### Wednesday Solution

Monday’s puzzle was open to various interpretations, but under what seems to me to be the most straightforward interpretation, if the number of runners you pass is the same as the number who pass you, you’re the mean runner, not the median.

You can find plenty of correct analysis in Monday’s comment section (see in particular Harold’s perfect comment #39), but here’s a more longwinded explanation:

First, suppose you randomly sample a large number of other runners and discover that half of them are faster than you and half are slower. Then you’re entitled to conclude that you’re the median runner (or, if we’re being careful, you’re entitled to conclude that you’re probably close to the median, since there’s always a chance your sample was unrepresentative).

Now in the problem as given it’s certainly true that half the runners you encounter are faster than you and half are slower. So you might be tempted to use the above reasoning and conclude that you’re the median runner. But that won’t work, because the runners you encounter are not a random sample.

So let’s start over. We might as well assume that you’re the center of the universe, so you’re completely motionless. Everyone who’s faster than you is running forward and everyone who’s slower than you is running backward. People “pass” you when they run past you in the forward direction, and you “pass” them when they run past you in the backward direction.

If there are, say, 32 10-mile-an-hour forward runners per mile, how many will pass you in an hour? Answer: 320. (Think about it). If there are 19 5-mile-an-hour forward runners per mile, how many will pass you in an hour? Answer: 95. (Think about it again.) In total, how many pass you in a given hour? Answer: 320+95=415.

And what’s the total speed of all the runners in a given mile? Answer: 32 at 10mph plus 19 at 5mph = 415.

In other words: The total speed of all the forward runners in a given mile is equal to the number who pass you per hour. Likeiwse, the total speed of all the backward runners in a given mile is equal to the number who you pass (i.e. who pass you going backward) per hour.

Now if we count forward speeds as positive and backward speeds as negative, then the total speed of all the runners in a given mile is equal to

(the total speed of all the forward runners in a mile)

minus

(the total speed of all the backward runners in a mile)

which in turn is equal (by the arguments we made above) to

(the number who pass you going forward each hour)

minus

(the number who pass you going backward each hour)

Since we’ve assumed that the number who pass you going forward is equal to the number going backward, this total comes to exactly zero.

And finally, the average speed of the other runners is equal to their total speed divided by the number of runners. Since their total speed is zero, so is their average speed. On average, the other runners are standing still — just like you!

You, in other words, are the average runner.

Lame hat tip: This was inspired by a question on MathOverflow which I no longer seem able to find. When I find it, I’ll replace this sentence with a proper link.

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#### 51 Responses to “Wednesday Solution”

1. 1 1 Keshav Srinivasan

Steve, I’m afraid I don’t quite understand what you mean by runners per mile.

2. 2 2 Keshav Srinivasan

Never mind Steve, I understand now. You’re assuming that the number of runners at any given speed are uniformly distributed throughout the line. So there are 32 10mph forward runners 10 miles behind you, 32 10mph forward runners 9 miles behind you, etc. so there’s 320 10mph forward runners within 10 miles behind you, all of whom will pass you in the next hour.

3. 3 3 Andy Wood

I haven’t read all the comments from Monday. Did anybody point out that this problem is very much like one in the kinetic theory of gases, where you calculate the net flux of molecules across a moving surface? The closest appears to be comment #3 by Ed.

4. 4 4 Robert

How is it that in your specific example you are not the median runner? It sounds like in a per mile scenario, you have 32 running 10 mph and 19 running 5 mph. But you also have 32 running -10 mph and 19 running -5 mph. The mean is 0 and thus you are standing still. It sounds like the median is also 0, no?

5. 5 5 Steve Landsburg

Robert: How do you know you have 32 running -10mph and 19 running -5mph? Maybe there are 415 running -1mph.

6. 6 6 ThomasBayes

Thanks for another interesting puzzle.

I’ll need to look over your answer again, but on first glance it appears that you are proving the following:

IF you are running at the average speed, THEN the number of people who pass you will be equal to the number of people you pass.

Did you also prove the following?

IF the number of people who pass you is equal to the number of people you pass, THEN you are running at the average speed.

7. 7 7 Ken B

Counterexample. 100 miles ahead of you is a very large group running faster than you in the same direction as you. You will never pass or be passed by them. You are running below the average speed.

8. 8 8 Jonathan Kariv

Ken B: In that case doesn’t the large group eventually lap you? Or are you talking about the case on an infinite track?

9. 9 9 Jonathan Kariv

@ThomasBayes: It goes both directions. If you where going faster than the average speed you’d pass more people than people passing you (slow everything down by the mean you’re still going forwards). Similiarly if you’re going below average speed you’ll be passed more often than you’ll pass.

10. 10 10 Steve Landsburg

Ken B: As I said in the post, it all depends on your assumptions. The analysis here uses the (to me) reasonable assumption that runners of any given speed are uniformly distributed along the (infinite straight) track.

11. 11 11 Alec Cawley

Are you not assuming that the runners are normally spread? Suppose most of the runners are your jogging neighbours, but one is Superman, chasing baddies at Mach 1. Then one single overtaker sways the mean way off: the mean runner is possibly the fastest non-superhero, while still leaving you just ahead of the median in this case.

Of course, there are no superheroes. But a sprinkling of visiting Olympic athletes superimposed on the out out-of-condition puffers will similarly skew the results.

And a couple of out of condition joggers having heart attacks which slow them way below the pack would produce an opposing spike of slow runners.

12. 12 12 Robert

Thanks Steve, that clears up my confusion. Great puzzle.

13. 13 13 Ken B

@Jonathan Kariv: I was responding to the amended puzzle, with a straight track. As I noted in an early comment Monday, the puzzle is underspecified.

14. 14 14 Ken B

The puzzle is interesting in that the answer, mean not median, is surprising. But the puzzle is also under specified. It should really refer to mean or median of the cohort of runners you pass or are passed by and not all the runners on the track. The Monday puzzle post refers to running with many others and asks about the mean and median of all the participants.

15. 15 15 David Grayson

Steve, I thought you were going to relate this problem back to relative wealth in society, and perhaps use it to refute a stupid comment by some politician or journalist. I was mildly surprised that it came from MathOverflow!

16. 16 16 Ken B

@Thomas Bayes re 6:
No he did not prove that. He proved it with the extra assumption that there is a uniform distribution of the runners at any given speed. He proved it also in the special case where the passers and passees make up all the runners.

17. 17 17 Ken

Ken B,

ed showed yesterday that you don’t need the uniform distribution assumption. ed proved the result for a general pdf f(x).

18. 18 18 Steve Landsburg

Ken b — you keep saying i’ve assumed that the passers and passees make up all the runners. Not at all. If that were true, you’d be the median runner.

19. 19 19 ed

Kens,

While it’s true that distribution of *speeds*, denoted f(x), doesn’t have to be uniform, I did assume that the distribution of runner *locations* (for which I used the parameter K) was both uniform and statistically independent of runners speed.

I think requiring this to be explicitly stated is a bit tedious. It was also not explicitly stated that we are modeling the runners as points moving along a one dimensional curve with constant speed, so that we can ignore the width of the runners and the track and resolve any ambiguity about what it means to “pass” a runner. The reader can generally fill in such assumptions for himself, since it is obvious that without them the puzzle is inelegant and probably unsolvable.

20. 20 20 ThomasBayes

I believe that my symmetry concern comes into play for anyone who runs for a finite time. I’d appreciate if someone could clear this up if it isn’t an issue.

Suppose I run at 1/10 miles per minute (mpm) for one hour (or 6 miles). Anyone that I pass has to be running slower than me, and has to have effectively began running at my starting point before me (but not too much before me). If, for instance, someone is running at (1/20) miles per minute (or 1/20 mpm slower than me), then I will catch them provided that they started no more than one hour before me.

(When I say that a runner “effectively” starts at a particular time, I just mean that they would be at my starting point at that time if they had been continuously running at their speed.)

Now, what about someone who is running 1/20 mpm faster than me with a speed of 3/20 mpm? They’ll run my 6 miles in 40 minutes, so to catch me they’d have to effectively be at my starting point within 40 minutes after my start time.

So, here’s my concern:

1. I’ll catch everyone who runs 1/20 miles per minute slower than me, provided they effectively start no more than one hour before me.

2. I’ll be caught by everyone who runs 1/20 miles per minute faster than me, provided they effective start no more than 40 minutes after me.

This issue goes away if everyone runs for ever, but won’t it matter if I only run for an hour or two hours or three hours? Won’t I have a sampling bias that favors people running at my speed plus dS relative to people running at my speed minus dS? Won’t this cause some small bias regardless of how long I run, provided it is a finite time?

21. 21 21 ed

Thomas Bayes,

I think I see the problem. You are imagining a finite track, with runners showing up at the starting end randomly according to some distribution of speeds, and then run until they finish the track.

What you need to realize, is that there are two different populations of runners you could consider. The distribution of speeds of runners who start is *not* the same the distribution of speeds of runners actually running on the finite track at any given time. The second population will have relatively more slower runners, since they stay on the track longer.

My interpretation of the question assumed that we were talking about the mean/median of runners actually running at a given time, not about runners starting. Although I now admit it’s a bit ambiguous.

22. 22 22 YGA

Here’s what I don’t get about the solution posted here.

In scenario A, I am passed by 5 runners and pass 5 runners. Let’s say the first guy who passes me is running a 5 minute mile.

In scenario B, nothing changes except that the first guy who passes me is running a 4 minute mile.

Both scenarios are effectively the same, except that I get passed a little sooner in the second. But clearly between the two the mean speed of the runners has changed!

(Maybe the assumption that runners of a given speed are uniformly distributed along the track means that I can’t just arbitrarily make one runner faster. But if that’s what it implies, then I think the restriction is very far out from the common-sense way of reading the question).

23. 23 23 Steve Landsburg

YGA:

Maybe the assumption that runners of a given speed are uniformly distributed along the track means that I can’t just arbitrarily make one runner faster.

Yes.

But if that’s what it implies, then I think the restriction is very far out from the common-sense way of reading the question

I don’t agree with this, but I can see where a reasonable person might agree with it entirely.

Edited to add: Even if you abandon the uniform-distribution assumption, I think one could quite reasonably argue that in the absence of further information about the other runners, the uniform-distribution assumption is your best guess about them, and hence the conclusion that you are the mean runner should be your best guess about your status.

24. 24 24 Mike H

You don’t need the uniform distribution assumption, you just need to assume (as Steve pointed out) that you’ve passed (and been passed by) a representative sample of those who will pass and be passed by you.

For example, you could be running for a sufficiently long time around a circuit, or for an infinite time (past and future) along a straight track.

As an example – if, on a 2 mile circuit, there’s a pack of runners running at 6mph, and another pack twice the size who are not running, there’s no uniform distribution of runners, but if I run at 2 mph for 1 hour, I will pass and be passed by equal numbers of runners – assuming I count Steve twice as he passes me twice.

25. 25 25 Steve Landsburg

Mike H: Yes, excellent comment (directly above this one). Others should read it!

26. 26 26 ThomasBayes

I think it may be that the “rate” (minutes per mile), not the speed (miles per minute), should be used to specify this problem.

Here’s an example using speed (miles per minute):

Let’s suppose that half the people run 4/40 miles per minute, a quarter of the people run 3/40 miles per minute, and a quarter of the people run 5/40 miles per minute:

Pr[S = 3/40 miles per minute] = 1/4 (13.333 minute miles)
Pr[S = 4/40 miles per minute] = 1/2 (10 minute miles)
Pr[S = 5/40 miles per minute] = 1/4 (8 minute miles)

Now suppose your speed is the mean speed, and a random runner starts every minute. After N minutes you will have run

Your distance = N(4/40) miles.

If a slower runner starts n minutes before you, after N minutes they will have run

Slower runner’s distance = (N + n)(3/40) miles, 0 (N + n)(3/40) ====> 0 < n < N/3.

That is, you will pass any slower runner after N minutes provided that they started more than 0 and less than N/3 minutes before you.

If a faster runner starts m minutes after you, they will have run

Faster runner's distance = (N – m)(5/40) miles, 0<m N(4/40) ====> 0 < m < N/5.

That is, a faster runner will pass you after N minutes provided that they started more than 0 and less than N/5 minutes after you.

On average, a quarter of the people who start after you will be faster than you, so, after N minutes, you can expect an average of N/20 people to have passed you. (One runner starts every minute.)

A quarter of the people who start before you will be slower than you, so, after N minutes, you can expect to pass an average of N/12 people.

I haven't put a restriction on N, so you can let N be as large as you like. (Start runners every minute of every hour and, once you start running, never stop.)

No matter how long you run,

the number you pass – the number who pass you = N/12 – N/20 = N/30.

So, you are running with the mean speed and, on average, you should never expect to be passed by the same number of people that you pass. The longer you run the bigger the difference.

Here's an example using "rate" (minutes per mile):

If you state the problem by specifying that you have the mean "rate" (minutes per mile) for the runners, then my example works out fine. That is, if

Pr[S = 12 minutes per mile] = 1/4, (10/120 miles per minute)
Pr[S = 10 minutes per mile] = 1/2, (12/120 miles per minute)
Pr[S = 8 minutes per mile] = 1/4, (15/120 miles per minute)

Then you will run N(12/120) miles in N minutes. You will pass a slow runner after N minutes provided that they start more than 0 and less than N/5 minutes after you. You will be passed by a faster runner provided that they start more than 0 and less than N/5 minutes before you. In this case, you can expect to pass and be passed by the same number of people.

27. 27 27 Ken B

@Ken and ed:
I think ed in 19 has clarified Ken’s error. You DO need an assumption about the distribution of runners. Some of us pointed this out early. I submit it’s easy to adjust the word problem: “you are part of a large evenly spread crowd running …”

@Steve re 18: No, I said in 16 that you proved it for the uniform distribution and ALSO for the case where the passers and passees make up all the runners. But now I see you are right: you did NOT also prove it for that special case :)

28. 28 28 ThomasBayes

I forgot that things can get messed up when you use the less-than or greater-than signs in a post. This section of my post:

If a slower runner starts n minutes before you, after N minutes they will have run

Slower runner’s distance = (N + n)(3/40) miles, 0 (N + n)(3/40) ====> 0 < n < N/3.

That is, you will pass any slower runner after N minutes provided that they started more than 0 and less than N/3 minutes before you.

If a faster runner starts m minutes after you, they will have run

Faster runner's distance = (N – m)(5/40) miles, 0 0 < m 0 .lt n .lt N/3.

That is, you will pass any slower runner after N minutes provided that they started more than 0 and less than N/3 minutes before you.

If a faster runner starts m minutes after you, they will have run

Faster runner’s distance = (N – m)(5/40) miles, 0 .lt m .lt N

For the faster runner to pass you, then, you need

(N – m)(5/40) .gt N(4/40) ====> 0 .lt m .lt N/5.

That is, a faster runner will pass you after N minutes provided that they started more than 0 and less than N/5 minutes after you.
——

Can someone remind me how to properly post an inequality sign?

29. 29 29 ThomasBayes

Okay, this is crazy. I left a less-than sign in the previous post, and the whole thing was goofed up again.

The bottom line of the example is this:

If you run at the average speed (miles per minute), the expected number of people you pass will not be equal to the expected number of people who pass you, regardless of how long you run.

If you run at the average rate (minutes per mile), the expected number of people you pass will be equal to the expected number of people who pass you, regardless of how long you run.

Unless I made a mistake, the proof for my example is simple, but my post is goofed up because I used several less-than and equal-than symbols, and those, evidently, mean something completely different than I intended.

30. 30 30 Ken B

@Thomas Bayes:
You should be able to use an escape sequence. This is, with no spaces, an ampersand followed by lt for less than, & then gt for greater than.

Here’s a test: &lt

31. 31 31 Ken

Ken B,

You DO need an assumption about the distribution of runners.

As Mike H points out “You don’t need the uniform distribution assumption, you just need to assume (as Steve pointed out) that you’ve passed (and been passed by) a representative sample of those who will pass and be passed by you.”

You don’t need to make any assumptions about distribution of speed or runners. The only assumption you’ve made is that you’ve run long enough that the law of large numbers has taken over and what you’ve observed hasn’t been skewed by over sampling fast (or slow) runners. The uniform distribution assumption just makes it easier for the math, as does ed’s K. Otherwise the math can get complicated and who wants that?

32. 32 32 ThomasBayes

Okay, here’s my example again. (Sorry for the sloppy posts with the inequality sign problems.)

We have an infinite stream of runners beginning at some starting point, and a new runner starts every minute. (There is nothing special about starting runners every minute. We could start them every second, or, I believe, we could start them at random times. The analysis is simpler if we start them at regular intervals.) The track is infinite, and everyone runs forever.

Scenario 1:

A quarter of the runners run at 3/40 miles per minute;
Half the runners run at 4/40 miles per minute;
A quarter of the runners run at 5/40 miles per minute.

You are the average (miles per minute) runner, so you run at 4/40 miles per minute. If you run long enough, you will eventually catch every slower runner who starts before you (sort of). If a slower runner starts n minutes before you, then you will pass them in 3n minutes. If you run for N minutes, then, you’ll pass every slower runner who started less than N/3 minutes before you. A runner starts every minute, and a quarter of those are slower than you, so, on average, you should pass N/12 runners in N minutes.

If a faster runner starts m minutes after you, then they will pass you in 5m minutes. If you run for N minutes, then, you’ll be passed by every faster runner who started less than N/5 minutes after you. A runner starts every minute, and a quarter of those are faster than you, so, on average, you should be passed by N/20 runners in N minutes.

Scenario 2:

A quarter of the runners run at 12 minutes per mile;
Half the runners run at 10 minutes per mile;
A quarter of the runners run at 8 minutes per mile.

You are the average (minutes per mile) runner, so you run at 10 minutes per mile. If you run long enough, you will eventually catch every slower runner who starts before you. If a slower runner starts n minutes before you, then you will pass them in 5n minutes. If you run for N minutes, then, you’ll pass every slower runner who started less than N/5 minutes before you. A runner starts every minute, and a quarter of those are slower than you, so, on average, you should pass N/20 runners in N minutes.

If a faster runner starts m minutes after you, then they will pass you in 5m minutes. If you run for N minutes, then, you’ll be passed by every faster runner who started less than N/5 minutes after you. A runner starts every minute, and a quarter of those are faster than you, so, on average, you should be passed by N/20 runners in N minutes.

Can someone explain why Scenario 1 is not a counter-example for the answer to this puzzle?

33. 33 33 Ken B

@Ken: What you describe IS an assumption about the distribution of runners. Look at my counter example in 7. Mike H’s comment widens the field from uniform to close enough to uniform your running period is a representative sample. Neither describes my counter example, so you must rule out such distributions. My example gains force if I stipulate 1 yard not 100 miles ahead. You need to eliminate the pig in the python.

34. 34 34 Ken B

@Thomas Bayes 32: Scenario 1 proves you might not get the passers = passees if you are the average, but that’s not what the puzzle is about. It’s about the reverse implication.

35. 35 35 Ken

Ken B,

What you describe IS an assumption about the distribution of runners. Look at my counter example in 7.

Yes I saw it. You’re also forgetting about the large clump of slow people who are behind you that you will never pass. This is why the LLN works, for any distribution, and why I explicitly mentioned the need for enough observations for LLN to justifiably be used.

36. 36 36 ThomasBayes

Ken B (34): But scenario 2 shows that if you are average for rate (minutes per mile), then you will get an equal number of passers and passees. But you can’t be average for rate AND average for speed (miles per minute), so if you see equal numbers you are not average for speed.

Being average for rate means that you are slower than average for speed, so you are slower than the average runner’s speed.

Does anyone else see that the issue might be rate (time per distance) versus speed (distance per time)?

37. 37 37 Ken B

@Ken: My point is that the initial distribution of runners matters. You rebut by … altering the initial distribution of runners?

38. 38 38 Ken B

@Thomas Bayes: But you asked why 1 isn’t a counter example. It isn’t because it addresses the reverse implication.

39. 39 39 ThomasBayes

Ken B (38): Fair enough. Do you think scenario 2 is a counter example, then? (Whereas no one in that scenario runs at the average speed, it should be simple to modify the scenario so that someone does.)

40. 40 40 Ken B

@ThomasBayes: Nope. Same reason. Steve has proven, that given some reasonable assumptions about the distribution of runners, passers = passees -> mean. You are looking at cases where you are the mean and looking to see if passers = passees. Affirming the consequent!

41. 41 41 Ken B

@ThomasBayes: OOPS. It now occurs to me that I might have missed what you asked. You claim in scenario 2 a case where passees = passers but not mean speed runner. Mean rate runner instead. Is that what you claim?

42. 42 42 ThomasBayes

Ken B (41): Yes. Unless I made a mistake, I showed that equal passers and passees occur for someone who is not running at the mean speed (in distance per time). Instead, they are running at the mean rate (in time per distance).

I think my scenario of starting a new runner every minute and letting this thing go on for as long as you want (hours, days, weeks, years) is a fair way to do the analysis. I allow for an unlimited number of runners ahead of me and a potentially unlimited number of runners behind me. Invoking some sort of large number limit won’t change my analysis.

43. 43 43 Ken B

OK, so let’s look at Steve’s example in a ThomasBayes way.

On my left is a stream of 10 mph runners, 32 to the mile
and a stream of 5 mph runners, 19 to the mile.
On my right, going the other way we have a stream of -1 mph runners, 415 to the mile.
In one hour I am passed by 23*10 + 19*5 = 415 on the left and 415*-1 on the right.
Equal.

Now SL wrote: “the average speed of the other runners is equal to their total speed divided by the number of runners. ”
Let’s figure this. 320 at 10 mph plus 95 at 5 mph plus 415 at -1 mph.
That’s Steve’s calculation right?
Hmmmm. That does not seem to come out to 0.

So perhaps ThomasBayes has trumped us all. Not for the first time either.

44. 44 44 Ken B

I forgot to finish the math in comment 43. 320*10 + 95*5 = 3675 for the “total speed”.

45. 45 45 ThomasBayes

Here’s another issue I’m struggling with. Again, here’s my distribution for runners:

1/4 of the runners run at 3/40 miles per minute; (slow)
1/2 of the runners run at 4/40 miles per minute; (average)
1/4 of the runners run at 5/40 miles per minute; (fast)

The average speed for runners is 4/40 miles per minute.

Now let’s use an infinitely long track and start randomly launching runners from the starting line at the rate of one per minute. The rate at which I put slow runners on the track is 1/4 per minute, and the rate at which I put fast runners on the track is 1/4 per minute. What is the runner density (runners per mile) on the track?

For a slow runner, we’ll have

(1/4 runners per minute) / (3/40 miles per minute) = 10/3 runners per mile.

For a fast runner, we’ll have

(1/4 runners per minute) / (5/40 miles per minute) = 2 runners per mile.

Now, let’s launch runners for a year, then go somewhere far down the track and tell everyone to subtract 4/40 from their speed. The fast runners will move forward at 1/40 miles per minute, the average runners will stop moving, and the slow runners will move backward at 1/40 miles per minute. If we observe this situation for 40 minutes, we’ll see one mile’s worth of runners from each direction. That would be an average of 2 runners moving forward and 10/3 runners moving backward. We won’t see an equal number because the runner densities won’t be equal. Slow runners will be more densely packed per mile than fast runners.

How could we fix this? I suppose we could have all the runners start at the same time, one at the starting line, the next 1/4 mile behind, the next 1/4 mile behind, the next 1/4 behind, etc. Then there would be 1 slow runner per mile, 1 fast runner per mile, and 2 average runners per mile. The density for fast and slow runners would be the same, and the analysis would work.

Why is one of these scenarios better than the other?

Here’s another way to think about this . . .

Suppose you stood somewhere on the track in the first scenario and tried to estimate the average speed. If you watched for one hour, you’d see an average of 15 slow runners, 15 fast runners, and 30 average runners. Based on their respective speeds, you’d make a pretty good estimate of the average speed.

In the second scenario, if you stood somewhere and tried to estimate the average speed, you’d see an average of 7.5 fast runners per hour, 4.5 slow runners per hour, and 6 average runners per hour. Based on their respective speeds, you’d overestimate their average speed.

46. 46 46 ThomasBayes

Suppose you have a “steady-state” situation with many runners. Enough runners so that you can make your favorite “large number” arguments. What does it mean to say that 10 percent of the runners are running at a rate of 1/10 miles per minute:

A) At any instant in time, you should expect 10 percent of the runners in a particular region of the route to be running at a rate of 1/10 miles per minute.

or

B) If you observe runners running by a particular point for some amount of time, you should expect 10 percent of the runners to be running at a rate of 1/10 miles per minute.

47. 47 47 Mike H

I just realised – this whole argument fails completely under special relativity. So, at normal running speeds, the best you can ever say is that I’m running at approximately the mean speed, but ever-so-slightly faster.

48. 48 48 ThomasBayes

I’d still like someone to help me understand the issue of linear density versus temporal flow.

If half the runners run with speed S1 and half run with speed S2, then why is it more reasonable to assume equal densities of runners per mile for the two speeds, rather than to assume equal flow rates of runners per minute? We have to pick one, so why–other than making this problem work out as desired–is it more reasonable to equalize runners-per-mile than to equalize runners-per-minute?

Thanks for any help on this.

49. 49 49 Mike H

@Thomas – what we are told is that as many runner pass you as are passed by you. Assumption : this means per second.

If someone says “half run with speed S1 and half with speed S2″, then the fact that as many pass you as are passed by you (per second) means you can calculate your speed. You’ll get different answers depending on what you assume about whether “half” means “half of those sampled per mile” or “half of those who pass the starting line”, but either way, your speed will be the mean of those of the runners that pass you or are passed by you.

50. 50 50 ThomasBayes

Mike H: (49) — I’m still confused.

I’m concerned about what it means when you describe the average speed of the other runners. If half the runners sampled per mile have speed S1, and half the runners sampled per mile have speed S2, then you might be tempted to say that the average speed is (S1 + S2)/2. But if this is true, then you won’t see an equal number of runners with speeds S1 and S2 cross a particular line during a particular time interval (on average).

So, the assumptions that lead to this answer imply that the average speed of runners who cross a particular line during any time interval will not be the speed of this story’s protagonist. Are you okay with that? That is, are you comfortable with a key assumption that implies that the average speed that a stationary observer computes will not be equal to your speed, even though you claim that your speed is the mean?

51. 51 51 Paul T

Mike H: “You don’t need the uniform distribution assumption, you just need to assume that you’ve passed (and been passed by) a representative sample of those who will pass and be passed by you.

For example, you could be running for a sufficiently long time around a circuit… ”

Right. If you run many laps around a loop, runners will be uniformly distributed around the loop, which will emulate the conditions of the original problem.

New idea: You perform a random telephone survey of local residents, to estimate population body weights. You notice that the number who weigh more than you, equals the number who weigh less, no matter how many you call. What can you infer about your weight, relative to the mean and median?

hmmmm……..

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