs and it’s easy to see that amortizing over a driver making $40/hr plus benefits is just not that big a deal, especially if you’re now adding a suite of sensors, electronics, and computing hardware.

Let's take this to it's logical conclusion, where each individual rider is given their own packetized vehicle that takes them right to their destination. We can amortize this cost by relying on the riders themselves as the drivers.

What's the cost of the driver vs fuel/maintenance/capital cost of bus?

Respectfully, this is a trick question. The driver isn't exclusively there to drive, they help tourists that ask "Can I get to X, or do I need another bus?". They help wheelchair users, they know to wait extra time when cyclists need to collect their bikes from the front bus rack. They know when to pull poles (and reattach them) on trolley buses in Vancouver (where I hail from). They report fare-dodgers, they radio control saying their bus is too full to pick up anyone waiting at the stop, and most importantly: they write down what goes wrong during their day, so maintenance knows what to fix when the bus is at the garage.

Seriously, Translink runs on a 28-hour day for bus service. If you want someone to test every single system, especially someone that knows how a system should work vs how it does work vs how it failed on this particular bus, you already have an employee for the job; one that's doing customer service and working around any shortcomings at the same time.

I would not count on bus drivers being replaced by AI any time soon.

I've seen a study done not that long ago for a small to medium sized German city. The idea was either taking the existing bus service and expanding it to a 10-minute-frequency all-day on all routes, or else keeping only the three most important (regional) bus routes and replacing everything else with a massive fleet of on-demand vehicles.

The surprising results:

- According to the results of the traffic modelling, the on-demand scenario wasn't substantially more successful in attracting additional passengers than the "classic buses running every 10 minutes" scenario (i.e. in both cases predicted passenger counts increased, but by approximately the same amount). This was because on-demand public transport has quite a bit of time-wise overhead, too: You need to order a vehicle instead of simply showing up at the bus stop, it takes a while for the next vehicle to arrive, due to ride sharing some detours might be incurred compared to a direct route, and the unpredictable journey time variation due to these factors is very disadvantageous when connecting to fixed-route fixed-timetable public transport, such as the remaining three bus routes, or railway services.

- The overhead of operating a large fleet of on-demand vehicles was high enough that even with driverless operation the on-demand scenario was more expensive to operate than the expanded every-ten-minutes bus service with drivers

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.

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

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.

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.

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.

Another simple strategy that I've seen is simply for the loaded bus to allow the empty bus to overtake it and go on ahead (and just stay ahead).

That sounds identical to what the Calgary busses do? You'd still need coordination between the busses to know when the loaded bus "wants" the empty one to overtake it.

The driver in the front sticks his hand out the window and waves to the one in the back

The difference is that it was a one-and-done thing rather than leapfrogging back and forth as it sounds like `theluketaylor` was describing.

And yes, the drivers would coordinate. (I've sometimes seen it done with a brief honk for attention followed by a hand wave.)

Why does the loaded bus’ preference matter? If you see a bus stopped ahead of you with the same route number and nobody requested that stop then just continue.

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/)

That's interesting.

Looking at the second link, it seems they implement it by having the buses pause at certain points. Does it do that with riders onboard? That seems like it could be a deterioration in experience for those riders that are on the bus pausing.

It does sometimes feel stupid for the bus to idle at the stop for 1-5 minutes, but on the flipside they are in time on the dot at the stop.

I guess the reality is, as a passenger, you either wait several minutes on the bus at a stop mid-route. Or you wait much longer at a bus stop in a crowd waiting for four buses to show up.

When possible, they pick "control points" at places where there are usually no passengers on board — for example, the ends of a linear route, or the bus depot.

But otherwise, they have a bunch of optimizations to spread the pauses out so they aren't too jarring. They also display the timer prominently so that riders are aware of what's going on.

In practice, riders seem happy with the tradeoff, since it has resulted in overall lower times to get from point A to point B.

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

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)_/¯

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

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.

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.

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.

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.

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

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)

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.

This makes no sense to me at all. The bus leaves when it is done unloading and loading. There is no question of fairness here, it is not like someone made the fries you were waiting for and gave them to the wrong person.

In a network metaphor, you do not halt all other TCP connection on a system because one TCP connection was stalled by a lost packet, in the name of fairness. No one would ever want that.

The people on the full bus can easily see that they are not moving because they are still unloading. Who would see another empty bus passing by and thinking "those people should be made to wait"?

In my experience on the subway, a train will start SKIPPING STATIONS when they are bunched together, forcing me to get off and wait for another train entirely. So I don't think fairness is the reasons buses keep their order.

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.

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.

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

``Then stop raising your arm.''

My favorite version:

A patient comes into a doctors office complaining of pain all over. The patient grabs their elbow and says: "It hurts when I touch my elbow like this." They then grabs onto their shin and says "When I grab my shin like this, it starts hurting too!" Finally, they massage their forehead and say "It even hurts when I rub my head! Doctor what's wrong with me!?"

The doctor runs a few tests and replies: "Your finger is broken."

My favorite doctor joke, semi related:

Patient: Doctor will I be able to play piano after the procedure?

Doctor: Yes, I don't see why not.

Patient: That's wonderful! I could never play piano before!

Simpsons did this one in the Planet of the Apes episode. Here's a musical rendition by Dankmus.

https://www.youtube.com/watch?v=sMRcIOjdojU

What’s wrong with me?

I think you’re crazy.

I want a second opinion!

You’re also lazy!

I am not crazy! (in the voice of chuck from Better Call Saul)

You just need to practice.

I’m fond of an exchange from the end of “The Doctor Dances”:

Mrs. Harcourt : My leg’s grown back. When I come to the ‘ospital I had one leg.

Dr. Constantine: Well, there is a war on. Is it possible you miscounted?

Patient: Every time I drink coffee I feel a pain in my right eye.

Doctor: Take the spoon out of the cup.

I feel this is my bones

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!

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.

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.

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.

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

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

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.

> 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.

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.

For anyone unconvinced, let's simulate this.

Instead of 1 in 14 million, we'll just do 1 in 2, and 8 players.

So we'll check how many bits are set in the average random byte:

    void Main()
     {
      byte[] buffer = new byte[256*1024];
      Random.Shared.NextBytes(buffer);
      var avg = buffer.Average(b =>     System.Runtime.Intrinsics.X86.Popcnt.PopCount(b));
      Console.WriteLine(avg);
     }

Okay, bounces around 3.998 to 4.001, seems normal.

Now let's check how many bits are set in the average random byte given that the low bit is 1 (i.e. player 1 has won!)

    void Main()
     {
      byte[] buffer = new byte[256*1024];
      Random.Shared.NextBytes(buffer);
      var avg = buffer
       .Where(b => (b & ((byte)0x01)) == 0x01)
       .Average(b =>     System.Runtime.Intrinsics.X86.Popcnt.PopCount(b));
      Console.WriteLine(avg); 
     }

Now ~=4.500

Which is 1+3.5

In this case, we're 1+ average from the 7 other players, so being an average of 7 others not 8 others is significant.

If we simulate with millions of players, you'll see that removing 1 person from the pool makes essentially no difference.

You are making an assumption you did not explain before and makes it confusing, you don’t know the number of winning tickets or the number of winning tickets affects the prize quantity, and not all lotteries work like that.

I'm essentially assuming the UK lottery (from 96-200x) rules, where players:

Pick 6 numbers from 49, so odds are independent, and roughly 1 in 14m.

The prize jackpot is determined from a set percentage from the ticket sales, and is shared between jackpot winners.

How much the jackpot prize is therefore determined by total sales and how many winners there are.

I'm also assuming your individual ticket contribution doesn't materially affect either the prize pool or the number of people playing. For a large N, small p, this holds true.

This is a variant of the Monty Hall problem, and the trick that makes it unintuitive is that you’ve snuck in the conditional of assuming you have already won.

If you have a ticket and haven’t checked that it is winning, you should expect two winners, regardless of whether you end up being one of them.

If you play the lottery trillions of times (always with the same odds, for simplicity) and build up a frequentist sample of winning events, you will on average be part of a pool of two winners in those instances where you win.

You snuck in the assumption that the first ticket checked (yours) is a winner, which screws up the statistics.

I haven't "snuck in" anything, I very explicitly stated that it's the conditional expectation I'm talking about.

Expected(Number of winners | Pop size N and You win) = 1 + Expected(Number winners | Pop size N-1)

For small p, large N, that's ~1 + Expected(number of winners).

The main problem with your argument is that the "average lottery winner" doesn't win JackpotPool/Average winners.

I think you forgot to mention the condition 'given that you already have a ticket', and whatever justifications are required to assume that two more winning tickets will be present (if each ticket has independent odds of being a winner then you end up with a distribution of other tickets yeah?). Otherwise your premise doesn't quite lead to your conclusion.

Right, I didn't spell out the format of the lottery. I'm assuming a "pick X numbers from Y" format, rather than a raffle style lottery.

This allows for multiple independent winners.

That doesn’t work very well when the average number of winners is much less than 1. The math might work out that the “expected value” is more than one winner but in a realistic lottery you should expect to be the only winner.

I'm not sure if this is true, as large jackpots see a higher than average number of tickets sold.

It is true. The average power ball does not have a winner.

Every lottery I know of has many winners. One big winner but many who match only one numbe, and so win a tiny amount.

Winners in this case means jackpot winners. This is especially relevant as unlike partial winners, the jackpot is shared, not duplicated.

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.

You said my conclusion was wrong (edit: Apologies, I confused the nesting level here), then proved it correct by calculating the expected number of winners given you win as 3/2.

(I think cortesoft was responding to pif, whom you also responded to, thus agreeing with you.)

Ah, thank you, navigating the nesting on here is difficult sometimes and this has proven a very contentious topic!

Nice. And, say the lottery jackpot is a constant 6$, then the average winning per player is 3$ (case 1) or 6$ (case 2) or 6$ (case 3), each equally likely (case 4 is not applicable), so $5.

However, if A wins, A wins either $3 (case 1) or $6 (case 2), so A's expected winnings are $4.5, which is indeed < $5, as GGP asserted.

The "average lottery winner" also wins $4.5 though. (The original claim was that "if you win the lottery jackpot, then you win less than the average lottery winner".)

If there are 100 draws with a $6 jackpot 25 will have no winners, 50 will have one ($6) winner and 25 will have two ($3 each) winners.

100 winners in total - half won $6 and half won $3.

You are correct. So, the statement should not be "if you win the lottery jackpot, then you win less than the average lottery winner", but "if you win the lottery jackpot, then you win less than lottery winners win on average"?

When you win the lottery you win on average what lottery winners win on average when they win the lottery.

One may find ways to define things differently so something is less than something else but I’m not sure what’s the point in doing so.

Maybe pif's comment

"The total number of expected winners (including you) is the same as the average number of winners"

means

"The total number of expected winners (including you) is the same as the average number of winners when there is at least one winner"

All the possible outcomes where zero people win are irrelevant when it comes to the calculation of how much "the average lottery winner" wins.

I mispoke a bit when saying it is ALL because of the case where zero people win.

It still holds for non-zero cases, too.

Since whether any individual wins is independent of other people winning, selecting only the situations where you win doesn't change the odds of other people winning, it simply adds a 100% chance of you winning. So it has all the same combination of winners, plus you.

I don't have time right now to type out a more full explanation, but I hope this somewhat makes sense given my previous comment.

> So it has all the same combination of winners, plus you.

And the same is true when you condition on having at least one winner. One winner doesn't change the odds of other people winning.

[edit: this may not be correct, never mind "In your example it doesn't matter whether you condition on A winning, on B winning or on at least one of A and B winning."]

Right, one winner doesn't change the odds... but we are choosing to throw out all the scenarios where that winner doesn't win, which DOES change the overall odds distribution. We are changing our selection criteria.

I think my previous comment was wrong. Anyway, the point is that the original claim

"if you win the lottery jackpot, then you win less than the average lottery winner"

seems wrong unless the winnings of "the average lottery winner" are defined in a quite unnatural way.

In your example the average lottery winner wins 3/4 of the jackpot. Half the winners take it all, the other half have to split it with someone else.

No, it it still the case that “if you win the lottery jackpot, then you will win less than the average lottery winner”. Let me see if I can explain in another way that might make this more clear… the example of only 2 people actually confuses the issue.

So in our example with 50% chance of winning, we know the average number of winners will be n/2, where n is the number of players. This means that the average lottery winner will win prize_pool / (n/2).

Now, let’s say we know I won. That means the average number of other winners is going to be (n-1) / 2. If you add in the known winner (me), we would have an average of 1 + (n-1)/2 winners… meaning the prize per person when I win is going to be prize_pool / (1 + (n-1)/2).

You can clearly see that the prize pool will be smaller when you know I am a winner. If it isn’t clear, just sub in 10 for N and solve it… the average winner will get prize_pool / (10/2) or prize_pool / 5. When I win, the average winner will get prize_pool / (1 + (10-1)/2), or prize_pool / 5.5. You can see that when I win, the average is lower.

This of course works whenever you start with the assumption that a particular person wins… you are turning the 1/2 chance for that person into a 100% chance, which increases the overall average number of winners.

> This means that the average lottery winner will win prize_pool / (n/2).

Does it?

Take the case n=2. Run the lottery a few times and take all the winners.

Half the winners win prize_pool.

Half the winners win prize_pool/2.

How do you define “average lottery winner” so the average lottery winner will win prize_pool / (n/2) = prize_pool ?

Taking the n=10 case, because you think n=2 is confusing.

> the average winner will get prize_pool / (10/2) or prize_pool / 5.

No, the average winner will get expected_prize_pool / expected_number_of_winners.

If 5 is the number of winners averaged over all draws - including those without winners - the (average) pot they share has also to take into account draws without winners.

The average prize shared in this case is not prize_pool, it’s 1023/1024 times prize_pool.

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!

> 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.

> 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.

Well, I simulated it and the numbers seem to agree with you. The example is interesting. I still have trouble seeing why my original reasoning doesn't hold though. I'll give an example, if anyone can clear up the issue that would be appreciated.

1/10 odds, 10 entrants, one winner expected on average.

Given a particular winner: Expect: 0.9 + 1 winners

Given the same particular loser: Expect: 0.9 winners

Over all cases we see: 0.1(1.9) + 0.9(0.9) = 1 winners

Checks out, but if the numbers are correct then any winner should be able to calculate the higher average and be right knowing only that there is at least one winner. So in cases where there is at least one winner: P(winners>=1)=1-(9/10)^10=~65% The expectation should work out to 1.9. The rest of the time we expect zero winners. However if I use those numbers I get an overall expected number of winners as 1.237, which has increased the overall number of winners across all cases. In order for that number to work out to one, the expected winners when there is at least one winner would have to be ~1.535. Which suggests that the expected outcome is different depending on if you check your own ticket, or someone else's, even if you see the same thing?

Am I just not on for math today? I thought the solution to the paradox would be that the higher expectation discounts outcomes with zero winners.

> 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).

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.

Conditioned on you winning, there are 1+N expected winners. The number of winners is always larger than zero [edit: the following is not correct "and the average prize is calculated diving the jackpot by the 1+N expected winners"].

Conditioned on you losing, there are N expected winners. The number of winners can be zero and the average prize cannot be calculated dividing the jackpot by the N expected winners [edit: not that you could before...]. [edit: the following is not correct "You have to divide the jackpot by a higher number: the expected winners conditional on having at least one."] No winner, no prize.

---

Let's say there are 2 people playing and the probability of winning is 50%. The number of expected winners is 1. If the jackpot is $1000 the "average lottery winner" doesn't get $1000.

Three outcomes are possible, with the following probabilities:

  1/4 zero winners
  1/2 one winner gets $1000
  1/4 two winners get $500 each

The "average lottery winner" gets less than $1000. The "average lottery winner" gets $750. (Imagine that a lot of draws have happened: for each split jackpot there were two jackpots going to a single winnner. All in all, half the winners got $1000 and the other half got $500.)

Consider now that you are one of the two players and you win. The other person will either win (you get $500) or not (you get $1000) with the same probability. Your expected prize? $750

What a coincidence!

> Consider now that you are one of the two players

This assumes that when you decide to buy a lottery ticket, you get to prevent someone else from buying one. If you decide to buy a lottery ticket, now there are three players.

Not sure what’s your point bur the original claim was about a constant (or constant-enough) number of players.

> I'm also assuming your individual ticket contribution doesn't materially affect either the prize pool or the number of people playing. For a large N, small p, this holds true.

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?

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.

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.)

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

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

Thanks for the tip!

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.

Filter tip only, I hope.

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

https://xkcd.com/583/

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.

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.

Eh, a bus ride is a bus ride. They're all outside of the 'comfort window' so to speak. When I was heavily dependent on nicotine I could still go a couple hours and not really mind it.

Maximizing the minutes doesn't track but people are irrational. When I got an e-cig my use went way, way higher.

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

I wish there were an r/AngryUpvote thing here

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

>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 ???

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.

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?

No, N means 10 in both parts of the sentence you quoted.

> when the average span between arrivals is [10] minutes, the average span experienced by riders is [20] minutes.

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

The average wait time is half the 'average span experienced by riders', since the rider arrives, on average, halfway through the average span.

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!

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

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.

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

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

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.

Well, they leave early all the time too.

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.

Here in the U.S. It probably happens to me personally about once a month as a commuter. Sometimes the bus just ends up a little faster than anticipated. Other times the bus driver is trying to get ahead of the schedule to buy them more time to run in to a mcdonalds and get food (I've only seen this once though). If they end up late its pretty bad for them as it eats into their break time on the bus layover.

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.

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.)

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.

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?

If the buses are perfectly at schedule and people arrive at random (uniform), the mean (and the median) waiting time will be half the scheduled interval.

If both are completely uniformly random, the mean waiting time will be the mean interval between buses. (In fact the distribution of the passenger arrival doesn't matter anymore.)

The real world is somewhere between those two.

Isn't that at odds with the article? Quoting, "The reason is that there are (of course) more students in the larger classes, and so you oversample large classes when computing the average experience of students." This leans on the experiences of the students, which necessarily leans on the population. (Granted, this is more clearly a different question/scenario.)

I'll try and play with the simulation some. And to be clear, I don't disagree with your statements. I just feel these are still different questions/statements. How long I expect to wait at a given trip/stop is not the same as how long I expect I have waited at that stop over the days.

My comment paraphrases what the article says.

You seem to be expecting something different from the inspection paradox, but all it says is that the waiting time will be higher than half of the interval between buses. And it only applies when the time between buses is random.

One of the things I've found far more often than makes sense, is that many paraphrases also change the statement they are paraphrasing.

The article goes out of its way to call out that sampling the average experience of students. This was obviously different in class sizes. Since there are more students than there are classes, it makes sense that the average size of classes that a student attends is different than the average size of classes offered. It isn't that people are giving you two different answers, they are answering two different questions.

Also, the article specifically says the waiting time paradox makes a stronger claim that the average will tend specifically to 2N. "But the waiting time paradox makes a stronger claim than this: when the average span between arrivals is N minutes, the average span experienced by riders is 2N minutes. Could this possibly be true?" I'm trying to explore/understand how that works out.

Oh, right. I was focused on the progression, I didn't notice you were talking about a factor of 2 that I forgot there.

The article says the following:

> If buses arrive exactly every ten minutes, it's true that your average wait time will be half that interval: 5 minutes.

Which means that "not arriving exactly on schedule" is indeed a requirement.

Ah, I clearly misread that spot. I don't think this changes too much of my questioning here. Will definitely be playing with this more.

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.

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?

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.

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?

(2018), but this article is timeless…

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

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?

> 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.

> 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.

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.

It would have had to have been back when NextBus existed and I still lived there, so 15+ years ago.

Well I can confirm it’s the same as of 2019. I stopped frequenting that part of town since.

(2018)

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

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.

The waiting time paradox is not only about the bus.

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

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.

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

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.

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

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...

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