I was a little surprised at the length of the comment thread in the post about the logic puzzle involving the monk Gaito going up and down a hill. On the one hand, I thought that there were some excellent explanations of why there had to be at least one instant where the monk was at the same location at the same time. These involved visualizing the situation in slightly different ways, such as instead of having one monk go up and down on two different days, having two monks going up and down on the same day or using graphs or films and so on.

But clearly these arguments were not persuasive enough for some and I have been trying to think why this might be so. In my teaching experience, it is often the case that what seems obvious to you as a teacher is by no means so to the student. It is no use repeating the same explanation more slowly or (worse) more loudly or (much worse) exasperatedly. There is clearly some opposing argument that the student finds persuasive that makes them reject your argument and yet they may not be able to identify and articulate what it is. Instead they feel that there must be some flaw in your reasoning that they cannot put their finger on. It is more fruitful as a teacher to try and figure out what *their* argument might be, rather than reiterating your own.

Comment #56 by larpar is the clue that there is something like that going on here when they say “But something in the back of my head, something I’ve heard before, tells me there is something wrong with comparing simultaneous and consecutive events. Sorry I can’t be clearer.” Also the comment #68: “I have considered being wrong. I even admitted being wrong about one aspect of my reasoning, but I think I have resolved that error. Maybe not. I could still be wrong. One other reason that I’m not yet ready to admit defeat is that this is a “logic problem”. Usually in these types of questions there is some kind of twist that makes the obvious answer wrong. I’m holding out for the twist. It may never come. : )”

In such situations one has to try and identify and put into words their argument. In this case, I think it may be because at the back of their minds is an argument along the following lines: If you take any given point on the path, there is no guarantee that the monk will be at that particular point at the same time on the two journeys. This is easy to see since all we have to do is vary the speeds of the monk during the two journeys so that that point is reached at different times. In other words, the probability that the two journeys will coincide at any given point at the same time is of measure zero. This can be repeated for *every point on the path*, because we can always postulate walking speed scenarios where the two journeys will reach that point at different times. It is a short step to go from there to think that there is no point at which this is guaranteed to happen. This argument is a powerful one that has significant plausibility. Before I present the subtle flaw with it, I would suggest that readers think about how they might counter it.

The flaw is that there is a difference between picking individual points in advance and arguing that the simultaneity of time and place cannot happen at any of them, to arguing that it cannot happen at any *unspecified* point. As an analogy, I am planning to leave my home sometime later today. If I specify an exact time, the probability of my leaving exactly at that time is zero, irrespective of what time I pick. But the probability that I will leave at *some* time is one. i.e., it is guaranteed. (Let’s leave out the possibility that something might cause me to change my plans about going out.)

This is a subtlety that arises with *continuous* distributions, like the monk’s position along the path or the time at which I leave. If we are dealing with *discrete* distributions, then the situation is different and their argument holds. For example, if instead of a path, the monk had to navigate 1000 steps while going up and down the the hill. Then there is no guarantee that he will be on the same step at the same time on the two journeys, because the time at which they cross could occur when they are each moving from one step to the next. i.e., the two paths could cross when the monk is in the process of going from one step to the next one up on the upward journey while taking a step down on the downward journey, so that they are not on the same step at the same time even though they are crossing. (This is the point that Deepak Shetty seems to be hinting at in comment #27.)

This long response is only partly about the monk puzzle. It is also about (a) the difference in probability estimates between continuous and discrete distributions of outcomes and (b) how important it is to not dismiss those who cannot seem to see what seems obvious to us, because understanding the reasons for their skepticism can reveal important insights about subtleties.

Matt G says

Creationists use an argument which invokes what you call “measure zero”. The likelihood of humans evolving from bacteria is so low, that it must have been aided by an intelligent agency. Of course, they have picked that outcome in advance; if you replayed evolution on Earth, you’d get something very different. I use the example of meeting an old friend by accident while on vacation. The chances are measure zero, but once you frame it “what are the chances that *someone* will see *some* old friend *somewhere* by accident, you see how it must happen from time to time. Or, think about how many old friends you *didn’t* see by accident while on vacation.

lanir says

My initial thought with the monk problem involved the flaw presented here. Obviously walking speeds vary so it seemed obvious he wouldn’t be at the same spot at any point. Then I realized that my mental model was mainly saying that the monk would be at the middle at the same time and if that weren’t so, he wouldn’t ever be at the same spot at the same time of day.

Figuring it out was pretty fast once I saw other people talking about it. Having the start and end of the journey be the same both trips is key as it removes outliers from consideration. Then I just had to realize that the spot could be anywhere along the path and my preoccupation with the speed issue disappeared. If he’s going double speed one trip, he crosses his path at the same time of day around 2/3rds of the way through on that trip. Triple speed and he’ll be at the same spot at the same time of day after walking about 3/4ths of the path.

larpar says

Darn, I guess I’m wrong.

I still working through things and don’t fully understand my error. One thing I don’t get is part of Mano’s leaving home analogy. To me, Mano seems like he would a pretty to be a punctual kind of person. If he says he’s going to leave at a specific time, then the probability of him leaving at that time would be pretty high. I don’t see how it could be zero. Maybe it’s the difference between a layman’s use of the word “probability” and a physicist/mathematicians use.

I’m also going to do some more studying on “continuous and discrete distributions”.

The probability me ever fully getting it is pretty low, but it’s not zero. : )

Thanks to all, and especially Mano, who tried to set me straight. I hope I haven’t wasted to much of your time.

Holms says

Larpar, if someone says they will leave at a certain time, what are the odds the person will leave at that

exacttime? Not just to the minute, but to the second, the microsecond, the nanosecond (etc.)?DonDueed says

larpar@3: I think Mano’s example may be confusing because the notion of “time I leave home” is rather fuzzy. What exactly does that mean, physically, and how accurately is it specified?

Instead, imagine Mano has a motion-activated switch to turn on the lights when he enters his closet. Now state the problem as: some time today he will enter the closet. What time does the light turn on? If you pick one particular microsecond, the probability of its being that exact time is nearly zero. But we still know that the light will turn on at some time today.

Microsecond timing still not convincing? You can make the time even more specific, and so on until you are satisfied that the chance of his triggering the light at any particular time to be zero.

DonDueed says

Holms: GMTA!

Matt G says

Speaking of low probabilities, I just remembered one of the weekly “science questions” I asked my students a few years back. It’s about bridge, so I’m sure you’ll appreciate it, Mano! I told them that the probability of being dealt a hand of all hearts is 635 billion to one. Now, you sit down for a game and get dealt a hand. What is the probability of being dealt that hand?

file thirteen says

Matt #7

Not being familiar with bridge, I don’t know which cards are used in the game or how many cards are in each hand, and consequently whether a hand of all hearts must contain all the hearts used in the game. But if it does, then 635 billion to 1 I guess (assuming you’re not talking about the probability that you’ve been dealt the hand you’ve just received, which should be 100% without any shenanigans going on)

John Morales says

file thirteen, the point is that all hands are equally likely (in this case, equally unlikely) given a random deal.

Matt G says

I use the bridge hand probability to illustrate why the creationist personal incredulity argument fails. There’s only one perfect hand (four actually), but there is a huge number of hands that guarantee a win for even a beginner. The chances of getting “humans” again if you replay evolution are minuscule, but there will be a huge number of ways of getting some self-aware life form.

Mano Singham says

larpar @#3,

It is not a matter of how punctual I am. As Holms has pointed out in #4, since time is a continuous variable, the probability of my leaving at an

exactly specified time(say 11:15) is zero.This fact is disguised by the way speak of time in real life, because we treat time as if it moves in

discretechunks, the way an analog clock’s second hand jerks between seconds or a digital clock jumps from one second to the next. So when I say I will leave at 11:15, we usually mean sometime in the one minute timeintervalbetween 11:14:30 and 11:15:30. Or if we are want to specify down to the second, we mean the one second timeintervalbetween 11:14:59.5 and 11:15:00.5. Then the probability is no longer zero. It is analogous to the difference between the monk walking along a continuous path and going up discrete steps.The probability will get smaller and smaller the more we make the time interval smaller because the

numberof time intervalwindowsin which I could leave becomes larger. In the limit that the interval becomes zero, the number of intervals becomes infinite and hence the probability of my leaving in any specific interval becomes zero.Ridana says

If the monk takes stairs like I do, he’s going to have one foot on one step and the other foot on the next step about half the time. 😀 So which step is he on when he’s not in the middle of moving one foot to the next step?

John Morales says

Ridana, “So which step is he on when he’s not in the middle of moving one foot to the next step?”

Exactly. Steps may be a good metaphor for discreteness, but traversing them is not a good metaphor for discreteness. Call it… um, a step function. 🙂

However, this particular monk is a true master of woo, and just teleports instantaneously from one step to the next.

Mano Singham says

Ridana,

We can overcome your problem by saying that the monk, as part of his ascetic practices,

jumpsfrom one step to the next, so that there is a time interval when he is in the air and not on any step.John Morales says

Mano, I think my response to Ridana is better than you.

The steps were supposed to make the journey consist of discrete steps, but being in the air reintroduces the continuity. Step function.

Mano Singham says

John @#15,

No, the question was whether the monk “will be on the same step at the same time on the two journeys”. If he is in the air between steps, then he is not on any specific step for that time interval. His trajectory remains continuous. It is the times that he is on a step that is discrete.

John Morales says

Mano, I get all that.

OK, this is what I got. You replaced the (continuous) path with (discrete steps), but you did not similarly replace the monk’s position on the [path|steps].

In short, the locus of the monk’s journey remains continuous, except that now only a portion of the path qualifies (the steps, not the distance between them) as a place where simultaneity may occur.

(And yes, I bear in mind what you wrote about pointless perseverance. Was good advice)

Rob Grigjanis says

Mano@11:A distinction without much difference. There is a built-in discreteness. It’s called ‘walking’. But maybe the monk has attained a level of enlightenment which allows him to glide along the path…

tuatara says

Sorry in advance for the long post.

I had doubts about the monk puzzle actually being one of logic. It doesn’t seem (to me) to come near to other logic puzzles, such as the joke about three logicians going into a bar and the bartender asking if they would all like a beer. This seemed more like just a maths puzzle.

What set me to thinking more deeply about the puzzle though was seeing the answer stated, without much explanation, that if the two journeys are plotted they will intersect with no explanation as to why, beyond that they will intersect.

I am sure we have all seen how data can be graphed in such a way as to be deceptive, so I like to know why things are as they are rather than just be given a graph and told it is so.

I also know that I am not a mathematician or physicist (I in fact gave up on school at 15 and only achieved a high-school level maths and general science education up to that point), and that most other commenters here are far more educated in these matters than I. So, I knew I was probably wrong but wanted a real reason why, which I wasn’t really getting from the other comments.

So I went away and approached the puzzle my own way. I thought about what information was necessary to be able to work it out and what was irrelevant.

I decided the start and end times are not relevant. It is sufficient that the duration of each journey is the same.

I also decided that any stops along the way or retracing of steps were also not relevant either.

I also realised that only two points are really essential for working it out. While the starting point is necessary for determining how far the first run has travelled in any given time, once that is established the only important points are the distances from the start of journey 2 either monk reaches at any given time.

The other necessary information is that of the velocity of each journey or part thereof, and that a difference in velocity for each journey is essential to recreate the original puzzle in this more simple way.

So, I imagined Usain Bolt running an olympic 100m track in precisely 20 seconds (but an Usain Bolt who can accelerate and decelerate instantly).

On the first run from point a (starting block) to point b (finish line), he instantly accelerated from 0 m/s to 9 m/s, maintaining this speed for exactly 10 seconds, then decelerated instantly and maintained 1 m/s for the remaining 10 seconds.

So at 10 seconds he was 90m from his start but more importantly was 10m from the point from which he will begin his second run.

On his second run from finish line to starting block, he maintained exactly 5 m/s for the entire 20 seconds. So 10 seconds in he has travelled 50m.

Obviously sometime in the first 10 seconds the two runs cross when comparing distance from point b.

Then I counted back.

At 9 seconds on the first run he was 81m into the journey or more importantly 19m from the beginning of the second run, at 8 seconds it was 28m from the beginning of the second run, at 7 seconds he was 37m from it, at six seconds 46m etc.

On the second run, at 9 seconds he had travelled 45m, at 8 seconds 40m, at 7 seconds 35m, at 6 seconds 30m etc.

It was immediately obvious to me that sometime between 7 and 8 seconds into either journey the time elapsed and distance from the same end (finish line of the 100m olympic track) for each journey converged.

Bingo!

I didn’t feel the need to calculate it exactly. In fact with my mathematical skills it looks fiendishly difficult to work out precisely, even with what appear to be simple starting assumptions.

More importantly though is that I had already felt that I was initially wrong in my answer to the original puzzle but now I had what to me was more proof than just a graph from a mathematician (respectfully of course).

Then I read where the idea was introduced of two films superimposed, which shows the same thing that I had concluded but in a very visual way. Confirmation!

That is what worked for me at least.

Two things stand out to me from my model.

1. The point of convergence will always occur before the slowest monk to travel half the distance has done so. I think this is also true of the original puzzle.

2. Not so sure about this one because it may be an artifact of my model, but perhaps the point of convergence will always be before half the time of the whole journey has elapsed.

I have been busy with work and other things so haven’t had a chance to properly think about the second one, but I am sure someone reading here will be able to tell me if it is the case.

I am just happy to be educated. Wish I had this attitude in my 15 year old self! I am 54 now.

Sorry again for the long post.

another stewart says

@3: I suspect that your intuition has been trained on finite and discrete sets of data, such as tossing coins, rolling dice, or dealing cards. Here we are dealing with continuous sets of data. Continuous sets of data imply an infinite number of possible values, so that chance of picking one particular value (such as the precise time* Mano leaves his house) is 1/infinity, or infinitesimal. (Infinitesimal means smaller than any possible non-zero number)

* Mano leaving his house is here modeled as an instantaneous event, rather than a process of non-zero duration. So that example, while having the advantage of being concrete, can be challenged as having unrealistic assumptions. A more abstract, but unchallengeable example, is to randomly select a real number between 0 and 1. There is an infinite number of real numbers in that interval. The chance of selecting any particular number is in layman’s terms zero. (Infinities are counterintuitive; in that range there is an infinite number of values expressible as x/y where x and y are integers, but the chance of picking one of them is still zero -- this is where the term measure zero mentioned in the prior thread comes in.)

Returning to the original problem:

It is clearly possible that there is a point which he is at at the same time of day on both days. It is clearly possible to adjust the schedules so that he is no longer at that point at the same time of day on both days. And that clearly applies to any possible point on the path. That shows that no particular point has to be a point of coincidence. The difference might be considered subtle, but that’s not the same as showing that there doesn’t have to be a point of coincidence. Whenever you adjust the schedules to eliminate one point of coincidence you create a different point of coincidence -- you can only move the point of coincidence -- you can’t eliminate it. (Allowing retracing of steps makes things more complicated -- you can have more than one point of coincidence -- but while you can reduce the number of points of coincidence you can’t eliminate the last one.)

A different formulation of the problem would be a locomotive on a single-track railway line. (Imagine a turntable at the end of the track to turn the locomotive round.) Assume the frontmost part of the locomotive is on its centre line. Then, can it be guaranteed that the frontmost part of the locomotive is at least one point at the same time of day on both days? The answer is that same, but many of the “loopholes” discussed in the previous thread look less reasonable, the potential variance of the path in three dimensions being much reduced, and the monk being replaced by something much closer, on a human scale, to a mathematical point.

another stewart says

I don’t know how precisely a perfect bridge hand is defined, but a hand of 13 hearts is not unbeatable. You could outbid it with a hand of thirteen spades. Twelve spades including the ace and a second ace is not a guaranteed win (if the first suite led is the same suite as the second ace, and the remaining spade is used to trump the first trick), but the chances are still good enough to justify outbidding a hand of 13 hearts. The unbeatable hands are those that are composed of consecutive top cards in all suites, or are long enough in a suite to guarantee being able to squeeze out any remaining cards, e.g. A,K,Q,J plus 5 other cards in a suite.

Rob Grigjanis says

tuatara@19:Not necessarily. Let’s change the problem slightly for simplicity. The duration of both trips is 5 hours, and the distance is 10 miles.

On the trip up, the monk walks a steady 2 mph. Due to an old soccer injury, his knee is buggered the next morning, and he can only hobble at 0.5 mph for 4 hours. So he’s only gone two miles by then. But that’s the same point that he got to after 4 hours the previous day (8 miles from the start). So that is the point of convergence, well past half the time.

By then, his knee feels better, and he can trot the last 8 miles in one hour.

another stewart says

@22: or more generally, run the journeys in reverse. If the original journeys had the point of convergence at less than half the time, the reversed journeys will have it at over half the time. (The sum of the convergence times for the original and reversed journeys is the full journey time.)

brucegee1962 says

I agree with John Morales @17 that taking into consideration the time when the monk is between steps, or with one foot on each, or hopping in the air and on neither step, negates the whole point of using steps in the model in the first place. We use steps to divide the journey into discrete steps — using any of the options above makes it into a continuous journey again. I assume the monk has gained transcendence and can teleport instantaneously from step to step.

So we start with one step on the journey. Or, going back to the original problem, we could ask “If the monk leaves at 8 and arrives at 10 each day, what are the chances that the monk will be somewhere on the entire path each day?” The answer is, trivially, 100%.

Now we go to two steps. Or, in the original problem, we put a marker halfway down the path. Maybe steps are easier to visualize at this stage, but again, it’s the same problem. Say that there are two steps between the top and the bottom, and the monk can teleport between steps. If we pick a step, what are the odds that the monk will be on that step at the same time? With some thought, we ought to see that the odds are 50%. If we make the steps smaller and smaller, the odds of their both being on any given step get smaller and smaller, but they still always add up to 100%. What Mano calls “measure zero” is not the same as actual zero.

I figured out the original problem pretty quickly once I realized that the problem was exactly the same as asking “If two monks who start on opposite ends of a trail meander towards one anothers’ starting points, what are the odds that at some point they will pass one another (thus occupying the same point at the same time)”? I was pleased to read the comments and see that others had made the same subsitution, so I didn’t bother to chime in.

But if you start thinking of movement as a series of discrete steps, like teleporting, then someone might ask “What if the two monks start on two different steps, and they both teleport to one another’s steps at the exact same instant?” This highlights exactly the difference that Mano brought up about continuous versus discrete movement. It is also why the various iterations of Zeno’s famous paradox initially seem puzzling. We can imagine time as being a set of infinitely divisible moments, but if we do so, then the notion of movement itself becomes impossible. Two monks walking towards each other can never meet, because the distance between them gets continuously smaller but can never reach 0. But of course, we know that it does reach 0, so the “infinitely divisible” time model must be wrong.

To put it another way — we may never be able to measure a “quantum time” unit — a fundamental bit of time which, like the frame rate of a video game, cannot be further subdivided. But to defang Zeno, we should act as if such a unit exists. In other words, we should act as if the universe has a pulse.

Deepak Shetty says

Im flattered! But I just was being pedantic -- A narrow path is still a rectangular-ish shape , not a line and so we dont have to occupy the same “point”. In normal terms we would have phrased it as Do they pass by each other at the same time -- But that sort of gives the game away.

I fluctuate between really liking logic puzzles (usually when I can solve them) and https://sites.ualberta.ca/~esimmt/haveyouheard/problems/sillyPuzzle.html when I cant.

Rob Grigjanis says

brucegee1962@24:Are you just presenting (a possible inference of) Zeno’s view, or do you actually believe this?

You can perform an infinite number of ‘steps’ (1,1/2,1/4,1/8…) in a finite time, because both the (highly contrived) steps and the time intervals for the steps add up to finite values.

larpar says

Mano @ 11

Thanks, that helped, but it leads me to this question. If the probability of you leaving on time is zero, then isn’t the probability of simultaneous events also zero? If probability of simultaneous events is zero, doesn’t that blow the monk problem out of the water? (this might be where the steps analogy comes in?)

another stewart @20

Thanks. Like Mano’s reply, that helps push me along the path to understanding. I’m not at the end yet but I’m getting there. I need lots of meditation, and even some backtracking, time. : )

Mano Singham says

larpar @#27,

You are right that the probability of my leaving at any specific time is zero, as is the probability of any specific point being the crossing point of the two journeys. But that is consistent with the probability of my leaving at

sometime being one, as is the probability of the two monks crossing atsomepoint at the same time.Perhaps this might help. Take a die with N sides that are equally likely. The probability of the die landing on one specific side is 1/N. Since there are N asides, the probability of it landing on some side is Nx(1/N) which is equal to 1. i.e., it is certain to land on some side. That is the discrete picture.

Now start making N larger and larger. The probability of the die falling on one specific side gets smaller but the probability of landing on some side remains one. When N becomes infinite (so that the die becomes a sphere), the probability of landing on any specific point becomes zero but the probability that it will land on some point is still one. That is the continuous picture.

Putting it in rather crude mathematical terms, in the limit that N tends to infinity, Nx(1/N) tends to (infinity)x(1/infinity) or (infinity)x(zero) but that product is still one.

Rob Grigjanis says

larpar@27:We think in terms of a probability

per unit timeP(t). So the probability of the event occurring in a time interval t to t+Δt is, if P(t) is smooth enough, and Δt is small enough, aboutP(t)Δt

So, yeah, if Δt is zero, this is zero, and you can read that as “the probability of the event occurring at exactly t is zero”. But that’s not very useful 😉

another stewart says

@27

As I understand, that’s one of the counterintuitive features of infinite sets -- zero probability events can and do occur.

larpar says

Mano @28 (and Rob @29)

Ok, I follow the dice example all the way through, and it makes a lot of sense. The only problem I see is that I’m going to have to look up my high school math teacher and chastise him for teaching me that anything x zero = zero. : )

One thing that still has me hung up is that it seems like extreme precision is being used for one aspect of the problem but not for other aspects.

Here’s another question that may be unrelated but I thought of after reading Mano’s dice example. Plank time (I think that’s the correct term), as I understand it, is the smallest amount of time that we can measure. Can plank time still be divided into infinite divisions.

Again, thanks for everyone’s time and patience.

larpar says

another stewart @30

Thanks! Stating it that way really helps.

brucegee1962 says

Of course, the all-time greatest example of the counterintuitive logic puzzle is the famous “Monte Hall” problem. https://en.wikipedia.org/wiki/Monty_Hall_problem The intelligence test there isn’t solving the problem — very very very few people get it right when they first consider it. The intelligence test is understanding the answer once it’s explained to you.

Mano Singham says

larger @#31,

In a post about five years ago, I discussed the possible meanings of Planck length, time, and mass. You may want to take a look at it.

Rob Grigjanis says

another stewart@30:Sure. Choose a random real number. The probability that you’d choose 1 is formally zero (maybe I’m misusing ‘formally’. I’m not a mathematician). But that doesn’t have much to do with the real world.

larpar@31:Yeah, that is a real issue in problems like this. There’s an assumption of ‘exact time’, or ‘exact position’, or both. Best to assume there is nothing exact. Everything is fuzzy; the position of the monk, the time that the up journey coincides with the down journey. The important point is that there will be time and position

intervalswhich will overlap.Also; Planck time has nothing to do with measurement. It’s just a unit (like the other Planck units) defined by the values of four universal constants.

tuatara says

@ 22 Rob Grigjanis & @23 another stewart.

Thanks. Seems bleeding obvious now!

larpar says

Mano @34

I read the article and I’ll reply the same way you ended the article; “Isn’t physics fun?”

: )

Rob Grigjanis says

Me @35:

To be clear, the Planck units don’t express constraints on measurability. For example, the Planck mass is on the order of a hundred thousandth of a gram. We can measure much smaller masses than that.

Holms says

#31 larpar

Planck time is actually far smaller than what we can measure with current technology, as is planck length. Contrast those with planck force or temperature…

ragove314 says

This problem is just the Intermediate Value Theorem from calculus. If f(t) is the difference in heights between going up and down, then f(0) is minus h. (h the mountain height). And if the end of the trip is t=1, f(1) is + h. The function f has to be 0 some place. ( It can not go from below 0 to above 0 without passing 0. ). At that time the monk is at the same place going up and down. We don’t need a monk,, it can be 2 rabbits going up and down. The paths just have to be continuous.

another stewart says

@31

In that context anything meant any number. Infinity is not a number, or at least not the type of number he was talking about. (There are such things as hyperreal, superreal and surreal numbers, but they’re not taught to high schoolers.)