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For homework I was asked to solve this classical problem "If you break a stick at two points chosen uniformly, the probability the three resulting sticks form a triangle is 1/4" and ok, it must result $1/4$. But I can't figure out why. This is my reasoning (obviously wrong somewhere!). Let $X,Y$ be the points (picked uniformy at random on the stick) There are $3$ case:

1) $0<X<Y<1$ with probability $\frac{1}{2}$ to happen

2) $0<Y<X<1$ with probability $\frac{1}{2}$ to happen

3) $0<Y=X<1$ which has probability $0$ to happen.

Consider the first case: ($X$ indicate now the length of the segment $[0,X]$) the triangle inequalities leads to:

$X<(Y-X)+(1-Y) = 1-X \longrightarrow X<\frac{1}{2}$

$Y-X < (X) + (1 - Y) \longrightarrow Y-X < \frac{1}{2}$

$1-Y < Y \longrightarrow Y > \frac{1}{2}$

which leads to a probability of "success" of $\frac{1}{8}$

and the "global" probability of this event is $\mathbb{P}(\text{case} \ 1)*\mathbb{P}(\text{ I can build a triangle in this case})=$ $\frac{1}{16}$. for the second case is the same. the third case has probability $0$, so it doesn't give any contribution to the probability. According to my reasoning the global probability is $\frac{1}{8} \neq \frac{1}{4}$.

I can't find why my reasoning is wrong. I don't want a solution of the problem, but WHERE and WHY I've made an error. I know there are a lot of questions about this problem, but I need a correction in this reasoning not a complete solution, and I didn't found an answer in the other questions :) Thank you in advance.

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sorry, $y-x < x +1-y $is equivalent to y-x<1/2$ right? But it doesn't change the final value i get i think –  Riccardo May 13 '13 at 21:09
That is one problem. Another problem is that you seem to assume certain things are independent when they are not. For example, when you assume that case 1 holds, the probability that $X < \frac 1 2$ is greater than $\frac 1 2$. You need to tease this apart a little more carefully. –  dfeuer May 13 '13 at 21:10
@dfeuer ok, this make sense, i start working on this observation! –  Riccardo May 13 '13 at 21:12
If $X$ is the smaller of the two, the probability that $X<1/2$ is more than $1/2$. This is not an unconditional probability... –  Thomas Andrews May 13 '13 at 21:19
@ThomasAndrews, i can't figure out why it is more than 1/2. It make sense ok, but i can't "prove" it let's say –  Riccardo May 13 '13 at 21:25
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1 Answer

up vote 1 down vote accepted

There's nothing wrong with your (corrected) calculation. The triangular region specified by the three inequalities $x<\frac 12$, $y<x+\frac 12$, and $y>\frac 12$, does indeed have area $\frac{1}{8}$. However, the total area under consideration is the portion of the $[0,1]$ square subject to $x<y$, which has area $\frac 12$. Hence the conditional probability is their ratio, or $\frac 14$.

You've made things more complicated by splitting into cases this way. Looking at the $[0,1]$ square, there are two triangular regions where the stick-breaking leads to a triangle; their total area is $\frac 14$.

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yes, so in the end the solution is $\frac{1}{2}*\frac{1}{4} + \frac{1}{2}*\frac{1}{4}$ right? i think you are right i made more complicated, but if i didn't split in cases i wasn't able to solve it... How do you find the 2 regions? –  Riccardo May 13 '13 at 21:19
The trouble with his original approach is that it seems to work only by accident. I haven't quite figured out why it works in this case. –  dfeuer May 13 '13 at 21:22
If $\max(x,y)<\frac 12$ it's not a triangle. If $\min(x,y)>\frac 12$ it's not a triangle. If $|x-y|> \frac 12$, it's not a triangle –  vadim123 May 13 '13 at 21:26
@defeuer, it's unclear what his "original approach" is. If it's to cube $\frac 12$, then of course it's coincidence that it works. But if it's to calculate the area of the triangle and (incorrectly) divide by 1 instead of $\frac 12$, it's fine. –  vadim123 May 13 '13 at 21:27
@vadim123 thanks for the explanation, I've understand the error and the easy way to do it :) –  Riccardo May 13 '13 at 21:30
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