A nice, elegant, puzzle this week.
I have a few thousand sticks. Each of them is 1 metre long. The other day I decided to break each of the sticks in two, with each of breaks happening at a random position along each stick. What is the average length of the shorter pieces?
As ever, please do NOT post your answers, but feel free to say if you think you have solved the puzzle and how long it took.
Richard Wiseman 
Two different answers immediately spring to mind, based on different interpretations of the statement.
After reviewing the updated wording, I’m satisfied that one of my answers has been thoroughly eliminated, which means I can comfortably go with the other one.
Got it. I think. ~45 seconds?
“Break the sticks in HALF”? I think you mean TWO.
I came up with an answer very quickly, which means it’s probably wrong…
Hmmm. I also have 2 answers, depending on interpretation. The method for working out the answer was the same either way, and took me a minute or so, using a computer to help with the maths bits.
I am assuming, as a commenter above, that you are breaking the sticks in TWO, not in HALF.
Will be interested to see the answer on Monday.
OK, Having played with a few more possibilities in Excel, I’m happy to commit to one of my interpretations, and therefore one of my answers.
I think we have to assume richard didn’t actually mean “half”.
And, additionally, we ignore the physics of breakage.
This one was painfully obvious and came to me immediately. Which means I’m probably wrong.
Done, about a minute or two, following a couple of hunches and convincing myself that one of them is right.
The question surely asks about the sticks being cleft in twain and the “obvious” answer won’t cut it, or there wouldn’t be much of a puzzle.
I think I have it – about 1 minute.
BTW – want to thank you for the puzzles each week – Look forward to them very much…..
Likewise I came up with an answer in about 2 minutes that seems straightforward and obvious to me. Great, I can relax now until Richard confirms how brilliant I am by revealing the answer I have come up with.
uh-huh…
Seconds, therefore almost certainly wrong.
In another – longer contemplated – interpretation, it’s too vague to reach an answer.
i got the answer after just completion of reading..and i have only one answer…
As he specifies that each stick is broken at a random position “along each stick”, there’s only one way of breaking them (ie, not lengthwise)
Come on, that is not a puzzle!
What are you trying to do? Confusing everybody by making them think “it’s so easy, it can’t be the answer”?
What is the random distribution of the breaking position? Gaussian? Evenly distributed? Other?
I did a Monte Carlo simulation for (i) 10,000 (ii) 100,000 runs… This is probabilistically speaking….
But breaking the sticks personally, I might have certain biases that would result in a different average than that I get from the calculations. My feeling is that this average would be different for different people.
Depending on your interpretation of the question (as usual, it’s badly worded) it doesn’t matter what the random distribution is.
I am pretty sure I have the answer wanted but I think there is too little information. If they are just trying to break the stick in two you might get that answer but if they are trying to break it in half you would get a different answer but there is not enough information to get the answer in that case.
Unless I am missing something.
I think I have it in 20 secs…
Assuming “randomly” means that you are choosing from a uniform distribution, I think I’ve got this one in under a minute. My argument persuades me at the moment but I’ve got these things wrong before!
And if we’re being pedantic, the question doesn’t say that you actually break the sticks, only that you have decided to do so. There are many things that I have decided to do that I haven’t actually done (like get down to work without spending time on Friday puzzles this morning).
Actually did LOL at that. 🙂
Taking the plain interpretation of the question, I think I have the answer by thinking about it for 30 seconds or so.
I suppose I could write something to simulate the question to double-check, or do a formal proof, but I think I’ll just wait until Monday to see if I’m right!
Well the obvious answer was in my head as I read the end of the sentence. Whether that answer is *correct* is another matter!
I got an answer in about 10 seconds. That seems far too fast so I am most likely wrong. Will wait impatiently for the answer.
less than ten seconds. Like everyone else, suspect it is too easy and i am wrong. (and I haven’t assumed it is a trick question with the answer in the question) who has time to break a thousand sticks and measure the results anyway?
Instantly as I was reading the question.
Like others, either this is really easy, or I am very wrong!
Like other peadants, my first thought was surely you mean they were broken in two. And to prove pedantry knows no limits or shame, either you have found a source of sticks that are uniform along their length, or the puzzle is impossible.
Very nice puzzle, I liked it very much. I have my almost instantaneous answer, but it’s the lind one were I’ll only be certain next monday…
@Joao Pedro …. Howdy! Long time no see…..lol….
Hey Joao Pedro! IT’S SNOWING HERE! OH JOY! …doing the HAPPY, HAPPY, JOY, JOY DANCE! IT’S SNOWING!
Just did a scaled down experiment with a couple of boxes of matchsticks and now have the answer.
Just kidding. Solution is obvious, isn’t it?
The obvious answer came to me as I was reading, but I’m thinking I must have missed something because it seems too easy…
Think i’m done. About five seconds
Really? this is what it has come to? Next week ‘John has one orange in each crate and 30 crates, how many oranges does John have?
I got the answer right away, and thinking further I can’t see how the answer could be anything else. I look forward to finding out the trick that renders my answer invalid on Monday. 🙂
I can’t see how the language is ambiguous, has it been rewritten since those complaints? If not, can someone explain the ambiguity in a way that doesn’t spoil the question?
the original post said ‘snap each one in half’ but so many people commented Richard has changed it.
Now, I though I had the right answer but I have just asked a friend who is a professional statistician and SHE says it’s a really complex question and she needs to work on it – apparently it depends on the type of randomness involved…
i got it in 20sec any ways i think am wrong
Very nice puzzle. No need to be pedantic or over-interpreting the words. Just read it in all it’s simplicity. I smiled when I figured out the answer, because my first guess was so clearly wrong.
The answer that first came to my mind was wrong as when I simulated with a few random examples on a spreadsheet the method became clear in minute or so
Wait, by ‘ shorter pieces’, do you mean all of the pieces broken from the 1m stick or only the shorter of the two?
I think I got it, in my second chance. 🙂
Hint: it isn’t 25cm.
Isn’t it?
I would like to know your reasoning, because I think you may be wrong.
Fascinating.
Read Chain Bear’s comment again.
@Simon H: As far as I know, it doesn’t change anything if you take the shorter one every time you break a stick, or if you sort all the sticks at the end and take the shorter half. Otherwise, you must define what you mean by “shorter pieces”.
Excellent Puzzle! I straightaway wrote up a simple simulation in Excel (six thousand sticks) to see how the average behaved. / Nary / New Delhi
My intuition told me an answer. And three minutes later a quick script running an emulation gave the same answer.
But can I prove that easily? Not even sure were to start 🙂
literally took me 2 sec. way too easy!
Hmm… my Excel test with 1000 randomly chosen breaks confirmed my first (obvious) guess at the answer.
Which is so obvious that I’m still convinced I’m missing something. So I will be intrigued to see what Monday’s post brings…
It is all “the shorter stick length sum” divided by “the number of shorter sticks”.
No idea why the obvious solution would be wrong. Must be missing something.
The term “shorter piece” might refer to the shorter part of each individual break. Or it might refer to all broken pieces. Two puzzles, one description. RW does it again, sigh…
After thinking it through I got a number, but after thinking a little more, figured why that number is probably wrong.
Now I’m anxious to discuss the solution, can’t wait untill monday…
In short another great friday puzzle 😀
a los 55 años me enteré que soy hija adoptiva, nunca se me dijo y ya habia adoptado un hijo por propia voluntad, no por infertilidad o falta de pareja ¿habra otros casos?
Sin duda es interesante, pero no entiendo porque lo has escrito aqui. Es que es una rompacabezas de la logica?
Porque se llama usted Juan: usted es una mujer, no?
Well it’s obvious isn’t it, I got it straight away and anyway who says i am wrong will have to fight me for it.
Like everybody else I have an obvious answer that is staring me in the face. I shall stick (ahem) with it.
Obviously the type of randomness matters, but I think we can safely assume that the breaks were randomly distributed along the full length of the sticks (including some itty bitty pieces consisting of a few tiny splinters) and the randomness was absolutely even, not clumpy like most random distributions.
I shall now do some actual work… oh sod it, I will test my answer instead.
A break at random positions suggests that a position of zero along the length of the stick is possible. In which case there is no shorter piece. That’s not unreasonable but if there must be a shorter piece then what is the minimum length it can be? If it’s less than the width of the stick it becomes more of a slice than a break…
Furthermore if they’re broken by hand then how wide is the hand holding the smaller piece? That would seem to be the minimum length of the shorter piece.
(Note to self: get a life))
I disagree. A break “at zero” means nothing has happened to the stick at all. Slices count, yes, but not breaks at zero, in my opinion.
I wasn’t sure what to do with the pieces that were exactly 50:50. Do we count both pieces, or take them out of the equation altogether?
This isn’t an interesting problem. When I first read this, I assumed he had set the interesting problem instead. The interesting problem is, if you break a stick at random in two places, what are the statistics. It is interesting because it is hard to know what “break a stick at random in two places” means. But I’m pretty sure we all know what breaking a stick at random in one place means.
I’m sort of pleased people did simulations to check their intuition, because it indicates a good attitude to checking the robustness of things. But I think this one is simple enough you should just write down the formula for calculating a mean by integration and observe by inspection it comes to the intuitive answer and be confident it is right. I usually reserve Monte Carlo simulations for things that are not amenable to simple calculation.
I had a quick think about what I thought the answer was, then ran a quick simulation to check this and was glad it came out the same.
How would you find the answer through integration?
The Friday puzzles are different each week and cater to a number of skill sets. This means there are bound to be puzzles where some people will find interesting or challenging whereas others, with the right skillset, will not even see how it is a puzzle at all. I’m sure there are plenty of intelligent people who don’t know how to “just write down the formula for calculating a mean by integration and observe by inspection” or know what a “Monte Carlo simulation” is. I would like to think I’m one of them.
@ivan after reading your post I went back and reread the question and I don’t see where it says to “break a stick at random in two places” but says “to break each of the sticks in two, with each of breaks happening at a random position along each stick.” which means to break the stick in to 2 pieces …if it said “two places” then the end result would be 3 pieces of sticks….
Or I could be wrong…..now reading more posts it seems there are a variety of ways to read the puzzle….so I asked my critters about it but they told me if I wasn’t going to play poker with ’em then I needed to go back inside as I was distracting them from the game…imagine being told what to do from a group of raccoons, opossums, squirrels and a sneaky stray cat….. I came back in….I know better than to play poker with a stray cat……lol…..
I guess I’m joining in on the line of “Wow I got it so fast it must the wrong answer”-ppl.
But I can’t possible think of another answer… I will reread the question until i understand all the mysteries of the universe.
I had a hunch, then filled an Excel column with random numbers and did a SUMIF() / COUNTIF() to test my hypothesis.
Assuming we’re talking about the mean (as opposed to median, mode, geometric mean, harmonic mean, or anything else), then my chosen number is correct, +/- 1cm.
Got the answer in about 20 seconds. I think it’s pretty intuative if you understand what the question is asking. I’ll wait for Monday to see if I’m right.
I think that my hunch was slightly wrong. My calculations indicate a slight but significant error in my initial assumption. If my calculation is right, then a variation of +/- 1cm (as mentioned by mittfh) is too big a variation. That is how wrong I was initially – by 0.5cm.
ooh, it tends towards that number, the more sticks you break, the closer you get…
Hope I am not the irritating spoiler again… bah humbug
No Rob. I’ve nit got a clue what you’re talking about!
I’ve got my answer but, after reading comments, becoming increasingly convinced it’s not correct!
Got an answer in a few seconds, checked it with Excell.
Ambiguous. It depends on the meaning of “shorter” pieces. ALL the resulting pieces will be shorter than 1m. That average? Or should the resulting pieces be sorted into two groups after each break? If so, then the “average length of the shorter pieces” has a different meaning, and the answer is different, 100% different.
…huh, four and a half hour to collect the thousands sticks in the wood. Tomorrow ill break them and measure. On Monday ill sale the whole packet on the market…
2 seconds. 2 different answers depending on how I read a single word in the last sentence.
This one is painfully obvious, which leads me to think it’s wrong, but thinking it through with two seconds of simple maths, I think I’ve got the right answer. When you do the maths to calculate the averages, even without definite numbers, it’s obvious.
Those of you (including me) who are doing this electronically, remember when you randomly break a stick, the length pair [1m 0m] does not count as a valid break in the real world — that would still be an unbroken stick.
Plus also, you’ll need an assumed actual real-world value for the shortest part you can break off. You assumption here may change your end result (I’ll leave the people writing simulations to find out).
I didn’t do a simulation, but two checked two different levels of granularity. Obviously, 1m/0m doesn’t count. However, if you also assume that 0.5/0.5 doesn’t count (because in this case there is no shorter stick), the answer is the same for all levels of granularity (i.e. assumptions of what the shortest distance is you can break off).
It doesn’t change anything if you don’t count the pair 0-1. There’s an infinite number of other combinations. Infinity – 1 is still infinity.
I found that if you include .5 once (I.e when you break each stick put one piece in the long pile and one in the short pile, and if you get an exact half you put one piece in each pile ) then it does change depending on the level of granularity.
Of course I want to go into detail, but can’t until Monday.
Here’s another similar puzzle:
If you break a stick randomly at two places, what is the probability that a triangle can be formed from the three pieces?
Hurrr durrr… What’s the definition of “average” (“mean”)? How many pieces are there in total? What’s the total length?
Yeah, I got it before I finished reading. WAY too easy.
hmmm… I think I got it, but this (as with all of us it seems) took me more time to convince myself I am right than I did to figure it out.
I think I got it very quickly. Which means I’m wrong. Tenterhooks…
If the shape of the sticks is what is shown in the photograph, it will not be the center always.
Obviously there are different answers because of some factors:
1) Distribution of randomness, a stick broken in 2 will not break evenly but at the weakest point ‘near’ the centre
2) Is a stick broken exactly in two able to have a ‘shorter piece’
If you ignore the above 2 then the answer came to me as I read the puzzle, plugging it into a spreadsheet and messing showed that 2) appears to be irrelevant whilst the distribution of the break point can make a *big* difference
I would like to say I know the answer but all I can think at the moment is…”how the hell do I know” !
Makes me laugh all the ‘i got it in 5 seconds’ answers. At least some of these will be wrong, but at least they’re wrong and quick! My logic said that, ignoring the 50/50 split and 100/0 split option there would always be a shorter stick, and this would range from 0(ish) to 50cm(ish) in an even distribution of numbers, giving an average of…(I’ve said too much, as has someone previously) :).
Instant. Then I re-read the question and I cannot see where the ‘trick’ is in the question, so I am assuming it really is easy-peasy this week (if you’re half-decent at maths).
I can only see one numeric answer which can be derived from the given data. This relies on ‘The Shorter Pieces’ referring to all the broken bits, as they are all shorter than the original sticks.
Anyone who thinks otherwise must be working for the UK Department for Work and Pensions, as they are the only people I know who don’t understand the ‘law of averages’.
I imagined breaking the sticks, lining them up in order, then eliminating 50% and got the “obvious” answer everyone else gets…is it simple or are we missing something?
As others. Did it on a spreadsheet and got the obvious answer. In the end it’s what’s the average of a bunch of random numbers between 0 and 50, surely?
spreadsheet why?? My answer most be wrong I dont see a problem.
bc it takes 2 mins and confirms the answer – please let me know what you think i should have been doing with my time and i shall obey. i shall obey.
well took me 10 secs to solve as I exclaimed shorter than what ? the original 1m stick ,lol, scratches head . Or perhaps I’m thick and missed the point
Well I came up with 2 answers right off and both are fun and therefore must be wrong! But they will keep me busy for awhile….lol….
From my experience your answers are rarely “wrong” and often interesting
@Rob J…..roflmao….. stop take deep breath and lol so more…..you must be easily impressed….. but thank you.
Your blog keeps me searchin for a solution of gold.
and I’m getting old.
I must say Strange Tributes……you are a deeply, deeply, deeeeeply disturbed puppy….. when I think of you I must put both hands over my eyes and both hands over my ears and say very loudly …”nanananananananananananananananananana” and think happy thoughts.
Got the obvious answer in 5 secs and then realised that was wrong. Took another couple of minutes to verify it was wrong. Haven’t quite got to a right answer though.
Easy peasy. Unless Admiral Ackbar is right.
Way too simple, got it just about immediately. It doesn’t matter where the break is or how many sticks there are, the average can never change
I haven’t solved it yet…
Any clue, please????
Well, there are two ways of reading the question, depending on how you interpret “the shorter sticks”. Either it means the two resulting sticks from the break, or it means “of each broken pair, the shorter of the two”.
In the first case, the result is static, doesn’t rely on probability at all, and always yields the exact same answer (modulo some micrometer lost when breaking the stick).
In the second case, the result can vary, but given enough broken sticks, the average should tend towards a particular answer, assuming the break is truly random
Take your pick 🙂
In the second case, in order to avoid misunderstandings it is then much better to ask for “the average length of the larger sticks”. As to what was the intention of the question one can only try to guess it: I lean towards “the two resulting sticks” interpretation. This is what it is all about, guessing what should have been correctly asked.
This was really not hard – I got it rather quickly and while it seems too obvious to be true, mathematically it works out.
If I’m correct, the randomness is a red herring and even if it was absurdly weighted towards one set of lengths, the average couldn’t be anything but the answer I got.
Um, does anyone else fall in the category of…
“Great – I love the Friday Puzzle, I’ll have a little think about it all weekend and then be (pleasantly surprised/a little confused/very smug) about the answer.”
Not –
“Great – I love the Friday Puzzle, I’ll (solve it in a split second/compose a colour coded spreadsheet on Excel/post patronising comments about how easy it all is on the blog).”
?
aaaaaah haaaaaa
Oh … I hadn’t even thought of the first possibility.
I suspect I could do the maths for the second one – which was my original interpretation – but I have little intuition for what the answer should be. Perhaps in the case of a uniform distribution, but I’d have to find the back of an envelope before committing to anything. In the case of a normal distribution, I’d need a table, which doesn’t sound right considering all the “2 seconds” claims.
it’s not. You don’t need a table. In a uniform distribution, the answer doesn’t really need any complicated mathematics.
Well, I’ve got an answer it may not be the right answer but never mind. Took me a few minutes of reading, re-reading and then thinking the problem through in stages. We’ll see tomorrow morning if my time was well spent!
I think the solution is the straight-forward one I came up with while still reading the puzzle.
Got it in perhaps 2 minutes using excel.
But since a thousand is a little too few I used 100 000 sticks 🙂
we don’t know, as the sticks can be any size i would guess 25cms
maybe its 1meter as it’s along the stick