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This question is similar to this but I cant find what goes wrong in my answer. The full statement of problem is this:

If a 1 meter rope is cut at two uniformly randomly chosen points (to give three pieces), what is the average length of the largest piece?

What I am doing:

|-------|------|------|
0       x      y      1

I assume rod as of being length 1 unit and divide it at distance $x$ from left and another cut is at distance $y$ from left ($y > x$). Then I take two cases:

  1. If $x \ge \frac{1}{2}$ then wherever we cut $y$ we always get $x$ as largest. So expectation is $$\int_{1/2}^{1}x(1-x) dx = \frac{1}{12}$$

  2. If cut is made $\frac{1}{3} \le x \le \frac{1}{2}$ then $y$ may lie only in region $1-x$ to $2x$ for leftmost part to be largest. Then we have $$\int_{1/3}^{1/2} x(3x-1) dx = \frac{1}{54}$$

So i thought expected length to be $1/12 + 1/54 = \frac{11}{108}$ but this is very wrong correct result is $11/18$

StubbornAtom
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jeea
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  • First of all, you shouldn't sum them as is, you need to multiply by the probability of each event. Second, I'd be happy to see more calculations and an explicit statement of the question, as I'm not sure where all the elements of the integral come from. – YJT Jul 30 '20 at 08:51
  • @YJT The statement is exactly similar to linked question just it is largest in this case. Ill edit question to add that. Regarding my calculation, for example the first one where $x \ge 1/2$, the integrand has $x$ to denote the length, and multiplied by $1-x$ which is probability that $y$ lies in the region $x$ to $1$. – jeea Jul 30 '20 at 09:33
  • It is indeed very wrong, considering that the minimum possible length of the longest piece is $\frac13$. – TonyK Jul 30 '20 at 23:44

1 Answers1

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Another approach: multivariable. Let the cuts be $x,y$ from the left, $y>x$. Then the pieces are $x,\,y-x,\,1-y$. The desired expected value is then $2$ times $$\frac12 E=\iint\limits_{x>y-x\\x>1-y} x\,\mathrm{d}x\,\mathrm{d}y+ \iint\limits_{y-x>x\\y-x>1-y} (y-x)\,\mathrm{d}x\,\mathrm{d}y+ \iint\limits_{1-y>y-x\\1-y>x} (1-y)\,\mathrm{d}x\,\mathrm{d}y.$$ 2 $$\frac12 E= \left(\int\limits_{1/3}^1 \int\limits_{1-x}^{2x} -\int\limits_{1/2}^1 \int\limits_{1}^{2x} -\int\limits_{1/2}^1 \int\limits_{1-x}^{x}\right) x\,\mathrm{d}y\,\mathrm{d}x+\\ \left(\int\limits_{0}^{1/2}\int\limits_{2x}^{1}- \int\limits_{0}^{1/3}\int\limits_{2x}^{(x+1)/2}\right) (y-x)\,\mathrm{d}y\,\mathrm{d}x+\\ \left(\int\limits_{0}^{1/2}\int\limits_{x}^{1-x}- \int\limits_{0}^{1/3}\int\limits_{(1+x)/2}^{1-x}\right) (1-y)\,\mathrm{d}y\,\mathrm{d}x$$ It appears that it suffices to take only the first integral, the other should be equal by symmetry. $$\frac16 E= \int\limits_{1/3}^1 (3x-1)x\,\mathrm{d}x- \int\limits_{1/2}^1 (2x-1)x\,\mathrm{d}x- \int\limits_{1/2}^1 (2x-1)x\,\mathrm{d}x=\\ \frac{14}{27}-2\cdot \frac{5}{24}=\frac{11}{108}.$$

Alexey Burdin
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    I understand one part that expected value of largest part is (Exp length of first part when it is largest) + (Exp length of second part when it is largest) + (Exp length of third part when it is largest) is this correct reasoning, and also why is desired expected value two times this as in your answers first paragraph? – jeea Jul 30 '20 at 10:11
  • Because of $y>x$ is half of the square. It might be $y<x$ as well. I'd rather say "sum from 1 to 3 (Exp length of n-th part when it is largest)*(probability when the n-th part when it is largest)" – Alexey Burdin Jul 30 '20 at 10:15
  • So actually what exactly is $11/108$ – jeea Jul 30 '20 at 10:22
  • $1/3$ of $1/2$ of the desired expected value. – Alexey Burdin Jul 30 '20 at 10:25
  • Can I say that in my answer I have not considered the values added to expected value of largest part by the second and third pieces and by $y < x$ – jeea Jul 30 '20 at 10:29
  • Yes, this reasoning like "let $x>y$, then multiply by $2$, then assume $x$ is the largest and multiply by $3$ by symmerty" sounds correct to me. Btw, the picture in this answer is so cute/neat. Far more than mine. – Alexey Burdin Jul 30 '20 at 10:42