A Puzzle

[lamford]

As this had some lively debate on Facebook, I shall put it on here:

 

Late at night Circle Line trains are now only three per hour (i.e. three in every hour, not an average of three per hour), at random intervals. I just miss one. What is my average wait for another, assuming that the last train has not left?

 

And what would my average wait have been if I had not just missed one?

[StevenG]

You cannot have three per hour at random intervals, for any unspecified hour. If you have three per hour for any unspecified hour, then the fourth must be exactly one hour after the first, the fifth exactly one hour after the second, etc..

 

If you divide the day into 24 hours and only consider those 24 hours, that is a different question.

 

Edit: If you have just missed one, you know there will only be two in the next 60 minutes, so the question is equivalent to "what is the expected value of the lowest of two random numbers in the range [0,60)." This is 20 minutes.

 

If you have not missed one you are now looking at three numbers in the range [0,60]. This gives 15 minutes.

 

(The expected value of the mth lowest number out of n in the range [0,1] is m/(n+1))

[barmar]

I was about to say that this is a trick question, because Circle Line is the name of the sightseeing cruises around Manhattan, so they're boats, not trains. Then I did some additional googling and learned that there's also a London Underground line with this name.

[barmar]

If you divide the day into 24 hours and only consider those 24 hours, that is a different question.

 

Yeah, I was wondering if he meant 3 between 11 and 12, 3 between 12 and 1, and so on.

 

But in this case, to answer you'd need to know what the current time is and how many of this hour's trains you missed. If it's 11:50 and you just missed the first train of the hour, you expect 2 trains in the next 10 minutes so your average wait would be 3 minutes 20 seconds. But if it's 11:30 and you just missed the second train, the average wait would be 15 minutes.

[hrothgar]

From my perspective, the most logical way to frame the problem is to assume a Poisson process with Lambda = 20 minutes

 

(it's silly to talk about three buses every hour at random intervals, because your analysis gets all higglety pigglety if you start measuring hours as running from 10:15 to 11:15 rather than 10:00 - 11:00)

[barmar]

From my perspective, the most logical way to frame the problem is to assume a Poisson process with Lambda = 20 minutes

Isn't that what it would be if the premise were "an average of 3 busses per hour"? But the question specifically says this is not the case.

 

So I think StevenG is right that they can't actually be at random intervals, they have to be every 20 minutes exactly, unless they go by clock hours as I described.

 

 

[broze]

Is the question basically this?:

 

"You are waiting at a train station; 3 trains will arrive randomly within the next hour; what is your expected wait?"

 

Because that seems relatively simple. (15 minutes; see Steven's formula above)

 

Or is that not what is meant by the OP? The puzzle as written seems badly formulated.

 

Also from a practical point of view you can't have randomly distributed trains because they can't realistically arrive within a minute or so of each others. I'm assuming light-fast trains and passengers. :P

[hrothgar]

Isn't that what it would be if the premise were "an average of 3 busses per hour"? But the question specifically says this is not the case.

 

So I think StevenG is right that they can't actually be at random intervals, they have to be every 20 minutes exactly, unless they go by clock hours as I described.

 

I don't think any of us are disagreeing

 

All three of us are indicating that the question, as framed, contains an inconsistency.

 

You prefer to relax the "at random" constraint.

I prefer to relax the constraint that all hours contain exactly three buses.

[Winstonm]

The interesting thing to me about this question is that each (or any) train represents all three types of trains, first, middle, and last within an hour. If I miss a train at 10:16, I can view that as the first train of the hour 10:16-11:16, the last train of the hour 9:16-10:16, or the middle train of some hour in between those.

 

It seems to me the missed train is irrelevant - the average wait for any of three trains appearing within any one hour would be 20 minutes.

[billw55]

Sounds a little like a game of Mornington Crescent.

[gszes]

Questions like this without sufficient details are a MENSA specialty so they

can claim their interpretation is the best:)))))) the questions seems to assume

we are dealing with one specific hour and that within that one hour there are going

to be three trains with no rhyme or reason as far as when those trains will depart

the station you are at. The question seems to further assume all three trains could

leave at the same time and that the timing of the departures during this hour in no

way shape or form affect when the trains will depart the following hour.

 

Note that we are also missing what time you arrived so your arrival time could be

anywhere from zero hour to 5959.

 

Longest wait??? assume the three trains all departed at zero hour and you arrived at 0001

the next hour all three trains could leave at 5959 so you would wait 1hour 59 min and 58

seconds.

 

Shortest zero since the train could depart at the precise moment you step into the train.

 

Average sighhhhhhhhhhhhhh it seems that the average overall time btn trains will average

out to 20 min btn departures. This means your average wait if you just missed a train

(presumably by 1 second) would be 19 min 59 seconds. If you arrived at any other random

time (and assuming any time after 1 second made further calculation impossible) your

average wait would be half that (unless you are a bridge player than my money is on

the full 1hour 59 min 58 seconds can trumps please stop breaking 50 please:)

[nige1]

Arithmetic isn't my long suit but these problems might be equivalent to:

A

• Place 3 dots (trains), randomly on a circle, circumference 60 units (mins)
• Designate another random dot as yourself.
• What is the expected distance along the cirumference, from yourself to a train in a clockwise direction.
• Arguably, the answer is (60 / 4 = ) 15 units (mins).

OK I guessed 10 minutes at first (I really can't do arithmetic) but SteveG's 15 minutes seems more sensible :)

B

• Place 3 trains as above.
• Designate yourself as (almost) coincident with any train.
• What is the average distance along the arc from yourself to the next train in a clockwise direction.
• Arguably, the answer is (60 / 3 =) 20 units (mins).

I suppose you need to subtract a train's platform-time from the above estimates.

The circle line really is a circle but that doesn't seem to be relevant :)

Edited June 13, 2014 by nige1

[helene_t]

20 mins is right, just divide the duration of the night by the number of trains in the night. There are three trains per hour so the average interval must be 20 mins.

[dave251164]

Some light relief needed after that?!

 

Tom is applying for a job as a signalman for the local railroad, and is told to meet the inspector at the signal box.

 

The inspector decides to give Tom a pop quiz, asking, "What would you do if you realized that two trains were heading towards each other on the same track?"

 

Tom says, "I would switch one train to another track."

 

"What if the lever broke?" asks the inspector.

 

"I'd run down to the tracks and use the manual lever," answers Tom.

 

"What if that had been struck by lightning?" challenges the inspector.

 

"Then," Tom continues, "I'd run back up here and use the phone to call the next signal box."

 

"What if the phone was busy?"

 

"In that case," Tom argues, "I'd run to the street level and use the public phone near the station".

 

"What if that had been vandalized?"

 

"Oh, well," says Tom, "in that case I'd run into town and get my Uncle Leo."

This puzzles the inspector, so he asks, "Why would you do that?"

 

"Because he's never seen a train crash."

[lamford]

It is quite hard to phrase a probability question unambiguously, but SteveG is right that if there are three trains in every hour, at random intervals, then one would wait 15 minutes. The trains do not have to be always 20 minutes apart for the requirement in the problem to be met, they just have to be at the same time past the hour. For example, someone uses a random-number device which generates, say, 7.29 minutes past, 28.41 minutes past and 29.11 minutes past for that day. A passenger arrives on the hour, say, and will wait an average of 15 minutes for a train. Of course this theoretical model is not what happens in practice. If the trains are "every 20 minutes on average" then the average waiting time is indeed 10 minutes.

 

I suspect that you have had enough of this problem, and will not want to prove that when there are n trains in every hour, at random intervals, the waiting time is 1/(n+1) hours!

[gwnn]

I think you are just double-counting trains, lamford. It's a bit like how there is only 1 vertex/volume element in a cubic grid, but each cube itself "has" 8 vertices. However, each vertex is shared by 8 different cubes so in fact cubes "have" only eight times one-eighth of a vertex. I suspect that's what you are missing, the train at 12:00 cannot be a part of both 11:00-12:00 and 12:00-13:00. As a sloppy physicist I often scoff at very carefully crafted definitions, but they can be very useful to avoid off-by-one errors. :)

[gwnn]

Are you really saying that if there is only one train per hour, then trains come at an average interval of 30 minutes between each other? Waiting for a train after just missing the previous one is equivalent to measuring the average time between trains.

[helene_t]

15 minutes is correct if you went to the tube station at a random time and weren't told how long time elapsed since the previous train. On average, there will be a 30 minutes interval and you will arrive 15 minutes after the last one, 15 minutes before the next one.

 

It may sound contraintuitive that the average interval length is 20 minutes if you arrive at the beginning of the interval while 30 minutes if you arrive at some random time. But think of it this way: some intervals will be longer than others. Arriving at the beginning of an interval, all intervals are equally likely, so you get the unbiased average which is 20 minutes. But if you arrive at a random time, you are more likely to arrive at a long interval.

 

Here is a similar (easier) classical puzzle: A man works in Manhattan and he has two lovers, one living in Brooklyn and one in Bronx. There are 6 trains per hour in each direction, at regular intervals. He goes to the tube station at random time points and take the train that comes first, in whichever direction it happens to go. He ends up visting his Bronx lover four times as frequently as his Brooklyn lover. Why?

[Mbodell]

Here is a similar (easier) classical puzzle: A man works in Manhattan and he has two lovers, one living in Brooklyn and one in Bronx. There are 6 trains per hour in each direction, at regular intervals. He goes to the tube station at random time points and take the train that comes first, in whichever direction it happens to go. He ends up visting his Bronx lover four times as frequently as his Brooklyn lover. Why?

 

The Bronx train runs through the station at 00, 10, 20, 30, 40, 50 minutes in the hour. The Brooklyn train runs through the station at 02, 12, 22, 32, 42, 52. Therefore 20% of the time he arrives after a Bronx train but before a Brooklyn train (00-02, 10-12, 20-22, 30-32, 40-42, 50-52) but 80% of the time he arrives after a Brooklyn but before a Bronx train (02-10, 12-20, 22-30, 32-40, 42-50, 52-00). So 80% of the time he goes to the Bronx which is 4 times the 20% he goes to Brooklyn.

[gwnn]

If you have not missed one you are now looking at three numbers in the range [0,60]. This gives 15 minutes.

 

(The expected value of the mth lowest number out of n in the range [0,1] is m/(n+1))

Shouldn't this be three numbers in (0,60]?

[helene_t]

The trains do not have to be always 20 minutes apart for the requirement in the problem to be met, they just have to be at the same time past the hour.

ok, I took it as meaning three times per clock hour. 0h20 0h40 0h50 1h10 1h15 1h35 2h... would be permissible

 

It doesn't matter, though. As long as the condition stands that we just missed a train, i.e. we are talking about average interval length, the answer will always be 20 minutes.

[StevenG]

Shouldn't this be three numbers in (0,60]?

Indeed it should. That radically changes the answer to 12.2 seconds :)

[kenberg]

Edit: It is clear that I have completely misunderstood the problem and I still do. Ignmore this post and all other posts by me on this.

 

 

15 minutes is correct if you went to the tube station at a random time and weren't told how long time elapsed since the previous train. On average, there will be a 30 minutes interval and you will arrive 15 minutes after the last one, 15 minutes before the next one.

 

It may sound contraintuitive that the average interval length is 20 minutes if you arrive at the beginning of the interval while 30 minutes if you arrive at some random time. But think of it this way: some intervals will be longer than others. Arriving at the beginning of an interval, all intervals are equally likely, so you get the unbiased average which is 20 minutes. But if you arrive at a random time, you are more likely to arrive at a long interval.

 

Here is a similar (easier) classical puzzle: A man works in Manhattan and he has two lovers, one living in Brooklyn and one in Bronx. There are 6 trains per hour in each direction, at regular intervals. He goes to the tube station at random time points and take the train that comes first, in whichever direction it happens to go. He ends up visting his Bronx lover four times as frequently as his Brooklyn lover. Why?

 

Note: From the post 25 below i guess I misunderstood the arrival time. He is assumed to arrive at the top of the hour. Then the answer is 15 minutes. I was thinking of a random arrival time, uniformly distributed over the hour. I'll leave this up anyway.

 

I need to think a bit about this unbiased average. Between 2 and 3 there are four intervals. If you arrive before the first intereval you have to wait on average half the length og the interva. Same with arriving between the first and second, or second and third. But if you arrive after the third, it's different. You certainly have to wait until 3, you have missed the last train in the 2-3 period, and then you have to woit some random time for the first train after.3. I take it that "random intervals", however this is to be understood, means that the arrival time of the first train after 3 is independent of the arrival time of the first train after 2.

 

So at least it doesn't seem to me that the various intervals are similar enough to look at an unbiased. average.

 

Simpler case: One train each hour at a random time, uniformly distributed throughout the hour.

 

X = arrival time of the train between 2 and 3

Y=arrival time of train between 3 and 4

Z = arrival time of the passenger between 2 and 3.

All uniform distributions.

 

F(X,Y,Z) = time he must wait. This is X-Z if Z<X and Y-Z if Z>X. We want the expected value of F

 

Is this the way you understand the (simplified) problem?

 

We can make the obvious modifcations

 

X1 = arrival time of the first train between 2 and 3

X 2= arrival time of the second train between 2 and 3

X3 = arrival time of the third train between 2 and 3

Y=arrival time of first train between 3 and 4

Z = arrival time of the passenger between 2 and 3.

Then modify F, and we need the expected value, is that right?

[Vampyr]

He ends up visting his Bronx lover four times as frequently as his Brooklyn lover. Why?

 

It is because girls who grow up in the Bronx turn out to be interesting, intelligent, witty and beautiful.

[lamford]

I think you are just double-counting trains, lamford. It's a bit like how there is only 1 vertex/volume element in a cubic grid, but each cube itself "has" 8 vertices. However, each vertex is shared by 8 different cubes so in fact cubes "have" only eight times one-eighth of a vertex. I suspect that's what you are missing, the train at 12:00 cannot be a part of both 11:00-12:00 and 12:00-13:00. As a sloppy physicist I often scoff at very carefully crafted definitions, but they can be very useful to avoid off-by-one errors. :)

Let us define the (edited: second part of the problem exactly). The person arrives at 11 pm, say (although it does not matter what time he arrives) and knows that between 11:00:00 pm and 11:59:59 pm inclusive there will be three trains, not an average of three trains, which is an entirely different problem. Their times from that station have indeed been set randomly, and the first in my simulation happens to be 29.78, 7.82, 49.75. The person will have 7.82 minutes to wait on that occasion, the minimum of three random numbers between 0 and 1 in hours. For completeness, I extended my Monte Carlo simulation to 1099509530625 trials, by looping through 1048575 sets of 1048575 tube journeys. The average for those was 14.99996 minutes waiting. I think I was rather lucky, in that I would have expected to wait 0.00004 minutes more, but not as lucky as in a backgammon game I once won where I was under 10^-27 to win a race. Consider the minutes being rounded down. Now those sets of three beginning with 0 are more frequent than those beginning with 1, which are more frequent than those beginning with 2. You may already know that three-card suits headed by the ace are more frequent than three-card suits headed by the king and these are more frequent than three-card suits headed by the queen, etc. The distribution of earliest departure times by minute rounded down is as follows:

0 0.049017953

1 0.047888801

2 0.046009108

3 0.044521374

4 0.042689364

5 0.040990869

6 0.039846458

7 0.038291014

8 0.036938703

9 0.035310779

10 0.034066233

11 0.032547982

12 0.031503707

13 0.03007987

14 0.028824834

15 0.027377155

16 0.026376749

17 0.02490809

18 0.023915314

19 0.022685073

20 0.021656057

21 0.020590802

22 0.019369144

23 0.018569964

24 0.017516153

25 0.016484276

26 0.015553489

27 0.014978423

28 0.013684286

29 0.01294328

30 0.011911404

31 0.01138116

32 0.010547648

33 0.009656915

34 0.008959779

35 0.008337983

36 0.007797249

37 0.007145888

38 0.006352431

39 0.005821234

40 0.005325322

41 0.004804616

42 0.004326824

43 0.00368214

44 0.003322604

45 0.002941134

46 0.002514842

47 0.002120974

48 0.001779558

49 0.001597406

50 0.001202584

51 0.001039506

52 0.000739098

53 0.000585557

54 0.00038147

55 0.000303269

56 0.00015831

57 9.72749E-05

58 2.67029E-05

59 3.8147E-06

We can trivially see that we are 51.9294% to wait 12 minutes, and if we were betting with someone at even money that a train would arrive within x minutes, we should choose a number of 12 or greater. This also answers a different puzzle, but I am now confident enough to bet on my (and SteveGs) original answer of 15 minutes for the problem as stated, that there are three trains per hour, randomly, not poissonly, distributed, but in any hour there are always exactly three trains, not three on average. The waiting time paradox deals with the "three on average" case correctly, but that assumes the times are distributed poissonly, not randomly. And I will have nothing to add to the above, which will be my last post on this subject.