Stick is broken randomly EDIT TWICE, what is expected length of small side?

[member 428835]

Homework Statement

A stick is broken randomly, what is expected length of small side?

Relevant Equations

Nothing comes to mind

Cut the stick twice, at locations $x$ and $y$. Assume $y>x$. The lengths of the stick are $x,y-x,1-y$. Assume $x < y-x \implies y> 2x$ and $x<1-y \implies y < 1-x$. These two boundaries intersect at $x = 1/3$. Thus the following integral should yield the expected value for $x$: $$\int_0^{1/3} \int_{2x}^{1-x} x \, dydx$$. But my answer is off, evidently by a factor of 6. I understand when $y>x$ there are three regions, each having same likelihood of being the small piece, and there are 3 regions when $x>y$ which is 6 total regions, so are we just multiplying by 6 from symmetry?

 

[hutchphd]

The problem as I read it is that the stick is cut once. I do not understand what you are doing.

 

[member 428835]

The problem as I read it is that the stick is cut once. I do not understand what you are doing.

Shoot, I meant to say the stick was broken twice.

 

[Nik_2213]

Having spent much of yesterday snapping dry sticks to fit into 'green bin', here's my calculus-free analysis...

Grasp stick about a cubit apart, flex: Bow usually fails some-where in middle third, so 1/3~~2/3 span, but there were outliers, probably due local stressors such as knots and side-branches. Also, several small 'keystone' fragments did detach sorta-radially at surprisingly high speed in plane of bow, startling watching cat(s)...

 

[PeroK]

Homework Statement:: A stick is broken randomly, what is expected length of small side?
Relevant Equations:: Nothing comes to mind

Cut the stick twice, at locations $x$ and $y$. Assume $y>x$. The lengths of the stick are $x,y-x,1-y$. Assume $x < y-x \implies y> 2x$ and $x<1-y \implies y < 1-x$. These two boundaries intersect at $x = 1/3$. Thus the following integral should yield the expected value for $x$: $$\int_0^{1/3} \int_{2x}^{1-x} x \, dydx$$. But my answer is off, evidently by a factor of 6. I understand when $y>x$ there are three regions, each having same likelihood of being the small piece, and there are 3 regions when $x>y$ which is 6 total regions, so are we just multiplying by 6 from symmetry?

This is tricky. You'll need to put your thinking cap on!

 

[PeroK]

Is the answer $\frac{31}{243}$ or thereabouts?

 

[member 428835]

Is the answer $\frac{31}{243}$ or thereabouts?

I mean, I think my work is correct (off by factor of 6). So the answer is 1/9. I'm just wondering if my logic on the 6 multiple is accurate.

 

[PeroK]

I mean, I think my work is correct (off by factor of 6). So the answer is 1/9. I'm just wondering if my logic on the 6 multiple is accurate.

Yes, I just did a simulation and it is $\frac 1 9$.

Your method is too simplistic. I have four integrals at least. I think I have the right method, but there are a lot of arithmetic calculations that can go wrong.

You really need to think about these things properly. If you want a symmetry argument you have to really make it watertight.

 

[PeroK]

Is it obvious that all three sticks have an equal likelihood of being the shortest? Maybe that's true, but you need a wateright argument. And, if shortest, have the same length distribution?

 

[member 428835]

Yes, I just did a simulation and it is $\frac 1 9$.

Your method is too simplistic. I have four integrals at least. I think I have the right method, but there are a lot of arithmetic calculations that can go wrong.

You really need to think about these things properly. If you want a symmetry argument you have to really make it watertight.

Too simplistic? I disagree (respectfully). The method makes sense. It also stands to reason that, given $x<y$, there are three distinct pieces: $x,y-x,1-y$ each of which could equally be the short stick. Then there's the case of $x>y$, for a total of 6 identical scenarios. Thus, we multiply the integral by 6 to arrive at the solution. Funny, after writing this down I'm now very confident the approach outlined is correct.

 

[PeroK]

Too simplistic? I disagree (respectfully). The method makes sense. It also stands to reason that, given $x<y$, there are three distinct pieces: $x,y-x,1-y$ each of which could equally be the short stick. Then there's the case of $x>y$, for a total of 6 identical scenarios. Thus, we multiply the integral by 6 to arrive at the solution. Funny, after writing this down I'm now very confident the approach outlined is correct.

Then why are you asking? If the symmetry was obvious then you wouldn't have to ask!

 

[PeroK]

Too simplistic? I disagree (respectfully). The method makes sense. It also stands to reason that, given $x<y$, there are three distinct pieces: $x,y-x,1-y$ each of which could equally be the short stick.

You need to prove that.

 

[member 428835]

Then why are you asking? If the symmetry was obvious then you wouldn't have to ask!

Funny, after writing this down I'm now very confident the approach outlined is correct.

Sometimes we don't see how clear something is until we write it down. I can't begin to tell you how many questions I've written down for PF and after writing it I delete the post because it becomes so clear. Good problem solving technique I guess.

 

[vela]

Too simplistic? I disagree (respectfully). The method makes sense. It also stands to reason that, given $x<y$, there are three distinct pieces: $x,y-x,1-y$ each of which could equally be the short stick. Then there's the case of $x>y$, for a total of 6 identical scenarios. Thus, we multiply the integral by 6 to arrive at the solution. Funny, after writing this down I'm now very confident the approach outlined is correct.

If you were calculating a probability, the factor of 6 from symmetry might make sense, but I'm left wondering why the symmetry would increase the expected value of the shortest length. If in each case, the average of the shortest length is 1/54, then the expected value taking into account all cases would still be 1/54, right?

 

[PeroK]

If you were calculating a probability, the factor of 6 from symmetry might make sense, but I'm left wondering why the symmetry would increase the expected value of the shortest length. If in each case, the average of the shortest length is 1/54, then the expected value taking into account all cases would still be 1/54, right?

It's only a calculation for one sixth of the cases, so effectively the calculation has a weighting of only one sixth. If the OP can make a totally convincing symmetry argument then it's valid.

Note that my simulation seems to give an expected length of $1/n^2$ for $n - 1$ cuts.

 

[member 428835]

You need to prove that.

Okay, how about this (not a rigorous proof): take a circle. One cut makes it a stick. Two cuts makes it two sticks. 3 cuts makes it 3 sticks. Denote the length stick of each piece $S_i : i \in \{1,2,3\}$. Given $k \in \{1,2,3\}$, is there any reason to think on average $S_{k\neq i} \neq S_i$? By symmetry, this seems obvious that the expected length of each stick would be equal. Which means the odds of $S_i$ being longest are equal for all $i \in \{1,2,3\}$.

 

[PeroK]

Okay, how about this (not a rigorous proof): take a circle. One cut makes it a stick. Two cuts makes it two sticks. 3 cuts makes it 3 sticks. Denote the length stick of each piece $S_i : i \in \{1,2,3\}$. Given $k \in \{1,2,3\}$, is there any reason to think on average $S_{k\neq i} \neq S_i$? By symmetry, this seems obvious that the expected length of each stick would be equal. Which means the odds of $S_i$ being longest are equal for all $i \in \{1,2,3\}$.

That works. What about the general case of $n$ cuts?

 

[member 428835]

That works. What about the general case of $n$ cuts?

Hmmmmm that integration would indeed pile up, and there would be a lot of bookkeeping since we'd have $x_1,x_2,...,x_n$. Tons of restrictions on our domain of interest. Do you know a better way?

 

[PeroK]

Hmmmmm that integration would indeed pile up, and there would be a lot of bookkeeping since we'd have $x_1,x_2,...,x_n$. Tons of restrictions on our domain of interest. Do you know a better way?

Let me think. It's getting later here.

 

[member 428835]

Let me think. It's getting later here.

In the meantime, if we have one cut at $x$, so that the stick is divided into two, then we have lengths $x,1-x$. Let's assume $x<1-x\implies x<1/2$. Finding the average length of $x$ is then $$\int_0^{1/2}x\, dx = \frac 1 8.$$ However, since either $x$ or $1-x$ has equal probability of being the shortest, multiply by 2 to get 1/4 as the answer.

So we know 1 cut = 1/4, 2 cuts = 1/9. I wrote the following python script (which may be very inefficient, open to criticism here)

import random
import statistics

class sticks:
    def length(self, n, cuts):
        short = []
        for i in range(n):
            cutLoc = [0,1]
            for i in range(cuts):
                cut = random.random()
                cutLoc.append(cut)

            sortCut = sorted(cutLoc, key = float)
            # print(sortCut)
            stickLengths = [j-i for i, j in zip(sortCut[:-1], sortCut[1:])]
            # print(stickLengths)
            small = min(stickLengths)
            short.append(small)

        mshort = statistics.mean(short)

        return(mshort)
if __name__ == "__main__":
    n = 1000000
    cuts = 3 # NUMBER OF NEW STICKS = cuts + 1
    ob1 = sticks()
    print(ob1.length(n, cuts))

As a sanity check, for 1 cut we get 1/4 and for 2 cuts we get 1/9. The output for 3 cuts is about $0.0625...\approx 1/16$, and 4 cuts is $0.0399...\approx 1/5^2$. I wonder what's happening here. Looks an awful lot like $n^{-2}$ for $n$ sticks. Could we try proof by induction here?

 

[PeroK]

In the meantime, if we have one cut at $x$, so that the stick is divided into two, then we have lengths $x,1-x$. Let's assume $x<1-x\implies x<1/2$. Finding the average length of $x$ is then $$\int_0^{1/2}x\, dx = \frac 1 8.$$ However, since either $x$ or $1-x$ has equal probability of being the shortest, multiply by 2 to get 1/4 as the answer.

So we know 1 cut = 1/4, 2 cuts = 1/9. I wrote the following python script (which may be very inefficient, open to criticism here)

import random
import statistics

class sticks:
    def length(self, n, cuts):
        short = []
        for i in range(n):
            cutLoc = [0,1]
            for i in range(cuts):
                cut = random.random()
                cutLoc.append(cut)

            sortCut = sorted(cutLoc, key = float)
            # print(sortCut)
            stickLengths = [j-i for i, j in zip(sortCut[:-1], sortCut[1:])]
            # print(stickLengths)
            small = min(stickLengths)
            short.append(small)

        mshort = statistics.mean(short)

        return(mshort)
if __name__ == "__main__":
    n = 1000000
    cuts = 3 # NUMBER OF NEW STICKS = cuts + 1
    ob1 = sticks()
    print(ob1.length(n, cuts))

As a sanity check, for 1 cut we get 1/4 and for 2 cuts we get 1/9. The output for 3 cuts is about $0.0625...\approx 1/16$, and 4 cuts is $0.0399...\approx 1/5^2$. I wonder what's happening here. Looks an awful lot like $n^{-2}$ for $n$ sticks. Could we try proof by induction here?

I doubt induction would work. The first step would be to find a way to do it for 3 cuts. Either extend the symmetry argument in some way or find some way to construct the integrals.

 

[PeroK]

For $n = 4$, i.e. three cuts:
$$E_4 = (4!)\int_0^{\frac 1 4} \int_{2x}^{1-2x} \int_{x + y}^{1-x} x \ dz \ dy \ dx = \frac 1 {16}$$
$$E_5 = (5!)\int_0^{\frac 1 5} \int_{2x}^{1-3x} \int_{x + y}^{1-2x} \int_{x + z}^{1-x} x \ dw \ dz \ dy \ dx = \frac 1 {25}$$

 

[PeroK]

We have a pattern for the integral for $E_n$, but how to relate one integral to the next? Something for tomorrow, perhaps.

 

[PeroK]

I couldn't find a way to generalise the integrals above, so I tried a different approach. For two cuts we want to calculate the probability that the shortest length is $\ge a$. There are two cases:

1) $a \le 1/4$

We assume the first cut is between $a$ and $1/2$ and this gives us half the total probability. Now we have three cases:$$a \le x \le 2a: \ \ x + a \le y \le 1-a$$$$2a \le x \le \frac 1 2: \ \ a \le y \le x - a \ \ \text{or} \ \ x + a \le y \le 1-a$$This gives us two integrals:
$$\int_{a}^{2a}\int_{x+1}{1-a} \ dy \ dx = a - \frac 9 2 a^2$$
$$ \int_{2a}^{\frac 1 2}\bigg (\int_{a}^{x-a} \ dy + \int_{x+a}^{1-a} \ dy \bigg ) \ dx = \frac 1 2 - 4a - 8a^2$$Adding these and doubling gives us:
$$p(l_{min} \ge a) = 1 - 6a - 9a^2 = (1-3a)^2$$

2) $1/4 \le a \le 1/3$

This time the first cut can only be between $a$ and $1- 2a$ or between $2a$ and $1-a$. Again we calculate the first of these and double the result:
$$\int_{a}^{1- 2a} \int_{x + a}^{1-a} \ dy \ dx = \frac 1 2 - 3a + \frac 9 2 a^2$$And as above we have:
$$p(l_{min} \ge a) = 1 - 6a + 9a^2 = (1-3a)^2$$
From this we can get the PDF for $l_{min}$:
$$f(x) = 6 - 18x$$And, the expected value of $l_{min}$ is:
$$E = \int_0^{\frac 1 3}(6 - 18x)x \ dx = \frac 1 9$$The next step is to try to calculate this for $n$ cuts, ideally using the result for two cuts inductively.

 

[PeroK]

Okay, I think I've got a solution.

First, consider the finite case of 10 points, labelled $0$ to $9$. We want to place two "cuts" that divide this line into three segments, with each segment at least two numbers long. For example, we may have:

0, 1, X, 3, 4, Y, 6, 7, 8, 9

This can be reduced to:

0, X, Y, 7, 8, 9

And, the number of solutions is $\binom 4 2$, which is the number of ways to arrange the X and Y in the four middle positions.

If we take $N$ numbers with $n$ cuts and $l$ numbers between each cut, then every solution is an arrangement of $n$ cuts in $N - (n+1)l$ positions.

Taking the limit as $N \rightarrow \infty$ we find that the probability of achieving this from random choice of cuts is approximately:$$p = (1 - (n+1)\frac l N)^n$$In the continuous case, we can take $\frac l N$ as the minimum length $a$, giving:
$$p(l_{min} \ge a) = (1 - (n+1)a)^n$$This generalises the formula for two cuts we found above.

Edit: to expand on this, the probability of the minimum length being $\ge l$ in the finite case is:
$$p(l_{min} \ge l) = \frac{\binom{N - (n+1)l}{n}}{\binom{N}{n}}$$And that transforms in the continuous case to the probability of choosing $n$ cuts in an interval of length $1 - (n+1)l$ when selecting uniformly from an interval of length $1$.

Finally, this gives us the pdf for $l_{min}$:$$f(x) = n(n+1)(1- (n+1)x)^{n-1}$$And, indeed:
$$\int_0^{\frac 1 {n+1}}f(x)x \ dx = \bigg(\frac 1 {n+1}\bigg )^2$$

 

[haruspex]

For two cuts, consider an equilateral triangle ABC, centre O. The sum of the distances to the sides for an arbitrarily chosen interior point is constant, so we can set this sum as the length of the stick. A point chosen from a uniform distribution on the triangle represents the breakup and the distances to the sides represent the fragment lengths.
The point is nearest to side AB if and only if it lies in the triangle OAB. The centroid of that is 1/3 of the way from AB to O, and O is 1/3 of the way from AB to C, so the average length of the shortest fragment is 1/9.

For three cuts we go tetrahedral, etc.

 

[Delta2]

Am I the only one that I can't understand a thing of post #2. I guess @Nik_2213 is a mean troll?!

 

[hutchphd]

Am I the only one that I can't understand a thing of post #2

You mean post #4?

 

[Delta2]

You mean post #4?

yes

 

[PeroK]

yes

I assumed it was a joke. Although I didn't understand it.