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Same question as here but I'm trying to work out what's wrong with my logic. The question is:

Take a stick and break it randomly into three pieces (i.e., two randomly placed breaks on the stick). What is the expected length of the longest piece?

Our stick has endpoints $0$ and $1$. Let the cuts be made at $x, y$. Now, I can choose $x\in [0,0.5]$ without loss of generality since I can always choose the first cut to be on the left half of the stick. Now, I consider four cases

  1. $1/3\leq x\leq 1/2$ and $y\geq x$
  2. $1/3\leq x\leq 1/2$ and $y< x$
  3. $x< 1/3$ and $y\geq x$
  4. $x< 1/3$ and $y< x$

I will work out the length of the longest part in each case and the probability of each case occurring. We have

$$P(1) = \frac{1}{3}(1-x), \quad L_{\rm longest}(1) = x$$ $$P(2) = \frac{1}{3}x, \quad L_{\rm longest}(2) = 1-x$$ $$P(3) = \frac{2}{3}(1-x), \quad L_{\rm longest}(3) = \frac{3}{4}(1-x)$$ $$P(4) = \frac{2}{3}x, \quad L_{\rm longest}(4) = 1-x$$

The probabilities are computed straightforwardly: $x$ is uniformly distributed in $[0,0.5]$ so $p(x<\frac{1}{3}) = \frac{2}{3}$. The probability of $y\geq x$ is just $(1-x)$. $L_{\rm longest}(3)$ is computed because the largest piece is obtained by taking a piece of length $(1-x)$ and breaking it in a random place. This gives us an expected length of $\frac{3}{4}(1-x)$. I now have to write each of these as integrals since $x$ is a continuous variable and I obtained

$$ I_1 = \int\limits_{1/3}^{1/2}\frac{dx}{1/2 - 1/3}\ \ \frac{1}{3}(1-x)x$$ $$ I_2 = \int\limits_{1/3}^{1/2}\frac{dx}{1/2 - 1/3}\ \ \frac{1}{3}x(1-x)$$ $$ I_3 = \int\limits_{0}^{1/3}\frac{dx}{1/3}\ \ \frac{2}{3}(1-x)\frac{3}{4}(1-x)$$ $$ I_4 = \int\limits_{0}^{1/3}\frac{dx}{1/3}\ \ \frac{2}{3}x(1-x)$$

But $\mathbb{E}(L_{\rm longest}) = I_1 + I_2 + I_3 + I_4 = \frac{97}{162} \neq \frac{11}{18}$. What is the error in my logic?

user1936752
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    Why is the longest length $x$ in the first case? It could be $y-x$ if $y$ is near $1$? – joriki Mar 22 '23 at 00:43
  • @joriki that's a good point! I made a mistake - let me try to see if I can fix Case 1 and obtain the correct result – user1936752 Mar 22 '23 at 00:59
  • These are not probabilities! What would it mean, for example, to claim that $P(y < x < \frac13) = \frac23x$? Would you say that $P(y < \frac14 < \frac13) = \frac16$?? – Misha Lavrov Mar 22 '23 at 01:17
  • @MishaLavrov I was trying to assign a probability to the case of two events $E_1 := x<\frac{1}{3}$ and $E_2:= y<x$ both occurring. Is there a problem with thinking in this way? – user1936752 Mar 22 '23 at 01:24
  • The problem is that the thing you're calling a probability is a function of $x$. Probabilities should be numbers. The probability that $x<\frac13$ and $y<x$ is $\frac1{18}$, or $\frac19$ if you're conditioning on $x<\frac12$. – Misha Lavrov Mar 22 '23 at 02:03

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