Later today I am giving a public talk at the University of Hertfordshire about the paranormal on film….and it is free! Hope you can make it – detailshere.

On Friday I posted this puzzle….

I have decided to hold a Mouse Clicking Competition. So far, there are only three entrants – John, James and Jack. John can click a mouse 10 times in 10 seconds. James can click a mouse 20 times in 20 seconds. Jack can click a mouse 5 times in 5 seconds. In my competition, the timing period begins with the first mouse click and ends with the final click. Which one of these entrants would be the first to complete 40 clicks?

If you have not tried to solve it, have a go now. For everyone else, the answer is after the break.

James wins. Making 10 clicks involves 9 pauses, and so John’s time per click is 10/9 (1.1 seconds). Similarly, the time per click for James is 20/19 (1.05 seconds) and for Jack it is 5/4 (1.25 seconds). Thus James is the quickest.

UPDATE: The original answer posted was not entirely correct.

Did you solve it?

I have produced an ebook containing 101 of the previous Friday Puzzles! It is called **PUZZLED** and is available for the**Kindle** (UK here and USA here) and on the **iBookstore** (UK here in the USAhere). You can try 101 of the puzzles for free here.

### Like this:

Like Loading...

*Related*

Didn’t get it and don’t understand why the actual time period is one second less…

I think the wording is still incorrect, it should be one click less, not one second less.

No, sorry.

Jack makes 5 clicks in 5 seconds:

t1 = 0s

t2 = 1.25s

t3 = 2.5s

t4 = 3.75s

t5 = 5s

Jack makes 1.25s per click, *not* 1.25 clicks per second, therefore he is the slowest clicker. The fastest clicker is James.

That makes more sense, although you’re assuming that a click is instantaneous, but there’s a refractory period afterwards, i.e. you assume click-pause. But a pause-click model (e.g. it take time to press the mouse) would have all 3 clicking at the same rate. The more realistic pause-click-pause model is between the two.

And then we have to start thinking about the reaction time between the gun going off and the first click being started.

Have I missed anything? If I’m going to over-think a problem, I want to do it properly.

If you ARE going with the pause-click model the original problem STILL states that time is measured from first click to last click, so regardless of your model, you are disregarding one entire sequence of pauses, which is where the arithmetic comes from.

And it works out the same regardless of where you put your pauses.

michiexile – The problem says that the timing for the competition starts at the first click, but is silent about the timing of the rates. One would not normally assume a rate like that starts with the first click. If Richard had said that these were what the J’s had acheived last year, it would be different, but for now there is ambiguity in what is meant.

Damn, why when there are problems with puzzles does it always seem to reduce to semantics?

“Damn, why when there are problems with puzzles does it always seem to reduce to semantics?”

I think this is a very interesting and important question. It seems to me that almost any puzzle, however carefully stated, can be interpreted in ways not foreseen by the author, given a sufficiently obtuse solver. When I’m reading Richard’s puzzles, I often have this running commentary in my head saying “Ah, there’s an ambiguity, that will cause a few comments… oops, another one! Still, I know what he MEANT to say…”

Perhaps almost any puzzle like this requires a certain amount of intuition about the intended method of solution, and a tacit agreement not to take the wording so literally that it becomes almost impossible to state the question clearly. But in that case, the solvers have to have some degree of collusion with the author, almost as if they are playing a game with unwritten rules.

People are fascinating🙂

Why is the time 1s less, though? Just declaring it so isn’t useful.

I’m with Ugo. Jack is the slowest.

The first click is on time zero so in 5 secs there are actually only 4 clicks which gives 1 click every 1.25 secs. Similar calculation for the others. So its not one sec less, its one click less!

You don’t actually mean 1 second less, do you? The actual COUNT is overstated by 1 in the problem statement: click-and-pause periods fit 9 in 10 seconds, 19 in 20 seconds or 4 in 5 seconds.

Now, using these to measure time per click gives us 10/9 ~= 1.1 seconds, 20/19 ~= 1.05 seconds and 5/4 = 1.25 seconds. Finally, multiplying up to measure 39 clicks-and-waits gives us the timing for 40 clicks as 43.33 seconds, 41.05 seconds and 48.75 seconds respectively.

Um, yep. What everyone else has already said, basically…

Missed it on Friday, didn’t get it today, but understand the answer. Monday’s brain getting a kick-start.

You got it backwards Wiseman… doing 2 clicks in 2 seconds the way you describe it would mean the clicker has to wait two seconds between the clicks… which would make him slower than someone that can manage 100 clicks in 100 seconds.

oh, this is one of those psychologist things isn’t it? We’re all supposed to think that it’s a mathematical problem and then you look at the way we react when you give what is clearly the wrong answer.

James is the winner. Comrade Wiseman is either a Stupidman or a troll. I vote for the latter.

Can anyone let us know what Richard’s original incorrect answer was?

The original answer still stands up there — it’s the one that claims the time decreases by 1 second.

Thank you Michiexile

The updated answer may be right, but the logic makes no sense. Where does the “1 second less” come from? The only thing that there’s exactly one less of than it seems is the number of *intervals* between clicks. If the question were phrased in (fractions of) minutes you obviously would not be saying “1 minute less”.

“Thus each second John completes 1.1 clicks, James = 1.05 clicks, and Jack = 1.25 clicks. Thus James is the quickest.”

The way the answer is now formulated, after the correction, makes even less sense. This is the correct answer, IMHO:

John has to wait 1.1s between clicks.

James has to wait 1.05s between clicks.

Jack has to wait 1.25s between clicks.

Therefore James is the quickest.

Agree with Goliath and Ugo.

Jack is slowest clicker @ 1.25 secs/click,

James fastest @ 1.0526 secs/click.

(He would make 40 clicks in 41.0526 secs; Jack takes 48.75 secs.)

RW states “Jack makes five clicks in five seconds.”, and

“Timer starts with first click, ends with last click.”

So Jack makes a click, which starts timer. That’s his first click.

Five seconds later he makes his fifth click, stopping timer.

FIFTH click. FIFTH! NOT SIXTH. Given that his first click started

the timer, he made FOUR MORE CLICKS in the next five seconds.

That’s 1.25 secs/click.

I too do not see where the missing second went. The time starts with

a click. JW can’t claim the timer starts a second earlier; it is stated in the puzzle to start with the first click.

How can someone who clicks 5 times in 5 seconds possibly take 48 seconds to do 40 clicks? It’s a click rate of One per Second so wouldn’t 40 clicks take 40 secs? What am I missing?

@mikekoz68 easy:

t1 = 0s

t2 = 1.25s

t3 = 2.5s

t4 = 3.75s

t5 = 5s

…

That’s it: 5 clicks in 5 seconds, spaced by 1.25s. The rate is not one per second, but one per 1.25s. The time it takes to do N clicks is (N-1)*1.25. It takes exactly 48.75s to do 40 clicks.

ok, so you’re telling me if someone can click 5 times in 5 seconds, if you were asked how long 40 clicks would take the answer is not 40 seconds? I’m lost here and I gotta leave for work! Oh, wait there is 7 spacings of 1.25 secs? Ok but still 5clicks/5secs = 40clicks/40secs Something is wrong with this

@mikekoz68 the math above is pretty simple. If you don’t believe the math, take a sheet of graph paper, mark one dot at 0, one at 5cm (or 5in), then three more equally spaced dots between those two. That’s 5 dots in 5cm (like 5 clicks in 5s), yet the distance between them is 1.25cm, not 1cm.

Sent from my old tin can connected to a piece of string…….

Begin forwarded message:

From: Stuart Hatwell

Date: 5 March 2012 22:37:21 GMT

To: stuart.hatwell@fsa.gov.uk, Stuart Hatwell

Subject: Youtubewannabe

@mikeoz68

So “Something is wrong with this” ya, the guy who writes books explaining the fallacies of the human mind has been corrected by a you-tube wannabe commentor lol The whole point of the puzzle is that people do not understand temporal sequencing – it is not intuitive, Richard posts the correct answer of James (20 clicks in 20 seconds) and the more people argue over it the more he is right- temporal sequencing is hard for us to

It is also incorrect to say that “each second John completes 1.1 clicks” without adding “on average”. There are no “tenths” of a click.

I agree with Ugo.

Hmm. i’m a little confused why it’s 10 clicks in 9 sec.

If the timing starts with the first click, that means that in sec0 there will be actually one click. So it will result in 9 click is 10 sec, 19 click in 20 sec, and 4 clicks in 5 seconds.

So.

Jack will take 50 sec to have 40 clicks

John will take 44 sec to have 40 clicks

James will take 42 sec to have 40 clicks

but as Beaky said, were is the missing second?…

You have the right idea, but you have it backwards. Just count the clicks, 0s, 1s, 2s, 3,s, 4s. You have 5 clicks in 4 seconds, extrapolate from that.

but the timer stopped at 5 sec. so my assumption is correct.

the first click is done outside the timer. Richard was wrong with his first answer…

|| —- 5 seconds

c1 | c2, c3, c4, c5| —- 4 clicks

The clicking speed for each can be fractionally under a second. So John would complete spend 9.99999 seconds for 10 clicks, etc. So while in theory James will always be fastest, there will be no clock precise enough to determine it.

I definitely agree that Richard’s solution is incorrect. David D hit the nail on the head. If the clock starts at the first click, then Jack has 5 full seconds (0 sec to 5 sec) to make 4 more clicks, John has 10 full seconds (0 sec to 10 sec) to make 9 more clicks, and James has 20 full seconds (0 sec to 20 sec) to make 19 more clicks. After the first click, the must all make 39 more clicks. Using their given rates:

5 sec / 4 clicks * 39 clicks = 48.75 sec

10 sec/ 9 clicks * 39 clicks = 43.33 sec

20 sec/ 19 clicks * 39 clicks = 40.0526 sec

That should say 41.0526…

I think Richard made the mistake of starting the clock at 1 sec instead of 0 sec

“UPDATE: The original answer posted was not entirely correct.”

With due respect, Richard, it was totally incorrect.

And the correction is incorrect, since it gives the correct answer with the still-wrong argument.

Get your act together, Richard!🙂

Admit you solved it wrong right from the start.

I’d like to submit that there is something wrong in the reasoning here.

Being able to click five times in five seconds does not mean being able to click four times or six times in one second, or any other different number. Either the basic data is correct, or it is wrong.

Therefore: if your first click happens on second zero, you have made your fifth click on second four, and you will then make your sixth click on second five, your seventh on second six, and so on,

UNLESS you assume a non-constant clicking speed.

If you are able to click 10 times in 10 seconds, and click your first click on second zero, your eleventh click will be on second ten, your twelfth on second eleven…

and so on.

You are assuming that each click takes one second, which it does not, and is the whole point of the puzzle. Nowhere does it state each click takes one second as your assert in saying the fifth click happens on second #4. It specifically states that it takes five seconds for the five clicks with the timer starting at the first click.

You are correct. My initial assumption, as probably many people have assumed, was that the first click occurs at second 1, not at second 0, i.e. at the start as said in the enunciation of the problem. I must have looked over that part, somehow. I don’t really understand how that’s possible, but evidence is evidence. I made a mistake.

Hogwash

Whoah! Has Richard edited his answer in the 50 minutes since he first wrote it? So he’s a dilly booliak who wrote a load of rubbish after all, and he’s not deliberately trolling us? What sort of devious mind-trick is this? What is this website coming to?

What! Now the update makes no sense! 5 clicks/ 4 seconds= 1.25 clicks/ 1 second, but this means it takes 0.8 seconds/ 1 click (not 1.25 seconds as Richard states). By this logic, Jack is STILL the fastest.

WHAT A MESS! The “corrected” explanation still manages to get the two key elements the wrong way round, as others have pointed out. Will further corrections be posted?

However, i suppose this still illustrates one of the larger points about these puzzles, which is (it seems to me) that apparently mathematical puzzles often require a different kind of thinking to be solved correctly.

I agree with that. However, the problem must also be enunciated correctly. Otherwise, thinking differently will only lead to nonsensical solutions.

Richard’s answers are incorrect, as many here already pointed out.

For the reasons stated, the click

ratesare(n – 1)/nclicks/s, so the player with the highestnfinishes a given numbermof clicks first, int = n (m – 1)/(n – 1)seconds.And that is of course why they teach to start counting at 0 when you measure the period of a pendulum with a stopwatch in the undergrad physics lab class.

Richard,

I know you’ve already updated the answer in resposne to some of the comments, but the answer as written is still not quite correct. Well, the answer is correct, but the explanation is not worded correctly. It should be…

“James wins. The actual number of clicks made by each, is 1 less than the number stated in the given period. John completes 9 clicks in 10 seconds. James completes 19 clicks in 20 seconds. Jack completes 4 clicks in 5 seconds (because the timing starts on their first click, click 1 occurs at time = 0, leaving the entire duration of the timing for the remaining clicks to be carried out). Thus, on average, each click takes John 1.1 seconds, James = 1.05 seconds, and Jack = 1.25 seconds. Thus James is the quickest.”

I looked at this totally differently – what if all the clicks happened really quickly in each time period and the rest of the time period was recovery time

So john last click ‘could’ be one second into the 35- 40 second period. So completed in say 36 seconds

James in the 30-40 second period does it in 31 seconds and

Jack the winner in the 20-40 second period does it in 21 seconds

Please don’t jump on me – I failed my maths A level – then the council pulled down my school to build houses .

Makes total sense. The problem should state that the rate is constant, not variable. In that case, there would be no difficulty to solve the problem.

Obviously the rate is constant, Bartholomew. You’re just being willfully pedantic and making life difficult for yourself.

That’s not nice. And no, I am not. I am merely a programmer who -in almost 35 years- has heard “this will never happen” and “this always happens” just a bit too often to still assume these are indeed true, and the only possibilities.

Came here (again) to say exactly that. I still believe all the J’s are equally fast… period. ;o)

Weird someone hasn’t mentioned this (possibly except wraakh if I understood him/her correctly).

Jack does 5 clicks in 4 seconds (s0, s1, s2, s3, s4), OK, but he still has to click 40 times. So, if he continues, he goes (s0, s1, s2, s3, s4, s5, s6, s7, s8, s9…) You can see here that he takes the same time as John to do 10 clicks, and if you keep going, same as James to do 40 clicks.

What am I missing?

Actually, that is exactly what I was suggesting in my comment a bit higher up. In my view, you are not missing anything. The problem has simply not been stated in a rigorous enough way to know what is meant.

Right, sorry I missed yours. It’s the same point.

No problem. Our solution is not the one desired, but given the way the problem was enunciated, it does make sense, at least to me.

I can’t see how taking the “rate” out of 5 clicks in 4 seconds and so on (i.e. 1.25 sec/click, etc.), and then extrapolating to 40 clicks makes any sense at all though, let alone be a desired explanation. Doesn’t take into account that after the 5th click there will still be a pause, and the first click of the next 5 clicks is not gonna be immediate.

That’s because it is just the other way around. In order to fill five seconds with 5 clicks, you start with click one at second 0, click two at second 1.25, click three at second 2.5, click four at second 3.75 and click five at second 5.

And you then continue from there: click 6 at second 6.25, click 7 at second 7.50 and so on…

But s0 could be at time 0. So if the time between 2 clicks is always t, then 5 clicks take 4t seconds (i.e. there are 4 gaps between clicks). Then for 10 clicks, there are 9 gaps, i.e. it takes 9t seconds.

It’s not clear, though, whether the times given for the J’s start with s0=0.

That’s why you must *assume* a constant gap. Otherwise, all guesses are valid.

But that doesn’t jibe with the established “answer” that it’s 5 clicks in 4 seconds. The 1.25 number was taken from that fact. That relationship between “5 in 4” and 1.25 is what I don’t get. If anything, it’s still 5 in 5 and James is quicker, but you can’t have both.

I agree completely. “5 clicks in 4 seconds” is wrong. No matter what. In order to reach that number you would have this:

This gives you 1 click per second:

Second 0 = click 1

Second 1 = click 2

Second 2 = click 3

Second 3 = click 4

Second 4 = click 5

But what you must get is

Second 0 = click 1

Second 1.25 = click 2

Second 2.5 = click 3

Second 3.75 = click 4

Second 5 = click 5

In other words: you get 1 click per 1.25 seconds

Ah, never mind. Was the answer changed to be more clear? I could have sworn there was a “5 clicks in 4 seconds” answer. Maybe just got confused.

The problem is in the way you look at things. That is why I like the pole example. You can easily do it by looking at a ruler: the first long line is at 0 cm (or 0 inches if you have an inched ruler), the second at 1 cm (or 1 inch), the third at 2 cm (or 2 inches), the fourth at 3 cm (or 3 inches)…

In other words, you always count 1 long line more than there are cm (or inches). The reason is that the cm (or inches) are the gaps between the lines.

Now, if you look at the clicks as though they were the lines, you’ll get exactly the same thing.

The crucial part in the problem enunciation is this:

Why is this crucial? Because it tells you that you don’t have to wait for the first click, just as you don’t have to pass 1 cm or 1 inch to encounter the first line on a ruler. As I’ve said elsewhere: you get the first click free.

The answer is James, the time is irrelevant.

There is a fundamental ambiguity in the wording related to how the contestants original click rates were expressed. Rather than being told that contestant A takes Y seconds to achieve X clicks, we are told that contestant A can achieve X clicks in Y seconds. These are not equivalent statements. The latter opens the possibility that contestant A can almost, but not quite achieve X+1 clicks in Y seconds. For instance, we are told John can achieve 10 clicks in 10 seconds, but for all we know he might almost, but not quite, have completed an 11th within that time.

Richard’s explanation is also not the clearest. The calculation is probably better thought of as the interval between two clicks. That makes the competition the time for 39 intervals. (This assumes clicks are “instantaneous”.

The “interval” periods for each contestant rounded to 3 decimal places are :-

John >1 1 1 <=1.026

All this makes an unambiguous answer impossible. What we know is a range of times that they may have been able to achieve.

John slightly over 39 to 43.3 seconds.

Jack slightly over 39 to 41.1 seconds

James slightly over 39 to 40 seconds

From this, it can be seen that it is possible for any of the contestants to have won. although clearly James has the narrowest of ranges and is the most likely from what we know.

Not quite true… You are told that ‘the timing period begins with the first mouse click and ends with the final click’, therefore Jacks 5th click occurs at exactly 5s, Johns 10th click occurs at exactly 10s and James’ 20th click occurs at exactly 20s

@Dribbling Sidney

We are only told about the timing for the competition.

“In my competition, the timing period begins with the first mouse click and ends with the final click.”

There is nothing that tells you how the three individual rates were produced.

Thank you for this Steve.

I would also note that the statements :-

“J can click a mouse 10 times in 10 seconds.”

“J can click a mouse 20 times in 20 seconds” and

“J can click a mouse 5 times in 5 seconds.”

are not mutually exclusive.

J could possibly do all of the above.

I’m sure you are aware of this, but do not think that everyone in this discussion appreciates it.

Regardless of any formulas, fences and timing issues the original puzzle states 5 clicks in 5 seconds not 5 clicks in 5-1 seconds.

The last click stops the clock, so there is no timed pause after the last click. at 4 pauses in 5 seconds, this would be 1.25 seconds per pause. The clicks are of negligible length, so it is only by measuring the pause lengths (or the number of pauses in a given time) that the solution can be found.

Although asking who will be first to complete 40 clicks, the time taken is not asked for, so the answwr to the question is the same as “Who is the fastest?”:

The speed of each person can be described as N clicks in n seconds. in the example above, N=n,but that does not have to be the case in other examples. The number of pauses in n seconds is N-1, in the same way that for a linear fence of length n with N posts would have N-1 gaps (for the people who, on Friday, noticed that this is similar to the Fence Post problem).

The formula to compare the rates in any finite time period ‘x’ where each person can do ‘N’ clicks in ‘n’ seconds is:

rate=((N-1)*(x/n))+1 clicks in ‘x’ seconds.

If x,N and n are the same, the rate would be as given in the question ie N-1+1 in n seconds

For a minute (ie x=60 seconds), this would give:

James = ((20-1)*(60/20))+1 = (19*3)+1 = 58 clicks

John = ((10-1)(60/10))+1 =(9*6)+1 = 55 clicks

Jack = ((5-1)(60/5))+1 = (4*12)+1 = 49 clicks

The person with the highest click rate is therefore James and he will reach 40 clicks first.

That duel does not make sense, 5 clicks in 5 seconds equals 60 clicks in 60 seconds, not 49.

The point here is that it is assumed that click 1 takes no time at all, something that is not assumed for any other click. It is as though the game master said: “Your first click is on the house!”

Another way of looking at it is if you were trying to measure distances by counting equidistant light poles: assume that there is 100 metres between two poles, and that this distance is constant for all subsequent poles:

Pole 1

100 metres

Pole 2

100 metres

Pole 3

100 metres

Pole 4

100 metres

Pole 5

100 metres

Pole 6

100 metres

Assume now that you have walked for 300 metres, how many poles have you seen? The answer is 4, but only if you started at a pole, not somewhere between 2 poles.

It really is 49 clicks, because the 1st click is at 0 seconds and the 5th click is at 5 seconds, a period of 1.25 seconds per click/pause. To get 60 clicks in 60 seconds the 6th click would be at 0 seconds in the 2nd group of 5 seconds, ie at 5 seconds, but that is when the 5th click happens and the 5th and 6th would therefore be simultaneous, when in the given rate, they should be 1.25 seconds apart.

@Bart: If the poles were in a dense forest that is – if they were on an open field I’ve seen them all before I even started😛

Only if there is no fog, of course. I happen to be from Flanders, where dense fog is the rule ^_^

The original answer posted was not entirely correct. hehehe…

My original solution was… heheheh

But it was easy ..

The solution is correct only in case when John, James and Jack are robots, predictably clicking mouse in exactly the same pattern – the author of the puzzle made an assumption that clicks are spaced evenly and that start and end of the time period coincides with first and last click.

Humans never click as predictably – and neither can they keep the tempo over longer periods – that’s why it is measured by counting the number of clicks made within a time period. Usually first click is made a bit after the start and last click a bit before the end is called. The averaged time per click was equal for all of them – 1 click per second, so we can say that in all probability they will all made 40 clicks in a similar amount of time – close to 40 seconds – but any predictions about who will make 40 clicks first are purely random.

He didn’t “make the assumption” that the “start and end of the time period coincides with first and last click”, he explicitly stated that as a tenet of the puzzle. If you decide that when solving a puzzle you can ignore a statement because it might not always be true, then surely it doesn’t matter what the puzzle states any more, because the answer will always be “we just can’t be sure”.

Not quite – it was explicitly stated that this was how the competition was judged, but there was no comment about whether this was also used in the measure of the rates of click.

What is the timing period? The period of 5, 10 or 20 seconds, or is the time needed to achieve 40 clicks? That’s a non-trivial question.

Has anyone thought about sampling error???

All this mouse arithmetic confuses me!

If I could click once click per second and you were TWICE AS FAST as me (two clicks every two seconds) why would that make you SLOWER!?

Once per second is one click then another 1 second later etc. This means that at 39 seconds you would have reached 40 clicks because click 1 = 0 seconds, click 2 = 1 second etc. However, teo clicks in 2 seconds means 1 click at time = 0 and the 2nd click at the end time i.e. at 2 seconds. This is only half the speed, NOT “twice as fast” as you suggest. It is, indeed, slower. The once per second rate would be described as 2 in 1 second, 3 in 2seconds, 4 in 3 seconds etc, as the time quoted includes the first click at time = 0 and the last click at the quoted time.

Jack will finish first coz those two waited 10-20 secs before they started clicking. Jack just clicked away from 5.

I don’t know why you people try to make this puzzle appear more complicated than it actually is. The answer is pretty obvious and one doesn’t need to make too many assumptions to find it.

First, when it says ” John can click a mouse 10 times in 10 seconds.” it is pretty obvious that this is the *fastest* John can click.

We do not know whether John can keep clicking at the same rate for 40 seconds, but we must assume he can. If he gets tired and slows down before the end of the test, we wouldn’t be able to determine the solution. If the puzzle has an answer, then we must conclude that the click rate is constant. This is a reasonable assumption that we must make.

The closing of the electrical circuit is an instantaneous event. You can model it as having a duration greater than zero, but in practice it would be so short as to be completely irrelevant (e.g. microseconds vs. seconds).

What is larger than zero is the time each player has to wait before being able to click again. This time can be the sume of the time it takes for the mouse button to travel up, plus the time the player has to rest before pressing the button again, plus the time it takes for the mouse button to travel down again. We don’t know how the waiting time is divided between these three activities, but we do know that this time is different for the three players. It doesn’t matter one bit whether we adopt a click-pause or a pause-click-pause method, all we need to know is that a given, constant interval of time elapses between clicks.

We also know that the clock starts when the first click is detected. We don’t know and don’t care how much time it took each player to reach that click before the clock started. One of them might have had their hand on the mouse button, while another might have had to order a new mouse on Amazon before starting. It doesn’t matter.

Now take a piece of graph paper, label the X axis ‘t’ and plot one point for each click that each player makes, putting the first point at t=0.

Put the 10th point for John at t=10, the 20th point for James at t=20, and the 5th point for Jack at t=5.

Put other 8 equally spaced points for John between the first two, 18 for James, and 3 for Jack.

Now measure the distance between points for each player. It will be ~1.1s for John, ~1.05s for James, and 1.25s for Jack.

Now plot more points along the t axis for each player, spacing each one by the distance you measured above, and count how many dots there are within the 40s line.

They will be 37 for John, 39 for James, and 33 for Jack. Therefore James wins. QED.

THANK YOU. That should pretty well cover all the non-existent and irrelevant ‘ambiguities’ brought up in the previous posts.It’s not that complicated, folks. And attempts to make it so seem pointlessly argumentative.

@Mervulon

Ah but there is an ambiguity. The problem states that contestant A can achieve Y clicks in X seconds. That means contestant A can’t achieve Y+1 clicks in X seconds, but it does mean that contestant A might be achieving any of a range of click speeds.

For instance, John achieved 10 clicks in 10 seconds. If the last click ended exactly at the end of 10 seconds, then the period between clicks would have been 1.25secs. However, if the 11th click would fall just after 10 seconds, then the period between clicks would have been fractionally over 1 second. I’d agree with @Ugo Cei that the question implies 10 is as many clicks as John can manage in 10 seconds, but that doesn’t mean all 10 seconds were used. The 10 clicks could have been completed in fractionally over 9 seconds. We simply aren’t told.

This ambiguity could have been avoided by stating that contestant A completed Y clicks in exactly X seconds. As it is, their is an ambiguity by just stating how many clicks could be achieved within a given number of seconds rather than stating eactly how long it took.

Now it is possible to specify unambiguous questions. The Enigma problems in New Scientist are generally much more rigorous (at the expense of length), but I rather suspect the studies ambiguity in these questions provides for a lot more discussion.

The text of the problem says: “the timing period begins with the first mouse click and ends with the final click.”

Now words can never be 100% unambiguos, but that sentence is as unambiguos as they come. In each of the three examples provided, the timer stops at the 5th, 10th, and 20th click exaclty, not some time after, and the time is 5s, 10s, and 20s exactly.

@Ugo Cei

You are being disengenuous by producing only a partial quote. Let me add the other relevant part.

“In my competition, the timing period begins with the first mouse click and ends with the final click.”

In other words, that timing rule applied to the competition. It does not tell us that the three individuals all ended on the exact tick of a second on the actual last click. As this condition explicitly states that this applies to the competition, it’s a stretch to say it applies in a different context.

My background in physics and IT has taught me only too well that ambiguities like this matter in real life. Sloppy wording has lead to many unfortunate results when people make unwarranted assumptions. It’s something I was trained to look out for. Note these are not physiological assumptions (a whole different game), but assumptions about the actual logic.

The use of unwarranted assumptions in problems solving is, in itself, is an interesting study in psychology.

So I’ll place a challenge. Where is the wording that forbids the possibility that in the 10 seconds available, John made his 10th click after 9.5 seconds?

OK Steve, I think you might have a point here. My opinion is that, if we have to make an assumption, it has to be the assumption that the problem is solvable and all my other assumptions derive from this.

For what is worth, I agree that we shouldn’t make any assumptions in real life, but this is a game and besides, it was fun catching the Prof with his pants down😉

That’s fair enough. I’m not disagreeing with the solution as such in that this is about the timing of the intervals. Just that as, is often the case, there’s some space for exploration.

even after reading all the comments, and that the esteemed prof has had a couple of goes at answering, I still don’t think there is a proper answer without assuming something that is as equally invalid as any other daft assumption I can apply…

eg

clicks take 0.75 sec recovery takes 0.18sec and smiling at the moon 0.07sec…

All 3Js click regularly / randomly

all 3Js are the same bloke with many aliases

etc

I have the answer! Jack is the fastest!

Ok, So Jack does the first 5 clicks in the first second so he can ponder the meaning of life for the next 4 seconds.

John clicks half his number in the first 2 seconds and the other half in the last 2 seconds because he was distracted by a beach ball.

James, being the lazy slob that he is, does all of his clicking at the last 3 seconds of his time. Still, he’s fast like that and puts woodpeckers to shame.

By the 36th second Jack is done, having gotten all his clicks in super fast. At that time John is busy clicking to get to 35 so he can wait for the 38th second to finish up. And James,that dozy cow, is still waiting for the 37th second to start so he can start clicking. So by that time the 40th second rolls around, they will all be done. But Jack did it first.

Just came back to check on the discussion, and it seems pretty clear there were various ambiguities/assumptions in the original problem, such that the “correct” solution could be disputed. The main assumption is that of a regular rate of clicking (the time starting with each candidate’s first click, and the uncanny ability of each candidate to complete their final click at the exact moment that time expired).

A rather unsatisfactory puzzle in and of itself, but interesting on a more general level in terms of the different ways that maths and language deal with logic and temporality.

P.S. I must admit i am really struggling with THIS week’s puzzle (about the people in odd- or even-numbered houses) whereas everyone else seems to find it ridiculously easy. I suppose i’ll have to wait and see whether there are once again ambiguities and assumptions at play (bothering fewer people on this occasion).

wonderful app many thanks for sharing

flagyl and azithromycin together azithromycin compared to penicillin cost of azithromycin treatment generic name for azithromycin azithromycin compared to clarithromycin azithromycin online uk how to buy azithromycin buy zithromax one time dose azithromy

hey buddy, this is a very interesting write-up